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;
 
typedef vector <int> vi;
typedef pair< int ,int > pii;
#define PB push_back
#define MP make_pair
#define FF first
#define SS second
#define ll long long
#define MOD 1000000007
#define OJ \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
int main(){
    FIO;
    //OJ;
    ll n,k,ne=0,cal=0;
    cin>>n>>k;
    vector<ll> v(n);
    for(ll &x:v){
        cin>>x;
        if(x<0){
            ne++;
        }
    }
    sort(v.begin(),v.end(),greater<int>());
    ll pro=1;
    if(ne==n){
        for(int i=0;i<k;i++){
            pro=(pro*v[i])%MOD;
        }
        cout<<pro+MOD;
        return 0;
    }
    if(ne%2==1)
        ne--;
    for(int i=n-1;i>=n-ne;i--)
        v[i]=abs(v[i]);
    sort(v.begin(),v.end(),greater<int>());
    for(int i=0;i<k;i++){
        if(v[i]==0)
            continue;
        pro=((pro*v[i])+MOD)%MOD;
    }
    cout<<pro;

}