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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), k = sc.nextInt();
        ArrayList<Long> p = new ArrayList<>();
        ArrayList<Long> m = new ArrayList<>();
        int plus = 0, minus = 0, zero = 0;
        int a;
        for(int i=0;i<n;i++){
            a = sc.nextInt();
            if(a == 0){
                zero++;
            }else if(a > 0){
                p.add((long)a);
                plus++;
            }else{
                m.add((long)a);
                minus++;
            }
        }
        Collections.sort(p);
        Collections.sort(m);
        if(zero != 0 && k > plus + minus) {   //0が絶対に入るとき
            System.out.println(0);
        }else if(k == plus + minus){
            if(minus % 2 == 1 && zero != 0){
                System.out.println(0);
            }else {     //全部掛けていいケース
                long ans = 1;
                for(Long pl : p){
                    ans *= pl;
                    ans %= 1000000007;
                }
                for(Long mn : m){
                    ans *= mn;
                    ans %= 1000000007;
                }
                ans += 1000000007;
                ans %= 1000000007;
                System.out.println(ans);
            }
        }else if(p.size() == 0){    //正の数がない時
            long ans = 1;
            if(k % 2 == 1) {    //奇数回掛けると負になるので小さい順に
                for (int i = 0; i < k; i++) {
                    ans *= m.get(m.size() - 1 - i);
                    ans %= 1000000007;
                }
            }else{      //偶数回掛けると正になるので大きい順に
                for (int i = 0; i < k; i++) {
                    ans *= m.get(i);
                    ans %= 1000000007;
                }
            }
            ans += 1000000007;
            ans %= 1000000007;
            System.out.println(ans);
        }else{
            long ans = 1;
            if(k % 2 == 1){     //奇数回掛ける時、正の数のうち最大のものは必ず使う
                ans *= p.get(p.size()-1);
                p.remove(p.size()-1);
                k--;
            }
            if(k != 0) {
            }
            System.out.println(ans);
        }
    }
}