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: Runtime Error)

#include<stdio.h>
#define rep(i,n)for(int i=0;i<(int)(n);i++)
#define MOD 1000000007
typedef long long LL;
LL n,k,p[200009],m[200009],pl=0,mn=0,ans=1,t;
int acsLL(const void*a,const void*b){return*(LL*)a<*(LL*)b?-1:*(LL*)a>*(LL*)b?1:0;}
void mr(int i){ans=ans*(MOD-m[i])%MOD;}
void pr(int i){ans=ans*p[i]%MOD;}
main(){
  scanf("%lld%lld",&n,&k);
  rep(i,n){
    scanf("%lld",&t);
    if(t>=0)p[pl++]=t;
    else m[mn++]=-t;
  }
  qsort(p,pl,8,acsLL);
  qsort(m,mn,8,acsLL);
  if(n==k){rep(i,mn)mr(i);rep(i,pl)pr(i);}
  else if(mn==n&&k%2)rep(i,k)mr(i);
  else{while(k){
    if(k>=2&&mn>=2&&(pl<=1||(pl>=2&&m[mn-1]*m[mn-2]>=p[pl-1]*p[pl-2]))){
      pr(--mn);pr(--mn);
      ans=ans*m[--mn]%MOD;
      k=k-2;
    }
    else{pr(--pl);k--;}
  }}
  printf("%lld",ans);
}