description: In a certain video game, the player controls a hero characterized by a single integer value: power. The hero will have to beat monsters that are also characterized by a single integer value: armor.
  
  On the current level, the hero is facing n caves. To pass the level, the hero must enter all the caves in some order, each cave exactly once, and exit every cave safe and sound. When the hero enters cave i, he will have to fight k_i monsters in a row: first a monster with armor a_{i, 1}, then a monster with armor a_{i, 2} and so on, finally, a monster with armor a_{i, k_i}.
  
  The hero can beat a monster if and only if the hero's power is strictly greater than the monster's armor. If the hero can't beat the monster he's fighting, the game ends and the player loses. Note that once the hero enters a cave, he can't exit it before he fights all the monsters in it, strictly in the given order.
  
  Each time the hero beats a monster, the hero's power increases by 1.
  
  Find the smallest possible power the hero must start the level with to be able to enter all the caves in some order and beat all the monsters.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of caves.
  
  The i-th of the next n lines contains an integer k_i (1 ≤ k_i ≤ 10^5) — the number of monsters in the i-th cave, followed by k_i integers a_{i, 1}, a_{i, 2}, …, a_{i, k_i} (1 ≤ a_{i, j} ≤ 10^9) — armor levels of the monsters in cave i in order the hero has to fight them.
  
  It is guaranteed that the sum of k_i over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the smallest possible power the hero must start the level with to be able to enter all the caves in some order and beat all the monsters.
  
  Example
  
  Input
  
  
  2
  1
  1 42
  2
  3 10 15 8
  2 12 11
  
  
  Output
  
  
  43
  13
  
  Note
  
  In the first test case, the hero has to beat a single monster with armor 42, it's enough to have power 43 to achieve that.
  
  In the second test case, the hero can pass the level with initial power 13 as follows: 
  
    * enter cave 2: 
      * beat a monster with armor 12, power increases to 14; 
      * beat a monster with armor 11, power increases to 15; 
    * enter cave 1: 
      * beat a monster with armor 10, power increases to 16; 
      * beat a monster with armor 15, power increases to 17; 
      * beat a monster with armor 8, power increases to 18. 
solution: import java.util.*;
  import java.lang.*;
  import java.io.*;
  import java.awt.*;
  
  /*AUTHOR - SHIVAM GUPTA */
  
  // U KNOW THAT IF THIS DAY WILL BE URS THEN NO ONE CAN DEFEAT U HERE ,JUst keep faith in ur strengths .................................................. 
  // ASCII = 48 + i ;// 2^28 = 268,435,456  > 2* 10^8 // log 10 base 2 = 3.3219 
  // odd:: (x^2+1)/2 , (x^2-1)/2 ; x>=3// even:: (x^2/4)+1 ,(x^2/4)-1  x >=4 
  // FOR ANY ODD NO N : N,N-1,N-2,ALL ARE PAIRWISE COPRIME,THEIR COMBINED LCM IS PRODUCT OF ALL THESE NOS
  // two consecutive odds are always coprime to each other, two consecutive even have always gcd  = 2 ;
  
  // Rectangle r = new Rectangle(int x , int y,int widht,int height) 
  //Creates a rect. with bottom left cordinates as (x, y) and top right as ((x+width),(y+height))
  
  //BY DEFAULT Priority Queue is MIN in nature in java
  //to use as max , just push with negative sign and change sign after removal 
  
  // We can make a sieve of max size 1e7 .(no time or space issue) 
  // In 1e7 starting nos we have about 66*1e4 prime nos
  // In 1e6 starting nos we have about 78,498 prime nos
  
   public class Main
   //public class Solution
  {
      static PrintWriter out;
      
  	static class FastReader{
  		BufferedReader br;
  		StringTokenizer st;
  		public FastReader(){
  			br=new BufferedReader(new InputStreamReader(System.in));
  			out=new PrintWriter(System.out);
  		}
  		String next(){
  			while(st==null || !st.hasMoreElements()){
  				try{
  					st= new StringTokenizer(br.readLine());
  				}
  				catch (IOException e){
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  		int nextInt(){
  			return Integer.parseInt(next());
  		}
  		long nextLong(){
  			return Long.parseLong(next());
  		}
  		double nextDouble(){
  			return Double.parseDouble(next());
  		}
  		String nextLine(){
  			String str = "";
  			try{
  				str=br.readLine();
  			}
  			catch(IOException e){
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  	
  ///////////////////////////////////////////////////////////////////////////////////////////////////////
   public static int sumOfDigits(long n)
   {
    if( n< 0)return -1 ;
    int sum = 0;
    while( n > 0){ sum = sum + (int)( n %10) ; n /= 10 ; } 
    return sum ;  
   }
   
