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;
long long n, m;
long long **a;
long long *bit;
bool *v;
long long rst_a = 0, rst_b = 0;
long long *master;
void func(long long left, long long right) {
  long long mid = (left + right) / 2;
  long long i, j;
  for (i = 0; i < (1 << m); i++) v[i] = false;
  for (i = 0; i < n; i++) bit[i] = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] >= mid) {
        bit[i] |= (1 << j);
      }
    }
    v[bit[i]] = true;
    master[bit[i]] = i;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 1 << m; j++) {
      if (v[j] && v[bit[i]] && (j | bit[i]) == (1 << m) - 1) {
        rst_a = i;
        rst_b = master[j];
        if (left == right) {
          return;
        }
        func(mid + 1, right);
        return;
      }
    }
  }
  if (left == right) {
    return;
  }
  func(left, mid);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j;
  cin >> n >> m;
  a = new long long *[n];
  for (i = 0; i < n; i++) a[i] = new long long[m];
  bit = new long long[n];
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) cin >> a[i][j];
  v = new bool[10000];
  master = new long long[10000];
  func(0, 10000000000LL);
  cout << rst_a + 1 << ' ' << rst_b + 1 << '\n';
  return 0;
}