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 ;
#define ll long long
#define pb push_back
#define __ ios_base::sync_with_stdio(false); cin.tie(NULL);
ll mod = 1000 * 1000 * 1000 + 7 ;
int main() {
 //   __

    ll  n , k ;
    cin >> n >> k ;
    vector<ll> v(n) ;
    ll i ;
    ll pos , neg , zr ;
    pos = 0 , neg = 0 , zr = 0 ;
    for (i = 0 ; i< n; i++) {
        cin >> v[i] ;
        if (v[i]==0) zr++ ;
        else if (v[i]<0) neg++ ;
        else pos++ ;
    }
    ll nneg = min(k,neg) ;
    if ( nneg - nneg%2 + pos>=k ) {
        ll ans = 1 ;
        sort(v.begin(),v.end()) ;
        i = 0 ;
        ll j = n-1 ;
        ll rem = k ;
        vector<ll>tmp ;
        while(rem) {
            if (rem>1) {
                if (v[i]*v[i+1]>=v[j]*v[j-1]  or (v[j-2]<0 and rem!=2)) {
                    tmp.pb(v[i]) , tmp.pb(v[i+1]) ;
                    i+=2 ;
                }
                else {
                    tmp.pb(v[j]) , tmp.pb(v[j-1]) ;
                    j-=2 ;
                }
                rem-=2 ;
            }
            else {
                rem-- ;
                tmp.pb(v[j]) ;
            }

        }
        for (i = 0 ; i< k ; i++) ans = (ans * abs(tmp[i]))%mod ;
        cout << ((ans%mod)+mod)%mod<< endl ;
        return 0 ;
    }
    ll ans = 1;
    for (i = 0 ; i< n ; i++) {
        v[i]= abs(v[i]) ;
    }
    sort(v.begin(),v.end()) ;
    for (i = 0 ; i< k ;i++) ans =( (( ans * (-v[i]) )%mod  ) + mod ) %mod ;
    cout << ((ans%mod)+mod)%mod << endl ;
   // cout << ((-ans%mod) + mod ) %mod << endl ;
    return 0 ;
}