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 ll long long int
#define fo(i, n) for (ll i = 0; i < n; i++)
#define fo1(i, n) for (ll i = 1; i <= n; i++)
#define pi (double)3.14159265359
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define vect vector<ll>
#define pb push_back
#define mod (ll) pow(10, 9) + 7
#define prq priority_queue <int, vector<int>, greater<in1t>>
#define scan(x) scanf("%lld",&x);
#define print(x) printf("%lld\n",x);
using namespace std;

long long mm(long long a,ll b) 
{ 
    long long ans = 0;
    a %= mod; 
    while (b) { 
        if (b & 1) 
            ans = (ans + a) % mod; 
        a = (2 * a) % mod; 
        b >>= 1;
    } 
    return ans; 
} 
void solve(){
	ll n,k,ans=1;
	cin>>n>>k;
	vect v(n);
	fo(i,n){
		cin>>v[i];
	}
	sort(v.begin(),v.end());
	if (v[n - 1] == 0 && (k & 1)) 
        cout<<"0"<<endl;
	else if (v[n - 1] <= 0 && (k & 1)) { 
        for (ll i = n - 1; i >= n - k; i--) 
            ans =ans*v[i]; 
        cout<<(ans%mod)<<endl; 
    } 
	else {
		ll i=0,j=n-1;
		if (k & 1) { 
        	ans *= v[j]; 
        	j--; 
        	k--; 
    	} 
		k >>= 1; 
		for (int itr = 0; itr < k; itr++) { 
        int left_product = (v[i] * v[i + 1]); 
        int right_product = (v[j] * v[j - 1]); 
        if (left_product > right_product) { 
            ans = (left_product*ans); 
            i += 2; 
        } 
        else { 
            ans = (right_product*ans); 
            j -= 2; 
        } 
    }
	cout<<(ans%mod)<<endl;
	}
} 
int main(){
	int t=1;
	//scanf("%d",&t);
	while (t--){
		solve();
	}
	return 0;
}