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 long long INF = 1e9;
const long long INFLL = 1e18;
const long long MOD = 1000000007;
const long double eps = 1e-6;
const long double eps2 = 1e-9;
const long long MOD2 = 998244353;
const long long dosz = 5e5;
const long long SZ = (1 << 18);
const long double PI = atan2l(0, -1);
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int maxn = 3e5 + 228;
int n, m;
int kek[maxn][10];
pair<int, int> check(int mid) {
  vector<int> mask((1 << m), -1);
  for (int i = 0; i < n; i++) {
    int cur_mask = 0;
    for (int j = 0; j < m; j++) {
      if (kek[i][j] >= mid) {
        cur_mask |= (1 << j);
      }
    }
    for (int j = 0; j < (1 << m); j++) {
      if ((j | cur_mask) == ((1 << m) - 1)) {
        if (mask[j] != -1) {
          return make_pair(mask[j], i);
        }
      }
    }
    mask[cur_mask] = i;
  }
  return make_pair(-1, -1);
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> kek[i][j];
    }
  }
  int l = 0, r = 1e9;
  pair<int, int> ans;
  while (l <= r) {
    int mid = (l + r) / 2;
    pair<int, int> lol = check(mid);
    if (lol.first == -1 && lol.second == -1) {
      r = mid - 1;
    } else {
      ans = lol;
      l = mid + 1;
    }
  }
  cout << ans.first + 1 << " " << ans.second + 1 << '\n';
}
signed main() {
  fast_io();
  int q = 1;
  while (q--) {
    solve();
  }
  return 0;
}