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>
const int N = 300000;
const int M = 8;
int n, m, a[N][M];
std::pair<bool, std::pair<int, int> > test(int v) {
  std::pair<int, int> x[N];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) x[i].first |= (a[i][j] >= v) << j;
    x[i].second = i;
  }
  std::sort(x, x + n);
  for (int i = 0; i < 1 << m; ++i)
    for (int j = 0; j < 1 << m; ++j)
      if ((i | j) == (1 << m) - 1) {
        int f = std::lower_bound(x, x + n, std::make_pair(i, 0)) - x,
            s = std::lower_bound(x, x + n, std::make_pair(j, 0)) - x;
        if (f != n && x[f].first == i && s != n && x[s].first == j)
          return std::make_pair(true, std::make_pair(x[f].second, x[s].second));
      }
  return std::make_pair(false, std::make_pair(0, 0));
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) scanf("%d", a[i] + j);
  int l = 0, r = 1000000000;
  std::pair<int, int> ans;
  while (l <= r) {
    int m = (l + r) >> 1;
    std::pair<bool, std::pair<int, int> > p;
    if ((p = test(m)).first) {
      ans = p.second;
      l = m + 1;
    } else
      r = m - 1;
  }
  printf("%d %d\n", ans.first + 1, ans.second + 1);
  return 0;
}