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 rep(i, n) for (int i = 0; i < (int)(n); i++)
#define PI 3.14159265358979323846
#define int long long
constexpr long long INF = numeric_limits<long long>::max() / 2;
constexpr int MOD = 1000000007;
using Graph = vector<vector<int>>;



signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int N,K;
    cin>>N>>K;

    int A[N];
    rep(i,N){
        cin>>A[i];
    }
    sort(A,A+N);
    int x=N;
    rep(i,N){
        if(A[i]>=0){
            x=i;
            break;
        }
    }
    int ans=1;
    int y=N-x;
    if(y==0&&K%2==1){
        int l=x-1;
        int r=x;
        while(r-l-1<K){
            if(r==N){
                ans*=A[l];
                l--;
                ans%=MOD;
            }
            else if(abs(A[r])<abs(A[l])){
                ans*=A[r];
                r++;
                ans%=MOD;
            }else{
                ans*=A[l];
                l--;
                ans%=MOD;
            }
        }
    }else{
        int l=0;
        int r=N-1;
        while(l+N-1-r<K){
            if(K-(l+N-1-r)==1){
                ans*=A[r];
                l++;
            }
            else if(A[l]*A[l+1]>A[r]*A[r-1]){
                ans*=A[l];
                ans%=MOD;
                ans*=A[l+1];
                ans%=MOD;
                l+=2;
            }else{
                ans*=A[r];
                ans%=MOD;
                ans*=A[r-1];
                ans%=MOD;
                r-=2;
            }
        }
    }
    


    while(ans<0)ans+=MOD;
    ans%=MOD;
    cout << ans << endl;




}