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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
    return *(int *)b - *(int *)a;
}

int main()
{

    int n;
    int num;
    long long A[200001];

    scanf("%d", &n);
    scanf("%d", &num);
    for (int i = 0; i < n; i++)
    {
        scanf("%lld", &A[i]);
    }

    qsort(A, n, sizeof(long long), compare_int);
    long long sum = 1;
    if (num % 2 == 1)
    { //奇数の時
        if (A[0] < 0)
        {
            //最大値が負　全部ふ
            for (int i = 0; i < num; i++)
            {
                sum = sum * A[n - i - 1]%1000000007;
            }
        }else if(A[0]>=0){
            sum = A[0];
            int front=1;
            int last = n;
            for (int i = 0; i < num - 1; i = i + 2)
            {
                if(A[front] * A[front + 1]>A[last - 1] * A[last - 2]){
                    sum = (sum * A[front] % 1000000007 )* A[front + 1] % 1000000007;
                    front += 2;
                }
                else
                {
                    sum = (sum * A[last - 1] % 1000000007) * A[last - 2] % 1000000007;
                    last -= 2;
                }
            }
        }
    }
    else
    {
        int front = 0;
        int last = n;
        for (int i = 0; i < num - 1; i = i + 2)
        {
            if (A[front] * A[front + 1] > A[last - 1] * A[last - 2])
            {
                sum = (sum * A[front] % 1000000007)* A[front + 1] % 1000000007;
                front += 2;
            }
            else
            {
                sum = (sum * A[last - 1] % 1000000007) * A[last - 2] % 1000000007;
                last -= 2;
            }
        }
    }
    if(sum<0){
        sum += 1000000007;
    }
    printf("%lld", sum);
    printf("\n");
}