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 = 1000000007;
const int N = 3 * (1e5) + 2;
int arr[N][8];
int n, m;
pair<int, int> pos(int val) {
  map<int, int> make_pair;
  for (int i = 0; i < n; ++i) {
    int mask = 0;
    for (int j = 0; j < m; ++j) {
      if (arr[i][j] >= val) mask |= 1 << j;
    }
    make_pair[mask] = i;
  }
  int need = 1;
  for (int i = 0; i < m; ++i) need *= 2;
  int mx = 0;
  for (auto in1 : make_pair) {
    for (auto in2 : make_pair) {
      mx = max(mx, in1.first | in2.first);
      if ((in1.first | in2.first) + 1 == need) return {in1.second, in2.second};
    }
  }
  return {-1, -1};
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> arr[i][j];
    }
  }
  pair<int, int> fn = {-1, -1};
  int st = 0, end = 1e9;
  while (st < end) {
    int mid = st + (end - st + 1) / 2;
    fn = pos(mid);
    if (fn.first != -1)
      st = mid;
    else
      end = mid - 1;
  }
  fn = pos(st);
  cout << fn.first + 1 << " " << fn.second + 1;
}