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: // package CF1561;
  
  
  import java.io.ByteArrayInputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.util.*;
  
  public class C {
  	static FastReader s;
  	static String INPUT = "2\n"
  			+ "1\n"
  			+ "1 42\n"
  			+ "2\n"
  			+ "3 10 15 8\n"
  			+ "2 12 11\n"
  			+ "";
  
  	public static void main(String[] args) {
  		long time = System.currentTimeMillis();
  		boolean oj = System.getProperty("ONLINE_JUDGE") != null;
  		s = new FastReader(oj);
  		int t = s.nextInt();
  		while (t-- > 0) {
  			ArrayList<Pair> list = new ArrayList<>();
  			int n = s.nextInt();
  			for (int i = 0; i < n; i++) {
  				int k = s.nextInt();
  				int[] arr = s.nextIntArray(k);
  				int entry = arr[0] + 1;
  				for (int j = 1; j < k; j++) {
  					entry = Math.max(entry, arr[j] + 1 - j);
  				}
  				list.add(new Pair(entry, entry + k));
  			}
  			
  			Collections.sort(list);
  			int currEntry = list.get(0).entry;
  			int ans = currEntry;
  			int gainedTillNow = 0;
  			for (int i = 0; i < list.size() - 1; i++) {
  				int currExit = currEntry + list.get(i).exit - list.get(i).entry;
  				gainedTillNow += list.get(i).exit - list.get(i).entry;
  				if (currExit < list.get(i + 1).entry) {
  					currEntry = list.get(i + 1).entry - gainedTillNow;
  					ans = Math.max(ans, currEntry);
  				} else {
  					currEntry = currExit;
  				}
  			}
  			
  			System.out.println(ans);
  			
  		}
  
  		if (!oj) {
  			System.out.println(Arrays.deepToString(new Object[] { System.currentTimeMillis() - time + " ms" }));
  		}
  	}
  	
  	private static class Pair implements Comparable<Pair>{
  		int entry, exit;
  		
  		public Pair(int entry, int exit) {
  			this.entry = entry;
  			this.exit = exit;
  		}
  
  		@Override
  		public int compareTo(Pair p) {
  			if (this.entry == p.entry) {
  				return Integer.compare(this.exit, p.exit);
  			}
  			
  			return Integer.compare(this.entry, p.entry);
  		}
  		
  		
  	}
  
  	private static class UnionFindDisjointSet {
  		int[] parent;
  		int[] size;
  		int n;
  		int size1;
  
  		public UnionFindDisjointSet(int n) {
  			this.n = n;
  			this.parent = new int[n];
  			this.size = new int[n];
  			for (int i = 0; i < n; i++) {
  				parent[i] = i;
  			}
  
  			for (int i = 0; i < n; i++) {
  				size[i] = 1;
  			}
  
  			this.size1 = n;
  		}
  
  		private int numDisjointSets() {
  			System.out.println(size1);
  			return size1;
  		}
  
  		private boolean find(int a, int b) {
  			int rootA = root(a);
  			int rootB = root(b);
  			if (rootA == rootB) {
  				return true;
  			}
  			return false;
  		}
  
  		private int root(int b) {
  			while (parent[b] != b) {
  				parent[b] = parent[parent[b]];
  				b = parent[b];
  			}
  			return b;
  		}
  
  		private void union(int a, int b) {
  			int rootA = root(a);
  			int rootB = root(b);
  			if (rootA == rootB) {
  				return;
  			}
  
  			if (size[rootA] < size[rootB]) {
  				parent[rootA] = parent[rootB];
  				size[rootB] += size[rootA];
  			} else {
  				parent[rootB] = parent[rootA];
  				size[rootA] += size[rootB];
  			}
  			size1--;
  			System.out.println(Arrays.toString(parent));
  		}
  	}
  
  	static ArrayList<Long> printDivisors(long n) {
  		// Note that this loop runs till square root
  		ArrayList<Long> list = new ArrayList<>();
  		for (long i = 1; i <= Math.sqrt(n); i++) {
  			if (n % i == 0) {
  				if (n / i == i) {
  					list.add(i);
  				} else {
  					list.add(i);
  					list.add(n / i);
  				}
  			}
  		}
  		return list;
  	}
  
