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 N = 3e5 + 5;
long long n, m, arr[N][9], Mask[1 << 8];
pair<long long, long long> ans;
bool cool(long long x) {
  memset(Mask, -1, sizeof(Mask));
  for (long long i = 0; i < n; i++) {
    long long mask = 0;
    for (long long w = 0; w < m; w++) {
      if (arr[i][w] >= x) {
        mask |= (1LL << w);
      }
    }
    Mask[mask] = i;
  }
  for (long long i = 0; i < (1LL << m); i++) {
    for (long long w = 0; w < (1LL << m); w++) {
      if (Mask[i] != -1 && Mask[w] != -1) {
        long long k = i | w;
        if (k == (1LL << m) - 1) {
          ans = make_pair(Mask[i] + 1, Mask[w] + 1);
          return true;
        }
      }
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long w = 0; w < m; w++) cin >> arr[i][w];
  }
  long long lo = 0, hi = 1e9, mid;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    if (cool(mid)) {
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  if (ans.first > ans.second) swap(ans.first, ans.second);
  cout << (long long)ans.first << " " << (long long)ans.second;
}