  //////////////////////////////////////////////////////////////////////////////////////////////////////
  public static void swapArray(int[] arr , int start , int end)
  {
      while(start<end){ int temp = arr[start] ;arr[start] = arr[end]; arr[end] = temp;  start++ ;end-- ;}
     
  }
  //////////////////////////////////////////////////////////////////////////////////
  
  static long factorial(long a)
  {
      if(a== 0L || a==1L)return 1L ;
     return a*factorial(a-1L) ;
  }
  ///////////////////////////////////////////////////////////////////////////////
  public static int[][] rotate(int[][] input){
  
  int n =input.length;int m = input[0].length ;
  int[][] output = new int [m][n];
  for (int i=0; i<n; i++)
  	for (int j=0;j<m; j++)
  		output [j][n-1-i] = input[i][j];
  return output;
  }
  ////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// ////////////////////////////////////////////////   
  public static String reverse(String input)
  {
    StringBuilder str  = new StringBuilder("") ;
     
      for(int i =input.length()-1 ; i >= 0  ; i-- )
      {
          str.append(input.charAt(i));
      }
      
  return str.toString() ;
  }
  ///////////////////////////////////////////////////////////////////////////////////////////
  
  public static int xorOfFirstN(int n)
  {
    if( n % 4 ==0)return n ; else if( n % 4 == 1)return 1 ;
    else if( n % 4 == 2)return n+1 ; else return 0 ;
   }
  //////////////////////////////////////////////////////////////////////////////////////////////
  
  public static int gcd(int a, int b )
  {
     if(b==0)return a ;else return gcd(b,a%b) ; 
  }
  
  public static long gcd(long a, long b )
  {
  if(b==0)return a ;else return gcd(b,a%b) ; 
  }
  ///////////////////////////////////////////////////////////////////////////////////////////
  
  public static int lcm(int a, int b ,int c )
  {
  int temp = lcm(a,b) ;int ans =  lcm(temp ,c) ;return ans  ;
  }
  
  ////////////////////////////////////////////////////////////////////////////////////////
      
  public static int lcm(int a , int b )
  {
     int gc = gcd(a,b);return (a/gc)*b ;
  }
  
  
  public static long lcm(long a , long b )
  {
      long gc = gcd(a,b);return (a/gc)*b;
  }
  
  
  ///////////////////////////////////////////////////////////////////////////////////////////
  static boolean isPrime(long n)
  { 
     if(n==1)return false  ;  
      for(long i = 2L; i*i <= n ;i++){ if(n% i ==0){return false; } }
      return true ;
  } 
  
  static boolean isPrime(int n)
  {
      if(n==1)return false  ;  
      for(int i = 2; i*i <= n ;i++){ if(n% i ==0){return false ; } }
      return true ;
  }      
  ////////////////////////////////////////////////////////////////////////////////////////////////////////
  
  ///////////////////////////////////////////////////////////////////////////
  
  static int sieve =  1000000 ;
  
   
  static boolean[] prime =  new boolean[sieve + 1] ;
  
  public static void sieveOfEratosthenes() 
      { 
          // FALSE == prime and 1 // TRUE ==  COMPOSITE
          
          // time complexity = 0(NlogLogN)== o(N)
          
          // gives prime nos bw 1 to N // size - 1e7(at max)
          
          for(int i = 4; i<= sieve ; i++)
          {
              prime[i] = true  ; i++ ;
             
          }
          
          for(int p = 3; p*p <= sieve; p++) 
          { 
             if(prime[p] == false) 
              { 
                      for(int i = p*p; i <= sieve; i += p) 
                      prime[i] = true; 
              } 
              
              p++ ;
          } 
            
      } 
   
  ///////////////////////////////////////////////////////////////////////////////////
  public static void sortD(int[] arr , int s  , int e)
  {
        sort(arr ,s , e) ;
        int i =s ; int j = e  ;
        while( i < j){  int temp = arr[i] ;arr[i] =arr[j] ; arr[j] = temp ; i++ ; j-- ;}
  }
  
