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;
vector<int> vec[300005];
int minv[2686];
int n, m;
int lst;
bool ok(int x) {
  for (int i = 0; i < n; ++i) {
    int k = 0;
    for (int j = 0; j < m; ++j) {
      if (vec[i][j] < x) {
        k |= (1 << j);
      }
    }
    if (!k || minv[k] >= x) {
      lst = i;
      return 1;
    }
  }
  return 0;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    vec[i].resize(m);
    for (int j = 0; j < m; ++j) {
      scanf("%d", &vec[i][j]);
    }
    for (int k = 0; k < (1 << m); ++k) {
      int tmp = 1e9;
      for (int j = 0; j < m; ++j) {
        if ((1 << j) & k) {
          tmp = min(tmp, vec[i][j]);
        }
      }
      minv[k] = max(tmp, minv[k]);
    }
  }
  int low = -1, high = 1e9 + 3;
  while (high - low > 1) {
    int mid = (low + high) >> 1;
    if (ok(mid)) {
      low = mid;
    } else {
      high = mid;
    }
  }
  int k = 0;
  for (int j = 0; j < m; ++j) {
    if (vec[lst][j] < low) {
      k |= (1 << j);
    }
  }
  for (int i = 0; i < n; ++i) {
    bool kd = 1;
    for (int j = 0; j < m; ++j) {
      if (k & (1 << j)) {
        if (vec[i][j] < low) {
          kd = 0;
          break;
        }
      }
    }
    if (kd) {
      cout << lst + 1 << " " << i + 1 << endl;
      return 0;
    }
  }
  cout << 1 << " " << 1;
  return 0;
}