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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > a(n, vector<int>(m));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  vector<pair<int, int> > ans(1 << m, pair<int, int>(-1, -1));
  for (int mask = 1; mask < (1 << m); mask++) {
    for (int i = 0; i < n; i++) {
      int mn = INT_MAX;
      for (int bit = 0; bit < m; bit++) {
        if ((mask >> bit) & 1) {
          mn = min(mn, a[i][bit]);
        }
      }
      ans[mask] = max(ans[mask], pair<int, int>(mn, i));
    }
  }
  int ones = (1 << m) - 1;
  int best = ans[ones].first;
  pair<int, int> cand(ans[ones].second, ans[ones].second);
  for (int mask1 = 1; mask1 < ones; mask1++) {
    int mask2 = ones - mask1;
    if (min(ans[mask1].first, ans[mask2].first) > best) {
      best = min(ans[mask1].first, ans[mask2].first);
      cand = pair<int, int>(ans[mask1].second, ans[mask2].second);
    }
  }
  cout << cand.first + 1 << ' ' << cand.second + 1 << '\n';
  return (0);
}