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;
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > a(n, vector<long long>(m, 0));
  long long l = 1000000007, r = -1;
  set<long long> s;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++) {
      cin >> a[i][j];
      l = min(l, a[i][j]);
      r = max(r, a[i][j]);
      s.insert(a[i][j]);
    }
  pair<long long, long long> am = {-1, -1};
  while (l <= r) {
    long long x = (l + r) >> 1;
    vector<long long> v(n, 0);
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        v[i] = v[i] << 1;
        if (a[i][j] >= x) v[i]++;
      }
    }
    long long dp[1 << 9];
    memset(dp, -1, sizeof dp);
    for (long long i = 0; i < n; i++) dp[v[i]] = i;
    for (long long i = 0; i < m; i++)
      for (long long mask = 0; mask < 1 << m; mask++)
        if (dp[mask] != -1 && (mask & (1 << i))) dp[mask ^ (1 << i)] = dp[mask];
    bool fl = false;
    for (long long i = 0; i < n; i++) {
      long long k = v[i];
      k = (1 << m) - 1 - k;
      if (dp[k] != -1) {
        fl = true;
        am.first = i;
        am.second = dp[k];
        break;
      }
    }
    if (fl) {
      l = x + 1;
    } else
      r = x - 1;
  }
  cout << am.first + 1 << " " << am.second + 1 << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}