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;
vector<vector<int> > a;
vector<int> vis;
int n, m;
int first, second;
bool check(int mid) {
  fill(vis.begin(), vis.end(), 0);
  for (int i = 1; i <= n; ++i) {
    int cur = 0;
    for (int j = 1; j <= m; ++j) {
      if (a[i][j] >= mid) cur |= (1 << (j - 1));
    }
    vis[cur] = i;
  }
  for (int i = 0; i < (1 << m); ++i) {
    for (int j = 0; j < (1 << m); ++j) {
      if (vis[i] && vis[j] && (i | j) == (1 << m) - 1) {
        first = vis[i];
        second = vis[j];
        return true;
      }
    }
  }
  return false;
}
void solve() {
  cin >> n >> m;
  a.resize(n + 1, vector<int>(m + 1));
  vis.resize((1 << m) + 10);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) cin >> a[i][j];
  }
  int l = 0, r = (int)1e9 + 10;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << first << " " << second << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}