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 mp make_pair
#define pb push_back
#define f first
#define s second
#define REP(i,a,b) for(int i=a; i<=b; i++)
#define LIM (1<<30)
using namespace std;

typedef long long int ll;
typedef pair<ll,ll> pi;

const ll mod = 1e9+7;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll n,k;
    cin >> n >> k;
    vector<ll> x(n);
    ll mx=-LIM;
    REP(i,0,n-1){
        cin >> x[i];
        mx=max(x[i],mx);
    }
    sort(x.begin(),x.end());
    ll ans=1;
    if(k&1){
        if(mx<=0){
            if(mx==0){
                cout << "0";
                return 0;
            }
            ll cur=1;
            REP(i,0,k-1){
                cur*=(x[n-i-1]);
            }
            cur=(cur+mod)%mod;
            cout << cur;
            return 0;
        }
        else{
            ans*=x[n-1];
          	n--;
        }
    }
    ll pairs=k/2;
    ll left=0,right=n-1;
    REP(i,0,pairs-1){
        ll c1=x[left]*x[left+1];
        ll c2=x[right]*x[right-1];
        if(c1>c2) left+=2;
        else right-=2;
        ans=(ans*max(c1,c2)%mod)%mod;
    }
    ans%=mod;
    if(ans<0) ans+=mod;
    ans%=mod;
    cout << ans;
    return 0;
}