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 int inf = 1e9;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 10;
int n, m;
int g[1 << 9];
int p[1 << 9];
int a[maxn][11];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
    }
    for (int j = 0; j < (1 << m); j++) {
      int mn = inf;
      for (int k = 0; k < m; k++)
        if (j >> k & 1) {
          mn = min(a[i][k], mn);
        }
      if (mn > g[j]) {
        g[j] = mn;
        p[j] = i;
      }
    }
  }
  int ans = 0, x, y;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      int s = 0;
      for (int k = 0; k < m; k++) {
        if (k != j && a[i][k] < a[i][j]) s |= (1 << k);
      }
      if (g[s] >= a[i][j]) {
        if (a[i][j] >= ans) {
          ans = a[i][j];
          x = i;
          y = p[s];
        }
      }
    }
  }
  printf("%d %d\n", x, y);
  return 0;
}