Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
  In the name of ALLAH
  Author : Raashid Anwar
*/
 
#include <bits/stdc++.h>
using namespace std;
 
#define int int64_t
const int M1 =  998244353;
const int M2 =  1000000007;

void solve() {
  int n, k;
  cin >> n >> k;
  vector <int> a(n);
  for(int &x : a)
    cin >> x;
  sort(a.begin(), a.end());
  int l = 0, r = n - 1, ans = 1;
  while(l <= r && k) {
    if(k == 1) {
      ans = (ans * a[r--]) % M2;
      k--;
    }
    else {
      int x = a[l] * a[l + 1];
      int y = a[r] * a[r - 1];
      if(x > y) {
        ans = (ans * (x % M2)) % M2;
        l += 2;
      }
      else {
        ans = (ans * (y % M2)) % M2;
        r -= 2;
      }
      k -= 2;
    }
  }
  cout << ans << "\n";
}
 
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}