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>
#define MOD 1000000007
long int PROD(long int a,long int b){return (a+MOD)%MOD*((b+MOD)%MOD)%MOD;}//a*b
void quicksort(long int a[],int s,int t){
  if(s<t){
    long int i=s,j=t,p=a[(s+t)/2];
    while(1){
      while(a[i]>p)i++;
      while(p>a[j])j--;
      if(i>=j)break;
      a[i]^=a[j]^=a[i]^=a[j];
      i++;j--;
    }
    quicksort(a,s,i-1);quicksort(a,j+1,t);
  }
}
int main(){
  long int p[200000],m[200000],i,j,k,n,x=0,y=0,r=1;
  scanf("%ld%ld",&n,&k);
  for(i=0;i<n;i++){scanf("%ld",&j);if(j>0){p[x]=j;x++;}else{m[y]=-j;y++;}}
  quicksort(p,0,x-1);quicksort(m,0,y-1);
  if(x==0&&k%2){for(i=1;i<=k;i++)r=PROD(r,-m[y-i]);printf("%ld",r);return 0;}
  j=k<y?k-k%2:y-y%2;
  while(j>1&&k-j+1<x&&p[k-j]*p[k-j+1]>m[j-1]*m[j-2])j-=2;
  for(i=0;i<k-j;i++)r=PROD(r,p[i]);
  for(i=0;i<j;i++)r=PROD(r,m[i]);
  printf("%ld",r);
}