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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;using ll=long long;typedef vector<ll> vl;typedef pair<ll, ll> pll;
typedef set<ll> sl;typedef map<ll,ll> mll;typedef vector<pll> vll;
#define pb push_back
#define F first
#define S second
#define mp make_pair
#define endl '\n'
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define sa(v) sort(v.begin(),v.end())
#define sd(v) sort(v.begin(),v.end(),greater<>())
//----------------------------------------------//

int main()
{
    int ts=1;//cin>>ts;
    for(int tss=1;tss<=ts;tss++)
    {
        ll n,k;
        cin>>n>>k;
        vl a(n);
        loop(i,0,n) cin>>a[i];
        sa(a);
        ll ans=1;
        for(int i=n-k;i<n;i++) {ans*=a[i];ans%=1000000007;}
        ll tmp=ans;
        for(int i=0;i<k;i++)
        {
            tmp*=a[i];
            tmp/=a[n-k+i];
            tmp%=1000000007;
            while(tmp<0) tmp+=1000000007;
            ans=max(tmp,ans);
        }
        ans%=1000000007;
        while(ans<0) ans+=1000000007;
        cout<<ans<<endl;
    }
}
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///Struggle goes here....