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 MAXN = 3e5 + 5, M = 8 + 2;
int a[MAXN][M], mark[1 << M];
pair<int, int> ans;
int n, m;
void input();
bool check(int);
void solve();
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input();
  solve();
  return 0;
}
void input() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
}
bool check(int x) {
  for (int i = 0; i < n; i++) {
    int ksh = 0;
    for (int j = 0; j < m; j++)
      if (a[i][j] >= x) ksh += 1 << j;
    mark[ksh] = i + 1;
  }
  bool p = false;
  for (int i = 0; i < 1 << m; i++)
    for (int j = i; j < 1 << m; j++)
      if ((i | j) == (1 << m) - 1 && mark[i] && mark[j]) {
        p = true;
        ans = {mark[i], mark[j]};
      }
  for (int i = 0; i < 1 << m; i++) mark[i] = 0;
  return p;
}
void solve() {
  int l = 0, r = 1e9 + 9, mid = (l + r) / 2;
  ans = {1, 1};
  while (r - l > 1) {
    check(mid) ? l = mid : r = mid;
    mid = (l + r) / 2;
  }
  cout << ans.first << " " << ans.second << endl;
}