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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
  public static void main(String[] args){
	Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    //long[] a = new long[n];
    int[] a = new int[n];
    long ans = 1;
    long MOD = 1000000007;
    boolean allminus = true;
    for(int i=0;i<n;i++){
      //a[i] = sc.nextLong();
      a[i] = sc.nextInt();
      if( a[i] > -1 ){
        allminus = false;
      }
    }
    Arrays.sort(a);
    
    int head = 0;
    int tail = n-1;
    int count = 0;
    while(true){
      if( count + 2 > k ){
        break;
      }
      if( a[tail] * a[tail-1] >= a[head] * a[head+1] ){
        ans = ( ans * a[tail] % MOD ) * a[tail-1] % MOD;
        tail -= 2;
      }else{
        ans = ( ans * a[head] % MOD ) * a[head+1] % MOD;
        head += 2;
      }
      count += 2;      
    }
    if( count + 1 == k ){
      ans = ans * a[tail] % MOD;
    }
    
    if( allminus && k % 2 == 1 ){
      ans = 1;
      for(int i=1;i<k+1;i++){
        ans =  ans * a[n-i] % MOD;
      }
    }
    if( ans < 0 ){
      ans += MOD;
    }
    System.out.println(ans);  
  }
}