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 A[303030][10];
int cnt[515];
int N, M;
int x1, x2;
bool solve(int m) {
  for (int i = 1; i <= N; i++) {
    int b = 0;
    for (int j = 1; j <= M; j++) {
      if (A[i][j] >= m) b += (1 << (j - 1));
    }
    cnt[b] = i;
  }
  bool ok = false;
  for (int i = 0; i < (1 << M); i++) {
    if (!cnt[i]) continue;
    int all = ((1 << M) - 1) ^ i;
    for (int j = 0; j < (1 << M); j++) {
      if ((all & j) == all) {
        if (cnt[j]) {
          ok = true;
          x1 = cnt[i];
          x2 = cnt[j];
        }
      }
    }
  }
  if (ok)
    return true;
  else
    return false;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= M; j++) scanf("%d", &A[i][j]);
  int l = 0, r = 1000000000;
  while (l <= r) {
    int m = l + r >> 1;
    if (solve(m))
      l = m + 1;
    else
      r = m - 1;
    for (int i = 0; i < (1 << M); i++) cnt[i] = 0;
  }
  solve(r);
  printf("%d %d\n", x1, x2);
  return 0;
}