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>
const int MAX = 3e5 + 5;
using namespace std;
int ara[MAX][10];
pair<int, int> rec[10000];
int n, m;
pair<int, int> solve(int tgt) {
  memset(rec, -1, sizeof rec);
  for (int i = 0; i < n; i++) {
    int mask = 0;
    for (int j = 0; j < m; j++) {
      if (ara[i][j] >= tgt) mask |= (1 << j);
    }
    rec[mask] = {1, i};
  }
  int mm = (1 << m);
  for (int i = 0; i < mm; i++) {
    if (rec[i] == pair<int, int>(-1, -1)) continue;
    for (int j = 0; j < mm; j++) {
      if (rec[j] == pair<int, int>(-1, -1)) continue;
      if ((i | j) != mm - 1) continue;
      return {rec[i].second + 1, rec[j].second + 1};
    }
  }
  return {-1, -1};
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cin >> ara[i][j];
  }
  int lo = 0, hi = 1e9 + 10;
  pair<int, int> ans = {-1, -1};
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    pair<int, int> ret = solve(mid);
    if (ret == pair<int, int>(-1, -1))
      hi = mid - 1;
    else {
      ans = ret;
      lo = mid + 1;
    }
  }
  cout << ans.first << " " << ans.second << '\n';
  return 0;
}