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 main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long ar[n + 1][m + 1];
  vector<long long> v;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      cin >> ar[i][j];
      v.push_back(ar[i][j]);
    }
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  long long L = 0;
  long long R = v.size() - 1;
  long long res = LONG_LONG_MIN;
  pair<long long, long long> fin;
  while (L <= R) {
    long long temp = v[(L + R) / 2];
    map<long long, long long> ma;
    long long val = 0;
    for (long long i = 1; i <= n; i++) {
      long long score = 0;
      for (long long j = 1; j <= m; j++) {
        if (ar[i][j] >= temp) {
          score += pow(2, j - 1);
        }
        if (i == 1) val += pow(2, j - 1);
      }
      ma[score]++;
    }
    map<long long, long long>::iterator it;
    vector<long long> sc;
    for (it = ma.begin(); it != ma.end(); it++) {
      sc.push_back((*it).first);
    }
    long long flag = 0;
    pair<long long, long long> want;
    for (long long i = 0; i < sc.size(); i++) {
      for (long long j = 0; j < sc.size(); j++) {
        if ((sc[i] | sc[j]) == val) {
          want.first = sc[i];
          want.second = sc[j];
          flag = 1;
          break;
        }
      }
      if (flag) {
        break;
      }
    }
    if (flag) {
      if (temp > res) {
        res = temp;
        for (long long i = 1; i <= n; i++) {
          long long score = 0;
          for (long long j = 1; j <= m; j++) {
            if (ar[i][j] >= temp) {
              score += pow(2, j - 1);
            }
          }
          if (score == want.first) {
            fin.first = i;
          }
          if (score == want.second) {
            fin.second = i;
          }
        }
      }
      L = (L + R) / 2 + 1;
    } else {
      R = (L + R) / 2 - 1;
    }
  }
  cout << fin.first << " " << fin.second << "\n";
}