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 N = 3e5 + 10;
const int mod = 1e9 + 7;
int n, m;
int a[N][10];
int h[N], g[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cin >> a[i][j];
    }
  }
  int l = 0, r = 1e9 + 1;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    bool fl = 0;
    for (int j = 0; j < (1 << m); ++j) g[j] = 0;
    for (int i = 1; i <= n; ++i) {
      h[i] = 0;
      for (int j = 1; j <= m; ++j) {
        if (a[i][j] >= mid) h[i] |= (1 << (j - 1));
      }
      g[h[i]] = i;
      for (int j = 0; j < (1 << m); ++j) {
        if ((h[i] | j) == ((1 << m) - 1) && g[j] != 0) {
          fl = 1;
          break;
        }
      }
    }
    if (!fl)
      r = mid;
    else
      l = mid;
  }
  int mid = l;
  for (int i = 1; i <= n; ++i) {
    h[i] = 0;
    for (int j = 1; j <= m; ++j) {
      if (a[i][j] >= mid) h[i] |= (1 << (j - 1));
    }
    g[h[i]] = i;
    for (int j = 0; j < (1 << m); ++j) {
      if ((h[i] | j) == ((1 << m) - 1) && g[j] != 0) {
        cout << i << ' ' << g[j];
        return 0;
      }
    }
  }
  return 0;
}