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;
vector<pair<long long, pair<long long, long long>>> v;
map<long long, long long> mp;
long long po[300005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, x;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> x;
      v.push_back({x, {i, pow(2, j)}});
    }
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  long long p = pow(2, m) - 1;
  int i1 = -1, i2 = -1;
  for (int i = 0; i < v.size(); i++) {
    po[v[i].second.first] ^= v[i].second.second;
    x = p ^ po[v[i].second.first];
    if (mp.find(x) != mp.end() || x == 0) {
      i1 = v[i].second.first;
      if (x == 0)
        i2 = i1;
      else
        i2 = mp[x];
      break;
    }
    mp[po[v[i].second.first]] = v[i].second.first;
  }
  cout << i1 + 1 << " " << i2 + 1;
}