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<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long
#define mod 1000000007

ll n,k;
ll a[212345];
char z;

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

int main(){
  scanf("%lld%lld",&n,&k);

  
  for(int i=0;i<n;i++){
    scanf("%lld",&a[i]);}
  
  
  sortup(a,n);
  
  ll right=n-1,left=0;
  ll count=k;
  ll ans=1;
  if(a[right]>=0 || k%2==0){
    //printf("right\n");
  while(count>0){
    if(count>=2){
      if(a[right]*a[right-1]>a[left]*a[left+1] && !(k%2==1 && a[right-2]>0)){
	ans = ans*a[right]%mod*a[right-1]%mod;
	right-=2;}
      else{
	ans = a[left]*a[left+1]%mod*ans%mod;
	left+=2;}
      count-=2;}
    else{
      ans = ans*a[right]%mod;
      right--;
      count--;}
  }
  }
  else{
    //printf("left\n");
    while(count>0){
      if(count>=2){
	ans = a[right]*a[right-1]%mod*ans%mod;
	right-=2;
	count-=2;}
      else{
	ans*=a[right];
	right--;
	ans=(ans+mod)%mod;
	count--;}
    }}
    
  printf("%lld\n",ans);
  
  return 0;
}