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 = 300050;
int n, m, fullst;
int ansi, ansj;
int a[N][10];
int vis[1000];
inline bool check(int x) {
  memset(vis, 0, sizeof(vis));
  int allst[N] = {0};
  for (int i = 1; i <= n; ++i) {
    int nowst = 0;
    for (int j = 1; j <= m; ++j) {
      if (a[i][j] >= x) nowst |= (1 << (j - 1));
    }
    allst[i] = nowst;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < fullst; ++j) {
      if ((j | allst[i]) == allst[i]) {
        vis[j] = i;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[fullst ^ allst[i]]) {
      ansi = i;
      ansj = vis[fullst ^ allst[i]];
      return true;
    }
  }
  return false;
}
int main() {
  scanf("%d%d", &n, &m);
  fullst = (1 << m) - 1;
  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 mid = (L + R) >> 1;
    if (check(mid))
      L = mid + 1;
    else
      R = mid - 1;
  }
  printf("%d %d\n", ansi, ansj);
  return 0;
}