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;
int ans1, ans2;
int m, n;
bool ok(long long int x, vector<vector<long long int>> &a) {
  vector<int> mask((1 << m) + 1, -1);
  for (int i = 0; i < n; i++) {
    int here = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= x) {
        here ^= (1 << j);
      }
    }
    mask[here] = i;
  }
  if (mask[(1 << m) - 1] != -1) {
    ans1 = mask[(1 << m) - 1], ans2 = mask[(1 << m) - 1];
    return true;
  }
  for (int i = 0; i < (1 << m); i++) {
    for (int j = 0; j < (1 << m); j++) {
      if (mask[i] != -1 && mask[j] != -1 && (i | j) == ((1 << m) - 1)) {
        ans1 = mask[i], ans2 = mask[j];
        return true;
      }
    }
  }
  return false;
}
void test_case() {
  cin >> n >> m;
  vector<vector<long long int>> a(n, vector<long long int>(m));
  for (vector<long long int> &tmp : a) {
    for (long long int &x : tmp) {
      cin >> x;
    }
  }
  long long int l = 0, r = 1e9 + 1;
  while (l + 1 < r) {
    long long int mid = (l + r) / 2;
    if (ok(mid, a)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << ans1 + 1 << ' ' << ans2 + 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    test_case();
  }
  return 0;
}