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 MOD 998244353
#define pb push_back
#define ll long long int
#define con 1000000007
using namespace std;



int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll i,j,m,n,k;
    ll t;
    cin>>n>>k;
    ll u=k;
    vector<pair<ll,ll>> a(n);
    for(i=0;i<n;i++)
    {
        cin>>m;
        if(m>=0)
            a[i].second=1;
        else
            a[i].second=-1;
        a[i].first=abs(m);
    }
    sort(a.begin(),a.end());
    reverse(a.begin(),a.end());
    ll ans=1;
    m=1;
    for(i=0;i<n;i++)
    {
        if(k==0)
            break;
        if(a[i].second==1)
        { ans=(ans*a[i].first)%con;
         k--;}
        else
        {
            if(m<=0)
            {
                if(k>1)
                {
                    k=k-2;
                    m=(abs(m)*a[i].first)%con;
                    ans=(ans*m)%con;
                    m=1;
                }
            }
            else
                m=0-a[i].first;
        }
    }
    if(k>0)
    {   k=u;
        sort(a.begin(),a.end());
        ans=1;
        for(i=0;i<k;i++)
        {
            ans=(ans*a[i].first*a[i].second)%con;
            ans=(ans+con)%con;
        }
    }
    
    cout<<ans;
}