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;
double PI = 3.141592653589793;
const double EPS = 1e-6;
const int N = 1e6 + 5;
const long long mod = 1e9 + 7;
int oo = 1e9;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int t[300005][10];
int n;
pair<int, int> a[N];
int m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m - 1; j++) scanf("%d", &t[i][j]);
  }
  a[0].first = 0;
  a[0].second = 1;
  for (int i = 1; i <= n; i++) {
    for (int mask = 0; mask < (1 << m); mask++) {
      int mn = 1e9;
      for (int j = 0; j <= m - 1; j++) {
        if (mask & (1 << j)) mn = min(t[i][j], mn);
      }
      if (mn >= a[mask].first) a[mask].first = mn, a[mask].second = i;
    }
  }
  int mx = -1, ans1, ans2;
  int all = (1 << m) - 1;
  for (int i = 1; i <= n; i++) {
    for (int mask = 0; mask < (1 << m); mask++) {
      int mn = 1e9;
      for (int j = 0; j <= m - 1; j++) {
        if (mask & (1 << j)) mn = min(t[i][j], mn);
      }
      int x = all ^ mask;
      if (min(mn, a[x].first) > mx) {
        mx = min(mn, a[x].first);
        ans1 = i;
        ans2 = a[x].second;
      }
    }
  }
  printf("%d %d", ans1, ans2);
}