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 n, m;
int ansx, ansy;
int a[300005][10];
int b[300005];
int check(long long x) {
  memset(b, 0, sizeof(b));
  for (int i = 1; i <= n; i++) {
    int p = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= x) p += (1 << (j - 1));
    }
    b[p] = i;
  }
  int k = (1 << m);
  int c = (1 << m) - 1;
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= k; j++) {
      if (b[i] && b[j] && ((i | j) == c)) {
        ansx = b[i];
        ansy = b[j];
        return 1;
      }
    }
  }
  return 0;
}
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]);
    }
  }
  long long l = 0, r = 1e9 + 7;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (check(mid)) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  printf("%d %d\n", ansx, ansy);
}