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<vector<long long>> a;
set<long long> masks;
map<long long, long long> b;
int main() {
  long long n, m, ans1 = 1, ans2 = 1;
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      long long x;
      cin >> x;
      a[i].push_back(x);
    }
  }
  long long left = 0;
  long long right = 2e18;
  while (left + 1 < right) {
    long long middle = (left + right) / 2;
    masks.clear();
    b.clear();
    for (int i = 0; i < n; i++) {
      long long num = 0;
      for (int j = 0; j < m; j++) {
        if (a[i][j] >= middle) num += (1 << (m - j - 1));
      }
      b[num] = i + 1;
      masks.insert(num);
    }
    bool f = false;
    for (auto i : masks) {
      for (auto j : masks) {
        bool f0 = true;
        long long x = i;
        long long y = j;
        for (int kk = 0; kk < m; kk++) {
          if (x % 2 == 1 || y % 2 == 1) {
          } else {
            f0 = false;
            break;
          }
          x /= 2;
          y /= 2;
        }
        if (f0) {
          f = true;
          ans1 = b[i];
          ans2 = b[j];
          break;
        }
      }
    }
    if (f) {
      left = middle;
    } else {
      right = middle;
    }
  }
  cout << ans1 << ' ' << ans2 << endl;
  ;
  return 0;
}