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 N = 3e5 + 5;
long long arr[N][9];
int n, m;
bool check(long long x) {
  set<int> tmpmsk;
  for (int i = 0; i < n; i++) {
    int msk = 0;
    for (int j = 0; j < m; j++) {
      msk <<= 1;
      msk |= (arr[i][j] >= x);
    }
    tmpmsk.insert(msk);
  }
  vector<int> tmp;
  for (auto i : tmpmsk) tmp.push_back(i);
  for (int i = 0; i < tmp.size(); i++)
    for (int j = 0; j < tmp.size(); j++)
      if ((tmp[i] | tmp[j]) == ((1 << m) - 1)) return 1;
  return 0;
}
long long bs(long long low, long long high, long long req) {
  if (low > high) return req;
  long long mid = (low + high) / 2;
  if (check(mid)) return bs(mid + 1, high, mid);
  return bs(low, mid - 1, req);
}
void solve() {
  vector<int> tmp[268];
  long long mx = bs(0, 1e9, 0);
  for (int i = 0; i < n; i++) {
    int msk = 0;
    for (int j = 0; j < m; j++) {
      msk <<= 1;
      msk |= (arr[i][j] >= mx);
    }
    tmp[msk].push_back(i);
  }
  for (int i = 0; i < 260; i++)
    for (int j = 0; j < 260; j++)
      if ((i | j) == ((1 << m) - 1) && tmp[i].size() && tmp[j].size()) {
        cout << tmp[i][0] + 1 << " " << tmp[j][0] + 1 << endl;
        return;
      }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> arr[i][j];
  solve();
  return 0;
}