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 int n, m, a, b;
long long int fn(vector<vector<long long int>> &A, long long int mid) {
  long long int i, j;
  vector<long long int> mask((1 << m), -1);
  for (i = 0; i < n; i++) {
    long long int ans = 0;
    for (j = 0; j < m; j++) {
      if (A[i][j] >= mid) ans = ans ^ (1 << j);
    }
    mask[ans] = i + 1;
  }
  if (mask[(1 << m) - 1] != -1) {
    a = mask[(1 << m) - 1];
    b = mask[(1 << m) - 1];
    return (1);
  }
  for (i = 0; i < (1 << m); i++) {
    for (j = 0; j < (1 << m); j++) {
      if (mask[i] != -1 && mask[j] != -1 && (i | j) == (1 << m) - 1) {
        a = mask[i];
        b = mask[j];
        return (1);
      }
    }
  }
  return (0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  long long int i, j, low = 1e11, mid, high = -1;
  cin >> n >> m;
  vector<vector<long long int>> A(n);
  vector<long long int> temp(m);
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> temp[j];
      high = max(high, temp[j]);
      low = min(low, temp[j]);
    }
    A[i] = temp;
  }
  while (low <= high) {
    mid = (low + high) / 2;
    if (fn(A, mid) == 1)
      low = mid + 1;
    else
      high = mid - 1;
  }
  cout << a << " " << b;
  return (0);
}