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 all(v) v.begin(), v.end()
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(), v.end()

using namespace std;
typedef long long LL;
const LL MOD = 1e9 + 7;
const double PI = 2 * acos(0);

bool cmp(int x, int y) {
	return abs(x) < abs(y);
}

int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	LL arr[n + 1];
	for (int i = 0; i < n; i++) {
		scanf("%lld", &arr[i]);
	}
	sort (arr, arr + n, cmp);
	stack<LL> pos, neg;
	if (arr[n - k] == 0) printf("0\n");
	else {
		for (int i = n - 1; i >= n - k; i--) {
			if (arr[i] > 0) pos.push(arr[i]);
			else neg.push(arr[i]);
		}
		int l1 = neg.size();
		if (l1 % 2) {
			LL x = INT_MAX;
			LL y = INT_MAX;
			for (int i = 0; i < n - k; i++) {
				if (arr[i] > 0) x = arr[i];
				else y = arr[i];
			}
			if (!pos.empty() && y != INT_MAX) {
				LL cnt1 = neg.top() * y;
				LL cnt2 = -1;
				if (x != INT_MAX) {
					cnt2 = x * pos.top();
				}
				if (cnt1 > cnt2) {
					pos.pop();
					LL ans = y;
					while (!pos.empty()) {
						ans *= pos.top();
						ans %= MOD;
						pos.pop();
					}
					while (!neg.empty()) {
						ans *= -neg.top();
						ans %= MOD;
						neg.pop();
					}
					printf("%lld\n", ans);
				} else {
					neg.pop();
					LL ans = x;
					while (!pos.empty()) {
						ans *= pos.top();
						ans %= MOD;
						pos.pop();
					}
					while (!neg.empty()) {
						ans *= -neg.top();
						ans %= MOD;
						neg.pop();
					}
					printf("%lld\n", ans);
				}
			} else if (x != INT_MAX) {
				neg.pop();
				LL ans = x;
				while (!pos.empty()) {
					ans *= pos.top();
					ans %= MOD;
					pos.pop();
				}
				while (!neg.empty()) {
					ans *= -neg.top();
					ans %= MOD;
					neg.pop();
				}
				printf("%lld\n", ans);
			} else {
				LL ans = 1;
				for (int i = 0; i < k; i++) {
					ans *= arr[i];
					while (ans < 0) ans += MOD;
					ans %= MOD;
				}
				printf("%lld\n", ans);
			}
		} else {
			LL ans = 1;
			while (!pos.empty()) {
				ans *= pos.top();
				ans %= MOD;
				pos.pop();
			}
			while (!neg.empty()) {
				ans *= -neg.top();
				ans %= MOD;
				neg.pop();
			}
			printf("%lld\n", ans);
		}
	}
}