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 power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int a[n + 1][m];
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < m; j++) cin >> a[i][j];
    long long l = 1;
    long long r = 1000000000;
    int x1, x2;
    x1 = x2 = 1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      vector<int> b1(256, 0);
      for (int i = 1; i <= n; i++) {
        long long val = 0ll;
        for (int j = 0; j < m; j++) {
          val *= 2ll;
          if (a[i][j] >= mid) val += 1;
        }
        b1[val] = i;
      }
      int check = 0;
      for (int i = 0; i < 256 && check == 0; i++) {
        if (b1[i] != 0) {
          for (int j = 0; j < 256; j++) {
            if (b1[j] != 0 && (i | j) == (1 << m) - 1) {
              x1 = b1[i];
              x2 = b1[j];
              check = 1;
              break;
            }
          }
        }
      }
      if (check == 1)
        l = mid + 1;
      else
        r = mid - 1;
    }
    cout << x1 << " " << x2 << "\n";
  }
  return 0;
}