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;
long long x[500], x2[500];
int n, m;
long long o[20];
long long oo = 1e15;
int pp, u;
void go(int cr, int sum, long long p) {
  if ((1 << cr) > u) {
    if (x[sum] < p && sum > 0) {
      x[sum] = p;
      x2[sum] = pp;
    }
    return;
  }
  if (x[sum] < p && sum > 0) {
    x[sum] = p;
    x2[sum] = pp;
  }
  go(cr + 1, sum, p);
  go(cr + 1, sum + (1 << cr), min(p, o[cr]));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  u = (1 << (m - 1));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> o[j];
    }
    pp = i;
    go(0, 0, oo);
  }
  u = (1 << m) - 1;
  long long mx = 0, idx = 1, idx2 = 1;
  for (int i = 0; i <= u; i++) {
    for (int j = 0; j <= u; j++) {
      if ((j | i) == u) {
        if (min(x[i], x[j]) > mx) {
          mx = min(x[i], x[j]);
          idx = x2[i];
          idx2 = x2[j];
        }
      }
    }
  }
  cout << idx << ' ' << idx2;
  return 0;
}