  /////////////////////////////////////////////////////////////////////////////////////////
  
  public static long countSubarraysSumToK(long[] arr ,long sum )
      {
        HashMap<Long,Long> map = new HashMap<>() ;
         
        int n = arr.length ; long prefixsum = 0 ; long count = 0L ;
          for(int i  = 0; i < n ; i++)
        {
            prefixsum  = prefixsum +  arr[i] ;
             
            if(sum == prefixsum)count = count+1 ;
             
            if(map.containsKey(prefixsum -sum))count = count + map.get(prefixsum -sum) ;
           
            if(map.containsKey(prefixsum )) map.put(prefixsum , map.get(prefixsum) +1 );
            else map.put(prefixsum , 1L );
        }
          
        return count  ;  
      }
  
  ///////////////////////////////////////////////////////////////////////////////////////////////
  
  
  // KMP ALGORITHM : TIME COMPL:O(N+M) 
  // FINDS THE OCCURENCES OF PATTERN AS A  SUBSTRING IN STRING
  //RETURN THE ARRAYLIST OF INDEXES 
  // IF SIZE OF LIST IS ZERO MEANS PATTERN IS NOT PRESENT IN STRING
  
  
  public static ArrayList<Integer> kmpAlgorithm(String str , String pat)
       {
          ArrayList<Integer> list =new ArrayList<>();
          
          int n = str.length() ;
          int m = pat.length() ;
          
          String q = pat + "#" + str ;
          
          int[] lps  =new int[n+m+1] ;
          
           longestPefixSuffix(lps, q,(n+m+1)) ;
           
           
           for(int i =m+1 ; i < (n+m+1) ; i++ )
           {
               if(lps[i] == m)
               {
                   list.add(i-2*m) ;
               }
           }
          
          return list ; 
           
          
      }
      
  
  public static void longestPefixSuffix(int[] lps ,String str , int n)
      {
          lps[0] =  0 ;
          
          for(int i = 1  ; i<= n-1; i++)
          {
            int l = lps[i-1] ;
             
            while( l > 0 && str.charAt(i) != str.charAt(l))
            {
                l = lps[l-1] ;
            }
             
            if(str.charAt(i) == str.charAt(l))
            {l++ ;}
                
                lps[i] = l ; 
          }
          
      }
      
   ///////////////////////////////////////////////////////////////////////////////////////////////////
  
   // CALCULATE TOTIENT Fn FOR ALL VALUES FROM 1 TO n
      // TOTIENT(N) = count of nos less than n and grater than 1 whose gcd with n is 1 
      // or n and the no  will be coprime in nature
      //time : O(n*(log(logn)))
      
      public static void eulerTotientFunction(int[] arr ,int n )
      {
        for(int i = 1; i <= n  ;i++)arr[i] =i  ;
        
         for(int i= 2 ; i<= n ;i++)
        {
            if(arr[i] == i)
            {
                arr[i] =i-1 ;
                
                for(int j =2*i ; j<= n  ; j+= i )
                {
                    arr[j] = (arr[j]*(i-1))/i ;
                }
                
            }
        }
          
      }
  	
  /////////////////////////////////////////////////////////////////////////////////////////////
  public static long nCr(int n,int k)
  {
      long ans=1L;
      k=k>n-k?n-k:k;
      int j=1;
      for(;j<=k;j++,n--)
      {
          if(n%j==0)
          {
              ans*=n/j;
          }else
          if(ans%j==0)
          {
              ans=ans/j*n;
          }else
          {
              ans=(ans*n)/j;
          }
      }
      return ans;
  }
  
  ///////////////////////////////////////////////////////////////////////////////////////////
  
  public static ArrayList<Integer> allFactors(int n)
  {   
        ArrayList<Integer> list = new ArrayList<>() ;
        
      for(int i = 1; i*i <= n ;i++)
      {
            if( n % i == 0)
            {
                if(i*i == n)list.add(i) ;
                else{ list.add(i) ;list.add(n/i) ; }
            }
      }
        
       return list ; 
  }
  
  
  public static ArrayList<Long> allFactors(long n)
  {   
        ArrayList<Long> list = new ArrayList<>() ;
        
      for(long i = 1L; i*i <= n ;i++)
      {
            if( n % i == 0)
            {
                if(i*i == n) list.add(i) ;
                else{  list.add(i) ; list.add(n/i) ; }
               
