description: The Olympic Games have just started and Federico is eager to watch the marathon race.
  
  There will be n athletes, numbered from 1 to n, competing in the marathon, and all of them have taken part in 5 important marathons, numbered from 1 to 5, in the past. For each 1≤ i≤ n and 1≤ j≤ 5, Federico remembers that athlete i ranked r_{i,j}-th in marathon j (e.g., r_{2,4}=3 means that athlete 2 was third in marathon 4).
  
  Federico considers athlete x superior to athlete y if athlete x ranked better than athlete y in at least 3 past marathons, i.e., r_{x,j}<r_{y,j} for at least 3 distinct values of j.
  
  Federico believes that an athlete is likely to get the gold medal at the Olympics if he is superior to all other athletes.
  
  Find any athlete who is likely to get the gold medal (that is, an athlete who is superior to all other athletes), or determine that there is no such athlete.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains a single integer n (1≤ n≤ 50 000) — the number of athletes.
  
  Then n lines follow, each describing the ranking positions of one athlete.
  
  The i-th of these lines contains the 5 integers r_{i,1},\,r_{i,2},\,r_{i,3},\,r_{i,4},  r_{i,5} (1≤ r_{i,j}≤ 50 000) — the ranking positions of athlete i in the past 5 marathons. It is guaranteed that, in each of the 5 past marathons, the n athletes have distinct ranking positions, i.e., for each 1≤ j≤ 5, the n values r_{1,j},  r_{2, j},  ...,  r_{n, j} are distinct.
  
  It is guaranteed that the sum of n over all test cases does not exceed 50 000.
  
  Output
  
  For each test case, print a single integer — the number of an athlete who is likely to get the gold medal (that is, an athlete who is superior to all other athletes). If there are no such athletes, print -1. If there is more than such one athlete, print any of them.
  
  Example
  
  Input
  
  
  4
  1
  50000 1 50000 50000 50000
  3
  10 10 20 30 30
  20 20 30 10 10
  30 30 10 20 20
  3
  1 1 1 1 1
  2 2 2 2 2
  3 3 3 3 3
  6
  9 5 3 7 1
  7 4 1 6 8
  5 6 7 3 2
  6 7 8 8 6
  4 2 2 4 5
  8 3 6 9 4
  
  
  Output
  
  
  1
  -1
  1
  5
  
  Note
  
  Explanation of the first test case: There is only one athlete, therefore he is superior to everyone else (since there is no one else), and thus he is likely to get the gold medal.
  
  Explanation of the second test case: There are n=3 athletes. 
  
    * Athlete 1 is superior to athlete 2. Indeed athlete 1 ranks better than athlete 2 in the marathons 1, 2 and 3. 
    * Athlete 2 is superior to athlete 3. Indeed athlete 2 ranks better than athlete 3 in the marathons 1, 2, 4 and 5. 
    * Athlete 3 is superior to athlete 1. Indeed athlete 3 ranks better than athlete 1 in the marathons 3, 4 and 5. 
  
  
  
  Explanation of the third test case: There are n=3 athletes. 
  
    * Athlete 1 is superior to athletes 2 and 3. Since he is superior to all other athletes, he is likely to get the gold medal. 
    * Athlete 2 is superior to athlete 3. 
    * Athlete 3 is not superior to any other athlete. 
  
  
  
  Explanation of the fourth test case: There are n=6 athletes. 
  
    * Athlete 1 is superior to athletes 3, 4, 6. 
    * Athlete 2 is superior to athletes 1, 4, 6. 
    * Athlete 3 is superior to athletes 2, 4, 6. 
    * Athlete 4 is not superior to any other athlete. 
    * Athlete 5 is superior to athletes 1, 2, 3, 4, 6. Since he is superior to all other athletes, he is likely to get the gold medal. 
    * Athlete 6 is only superior to athlete 4. 
solution: 
  ///package com.company;
  
  
  
  import  java.math.*;
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  
  
  public final class Main {
  
      FastReader s;
  
  
      public static void main (String[] args) throws java.lang.Exception
      {
  
          new Main().run();
  
      }
      void run()
      {
          s = new FastReader();
          solve();
      }
     StringBuffer sb;
     // int counter;
      void solve()
      {
           sb = new StringBuffer();
  
          for(int T = s.nextInt();T > 0;T--)
          {
  
  
              start();
  
  
          }
  
         // System.out.print(sb);
  
