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 + 10;
int n, m;
int a[maxn][10];
int id[maxn];
int ans1, ans2;
bool check(int mid) {
  memset(id, 0, sizeof(id));
  for (int i = 1; i <= n; i++) {
    int s = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= mid) s |= (1 << j);
    }
    id[s] = i;
  }
  for (int i = (1 << m) - 1; i; i--) {
    for (int j = 0; j < m; j++)
      if (id[i] && (i & (1 << j))) id[i ^ (1 << j)] = id[i];
  }
  for (int i = 0; i < (1 << m); i++)
    if (id[i] && id[((1 << m) - 1) ^ i]) {
      ans1 = id[i];
      ans2 = id[((1 << m) - 1) ^ i];
      return 1;
    }
  return 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) scanf("%d", &a[i][j]);
  }
  int L = 0, R = 1e9;
  while (L < R) {
    int mid = (L + R + 1) / 2;
    if (check(mid))
      L = mid;
    else
      R = mid - 1;
  }
  check((L + R) / 2);
  printf("%d %d\n", ans1, ans2);
  return 0;
}