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;
const int mod = int(1e9) + 7;
const int MOD = int(1e9) + 7;
const int MAX = int(3e5) + 5;
const int NN = int(1e6) + 1;
long long a[MAX][9];
long long N, M;
long long cx = 0;
long long cy = 0;
long long ans = 0;
bool check(long long x) {
  long long mask[N];
  memset(mask, 0, sizeof(mask));
  map<long long, long long> freq;
  map<long long, long long> pos;
  for (long long i = 0; i < N; ++i) {
    for (long long j = 0; j < M; ++j) {
      mask[i] += (1 << j) * (a[i][j] >= x);
    }
    freq[mask[i]]++;
    pos[mask[i]] = i + 1;
  }
  long long MM = (1 << M) - 1;
  for (long long i = 0; i <= MM; ++i) {
    for (long long j = 0; j <= MM; ++j) {
      if (((i | j) == MM) and freq[i] and freq[j]) {
        if (x > ans) {
          cx = pos[i];
          cy = pos[j];
        }
        return true;
      }
    }
  }
  return false;
}
void solve() {
  cin >> N >> M;
  cx = cy = 1;
  ans = 0;
  for (long long i = 0; i < N; ++i) {
    for (long long j = 0; j < M; ++j) {
      cin >> a[i][j];
    }
  }
  long long mini = 0;
  long long maxi = mod;
  while (maxi >= mini) {
    long long mid = (mini + maxi) >> 1;
    if (check(mid)) {
      ans = max(ans, mid);
      mini = mid + 1;
    } else {
      maxi = mid - 1;
    }
  }
  cout << cx << " " << cy << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}