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 maxn = 3e5 + 10;
int n, m;
int a[maxn][10];
int mask[maxn];
int valid[1024];
pair<int, int> ans;
void print(int x) {
  for (int i = 7; i >= 0; i--) cout << ((x >> i) & 1);
  puts("");
}
bool check(int x) {
  memset(valid, -1, sizeof(valid));
  memset(mask, 0, sizeof(mask));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++)
      if (a[i][j] >= x) mask[i] |= (1 << j);
    valid[mask[i]] = i;
  }
  for (int i = 0; i < (1 << m); i++) {
    if (valid[i] == -1) continue;
    for (int j = 0; j < (1 << m); j++)
      if (((i | j) == ((1 << m) - 1)) && valid[j] != -1) {
        ans = make_pair(valid[i], valid[j]);
        return true;
      }
  }
  return false;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; 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 >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  check(l);
  printf("%d %d", ans.first + 1, ans.second + 1);
  return 0;
}