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;
const int MOD = 1e9 + 7;
const int N = 300005;
int n, m;
int a[N][10];
int b[300], res = 0, luu[300];
bool getbit(int mask, int i) { return (mask >> (i - 1) & 1); }
int get(int pos, int mask) {
  int kq = MOD;
  for (int i = 1; i <= m; i++) {
    if (getbit(mask, i)) {
      kq = min(kq, a[pos][i]);
    }
  }
  return kq;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
    for (int mask = 0; mask < (1 << m); mask++) {
      int x = get(i, mask);
      if (b[mask] < x) {
        b[mask] = x;
        luu[mask] = i;
      }
    }
  }
  int kq1 = 1, kq2 = 1;
  for (int mask1 = 0; mask1 < (1 << m); mask1++) {
    int mask2 = ((1 << m) - 1) ^ mask1;
    if (res < min(b[mask1], b[mask2])) {
      res = min(b[mask1], b[mask2]);
      kq1 = luu[mask1];
      kq2 = luu[mask2];
    }
  }
  cout << kq1 << " " << kq2;
  return 0;
}