            }
      }
        
       return list ; 
  }
  ////////////////////////////////////////////////////////////////////////////////////////////////////
  
    static final int MAXN = 1000001; 
         
      
      static int spf[] = new int[MAXN]; 
     
      static void sieve() 
      { 
         for (int i=1; i<MAXN; i++) spf[i] = i; // ALL NOS HAVE SPF[i] =i AT START
         
          for (int i=4; i<MAXN; i+=2) //ALL EVEN NOS HAVE SPF[i] = 2 
              spf[i] = 2; 
         
          for (int i=3; i*i<MAXN; i++) 
          { 
             if (spf[i] == i) 
              { 
                 for (int j=i*i; j<MAXN; j+=i) 
         
                      if (spf[j]==j) spf[j] = i;  
              } 
          } 
      } 
         
  
      static ArrayList<Integer> getPrimeFactorization(int x) 
      { 
          ArrayList<Integer> ret = new ArrayList<Integer>(); 
          while (x != 1) 
          { 
              ret.add(spf[x]); 
              x = x / spf[x]; 
          } 
          return ret; 
      } 
         
   //////////////////////////////////////////////////////////////////////////////////////////////////
   //////////////////////////////////////////////////////////////////////////////////////////////////
     
  public static void merge(int arr[], int l, int m, int r)
      {
          // Find sizes of two subarrays to be merged
          int n1 = m - l + 1;
          int n2 = r - m;
   
          /* Create temp arrays */
          int L[] = new int[n1];
          int R[] = new int[n2];
   
         //Copy data to temp arrays
          for (int i=0; i<n1; ++i)
              L[i] = arr[l + i];
          for (int j=0; j<n2; ++j)
              R[j] = arr[m + 1+ j];
   
   
          /* Merge the temp arrays */
   
          // Initial indexes of first and second subarrays
          int i = 0, j = 0;
   
          // Initial index of merged subarry array
          int k = l;
          while (i < n1 && j < n2)
          {
              if (L[i] <= R[j])
              {
                  arr[k] = L[i];
                  i++;
              }
              else
              {
                  arr[k] = R[j];
                  j++;
              }
              k++;
          }
   
          /* Copy remaining elements of L[] if any */
          while (i < n1)
          {
              arr[k] = L[i];
              i++;
              k++;
          }
   
          /* Copy remaining elements of R[] if any */
          while (j < n2)
          {
              arr[k] = R[j];
              j++;
              k++;
          }
      }
   
      // Main function that sorts arr[l..r] using
      // merge()
    public static void sort(int arr[], int l, int r)
      {
          if (l < r)
          {
              // Find the middle point
              int m = (l+r)/2;
   
              // Sort first and second halves
              sort(arr, l, m);
              sort(arr , m+1, r);
   
              // Merge the sorted halves
              merge(arr, l, m, r);
          }
      }
  
  public static void sort(long arr[], int l, int r)
      {
          if (l < r)
          {
              // Find the middle point
              int m = (l+r)/2;
   
              // Sort first and second halves
              sort(arr, l, m);
              sort(arr , m+1, r);
   
              // Merge the sorted halves
              merge(arr, l, m, r);
          }
      }
  
  
  public static void merge(long arr[], int l, int m, int r)
      {
          // Find sizes of two subarrays to be merged
          int n1 = m - l + 1;
          int n2 = r - m;
   
          /* Create temp arrays */
          long L[] = new long[n1];
          long R[] = new long[n2];
   
          //Copy data to temp arrays
          for (int i=0; i<n1; ++i)
              L[i] = arr[l + i];
          for (int j=0; j<n2; ++j)
              R[j] = arr[m + 1+ j];
   
   
          /* Merge the temp arrays */
   
          // Initial indexes of first and second subarrays
          int i = 0, j = 0;
   
          // Initial index of merged subarry array
          int k = l;
          while (i < n1 && j < n2)
          {
              if (L[i] <= R[j])
              {
                  arr[k] = L[i];
                  i++;
              }
              else
              {
                  arr[k] = R[j];
                  j++;
              }
              k++;
          }
   
          /* Copy remaining elements of L[] if any */
          while (i < n1)
          {
              arr[k] = L[i];
              i++;
              k++;
          }
   
          /* Copy remaining elements of R[] if any */
          while (j < n2)
          {
              arr[k] = R[j];
              j++;
              k++;
          }
      }
   
