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)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MOD = 1e9 + 7;

void fn(){
	fn();
}
int main(){
	int n, k;
	cin >> n >> k;
	int a[n];
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	sort(a, a + n);
	if(a[n - 1] == 0 && k % 2){
		cout << 0 << '\n';
		return 0;
	}
	ll sol = 1;
	if(a[n - 1] <= 0 && k % 2){
		for(int i = n - 1;  i >= n - k; i--){
			sol = (sol * 1LL * a[i] + MOD) % MOD;
		}
		cout << sol << '\n';
		return 0;
	}
	int i = 0, j = n - 1;
	if(k % 2){
		sol = (sol * 1LL *  a[j]) % MOD;
		j--;
		k--;
	}
	k /= 2;
	for(int p = 0; p < k; p++){
		ll neg = (a[i] * 1LL * a[i + 1]);
		ll pos = (a[j] * 1LL * a[j - 1]);
		if(neg > pos){
			sol = ((sol * 1LL * (neg % MOD)) + MOD) % MOD;
			i += 2;
		}else{
			sol = ((sol * 1LL * (pos % MOD)) + MOD) % MOD;
			j -= 2;
		}
	}
	cout << sol << '\n';
}