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;
void online_judge() {}
pair<long long, long long> res;
bool pos(vector<vector<long long>> &v, long long val, long long n,
         long long m) {
  set<long long> s;
  vector<pair<long long, long long>> v1;
  for (int i = 0; i < n; ++i) {
    long long mask = 0;
    for (int j = 0; j < m; ++j) {
      if (v[i][j] >= val) {
        mask |= (1 << j);
      }
    }
    if (!s.count(mask)) {
      s.insert(mask);
      v1.push_back({mask, i + 1});
    }
  }
  for (int i = 0; i < v1.size(); ++i) {
    for (int j = i; j < v1.size(); ++j) {
      if ((v1[i].first | v1[j].first) == (1 << m) - 1) {
        res.first = v1[i].second;
        res.second = v1[j].second;
        return 1;
      }
    }
  }
  return 0;
}
void go() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> v(n, vector<long long>(m, 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) cin >> v[i][j];
  }
  long long be = 0;
  long long en = 1e18;
  pair<long long, long long> ans = {-1, -1};
  while (be <= en) {
    long long mid = (be + en) / 2;
    res = {-1, -1};
    if (pos(v, mid, n, m)) {
      ans = res;
      be = mid + 1;
    } else
      en = mid - 1;
  }
  cout << ans.first << " " << ans.second;
}
int main() {
  cin.tie(0), ios_base ::sync_with_stdio(0);
  online_judge();
  go();
  return 0;
}