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;
int arr[300000][8];
int n, m;
int a, b;
void bin_search(int lb, int ub) {
  int pos[256];
  if (ub - lb < 2) {
    return;
  }
  int next;
  next = (lb + ub) / 2;
  memset(pos, -1, sizeof(pos));
  for (int i = 0; i < n; i++) {
    int mask = 0;
    for (int j = 0; j < m; j++) {
      if (arr[i][j] >= next) {
        mask |= 1 << j;
      }
    }
    pos[mask] = i;
  }
  for (int i = (1 << m) - 1; i >= 0; i--) {
    if (pos[i] != -1) {
      for (int j = 0; j < m; j++) {
        pos[i & (~(1 << j))] = pos[i];
      }
    }
  }
  bool valid = false;
  for (int i = 0; i < (1 << m); i++) {
    if (pos[i] != -1) {
      int another;
      another = ~i & ((1 << m) - 1);
      if (pos[~i & ((1 << m) - 1)] != -1) {
        a = pos[i];
        b = pos[another];
        valid = true;
        break;
      }
    }
  }
  if (valid) {
    bin_search(next, ub);
  } else {
    bin_search(lb, next);
  }
}
int main() {
  scanf("%lld %lld", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &arr[i][j]);
    }
  }
  if (n == 1) {
    printf("1 1\n");
    return 0;
  }
  bin_search(0, 1000000001);
  printf("%d %d\n", a + 1, b + 1);
  return 0;
}