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: Runtime Error)

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 200005;
const ll mod = 1e9 + 7;
int n, k;
ll a[maxn], b[maxn], tot;

ll quickpow(ll a, ll x) {
	ll ret = 1;
	while (x) {
		if (x & 1) ret = (ret * a) % mod;
		a = (a * a) % mod;
		x >>= 1;
	}
	return ret;
}

int main() {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	if (k == n) {
		ll ans = 1;
		for (int i = 1; i <= n; i++) ans = (ans * a[i]) % mod;
		printf("%lld\n", ((ans % mod) + mod) % mod);
		return 0;
	}
	sort(a + 1, a + n + 1);
	int fir = 0;
	for (int i = n; i >= 1; i--) {
		if (a[i] < 0) b[++tot] = a[i];
	}
	fir = tot;
	for (int i = n; i >= 1; i--) {
		if (a[i] >= 0) b[++tot] = a[i];
	}
	ll ans = 1;
	for (int i = 1; i <= k; i++) ans = (ans * b[i]) % mod;
	if (!fir || fir == n) { printf("%lld\n", ((ans % mod) + mod) % mod); return 0; }
	ans = 1;
	for (int i = fir - k + 1; i <= fir; i++) ans = ((ans * b[i]) % mod + mod) % mod;
	for (int i = fir + 2; i <= n; i += 2) {
		if (b[i - k - 1] * b[i - k] > b[i - 1] * b[i]) break;
		ans = ans * quickpow((b[i - k - 1] * b[i - k]) % mod, mod - 2) * b[i - 1] % mod * b[i] % mod;
	}
	printf("%lld\n", ((ans % mod) + mod) % mod);
	return 0;
}