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::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> table(n, vector<int>(m));
  for (auto& vec : table) {
    for (auto& val : vec) cin >> val;
  }
  vector<int> min_for_mask(1 << m, 0);
  vector<int> value_for_mask(1 << m, 0);
  value_for_mask[0] = 1000000000;
  for (int mask = 1; mask < (1 << m); ++mask) {
    int cur_min = 0;
    int i = 0;
    vector<int> pos;
    pos.reserve(__builtin_popcount(mask));
    for (int p = 0; p < m; ++p) {
      if (1 << p & mask) pos.push_back(p);
    }
    int ind = 0;
    for (const auto& vec : table) {
      int mn = 1000000000;
      for (int p : pos) {
        if (vec[p] < mn) mn = vec[p];
      }
      if (mn > cur_min) {
        cur_min = mn;
        i = ind;
      }
      ++ind;
    }
    min_for_mask[mask] = i;
    value_for_mask[mask] = cur_min;
  }
  int cur_res = 0;
  pair<int, int> an = {0, 0};
  int full = (1 << m) - 1;
  for (int i = 0; i < (1 << (m - 1)); ++i) {
    int rest = full ^ i;
    int ans = min(value_for_mask[i], value_for_mask[rest]);
    if (ans > cur_res) {
      cur_res = ans;
      an = {min_for_mask[i], min_for_mask[rest]};
    }
  }
  cout << an.first + 1 << ' ' << an.second + 1;
  return 0;
}