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;
using ll = long long;
const int maxn = 3e5 + 5;
int n, m;
int a[maxn][10];
int b[maxn];
int valid[1 << 10];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  int lo = 0, hi = 1e9 + 9;
  pair<int, int> ret;
  while (lo + 1 < hi) {
    int mid = (lo + hi) / 2;
    bool poss = false;
    pair<int, int> ans;
    for (int j = 0; j < (1 << m); j++) valid[j] = -1;
    for (int i = 0; i < n; i++) {
      b[i] = 0;
      for (int j = 0; j < m; j++) {
        b[i] |= (a[i][j] < mid ? 0 : 1) << j;
      }
      if (valid[b[i]] == -1) {
        for (int s = b[i];; s = (s - 1) & b[i]) {
          valid[s] = i;
          if (s == 0) break;
        }
        int f = ((1 << m) - 1) ^ b[i];
        if (valid[f] != -1) {
          poss = true;
          ans = {i, valid[f]};
          break;
        }
      }
    }
    if (poss) {
      ret = ans;
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << ret.first + 1 << ' ' << ret.second + 1 << '\n';
}