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;
long long int power(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long int a[300009][10];
pair<long long int, long long int> ans;
long long int n, m;
bool check(long long int num) {
  vector<long long int> v[1024];
  long long int mask[n + 2];
  for (long long int i = 1; i < n + 1; i++) {
    mask[i] = 0;
    for (long long int j = 1; j < m + 1; j++) {
      long long int lol = m - j;
      if (a[i][j] >= num) mask[i] += (1LL << lol);
    }
    v[mask[i]].push_back(i);
  }
  for (long long int i = 0; i < 1LL << m; i++) {
    for (long long int j = 0; j < 1LL << m; j++) {
      if (v[i].size() == 0 or v[j].size() == 0) continue;
      if (i == j) {
        if (v[i].size() < 1) continue;
      }
      long long int lol = i | j;
      if (lol == ((1LL << m) - 1)) {
        if (i == j)
          ans = {v[i][0], v[i][0]};
        else
          ans = {v[i][0], v[j][0]};
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < m + 1; j++) cin >> a[i][j];
  long long int low = 0, high = 1e9;
  while (low <= high) {
    long long int mid = (low + high) / 2;
    if (check(mid))
      low = mid + 1;
    else
      high = mid - 1;
  }
  cout << ans.first << " " << ans.second << endl;
}