  	//    GCD - Using Euclid theorem.
  	private static long gcd(long a, long b) {
  		if (b == 0) {
  			return a;
  		}
  
  		return gcd(b, a % b);
  	}
  
  	//    Extended euclidean algorithm
  	//    Used to solve equations of the form ax + by = gcd(a,b)
  	//    return array [d, a, b] such that d = gcd(p, q), ap + bq = d
  	static long[] extendedEuclidean(long p, long q) {
  		if (q == 0)
  			return new long[] { p, 1, 0 };
  
  		long[] vals = extendedEuclidean(q, p % q);
  		long d = vals[0];
  		long a = vals[2];
  		long b = vals[1] - (p / q) * vals[2];
  		return new long[] { d, a, b };
  	}
  
  	// X ^ y mod p
  	static long power(long x, long y, long p) {
  		long res = 1;
  		x = x % p;
  
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y = y >> 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	// Returns modulo inverse of a
  	// with respect to m using extended
  	// Euclid Algorithm. Refer below post for details:
  	// https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/
  	static long inv(long a, long m) {
  		long m0 = m, t, q;
  		long x0 = 0, x1 = 1;
  
  		if (m == 1)
  			return 0;
  
  		// Apply extended Euclid Algorithm
  		while (a > 1) {
  			q = a / m;
  			t = m;
  			m = a % m;
  			a = t;
  			t = x0;
  			x0 = x1 - q * x0;
  			x1 = t;
  		}
  
  		// Make x1 positive
  		if (x1 < 0)
  			x1 += m0;
  
  		return x1;
  	}
  
  	// k is size of num[] and rem[].
  	// Returns the smallest number
  	// x such that:
  	// x % num[0] = rem[0],
  	// x % num[1] = rem[1],
  	// ..................
  	// x % num[k-2] = rem[k-1]
  	// Assumption: Numbers in num[] are pairwise
  	// coprime (gcd for every pair is 1)
  	static long findMinX(long num[], long rem[], long k) {
  		int prod = 1;
  		for (int i = 0; i < k; i++)
  			prod *= num[i];
  
  		int result = 0;
  		for (int i = 0; i < k; i++) {
  			long pp = prod / num[i];
  			result += rem[i] * inv(pp, num[i]) * pp;
  		}
  
  		return result % prod;
  	}
  
  	//  Binary search
  	private static int binarySearch(int[] arr, int ele) {
  		int low = 0;
  		int high = arr.length - 1;
  		while (low <= high) {
  			int mid = (low + high) / 2;
  			if (arr[mid] == ele) {
  				return mid;
  			} else if (ele < arr[mid]) {
  				high = mid - 1;
  			} else {
  				low = mid + 1;
  			}
  		}
  		return -1;
  	}
  
  	//    First occurence using binary search
  	private static int binarySearchFirstOccurence(int[] arr, int ele) {
  		int low = 0;
  		int high = arr.length - 1;
  		int ans = -1;
  		while (low <= high) {
  			int mid = (low + high) / 2;
  			if (arr[mid] == ele) {
  				ans = mid;
  				high = mid - 1;
  			} else if (ele < arr[mid]) {
  				high = mid - 1;
  			} else {
  				low = mid + 1;
  			}
  		}
  		return ans;
  	}
  
  	//    Last occurenece using binary search
  	private static int binarySearchLastOccurence(int[] arr, int ele) {
  		int low = 0;
  		int high = arr.length - 1;
  		int ans = -1;
  		while (low <= high) {
  			int mid = (low + high) / 2;
  			if (arr[mid] == ele) {
  				ans = mid;
  				low = mid + 1;
  			} else if (ele < arr[mid]) {
  				high = mid - 1;
  			} else {
  				low = mid + 1;
  			}
  		}
  		return ans;
  	}
  
  	// Merge sort
  	static void merge(int arr[], int l, int m, int r) {
  		int n1 = m - l + 1;
  		int n2 = r - m;
  		int L[] = new int[n1];
  		int R[] = new int[n2];
  		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];
  		int i = 0, j = 0;
  		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++;
  		}
  
  		while (i < n1) {
  			arr[k] = L[i];
  			i++;
  			k++;
  		}
  
  		while (j < n2) {
  			arr[k] = R[j];
  			j++;
  			k++;
  		}
  	}
  
  	static void sort(int arr[], int l, int r) {
  		if (l < r) {
  			int m = (l + r) / 2;
  			sort(arr, l, m);
  			sort(arr, m + 1, r);
  			merge(arr, l, m, r);
  		}
  	}
  
