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 n, m, a[300000 + 100][10];
pair<long long, long long> dp[300];
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (long long i = 1; i < n + 1; i++) {
    for (long long j = 0; j < m; j++) cin >> a[i][j];
    for (long long mask = 0; mask < (1 << m); mask++) {
      long long mn = 1000000000000000000;
      for (long long k = 0; k < m; k++) {
        if (mask & (1 << k)) {
          mn = min(mn, a[i][k]);
        }
      }
      if (dp[mask].first <= mn) {
        dp[mask] = {mn, i};
      }
    }
  }
  long long ans = 0, ind1, ind2;
  for (long long mask = 0; mask < 1 << m; mask++) {
    long long rev = 0;
    for (long long j = 0; j < m; j++)
      if (!(mask & (1 << j))) rev += (1 << j);
    if (min(dp[mask].first, dp[rev].first) >= ans) {
      ans = min(dp[mask].first, dp[rev].first);
      ind1 = dp[mask].second;
      ind2 = dp[rev].second;
    }
  }
  cout << ind1 << " " << ind2;
}