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;
pair<int, int> prop(int val, vector<vector<int> >& mat) {
  vector<pair<int, int> > V;
  map<int, int> mapa;
  for (int i = 0; i < n; i++) {
    int val1 = 0, val2 = 0;
    for (int j = 0; j < m; j++) {
      if (mat[i][j] < val) {
        val1 = val1 | (1 << j);
      }
    }
    if (mapa[val1] == 0) {
      mapa[val1] = 1;
      V.push_back(make_pair(val1, i + 1));
    }
  }
  for (int i = 0; i < V.size(); i++) {
    int valor = V[i].first;
    for (int j = 0; j < V.size(); j++) {
      int valor2 = V[j].first;
      if ((valor & valor2) == 0) {
        pair<int, int> ans = make_pair(V[i].second, V[j].second);
        return ans;
      }
    }
  }
  pair<int, int> ans = make_pair(-1, -1);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  vector<vector<int> > mat(n, vector<int>(m));
  int maximo = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mat[i][j];
      maximo = max(mat[i][j], maximo);
    }
  }
  int left = 0;
  int right = maximo + 1;
  pair<int, int> cumple;
  for (int mid = (left + right) / 2; left < right; mid = (left + right) / 2) {
    cumple = prop(mid, mat);
    if (cumple.first == -1) {
      right = mid;
    } else
      left = mid + 1;
  }
  left--;
  if (left < 0) {
    cout << "1 1\n";
    return 0;
  }
  cumple = prop(left, mat);
  pair<int, int> cumple2 = prop(left + 1, mat);
  if (cumple2.first != -1) cumple = cumple2;
  cout << cumple.first << " " << cumple.second << "\n";
  return 0;
}