  	static void sort(int[] arr) {
  		sort(arr, 0, arr.length - 1);
  	}
  
  	private static class FastReader {
  		InputStream is;
  
  		public FastReader(boolean onlineJudge) {
  			is = onlineJudge ? System.in : new ByteArrayInputStream(INPUT.getBytes());
  		}
  
  		byte[] inbuf = new byte[1024];
  		public int lenbuf = 0, ptrbuf = 0;
  
  		int readByte() {
  			if (lenbuf == -1)
  				throw new InputMismatchException();
  			if (ptrbuf >= lenbuf) {
  				ptrbuf = 0;
  				try {
  					lenbuf = is.read(inbuf);
  				} catch (IOException e) {
  					throw new InputMismatchException();
  				}
  				if (lenbuf <= 0)
  					return -1;
  			}
  			return inbuf[ptrbuf++];
  		}
  
  		boolean isSpaceChar(int c) {
  			return !(c >= 33 && c <= 126);
  		}
  
  		int skip() {
  			int b;
  			while ((b = readByte()) != -1 && isSpaceChar(b))
  				;
  			return b;
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		char nextChar() {
  			return (char) skip();
  		}
  
  		String next() {
  			int b = skip();
  			StringBuilder sb = new StringBuilder();
  			while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
  				sb.appendCodePoint(b);
  				b = readByte();
  			}
  			return sb.toString();
  		}
  
  		String nextLine() {
  			int b = skip();
  			StringBuilder sb = new StringBuilder();
  			while ((!isSpaceChar(b) || b == ' ')) { // when nextLine, (isSpaceChar(b) && b != ' ')
  				sb.appendCodePoint(b);
  				b = readByte();
  			}
  			return sb.toString();
  		}
  
  		char[] next(int n) {
  			char[] buf = new char[n];
  			int b = skip(), p = 0;
  			while (p < n && !(isSpaceChar(b))) {
  				buf[p++] = (char) b;
  				b = readByte();
  			}
  			return n == p ? buf : Arrays.copyOf(buf, p);
  		}
  
  		int nextInt() {
  			int num = 0, b;
  			boolean minus = false;
  			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  				;
  			if (b == '-') {
  				minus = true;
  				b = readByte();
  			}
  
  			while (true) {
  				if (b >= '0' && b <= '9') {
  					num = num * 10 + (b - '0');
  				} else {
  					return minus ? -num : num;
  				}
  				b = readByte();
  			}
  		}
  
  		long nextLong() {
  			long num = 0;
  			int b;
  			boolean minus = false;
  			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  				;
  			if (b == '-') {
  				minus = true;
  				b = readByte();
  			}
  
  			while (true) {
  				if (b >= '0' && b <= '9') {
  					num = num * 10 + (b - '0');
  				} else {
  					return minus ? -num : num;
  				}
  				b = readByte();
  			}
  		}
  
  		char[][] nextMatrix(int n, int m) {
  			char[][] map = new char[n][];
  			for (int i = 0; i < n; i++)
  				map[i] = next(m);
  			return map;
  		}
  
  		int[] nextIntArray(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		long[] nextLongArray(int n) {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		int[][] next2DInt(int n, int m) {
  			int[][] arr = new int[n][];
  			for (int i = 0; i < n; i++) {
  				arr[i] = nextIntArray(m);
  			}
  			return arr;
  		}
  
  		long[][] next2DLong(int n, int m) {
  			long[][] arr = new long[n][];
  			for (int i = 0; i < n; i++) {
  				arr[i] = nextLongArray(m);
  			}
  			return arr;
  		}
  
  		int[] shuffle(int[] arr) {
  			Random r = new Random();
  			for (int i = 1, j; i < arr.length; i++) {
  				j = r.nextInt(i);
  				arr[i] = arr[i] ^ arr[j];
  				arr[j] = arr[i] ^ arr[j];
  				arr[i] = arr[i] ^ arr[j];
  			}
  			return arr;
  		}
  
  		int[] uniq(int[] arr) {
  			Arrays.sort(arr);
  			int[] rv = new int[arr.length];
  			int pos = 0;
  			rv[pos++] = arr[0];
  			for (int i = 1; i < arr.length; i++) {
  				if (arr[i] != arr[i - 1]) {
  					rv[pos++] = arr[i];
  				}
  			}
  			return Arrays.copyOf(rv, pos);
  		}
  	}
  }
  
  