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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MAXN = 3e5 + 7;
int n, m, mat[MAXN][10];
bool check(int mid) {
  unordered_set<int> S;
  for (int i = 1; i <= n; i++) {
    int msk = 0;
    for (int j = 1; j <= m; j++)
      if (mat[i][j] >= mid) msk |= (1 << (j - 1));
    S.insert(msk);
    for (auto x : S)
      if ((x | msk) == ((1 << m) - 1)) return true;
  }
  return false;
}
int main() {
  scanf("%d %d", &n, &m);
  int l = 0, r = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      scanf("%d", &mat[i][j]);
      r = max(r, mat[i][j]);
    }
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  unordered_map<int, int> mp;
  for (int i = 1; i <= n; i++) {
    int msk = 0;
    for (int j = 1; j <= m; j++)
      if (mat[i][j] >= r) msk |= (1 << (j - 1));
    mp.insert(make_pair(msk, i));
    for (auto x : mp)
      if ((x.first | msk) == ((1 << m) - 1)) {
        cout << x.second << ' ' << i << endl;
        return 0;
      }
  }
  return 0;
}