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;
void minimax(int n, int m, vector<vector<int> >& a) {
  int p1, p2, low, mid, high;
  low = 0;
  high = 1e9;
  function<bool(int)> is_valid = [&](int x) {
    int valid = (1 << m) - 1;
    unordered_set<int> masks;
    map<int, int> pos;
    for (int i = 0; i < n; i++) {
      int mask = 0;
      for (int j = 0; j < m; j++) {
        mask <<= 1;
        mask |= a[i][j] >= x;
      }
      pos[mask] = i;
      masks.insert(mask);
    }
    for (auto it1 = masks.begin(); it1 != masks.end(); it1++)
      for (auto it2 = masks.begin(); it2 != masks.end(); it2++)
        if ((*it1 | *it2) == valid) {
          p1 = pos[*it1] + 1;
          p2 = pos[*it2] + 1;
          return true;
        }
    return false;
  };
  while (low <= high) {
    mid = (low + high) / 2;
    if (is_valid(mid))
      low = mid + 1;
    else
      high = mid - 1;
  }
  cout << p1 << " " << p2 << endl;
}
int main() {
  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];
  minimax(n, m, a);
  return 0;
}