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 INF = 0x3f3f3f3f;
const int MAXN = 3e5 + 5;
int n, m;
int a[MAXN][10];
unordered_map<int, int> mp;
void _out(int x) {
  for (int i = 0; i < m; i++) {
    if ((1 << i) & x)
      cout << 1;
    else
      cout << 0;
  }
  cout << endl;
}
bool check(int x) {
  mp.clear();
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= x) {
        now |= (1 << (j - 1));
      }
    }
    mp[now] = i;
    for (int j = 0; j < (1 << m); j++) {
      if ((now | j) == (1 << m) - 1 && mp.count(j)) {
        return true;
      }
    }
  }
  return false;
}
void check_2(int x) {
  mp.clear();
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= x) {
        now |= (1 << (j - 1));
      }
    }
    mp[now] = i;
    for (int j = 0; j < (1 << m); j++) {
      if ((now | j) == (1 << m) - 1 && mp.count(j)) {
        int _min = INF, tmp;
        for (int k = 0; k < m; k++) {
          tmp = -1;
          if (now & (1 << k)) tmp = max(tmp, a[i][k + 1]);
          if (j & (1 << k)) tmp = max(tmp, a[mp[j]][k + 1]);
          _min = min(_min, tmp);
        }
        if (_min == x) {
          cout << i << " " << mp[j] << endl;
          return;
        }
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  int l = 0, r = 1e9, mid, ans = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  check_2(ans);
  return 0;
}