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(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  int a[n + 5][m + 5];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  vector<pair<int, int> > v;
  int x = (1 << m) - 1;
  for (int i = 1; i <= x; i++) {
    for (int j = 1; j <= x; j++) {
      if ((i | j) == x) {
        v.push_back(make_pair(i, j));
      }
    }
  }
  int l = 0, h = 1e9, ans = 0, a1 = 0, a2 = 0;
  while (l <= h) {
    int mid = (l + h) / 2;
    bool y = 0;
    int b[n + 5];
    vector<int> k[x + 5];
    for (int i = 1; i <= n; i++) {
      int p = 0;
      for (int j = 1; j <= m; j++) {
        if (a[i][j] >= mid) p++;
        p *= 2;
      }
      p /= 2;
      b[i] = p;
      k[p].push_back(i);
    }
    for (int i = 0; i < v.size(); i++) {
      if (k[v[i].first].size() >= 1 && k[v[i].second].size() >= 1) {
        y = 1;
        a1 = k[v[i].first][0];
        a2 = k[v[i].second][0];
        break;
      }
    }
    if (k[x].size() >= 2) {
      y = 1;
      a1 = k[x][0];
      a2 = k[x][1];
    }
    if (y) {
      ans = mid;
      l = mid + 1;
    } else {
      h = mid - 1;
    }
  }
  cout << a1 << " " << a2 << endl;
  return 0;
}