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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 3e5 + 10;
int n, m;
int dp[1 << 9];
int id[1 << 9];
int g[maxn][10];
int ans = -1;
pair<int, int> inds;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int lo = 1e9 + 10;
    for (int j = 0; j < m; j++) {
      cin >> g[i][j];
      lo = min(g[i][j], lo);
    }
    if (lo > ans) {
      ans = lo;
      inds = make_pair(i, i);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int mask = 1; mask < (1 << m); mask++) {
      int lo = 1e9 + 10;
      for (int j = 0; j < m; j++) {
        if (mask >> j & 1) {
          lo = min(g[i][j], lo);
        }
      }
      int o = ((1 << m) - 1) & (~mask);
      lo = min(lo, dp[o]);
      if (id[o] && lo > ans) {
        ans = lo;
        inds = make_pair(id[o], i);
      }
    }
    for (int mask = 1; mask < (1 << m); mask++) {
      int lo = 1e9 + 10;
      for (int j = 0; j < m; j++) {
        if (mask >> j & 1) {
          lo = min(g[i][j], lo);
        }
      }
      if (lo > dp[mask]) {
        dp[mask] = lo;
        id[mask] = i;
      }
    }
  }
  cerr << ans << endl;
  cout << inds.first << " " << inds.second << endl;
  return 0;
}