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 inf = 1e9;
const long long maxn = 3e5 + 10;
const long long maxm = 10;
long long a[maxn][maxm];
long long n, m;
pair<long long, long long> f(long long first) {
  map<long long, long long> mp;
  long long ones = (1 << m) - 1;
  for (long long i = 0; i < n; i++) {
    long long mask = 0;
    for (long long j = 0; j < m; j++) {
      mask |= ((1 << j) * (a[i][j] >= first));
    }
    if (!mp[mask]) {
      for (long long s = mask;; s = (s - 1) & mask) {
        mp[s] = i + 1;
        if (s == 0) break;
      }
      mp[mask] = i + 1;
    }
    if (mp[mask ^ ones]) {
      return {mp[mask ^ ones], i + 1};
    }
  }
  return {-1, -1};
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long l = 0, r = inf + 1, mid;
  while (r - l > 1) {
    mid = l + (r - l) / 2;
    if (f(mid).first != -1) {
      l = mid;
    } else {
      r = mid;
    }
  }
  auto ans = f(l);
  cout << ans.first << ' ' << ans.second << '\n';
  return 0;
}