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<stdio.h>
#include<iostream>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<math.h>
#define pans ((long long)(pow(10,9))+7)
#define rep(i,n) for(int i = 0;i < (n);++i)
using namespace std;

int main(){
    long long n,k;
    long long ans = 1;
    cin >> n >> k;
    int s = 0,e = n-1;
    bool f = false;
    vector<long long> A(n);
    rep(i,n) cin >> A[i];
    sort(A.rbegin(),A.rend());
    if(k%2 != 0){
        if(A[s] < 0) f = true;
        ans *= A[s];
        s++;
    }
    rep(i,int(k/2)){
        if(k == 1)break;
        if(A[s]*A[s+1] > A[e]*A[e-1] || f == true){
            ans *= A[s]*A[s+1];
            s += 2;
        }else{
            ans *= A[e]*A[e-1];
            e -= 2;
        }
        if(ans > pans)ans %= pans;
        printf("%lld\n",ans);
    }
    if(ans < 0){
        ans = pans - labs(ans)%pans;
    }else{
        ans = ans%pans;
    }
    printf("%lld",ans);
}