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 = 1e9;
const int MOD = 1e9 + 7;
int n, m;
int arr[301010][8];
int ansx, ansy;
int cand[256];
int solve(long long x) {
  memset(cand, -1, sizeof(cand));
  for (int i = 0; i < n; i++) {
    int bit = 1;
    int mask = 0;
    for (int j = 0; j < m; j++) {
      if (arr[i][j] >= x) mask += bit;
      bit <<= 1;
    }
    cand[mask] = i;
  }
  for (int i = 0; i < (1 << m); i++) {
    for (int j = 0; j < (1 << m); j++) {
      if ((i | j) == (1 << m) - 1) {
        if (cand[i] != -1 && cand[j] != -1) {
          ansx = cand[i];
          ansy = cand[j];
          return 1;
        }
      }
    }
  }
  return 0;
}
long long binary_search(long long x, long long y) {
  long long lo = x - 1;
  long long hi = y + 1;
  while (lo + 1 < hi) {
    long long mid = (lo + hi) / 2;
    if (solve(mid))
      lo = mid;
    else
      hi = mid;
  }
  return lo;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) scanf("%d", &arr[i][j]);
  }
  long long ans = binary_search(0, (long long)1e9);
  solve(ans);
  printf("%d %d\n", ansx + 1, ansy + 1);
  return 0;
}