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: #include <bits/stdc++.h>
  using namespace std;
  const int32_t M = 1e9 + 7;
  int32_t mod_inv(int32_t a, int32_t m) {
    int32_t g = m, r = a, x = 0, y = 1;
    while (r != 0) {
      int32_t q = g / r;
      g %= r;
      swap(g, r);
      x -= q * y;
      swap(x, y);
    }
    return x < 0 ? x + m : x;
  }
  int gcdi(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcdi(b, a % b);
  }
  long long gcdl(long long a, long long b) {
    if (b == 0) {
      return a;
    }
    return gcdl(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a / gcdl(a, b)) * b; }
  int factorial(int n) {
    if (n == 0) return 1;
    return n * factorial(n - 1);
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  void FAST() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  void swap_ll(long long *a, long long *b) {
    long long c;
    c = *a;
    *a = *b;
    *b = c;
  }
  void swap_int(int *a, int *b) {
    int c;
    c = *a;
    *a = *b;
    *b = c;
  }
  void swap_char(char *a, char *b) {
    char c;
    c = *a;
    *a = *b;
    *b = c;
  }
  bool mycomp(pair<int, int> &a, pair<int, int> &b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first < b.first;
  }
  long long fast_pow(long long a, long long p) {
    long long res = 1;
    while (p) {
      if (p % 2 == 0) {
        a = a * 1ll * a % 000000007;
        p /= 2;
      } else {
        res = res * 1ll * a % 000000007;
        p--;
      }
    }
    return res;
  }
  long long fact(long long n) {
    long long res = 1;
    for (long long i = 1; i <= n; i++) {
      res = res * 1ll * i % 000000007;
    }
    return res;
  }
  long long nCk(long long n, long long k) {
    return fact(n) * 1ll * fast_pow(fact(k), 000000007 - 2) % 000000007 * 1ll *
           fast_pow(fact(n - k), 000000007 - 2) % 000000007;
  }
  long long firstDigit(long long n) {
    long long digits = (long long)log10(n);
    n = (long long)(n / pow(10, digits));
    return n;
  }
  long long find(long long n, long long d) {
    while (n > 0) {
      if (n % 10 == d) return 1;
      n /= 10;
    }
    return 0;
  }
  struct node {
    int data;
    node *next;
    node(int d) {
      data = d;
      next = NULL;
    }
  };
  void ffin() { FAST(); }
  long long add(vector<long long> a, long long b, long long e) {
    long long sum = 0;
    for (long long i = b; i < e; i++) {
      sum += a[i];
    }
    return sum;
  }
  void decToBinary(int n) {
    int binaryNum[32];
    int i = 0;
    while (n > 0) {
      binaryNum[i] = n % 2;
      n = n / 2;
      i++;
    }
    for (int j = i - 1; j >= 0; j--) cout << binaryNum[j];
  }
  bool isInteger(double N) {
    int X = N;
    double temp2 = N - X;
    if (temp2 > 0) {
      return false;
    }
    return true;
  }
  bool check(vector<int> &v) {
    int n = (int)((v).size());
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        for (int k = j + 1; k < n; k++) {
          if ((v[i] >= v[j] && v[j] >= v[k]) || (v[i] <= v[j] && v[j] <= v[k])) {
            return true;
          }
        }
      }
    }
    return false;
  }
  bool digitscheck(long long n) {
    while (n != 0) {
      long long y = n % 10;
      if (y != 0 and y != 1) {
        return false;
      }
      n = n / 10;
    }
    return true;
  }
  const long long N = 5e+5;
  vector<vector<long long>> a(N, vector<long long>(5));
  long long better(long long i, long long j) {
    long long cnt = 0;
    for (int k = 0; k < 5; k++) {
      cnt += (a[i][k] < a[j][k]);
    }
    return (cnt >= 3);
  }
  void solve() {
    long long n;
    cin >> n;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 5; j++) cin >> a[i][j];
    long long best = 0;
    for (int i = 1; i < n; i++) {
      if (better(i, best)) best = i;
    }
    bool ok = true;
    for (int i = 0; i < n; i++) {
      if (i == best) continue;
      ok &= better(best, i);
    }
    if (!ok) best = -2;
    cout << best + 1 << "\n";
  }
  int main() {
    ffin();
    long long tests = 1;
    cin >> tests;
    while (tests--) {
      solve();
    }
  }
  