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 MAX_N = 3e5 + 5;
const int MAX_M = 8;
const long long INF = 1e9 + 5;
void solve();
int main() {
  solve();
  return 0;
}
int n, m;
long long a[MAX_N][MAX_M];
int vec[1 << MAX_M];
pair<int, int> check(long long x) {
  for (int i = 0; i < (1 << m); ++i) vec[i] = -1;
  for (int i = 0; i < n; ++i) {
    int cur = 0;
    for (int j = 0; j < m; ++j)
      if (a[i][j] >= x) cur |= (1 << j);
    for (int s = cur;; s = ((s - 1) & cur)) {
      vec[s] = i;
      if (s == 0) break;
    }
    int c = (1 << m) - 1 - cur;
    if (vec[c] != -1) {
      return make_pair(i + 1, vec[c] + 1);
    }
  }
  return make_pair(-1, -1);
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) cin >> a[i][j];
  long long l = 0;
  long long r = INF;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    pair<int, int> cur = check(m);
    if (cur.first != -1)
      l = m;
    else
      r = m;
  }
  pair<int, int> ans = check(l);
  cout << ans.first << ' ' << ans.second << '\n';
  return;
}