      }
  
  
      void start()
      {
          int n = s.nextInt();
            int mat[][] = new int[n][5];
          for (int i = 0; i < n; i++)
          {
              for (int j = 0; j < 5; j++)
              {
                  mat[i][j] = s.nextInt();
              }
          }
  
          ArrayList<Integer> arr  = new ArrayList<>();
          int ans = 1;
  
          for (int i = 0; i < 5; i++)
          {
              arr.add(mat[0][i]);
          }
  
  
          for (int i = 1; i < n; i++)
          {
              int cnt = 0;
              for (int j = 0; j < 5; j++)
              {
                  if (mat[i][j] > arr.get(j))
                      cnt++;
                  else
                      cnt--;
              }
  
  
              if (cnt < 0)
              {
                  arr.clear();
                  for (int j = 0; j < 5; j++)
                  {
                      arr.add(mat[i][j]);
                  }
                  ans = i + 1;
              }
          }
  
          for (int i = 0; i < n; i++)
          {
              int cnt = 0;
              for (int j = 0; j < 5; j++)
              {
                  if (arr.get(j) <= mat[i][j])
                      cnt++;
                  else
                      cnt--;
              }
              if (cnt <= 0)
              {
                  ans = -1;
                  break;
              }
  
          }
          System.out.println(ans);
  
  
  
      }
  
  
  
  
  
  
      long gcd(long a, long b)
      {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
      long power(long x, long y, long p)
      {
          long res = 1; // Initialize result
  
          // Update x if it is more
          // than or equal to p
          x = x % p;
  
          while (y > 0)
          {
              // If y is odd, multiply
              // x with the result
              if ((y & 1) > 0)
                  res = (res * x) % p;
  
              // y must be even now
              y = y >> 1; // y = y/2
              x = (x * x) % p;
          }
          return res;
      }
  
      int lower_bound(int [] arr , int key)
      {
  
          int i = 0;
          int j = arr.length-1;
            //if(arr[i] > key)return -1;
           if(arr[j] < key)return -1;
  
          while(i<j)
          {
              int mid = (i+j)/2;
  
              if(arr[mid] == key)
              {
                  j = mid;
              }
  
              else if(arr[mid] < key)
              {
                  i = mid+1;
              }
              else
                  j = mid-1;
        }
  
          return i;
      }
  
      int upper_bound(int [] arr , int key)
      {
  
          int i = 0;
          int j = arr.length-1;
          if(arr[j] <= key)return j+1;
  
          while(i<j)
          {
              int mid = (i+j)/2;
  
  
  
               if(arr[mid] <= key)
              {
                  i = mid+1;
              }
              else
                  j = mid;
          }
  
          return i;
      }
  
  
      static void sort(int[] A){
          int n = A.length;
          Random rnd = new Random();
          for(int i=0; i<n; ++i){
              int tmp = A[i];
              int randomPos = i + rnd.nextInt(n-i);
              A[i] = A[randomPos];
              A[randomPos] = tmp;
          }
          Arrays.sort(A);
      }
      static void sort(long[] A){
          int n = A.length;
          Random rnd = new Random();
          for(int i=0; i<n; ++i){
              long tmp = A[i];
              int randomPos = i + rnd.nextInt(n-i);
              A[i] = A[randomPos];
              A[randomPos] = tmp;
          }
          Arrays.sort(A);
      }
      static String sort(String s) {
          Character ch[]=new Character[s.length()];
          for(int i=0;i<s.length();i++) {
              ch[i]=s.charAt(i);
          }
          Arrays.sort(ch);
          StringBuffer st=new StringBuffer("");
          for(int i=0;i<s.length();i++) {
              st.append(ch[i]);
          }
          return st.toString();
      }
  
      //long array input
      public long [] longArr(int len)
      {
          // long arr input
          long [] arr = new long[len];
          String [] strs =  s.nextLine().trim().split("\\s+");
          for(int i =0; i<len; i++)
          {
              arr[i] = Long.parseLong(strs[i]);
  
          }
          return arr;
      }
      // int arr input
      public int [] intArr(int len)
      {
          // long arr input
          int [] arr = new int[len];
          String [] strs =  s.nextLine().trim().split("\\s+");
          for(int i =0; i<len; i++)
          {
              arr[i] = Integer.parseInt(strs[i]);
  
          }
          return arr;
      }
  
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          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;
          }
      }
  
  
  
  }
  