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>
using namespace std;
int main(){
    int n,k;
    cin >> n >> k;
    long long int a[n];
    for(int i=0;i<n;i++)cin >> a[i];
    sort(a,a+n);
    long long int mod=1e9+7;
    long long int ans=1;
    vector<long long int> plus;
    vector<long long int> minus;
    for(int i=0;i<n;i++){
        if(a[i]<=0)minus.push_back(a[i]);
        else plus.push_back(a[i]);
    }
    if(plus.size()<(long unsigned int)k && (k-plus.size())%2==1){
        long long int b[n];
        for(int i=0;i<n;i++)b[i]=abs(a[i]);
        sort(b,b+n);
        for(int i=0;i<k;i++){
            ans *= b[i];
            ans %= mod;
        }
        if(ans!=0)cout << mod - ans;
        else cout << 0;
    }
    else{
        int p;
        long unsigned int m;
        int count=0;
        p=plus.size()-1; m=0;
        while(count<k){
            while((m+1<minus.size() && p-1>=0) && count+2<=k){
                if(minus[m]*minus[m+1]>=plus[p]*plus[p-1]){
                ans *= ((minus[m]*minus[m+1])%mod);
                ans %= mod;
                m+=2;
                count+=2;
                }
                else break;
            }
            if((p<0 && m+1<minus.size()) && count+2<=k){
                ans *= ((minus[m]*minus[m+1])%mod);
                ans %= mod;
                m+=2;
                count+=2;                
            }
            if(count+1<=k && p>=0){
                ans *= ((plus[p])%mod);
                ans %= mod;
                p--;
                count++;                
            }
        }
        cout << ans;
    }
}