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 + 20, M = 10;
int n, m, a[N][M], vis[1 << M], w[M], ans1, ans2;
bool fun(int v) {
  int en = 1 << m;
  for (int i = 0; i < en; i++) vis[i] = -1;
  for (int i = 1; i <= n; i++) {
    int t = 0;
    for (int j = 1; j <= m; j++)
      if (a[i][j] >= v) t += w[j];
    vis[t] = i;
  }
  for (int i = 0; i < en; i++)
    for (int j = i; j < en; j++)
      if (vis[i] != -1 && vis[j] != -1 && (i | j) == en - 1) {
        ans1 = vis[i], ans2 = vis[j];
        return 1;
      }
  return 0;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  w[1] = 1;
  for (int i = 2; i < M; i++) w[i] = w[i - 1] << 1;
  int l = 0, r = 1e9 + 10;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (fun(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << ans1 << ' ' << ans2 << endl;
  return 0;
}