  
   /////////////////////////////////////////////////////////////////////////////////////////
   
  	public static long knapsack(int[] weight,long value[],int maxWeight){
  
          int n=  value.length ;
  
  	//dp[i] stores the profit with KnapSack capacity "i" 
    long []dp = new long[maxWeight+1]; 
        
      //initially profit with 0 to W KnapSack capacity is 0 
      Arrays.fill(dp, 0); 
    
      // iterate through all items 
      for(int i=0; i < n; i++)  
        
          //traverse dp array from right to left 
          for(int j = maxWeight; j >= weight[i]; j--) 
              dp[j] = Math.max(dp[j] , value[i] + dp[j - weight[i]]); 
                
      /*above line finds out maximum of dp[j](excluding ith element value) 
      and val[i] + dp[j-wt[i]] (including ith element value and the 
      profit with "KnapSack capacity - ith element weight") */
      return dp[maxWeight]; 
  	}
  
  ///////////////////////////////////////////////////////////////////////////////////////////
  //////////////////////////////////////////////////////////////////////////////////////////
  
  
  // to return max sum of any subarray in given array
  public static long kadanesAlgorithm(long[] arr)
  {
      if(arr.length == 0)return  0 ;    
      
      long[] dp = new long[arr.length] ; dp[0] = arr[0] ;long max =  dp[0] ;
        
      for(int i = 1; i <  arr.length ; i++)
        {
              if(dp[i-1] > 0) dp[i] = dp[i-1] + arr[i] ; else  dp[i] = arr[i] ;
              if(dp[i] >  max)max = dp[i] ;
      }
         return max  ;
        
  }
  /////////////////////////////////////////////////////////////////////////////////////////////
  public static long kadanesAlgorithm(int[] arr)
  {    if(arr.length == 0)return  0 ;
      
        long[] dp = new long[arr.length] ;dp[0] = arr[0] ; long max =  dp[0] ;
        for(int i = 1; i <  arr.length ; i++)
        {
              if(dp[i-1] > 0) dp[i] = dp[i-1] + arr[i] ;  else  dp[i] = arr[i] ;
             if(dp[i] >  max)max = dp[i] ;
        }
        
        return max  ;
  }
  
  /////////////////////////////////////////////////////////////////////////////////////////////////////////
      public static long power(int a  ,int b)
      {
          //time comp : o(logn) 
          long x = (long)(a) ; long n = (long)(b) ;
          if(n==0)return 1 ;if(n==1)return x;
           long ans =1L  ;
   
        while(n>0)
        {
            if(n % 2 ==1)ans = ans *x ;
           n = n/2L ;x =  x*x ; 
        }    
        return ans ;
      }
      
      public static long power(long a  ,long b)
      {
          //time comp : o(logn) 
          long x = (a) ;long n = (b) ;
           if(n==0)return 1L ;if(n==1)return x;
           long ans =1L  ;
          
        while(n>0)
        {
            if(n % 2 ==1)ans = ans *x ;
          
            n = n/2L ; x =  x*x ;
        }
         
        return ans ;
      }
      
  //////////////////////////////////////////////////////////////////////////////////
  
  /*
  lowerBound - finds largest element equal or less than value paased
  upperBound - finds smallest element equal or more than value passed
  if not present return -1;
  */
  
  public static long lowerBound(long[] arr,long k)
  	{
  		long ans=-1;
  		
  		int start=0;
  		int end=arr.length-1;
  		
  		while(start<=end)
  		{
  			int mid=(start+end)/2;
  			
  			if(arr[mid]<=k)
  			{
  				ans=arr[mid];
  				start=mid+1;
  			}
  			else
  			{
  				end=mid-1;
  			}
  			
  		}
  		
  		return ans;
  		
  	}
  	
  	public static int lowerBound(int[] arr,int k)
  	{
  		int ans=-1;int start=0;int end=arr.length-1;
  	
  		while(start<=end)
  		{
  			int mid=(start+end)/2;
  			
  			if(arr[mid]<=k)
  			{
  				ans=arr[mid];
  				start=mid+1;
  			}
  			else
  			{
  				end=mid-1;
  			}
  			
  		}
  		
  		return ans;
  		
