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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*
import kotlin.math.abs

fun main() {
    var (n,k) = readLine()!!.split(" ").map { it.toInt() }
    val arr = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
    val mod = (1e9+7).toLong()
    val po = arr.count { it >= 0 }
    var ans = 1L

    if((po == 0 && k % 2 != 0) || k == n) {
        val list = arr.sortedBy { abs(it) }
        for(i in 0 until k) {
            ans *= (mod + list[i])
            ans %= mod
        }
    } else {
        var poArr = arr.filter { it >= 0 }.sortedDescending()
        val negArr = arr.filter { it < 0 }.sorted()
        var pp = 0
        var np = 0
        var cnt = 0

        if(k % 2 != 0) {
            //poArr 에서 하나 뽑는다.
            ans *= poArr[0]
            poArr = poArr.subList(1, poArr.size)
            k--
        }

        val pq = PriorityQueue<Long>(compareByDescending { it })

        while ( poArr.size - pp >= 2) {
            val v = (poArr[pp] * poArr[pp + 1]) % mod
            pq.add(v)
            pp+=2
        }

        while ( negArr.size - np >= 2) {
            val v = (negArr[np] * negArr[np + 1]) % mod
            pq.add(v)
            np+=2
        }

        while(cnt * 2 < k) {
            val v = pq.poll()
            ans *= v
            ans %= mod
            ++cnt
        }
    }
    print(ans)

}