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;
vector<vector<int> > v;
int A = -1, B = -1;
bool ok(int val) {
  int MaxMask = (1 << m);
  int off[(1 << m)];
  for (int i = 0; i < (1 << m); ++i) {
    off[i] = -1;
  }
  for (int i = 0; i < n; ++i) {
    int mask = 0;
    for (int j = 0; j < m; ++j) {
      if (v[i][j] >= val) {
        mask |= (1 << j);
      }
    }
    off[mask] = i;
  }
  if (off[(1 << m) - 1] != -1) {
    A = off[(1 << m) - 1];
    B = off[(1 << m) - 1];
    return true;
  }
  for (int mask = 0; mask < (1 << m); ++mask) {
    if (off[mask] == -1) continue;
    for (int mask2 = 0; mask2 < (1 << m); ++mask2) {
      if (off[mask2] == -1 || (mask | mask2) != ((1 << m) - 1)) continue;
      A = off[mask];
      B = off[mask2];
      return true;
    }
  }
  return false;
}
void solve() {
  cin >> n >> m;
  v.assign(n, vector<int>(m, 0));
  int Max = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> v[i][j];
      Max = max(Max, v[i][j]);
    }
  }
  int L = 0, R = Max + 1;
  while (L + 1 < R) {
    int M = (L + R) / 2;
    if (ok(M)) {
      L = M;
    } else {
      R = M;
    }
  }
  ok(L);
  cout << A + 1 << " " << B + 1 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}