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;
long long masks[1 << 10];
long long a[300005][10];
pair<bool, pair<long long, long long> > solve(long long x, int m) {
  for (int i = (0); i < (int)(1 << m); i++) masks[i] = -1;
  for (int i = (0); i < (int)(n); i++) {
    int mask = 0;
    for (int j = m - 1; j >= 0; --j)
      mask += (1 << (m - 1 - j)) * (a[i][j] >= x);
    masks[mask] = i;
  }
  for (int j = (0); j < (int)(1 << m); j++) {
    if (masks[j] == -1) continue;
    for (int i = (0); i < (int)(1 << m); i++) {
      if (masks[i] != -1 && ((j | i) == (1 << m) - 1))
        return make_pair(true, make_pair(masks[i], masks[j]));
    }
  }
  return make_pair(false, make_pair(-1, -1));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long maxA = 0LL;
  for (int i = (0); i < (int)(n); i++)
    for (int j = (0); j < (int)(m); j++) {
      cin >> a[i][j];
      maxA = max(maxA, a[i][j]);
    }
  long long l = 0, r = maxA;
  while (l <= r) {
    long long med = (l + r) / 2;
    pair<bool, pair<long long, long long> > res = solve(med, m);
    if (res.first) {
      pair<bool, pair<long long, long long> > res1 = solve(med + 1, m);
      if (!res1.first) {
        cout << res.second.first + 1 << " " << res.second.second + 1 << endl;
        return 0;
      }
      l = med + 1;
    } else
      r = med - 1;
  }
  return 0;
}