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 s[260];
int n, m, i, j, maxi, st, dr, mij, ans, ansst, ansdr, mini;
int a[300005][10];
inline bool ok(int nr) {
  for (int i = 0; i <= (1 << m) - 1; i++) s[i] = 0;
  int nfr;
  for (int i = 1; i <= n; i++) {
    nfr = 0;
    for (int j = 1; j <= m; j++) nfr = (nfr << 1) + (a[i][j] >= nr);
    s[nfr] = i;
  }
  for (int i = 1; i <= (1 << m) - 1; i++) {
    if (!s[i]) continue;
    for (int j = 1; j <= (1 << m) - 1; j++) {
      if (!s[j]) continue;
      if ((i | j) == (1 << m) - 1) {
        ansst = s[i];
        ansdr = s[j];
        return true;
      }
    }
  }
  return false;
}
int main() {
  scanf("%d %d", &n, &m);
  mini = 1e9;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      scanf("%d", &a[i][j]);
      maxi = max(maxi, a[i][j]);
      mini = min(mini, a[i][j]);
    }
  }
  if (ok(1))
    ans = 1;
  else {
    cout << "1 1";
    return 0;
  }
  st = 1;
  dr = maxi;
  while (st < dr) {
    mij = (st + dr) / 2;
    if (ok(mij)) {
      ans = mij;
      st = mij + 1;
    } else {
      dr = mij - 1;
    }
  }
  if (ans != maxi) bool nans = ok(ans + 1);
  printf("%d %d\n", ansst, ansdr);
  return 0;
}