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[300005][10];
long long mask[1 << 10];
pair<long long int, long long int> check(long long x) {
  pair<long long int, long long int> p = {-1, -1};
  memset(mask, -1, sizeof(mask));
  for (long long int i = 1; i < n + 1; i++) {
    long long c = 0;
    for (long long int j = 1; j < m + 1; j++) {
      if (arr[i][j] >= x) c += (1 << (m - j));
    }
    mask[c] = i;
  }
  for (long long int i = 0; i < (1 << m); i++)
    for (long long int j = 0; j < (1 << m); j++) {
      if (mask[i] != -1 && mask[j] != -1 && ((i | j) == ((1 << m) - 1))) {
        p.first = mask[i];
        p.second = mask[j];
        return p;
      }
    }
  return p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    cin >> n >> m;
    long long mini = 1000000007;
    long long maxi = -1;
    for (long long int i = 0; i < n; i++)
      for (long long int j = 0; j < m; j++) {
        cin >> arr[i + 1][j + 1];
        mini = min(mini, arr[i + 1][j + 1]);
        maxi = max(maxi, arr[i + 1][j + 1]);
      }
    long long l = mini, r = maxi, m;
    long long x = -1, y = -1;
    while (l <= r) {
      m = (l + r) >> 1;
      pair<long long int, long long int> p = check(m);
      if (p.first != -1 && p.second != -1) {
        x = p.first;
        y = p.second;
        l = m + 1;
      } else
        r = m - 1;
    }
    cout << x << " " << y << '\n';
  }
  return 0;
}