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 a[300005][10];
int vis[300];
int n, m;
vector<unsigned long> v, id;
pair<int, int> solve(int p) {
  v.clear();
  id.clear();
  for (int i = 0; i < (1 << m); i++) vis[i] = 0;
  for (int i = 0; i < n; i++) {
    bitset<8> bs;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= p)
        bs[j] = 1;
      else
        bs[j] = 0;
    }
    if (!vis[bs.to_ulong()]) {
      vis[bs.to_ulong()] = 1;
      id.push_back(i);
      v.push_back(bs.to_ulong());
    }
  }
  for (int i = 0; i < v.size(); i++) {
    for (int j = i; j < v.size(); j++) {
      if ((v[i] | v[j]) == ((1 << m) - 1)) return pair<int, int>(i, j);
    }
  }
  return pair<int, int>(-1, -1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  int l = 0, r = 1000000000;
  pair<int, int> p;
  while (l < r) {
    int m = r + (l - r) / 2;
    p = solve(m);
    if (p.first != -1)
      l = m;
    else
      r = m - 1;
  }
  p = solve(l);
  printf("%lu %lu\n", id[p.first] + 1, id[p.second] + 1);
}