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>
#define int long long int
using namespace std;

#undef int
int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
#define int long long int
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t = 1;
	while (t--) {
		int n, k;
		cin >> n >> k;
		int arr[n], brr[n], crr[n];
		int a = 0, b = 0;
		for (int i = 0; i < n; i++)
			cin >> arr[i];
		vector<int> v1, v2, v3;
		for (int i = 0; i < n; i++) {
			if (arr[i] >= 0) {
				brr[a] = arr[i];
				a++;
			}
			else if (arr[i] < 0) {
				crr[b] = arr[i];
				b++;
			}
		}
		int ans = 0;
		sort(brr, brr + a);
		sort(crr, crr + b);
		if (k % 2 == 1) {
			ans = brr[a - 1];
			a--;
			k--;
		}
		while (k > 0) {
			int x = 0, y = 0, flag = 0, flag1 = 0;
			if (a == 0) {
				x = brr[a];
				a--;
				flag = 1;
			}
			if (b == 0) {
				y = brr[b];
				b--;
				flag1 = 1;
			}
			if (a >= 0 && a - 1 >= 0) {
				x = brr[a] * brr[a - 1];
				a = a - 2;
			}
			if (b >= 0 && b - 1 >= 0) {
				y = crr[b] * crr[b - 1];
				b = b - 2;
			}
			if (x > y) {
				ans = (ans * x)%1000000007;
				if (flag == 1) k = k - 1;
				else k = k - 2;
			}
			else {
				ans = (ans * y)%1000000007;
				if (flag1 == 1) k = k - 1;
				else k = k - 2;
			}
		}
		cout << ans << endl;
	}
	return 0;
}