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 mat[300005][10];
int maskExists[305];
pair<int, int> possible(int v) {
  for (int i = 0; i < 305; i++) maskExists[i] = 0;
  for (int i = 1; i <= n; i++) {
    int mask = 0;
    for (int j = 1; j <= m; j++) mask = (mask << 1) + (mat[i][j] >= v);
    maskExists[mask] = i;
  }
  for (int mask1 = 0; mask1 < (1 << m); mask1++) {
    if (maskExists[mask1]) {
      for (int mask2 = 0; mask2 < (1 << m); mask2++) {
        if (maskExists[mask2]) {
          if ((mask1 | mask2) == (1 << m) - 1)
            return {maskExists[mask1], maskExists[mask2]};
        }
      }
    }
  }
  return {0, 0};
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> mat[i][j];
  int ans = 0;
  pair<int, int> ansIdx = {1, 1};
  for (int step = (1 << 30); step; step >>= 1) {
    pair<int, int> tmp = possible(ans + step);
    if (tmp.first != 0) {
      ans += step;
      ansIdx = tmp;
    }
  }
  cout << ansIdx.first << ' ' << ansIdx.second;
}