Problem: You are given n arrays a_1, a_2, ..., a_n; each array consists of exactly m integers. We denote the y-th element of the x-th array as a_{x, y}.

You have to choose two arrays a_i and a_j (1 ≤ i, j ≤ n, it is possible that i = j). After that, you will obtain a new array b consisting of m integers, such that for every k ∈ [1, m] b_k = max(a_{i, k}, a_{j, k}).

Your goal is to choose i and j so that the value of min _{k = 1}^{m} b_k is maximum possible.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 8) — the number of arrays and the number of elements in each array, respectively.

Then n lines follow, the x-th line contains the array a_x represented by m integers a_{x, 1}, a_{x, 2}, ..., a_{x, m} (0 ≤ a_{x, y} ≤ 10^9).

Output

Print two integers i and j (1 ≤ i, j ≤ n, it is possible that i = j) — the indices of the two arrays you have to choose so that the value of min _{k = 1}^{m} b_k is maximum possible. If there are multiple answers, print any of them.

Example

Input


6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0


Output


1 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long oo = 1e18;
const long long MAXN = 3e5 + 100;
long long N, M;
long long A[MAXN][10];
bool mk[MAXN][10];
long long arr[300];
long long tipo[300];
long long ta, tb;
bool f(long long mid) {
  for (long long i = 0; i < 300; i++) arr[i] = false;
  for (long long i = 0; i < N; i++) {
    long long num = 0;
    for (long long j = 0; j < M; j++) {
      if (A[i][j] >= mid)
        mk[i][j] = true;
      else
        mk[i][j] = false;
      num += (1 << j) * mk[i][j];
    }
    arr[num] = true;
    tipo[num] = i;
  }
  for (long long i = 0; i < 300; i++) {
    for (long long j = 0; j < 300; j++) {
      if (arr[j] && ((i & j) == i)) arr[i] = true, tipo[i] = tipo[j];
    }
  }
  bool flag = false;
  for (long long i = 0; i < N; i++) {
    long long num = 0;
    long long temp = 0;
    for (long long j = 0; j < M; j++) {
      temp += (1 << j) * mk[i][j];
      num += (1 << j) * (!mk[i][j]);
    }
    if (arr[num]) flag = true, ta = tipo[temp], tb = tipo[num];
  }
  return flag;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < M; j++) {
      cin >> A[i][j];
    }
  }
  long long a = 0, b = 1e9;
  while (a <= b) {
    long long mid = (a + b) / 2;
    if (f(mid)) {
      a = mid + 1;
    } else {
      b = mid - 1;
    }
  }
  cout << ta + 1 << " " << tb + 1 << '\n';
  return 0;
}