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 = (int)3e5;
int n, m;
long long val[(1 << 8) + 5], koji[(1 << 8) + 5];
long long sol = -1;
pair<int, int> res = make_pair(1, 1);
vector<int> v;
const long long inf = (long long)1e9 + 5;
void solve(long long mask, long long first, long long sad, int idx) {
  if (first == m) {
    sad = min(sad, val[mask]);
    if (koji[mask] && sad > sol) {
      sol = sad;
      res = make_pair(idx, koji[mask]);
    }
    return;
  }
  solve(mask, first + 1, min(sad, (long long)v[first]), idx);
  solve(mask | (1 << first), first + 1, sad, idx);
}
void procesiraj(long long mask, long long first, long long sad, int idx) {
  if (first == m) {
    if (sad >= val[mask]) {
      val[mask] = sad;
      koji[mask] = idx;
    }
    return;
  }
  procesiraj(mask, first + 1, sad, idx);
  procesiraj(mask | (1 << first), first + 1, min(sad, (long long)v[first]),
             idx);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    v.clear();
    for (int j = 0; j < m; ++j) {
      int first;
      scanf("%d", &first);
      v.push_back(first);
    }
    solve(0, 0, inf, i);
    procesiraj(0, 0, inf, i);
  }
  printf("%d %d", res.first, res.second);
  return 0;
}