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;
const int MAXN = 300005, INF = 1e9 + 2;
int a[MAXN][10];
int mark[305];
int n, m;
int ans1 = 1, ans2 = 1;
bool check(int x) {
  fill(mark, mark + 300, 0);
  for (int i = 0; i < n; i++) {
    int b = 0;
    for (int j = 0; j < m; j++)
      if (a[i][j] >= x)
        b = b * 2 + 1;
      else
        b *= 2;
    mark[b] = i + 1;
  }
  int pm = (1 << m);
  for (int i = 0; i < pm; i++)
    for (int j = i; j < pm; j++)
      if (mark[i] && mark[j] && (i | j) == pm - 1) {
        ans1 = mark[i];
        ans2 = mark[j];
        return true;
      }
  return false;
}
inline int in() {
  int x, y;
  y = scanf("%d", &x);
  return x;
}
int main() {
  n = in(), m = in();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) a[i][j] = in();
  int lo = 0, hi = INF;
  while (hi - lo > 1) {
    int mid = (lo + hi) / 2;
    if (check(mid))
      lo = mid;
    else
      hi = mid;
  }
  cout << ans1 << ' ' << ans2 << endl;
  return 0;
}