  	}
  	
  	
  	public static long upperBound(long[] arr,long k)
  	{
  		long ans=-1;int start=0;int end=arr.length-1;
  	
  		while(start<=end)
  		{
  			int mid=(start+end)/2;
  			
  			if(arr[mid]>=k)
  			{
  				ans=arr[mid];
  				end=mid-1;
  			}
  			else
  			{
  				start=mid+1;
  			}
  			
  		}
  		
  		return ans;
  	}
  	
  	
  	public static int upperBound(int[] arr,int k)
  	{
  		int ans=-1; int start=0; 	int end=arr.length-1;
  
  		while(start<=end)
  		{
  			int mid=(start+end)/2;
  			
  			if(arr[mid]>=k)
  			{
  				ans=arr[mid];
  				end=mid-1;
  			}
  			else
  			{
  				start=mid+1;
  			}
  			
  		}
  		
  		return ans;
  	}
  	
  ////////////////////////////////////////////////////////////////////////////////////////
  //////////////////////////////////////////////////////////////////////////////////////////
  
  public static void printArray(int[] arr , int si ,int ei, int f)
  {
      for(int i = si  ; i  <= ei ; i++){ out.print(arr[i] +" ") ; }  if(f==1)out.println() ;
  }
  
  public static void printArray(long[] arr , int si ,int ei, int f)
  {
      for(int i = si  ; i  <= ei ; i++){ out.print(arr[i] +" ") ; }  if(f==1)out.println() ;
  }
  
  
  public static void printtwodArray(int[][] ans)
  {
      for(int i = 0; i< ans.length ; i++)
      {
          for(int j  = 0 ; j <  ans[0].length ; j++)out.print(ans[i][j] +" "); out.println() ;
      }
      out.println() ;
  }
  /////////////////////////////////////////////////////////////////////////////////////////////////////
   ////////////////////////////////////////////////////////////////////////////////////////////////////
   
     static long modPow(long a, long x, long p) {
      //calculates a^x mod p in logarithmic time.
      
      a = a % p ;
      
      if(a == 0)return 0L ;
      if(x==0)return 1L;
      if(x==1)return a;
      
      long res = 1L;
      while(x > 0) {
          if( x % 2 != 0) {
              res = (res * a) % p;
          }
          a = (a * a) % p;
          x =x/2;
      }
      return res;
  }
   
   
    static long modInverse(long a, long p) {
      //calculates the modular multiplicative of a mod p. //(assuming p is prime).
      return modPow(a, p-2, p);
  }
   
   static long[] factorial =  new long[1000001] ;
   
   static void modfac(long mod)
   {
       factorial[0]=1L ; factorial[1]=1L ;
       
       for(int i = 2; i<= 1000000 ;i++)
       {
           factorial[i] = factorial[i-1] *(long)(i) ;
            factorial[i] = factorial[i] % mod ;
       }
      
       
   }
   
  static long modBinomial(long n, long r, long p) {
  // calculates C(n,r) mod p (assuming p is prime).
   
    if(n < r) return 0L ; 
   
      long num = factorial[(int)(n)] ;
      
      long den = (factorial[(int)(r)]*factorial[(int)(n-r)]) %  p ;
      
      
      long ans  = num*(modInverse(den,p)) ;
      
      ans = ans % p ;
      
      return ans  ;
   
      
  }
   
   
   static void update(int val , long[] bit ,int n)
   {
       for( ; val <= n ; val += (val &(-val)) )
       {
           bit[val]++ ;
       }
      
       
   }
   
   
   static long query(int val , long[] bit , int n)
   {
       long ans = 0L; 
       for( ; val >=1 ; val-=(val&(-val))  )ans += bit[val];
       
       return ans ;
   }
  
  
  
   static int countSetBits(long n) 
      { 
          int count = 0; 
          while (n > 0) { 
              n = (n) & (n - 1L);  count++; 
              
          } 
          return count; 
      } 
  
  
  static int abs(int x)
  { if(x<0)x=-x ; return x ; }
      
  static long abs(long x)
  {  if(x<0)x=-x ; return x ; }
  
  ////////////////////////////////////////////////////////////////////////////////////////////
  
   // calculate total no of nos greater than or equal to key in  sorted array arr
   
  static  int bs(int[] arr, int s ,int e  ,int key)
  {
          if( s> e)return 0 ;
          int mid = (s+e)/2 ;
          
          if(arr[mid] <key) return bs(arr ,mid+1,e , key) ;
          else  {return bs(arr ,s ,mid-1, key)  + e-mid+1;}
     
