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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void solve() {
  int n, m;
  cin >> n >> m;
  vi mel(1000, -1);
  vi pos_mel(1000, -1);
  vector<int> aux(m, 0);
  vector<vi> vet(n, aux);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> vet[i][j];
    }
    for (int k = 1; k < (1 << m); k++) {
      int minn = INF;
      for (int j = 0; j < m; j++) {
        if (k & (1 << j)) {
          minn = min(minn, vet[i][j]);
        }
      }
      if (minn == INF) minn = 0;
      if (mel[k] < minn) {
        mel[k] = minn;
        pos_mel[k] = i;
      }
    }
  }
  int ans = -1;
  pii par_ans = make_pair(-1, -1);
  for (int kk = 0; kk < (1 << m); kk++) {
    int k = kk;
    vi bits(m, 0);
    for (int j = 0; j < m; j++) {
      if (k & (1 << j)) {
        bits[j] = 0;
      } else {
        bits[j] = 1;
      }
    }
    int x = 0;
    for (int j = m - 1; j >= 0; j--) {
      x = x * 2;
      if (bits[j]) {
        x += 1;
      }
    }
    if (k == 0) {
      k = x;
    }
    if (x == 0) {
      x = k;
    }
    int at = min(mel[k], mel[x]);
    if (at > ans) {
      ans = at;
      par_ans.first = pos_mel[k];
      par_ans.second = pos_mel[x];
    }
  }
  cout << par_ans.first + 1 << " " << par_ans.second + 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}