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 n, m, rez, a, b;
int ma[300001][9];
inline bool can(int val) {
  vector<int> msk(1 << m, -1);
  for (int i = 0; i < n; ++i) {
    int cur = 0;
    for (int j = 0; j < m; ++j) {
      if (ma[i][j] >= val) cur |= (1 << j);
    }
    msk[cur] = i;
  }
  if (msk[(1 << m) - 1] != -1) {
    a = b = msk[(1 << m) - 1];
    return 1;
  }
  for (int i = 0; i < (1 << m); ++i) {
    for (int j = 0; j < (1 << m); ++j) {
      if (msk[i] != -1 && msk[j] != -1 && (i | j) == (1 << m) - 1) {
        a = msk[i];
        b = msk[j];
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) cin >> ma[i][j];
  }
  int lg;
  for (lg = 1; lg <= (int)(1e9 + 1); lg <<= 1)
    ;
  for (int i = -1; lg; lg >>= 1) {
    if (i + lg <= (int)(1e9 + 1) && can(i + lg)) i += lg;
  }
  return cout << a + 1 << " " << b + 1, 0;
}