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 inf = 1 << 30;
long long linf = (long long)1 << 62;
pair<int, int> verify(int val, vector<vector<int> >& data) {
  vector<int> idx(1 << data[0].size(), -1);
  int mask = 1 << data[0].size();
  mask--;
  for (int k = 0; k < data.size(); k++) {
    auto v = data[k];
    int currentmask = 0;
    for (int i = 0; i < v.size(); i++) {
      currentmask |= (v[i] >= val) << i;
    }
    idx[currentmask] = k + 1;
  }
  for (int i = 0; i < idx.size(); i++) {
    for (int j = 0; j < idx.size(); j++) {
      if ((i | j) == mask && idx[i] != -1 && idx[j] != -1) {
        return {idx[i], idx[j]};
      }
    }
  }
  return {-1, -1};
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > data(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> data[i][j];
    }
  }
  int lo = 0, hi = inf;
  int mid;
  while (lo < hi) {
    mid = lo + (hi - lo + 1) / 2;
    auto res = verify(mid, data);
    ;
    if (res.first != -1) {
      lo = mid;
      ;
    } else {
      ;
      hi = mid - 1;
    }
  }
  auto res = verify(lo, data);
  cout << res.first << " " << res.second << endl;
}