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 A[300010][9];
pair<int, int> mx[257];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cin >> A[i][j];
  }
  mx[0] = make_pair(2 * 1e9, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (1 << m); j++) {
      pair<int, int> maxi = make_pair(2 * 1e9, -1);
      for (int k = 0; k < m; k++) {
        if (j & (1 << k)) {
          maxi = min(maxi, make_pair(A[i][k], i));
        }
      }
      mx[j] = max(mx[j], maxi);
    }
  }
  pair<int, pair<int, int> > res = make_pair(-1, make_pair(-1, -1));
  for (int i = 0; i < (1 << m); i++) {
    res = max(
        res, make_pair(min(mx[i].first, mx[i ^ ((1 << m) - 1)].first),
                       make_pair(mx[i].second, mx[i ^ ((1 << m) - 1)].second)));
  }
  cout << res.second.first + 1 << " " << res.second.second + 1 << '\n';
}