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 maxn = 3e5 + 5;
int n, m;
int cnt[maxn];
int a[maxn][10];
int num[270];
int ans1, ans2;
bool check(int cur) {
  memset(num, 0, sizeof(num));
  for (int i = 1; i <= n; i++) {
    int x = 0;
    for (int j = m - 1; j >= 0; j--) {
      if (a[i][j] >= cur) x += (1 << j);
    }
    num[x] = i;
  }
  for (int i = 0; i < (1 << m); i++) {
    for (int j = 0; j < (1 << m); j++) {
      if (num[i] != 0 && num[j] != 0 && (j | i) == (1 << m) - 1) {
        ans1 = num[i];
        ans2 = num[j];
        return true;
      }
    }
  }
  return false;
}
int main() {
  int r = -1, l = 1e9 + 10;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
      r = max(r, a[i][j]);
      l = min(l, a[i][j]);
    }
  }
  int mid;
  while (l < r) {
    mid = (l + r + 1) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  check(l);
  printf("%d %d", ans1, ans2);
  return 0;
}