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 n, m, x, mx, p1, p2;
map<int, int> mp;
vector<int> v[300005];
bool check(int pos) {
  mp.clear();
  for (int i = 1; i <= n; i++) {
    int val = 0;
    for (int j = 0; j < m; j++) {
      if (v[i][j] >= pos) val += (1 << j);
    }
    if (mp.find(val) == mp.end()) {
      for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
        if ((it->first) + val - ((it->first) & val) == (1 << m) - 1) {
          p1 = it->second;
          p2 = i;
          return true;
        }
      }
      if (val == (1 << m) - 1) {
        p1 = i;
        p2 = i;
        return true;
      }
      mp[val] = i;
    }
  }
  return false;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      cin >> x;
      v[i].push_back(x);
      mx = max(mx, x);
    }
  int st = 0;
  int dr = mx + 1;
  while (st < dr) {
    int mid = (st + dr) / 2;
    if (check(mid))
      st = mid + 1;
    else
      dr = mid;
  }
  cout << p1 << " " << p2;
  return 0;
}