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;
int n, m, index1 = 1, index2 = 1, a[300000 + 2][9];
map<int, int> o;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  int x1 = 0, x2 = 1e9;
  while (x1 <= x2) {
    int v = (x1 + x2) / 2;
    while (o.size()) o.erase(o.begin());
    for (int i = 0; i < n; i++) {
      int s = 1;
      int t = 0;
      for (int j = 0; j < m; j++) {
        if (a[i][j] >= v) {
          t += s;
        }
        s *= 2;
      }
      o[t] = i + 1;
    }
    int lam = 0;
    while (o.size()) {
      int y = (*o.begin()).first;
      for (int k = 0; k <= 300; k++) {
        if ((y | k) == pow(2, m) - 1 && (o.count(k))) {
          index1 = (*o.begin()).second;
          index2 = (*o.find(k)).second;
          lam = 1;
        }
      }
      o.erase(o.begin());
    }
    if (lam)
      x1 = v + 1;
    else
      x2 = v - 1;
  }
  cout << index1 << ' ' << index2;
}