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;
long long mod = 1000000000 + 7;
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
void fun() {}
long long n, m;
long long a[300001][9];
long long a1, a2;
long long fun(long long x) {
  long long mask[1 << m];
  memset(mask, -1, sizeof(mask));
  for (long long i = 0; i < n; i++) {
    long long num = 0;
    for (long long j = 0; j < m; j++) {
      if (a[i][j] >= x) num += (1 << j);
    }
    mask[num] = i;
  }
  if (mask[(1 << m) - 1] != -1) {
    a1 = a2 = mask[(1 << m) - 1];
    return 1;
  }
  for (long long i = 0; i < (1 << m); i++) {
    for (long long j = i; j < (1 << m); j++) {
      if (mask[i] != -1 && mask[j] != -1 && ((i | j) == ((1 << m) - 1))) {
        a1 = mask[i];
        a2 = mask[j];
        return 1;
      }
    }
  }
  return 0;
}
signed main() {
  fun();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long l = 0, r = INT_MAX;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (fun(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << a1 + 1 << " " << a2 + 1;
}