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 long long NN = 2e5 + 5;
long long n, m;
long long a[303030][9], K, x[303030];
long long jaw1, jaw2;
long long cek(long long aa) {
  for (long long i = 0; i <= K; i++) {
    x[i] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    long long tem = 0;
    for (long long j = 0; j < m; j++)
      if (a[i][j + 1] >= aa) tem += (1 << j);
    x[tem] = i;
  }
  for (long long i = 0; i <= K; i++)
    for (long long j = i; j <= K; j++) {
      if (((i | j) == K) && (x[i] != 0) && (x[j] != 0)) {
        jaw1 = x[i];
        jaw2 = x[j];
        return 1;
      }
    }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) cin >> a[i][j];
  K = (1 << m) - 1;
  long long L = 0, R = 1e9, H;
  while (L <= R) {
    long long C = (L + R) / 2;
    if (cek(C)) {
      H = C;
      L = C + 1;
    } else
      R = C - 1;
  }
  cek(H);
  cout << jaw1 << " " << jaw2 << "\n";
}