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 arr[300000][8];
long long a, b;
bool f(long long x, long long n, long long m) {
  vector<int> bitmasks(1 << m, -1);
  for (int i = 0; i < n; i++) {
    int bitmask = 0;
    for (int j = 0; j < m; j++) {
      if (arr[i][j] >= x) bitmask += (1 << (m - (j + 1)));
    }
    bitmasks[bitmask] = i + 1;
  }
  if (bitmasks.back() != -1) {
    a = b = bitmasks[(1 << m) - 1];
    return 1;
  }
  for (int i = 0; i < (1 << m); i++)
    for (int j = 0; j < (1 << m); j++)
      if (bitmasks[i] != -1 && bitmasks[j] != -1 &&
          ((i | j) == ((1 << m) - 1))) {
        a = bitmasks[i];
        b = bitmasks[j];
        return 1;
      }
  return 0;
}
int main() {
  cin >> n >> m;
  long long maxi = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
      maxi = max(maxi, arr[i][j]);
    }
  long long lo = 0;
  long long hi = maxi;
  a = n;
  b = n;
  while (lo < hi) {
    long long x = lo + (hi - lo + 1) / 2;
    if (f(x, n, m))
      lo = x;
    else
      hi = x - 1;
  }
  cout << a << " " << b << endl;
}