  }
   
  // static ArrayList<Integer>[] adj ;
  // static int mod= 1000000007 ;
  
  
  static class pair {
  		int u;
  	     int v;
   
  	public	pair(int u,int v) {
  			this.u = u ; this.v=v;
  		
  		}
  	}
   
  static class Dsu
  {
      int n ;int[] par ; int[] size ;
      
      public Dsu(int n)
      {
          this.n = n ;
          par = new int[n] ;
          size = new int[n] ;
          
          for(int i = 0 ; i< n ;i++)
          {
              par[i] =i ;size[i] =1 ;
          }
      }
      
      // cheack if given element is root element of its set
      public boolean isRoot(int x)
      {
          return par[x] == x ;
      }
      
      // find root of given set
     public int findRoot(int x)
     {
         if(par[x] == x)
         {
            return par[x] ; 
         }
         
         par[x] = findRoot(par[x]) ; //path compression
         return par[x] ;
     }
     
     
     public boolean unionSet(int a, int b)
     {
         int rootA = findRoot(a);
         int rootB=findRoot(b) ;
         
         if(rootA==rootB)return true ;//they are alraedy present in same set/connected already(in graph)
         
         if(size[rootA] > size[rootB])
         {
             size[rootA] += size[rootB] ;
             par[rootB] =rootA ;
         }
         
         else{
             
             size[rootB] += size[rootA] ;
             par[rootA] =rootB ;
         }
         
         return false ;     // means we union both the sets now
     }
     
  }
  
  
  static int[][] arr;
  
  static boolean check(long mid)
  {
      long cur =mid;
      
      int i = 0 ; 
      
      while( i < arr.length)
      {
          if(cur > arr[i][0])
          {
              cur =cur + arr[i][1] ; i++;
          }
          else{
              return false ;
          }
          
         
      }
      
      return true ;
  }
  
  ////////////////////////////////////////////////////////////////////////////////////////////
  
  public static void solve()
  {
  FastReader scn = new FastReader() ;
  
  
  int testcase = 1;
   testcase = scn.nextInt() ;
  for(int testcases =1  ; testcases <= testcase ;testcases++)
  {
   
   
  int n= scn.nextInt() ;
  
  arr =  new int[n][2] ;
  
  for(int i = 0 ;  i < n ;i++)
  {
      int  c =scn.nextInt() ;
           
       int max = 0 ;
       int j =  0 ;
       while( j < c)
       {
             int t = scn.nextInt()-j ;
             
             if(t > max)max=t;
             
              j++;
       }
       
       arr[i][0] =max;arr[i][1]=c;
      
  }
  
  
  Arrays.sort(arr, new Comparator<int[]>(){
      
     public int compare(int[] a, int[] b)
     {
         if(a[0] != b[0])
         {
             return Integer.compare(a[0],b[0]);
         }
         
         else{
             
              return Integer.compare(a[1],b[1]);
         }
     }
      
  });
  
  
  long s =1 ;long e = 100000000000L;
  long ans  = e;
  
  while(s<=e)
  {
      long mid =(s+e)/2 ;
      
      if(check(mid))
      {
          ans =mid;e=mid-1;
      }
      
      else{
          s=mid+1;
      }
  }
  
  out.println(ans);
  
  
  } // test case  end loop
  
  
  out.flush() ;  
  } // solve fn ends
  
  
  public static void main (String[] args) throws java.lang.Exception
  {     solve() ;    }
        
  }
    
   class Pair 
  {
    int first ;int second  ;
   
    public  Pair(int x, int y)
    {
        this.first = x ;this.second = y ;
    }
    
    @Override
   public boolean equals(Object obj)
   {
       if(obj == this)return true  ;
       
       if(obj == null)return false  ;
       
       if(this.getClass() != obj.getClass()) return false ;
       
       Pair other  = (Pair)(obj) ;
       
       if(this.first != other.first)return false  ;
       if(this.second != other.second)return false  ;
       
       return true  ;
       
   }
      
      @Override
      
      public int hashCode()
      {
          return this.first^this.second  ;
      }
        
  @Override
  	public String toString() {
  	
  	String ans = "" ;ans  += this.first ;	ans += " ";	ans += this.second ;
  	return ans  ;
  	}
  
  }
  
  
  
  