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(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define ff first
#define ss second
#define P 1000000007
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)

using namespace std;
using ll = long long;
typedef pair<int, int> ii;
typedef vector<ii> vii; 
typedef vector<int> vi;

void solve() {
	int n, k;
	cin >> n >> k;
	int kk = k;
	
	vector<pair<long double, ii>> a; vector<long double> b; vector<int> inc(n);
	for(int i = 0; i < n; i++) {
		long double c; cin >> c; inc[i] = abs(c);
		if(c >= 0) a.pub({c, {c, 1}}); else b.pub(c);
	}
	
	sort(all(b));
	for(int i = 0; i < b.size(); i += 2) {
		if(i + 1 >= b.size()) continue;
		a.pub({(long double)sqrt(b[i] * b[i + 1]), {b[i] * b[i + 1], 2}});
	}
	
	ll ans = 1;
	sort(rall(a));
	for(int i = 0; i < a.size() && k; i++) {
		if(k - a[i].ss.ss < 0) continue;
		k -= a[i].ss.ss;
		ans = ((ans * a[i].ss.ff) % P + P) % P;
	}
	
	if(k) {
		ans = 1;
		sort(all(inc));
		for(int i = 0; i < kk; i++) ans = ((ans * inc[i]) % P + P) % P;
		ans = (ans * (-1) % P + P) % P;
		cout << ans << endl;
	}

	cout << ans << endl;
}

int main(){
#define MULTITEST 0
#if MULTITEST   
 
    CASET {
        solve();
    }
    
#else

    solve();
    
#endif

    return 0;
}