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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  int a[N][M];
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      cin >> a[i][j];
    }
  }
  long long l = 0, r = 1e9 + 1;
  int i1 = 1, i2 = 1;
  while (r - l > 1) {
    long long m = (l + r) >> 1;
    bool check = false;
    vector<int> vis((1 << M), 0);
    for (int i = 0; i < N; i++) {
      int bit = 0;
      for (int j = 0; j < M; j++) {
        if (a[i][j] >= m) {
          bit += (1 << j);
        }
      }
      vis[bit] = i + 1;
    }
    for (int i = 0; i < (1 << M); i++) {
      for (int j = 0; j < (1 << M); j++) {
        if (vis[i] && vis[j] && ((i | j) + 1) == (1 << M)) {
          i1 = vis[i], i2 = vis[j];
          check = true;
          break;
        }
      }
    }
    if (check)
      l = m;
    else
      r = m;
  }
  cout << i1 << " " << i2;
}