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 INF = 1e9 + 5;
int N, M;
vector<vector<int>> A;
pair<int, int> solve(int threshold) {
  vector<int> exists(1 << M, -1);
  for (int i = 0; i < N; i++) {
    int mask = 0;
    for (int j = 0; j < M; j++) mask |= (A[i][j] >= threshold) << j;
    exists[mask] = i;
  }
  for (int mask = 0; mask < 1 << M; mask++)
    for (int other = 0; other < 1 << M; other++)
      if ((mask | other) == (1 << M) - 1 && exists[mask] >= 0 &&
          exists[other] >= 0)
        return {exists[mask], exists[other]};
  return {-1, -1};
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> M;
  A.assign(N, vector<int>(M));
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) cin >> A[i][j];
  int low = 0, high = INF;
  while (low < high) {
    int mid = (low + high + 1) / 2;
    if (solve(mid).first >= 0)
      low = mid;
    else
      high = mid - 1;
  }
  pair<int, int> answer = solve(low);
  cout << answer.first + 1 << ' ' << answer.second + 1 << '\n';
}