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;
long long dx[] = {-1, 1, 0, 0};
long long dy[] = {0, 0, -1, 1};
pair<long long, long long> dp[256];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> v(n, vector<long long>(m, 0));
  long long t = 1ll << m;
  for (long long i = 1; i < t; ++i) dp[i] = {-1, -1};
  for (auto &k : v)
    for (auto &i : k) cin >> i;
  for (long long i = 0; i < n; ++i) {
    for (long long j = 1; j < t; ++j) {
      long long mn = INT_MAX;
      for (long long k = 0; k < m; ++k) {
        if (j & (1 << k)) {
          mn = min(mn, v[i][k]);
        }
      }
      if (dp[j].first <= mn) {
        dp[j].first = mn;
        dp[j].second = i;
      }
    }
  }
  pair<long long, long long> ans = {-1, -1};
  long long mx = -1;
  for (long long i = 1; i < t; ++i) {
    for (long long j = i; j < t; ++j) {
      long long p = i | j;
      if (p == t - 1 and min(dp[i].first, dp[j].first) >= mx) {
        mx = min(dp[i].first, dp[j].first);
        ans.first = dp[i].second;
        ans.second = dp[j].second;
      }
    }
  }
  cout << ans.first + 1 << " " << ans.second + 1;
}