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;
int f[1005];
int a[300005][15];
void mo() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void nhap() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
}
pair<int, int> check(int g) {
  for (int i = 0; i < (1 << m); i++) {
    f[i] = 0;
  }
  f[0] = n + 1;
  for (int i = 1; i <= n; i++) {
    int state = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= g) state = state ^ (1 << (j - 1));
    }
    f[state] = i;
  }
  for (int state = 0; state < (1 << m); state++) {
    if (f[state] != 0)
      for (int prestate = state; prestate; prestate = (prestate - 1) & state) {
        f[prestate] = f[state];
      }
  }
  for (int state = 0; state < (1 << m); state++) {
    int prestate = state ^ ((1 << m) - 1);
    if (f[state] != 0 && f[prestate] != 0) {
      return pair<int, int>(f[state], f[prestate]);
    }
  }
  return pair<int, int>(-1, -1);
}
void xuli() {
  int l = 0;
  int r = round(1e9);
  pair<int, int> res;
  while (l <= r) {
    int mid = (l + r) >> 1;
    pair<int, int> tmp = check(mid);
    if (tmp.first != -1) {
      res = tmp;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  if (res.first == n + 1) res.first = res.second;
  if (res.second == n + 1) res.second = res.first;
  cout << res.first << " " << res.second;
}
int main() {
  mo();
  nhap();
  xuli();
}