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 a[300009][9], k, i, n, m, ct, t;
pair<long long, long long> f(long long x) {
  k = 0;
  long long q[600] = {};
  vector<long long> b;
  while (k < n) {
    i = m - 1;
    ct = 0;
    t = 1;
    while (i > -1) {
      if (a[k][i] >= x) ct += t;
      t *= 2;
      i--;
    }
    if (q[ct] == 0) {
      q[ct] = k + 1;
      b.push_back(ct);
    }
    k++;
  }
  t--;
  i = 0;
  while (i < b.size()) {
    k = i;
    while (k < b.size()) {
      if ((b[i] | b[k]) == t) {
        return {q[b[i]], q[b[k]]};
      }
      k++;
    }
    i++;
  }
  return {-1, -1};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i = 0, j;
  cin >> n >> m;
  while (i < n) {
    j = 0;
    while (j < m) {
      cin >> a[i][j];
      j++;
    }
    i++;
  }
  pair<long long, long long> w;
  long long s = 0, e = 1e10, mid, ans, an;
  while (s <= e) {
    mid = (s + e) / 2;
    w = f(mid);
    if (w.first != -1) {
      ans = w.first;
      an = w.second;
      s = mid + 1;
    } else
      e = mid - 1;
  }
  cout << ans << " " << an << "\n";
}