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 a[2000010][9], n, m, d[2000010], p, q;
bool kt(long long gt) {
  for (int i = 0; i <= 300; ++i) d[i] = 0;
  for (int i = 1; i <= n; ++i) {
    long long s = 0;
    for (int j = 1; j <= m; ++j)
      if (a[i][j] >= gt) s += (1 << (j - 1));
    d[s] = i;
  }
  for (int i = 0; i <= 300; ++i)
    for (int j = 0; j <= 300; ++j) {
      bool c = true;
      for (int z = 0; z < m; ++z)
        if (((i >> z) & 1) == 0 && ((j >> z) & 1) == 0) c = false;
      if (c == true && d[i] > 0 && d[j] > 0) {
        p = d[i];
        q = d[j];
        return true;
      }
    }
  if (d[(1 << m)] > 0) {
    p = q = d[(1 << m)];
    return true;
  }
  return false;
}
void nhap() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> a[i][j];
  long long l = 0, r = 1e9;
  while (l <= r) {
    long long mid = (l + r) / 2;
    bool o = kt(mid);
    if (o == true && (mid == 1e9 || kt(mid + 1) == false)) {
      cout << p << " " << q;
      return;
    }
    if (o == true)
      l = mid + 1;
    else
      r = mid - 1;
  }
}
int main(int argc, const char* argv[]) {
  nhap();
  return 0;
}