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: Accepted)

import kotlin.math.abs

fun main(args: Array<String>) {
    val (N, K) = readLine()!!.split(' ').map { it.toInt() }
    val A = readLine()!!.split(' ').map { it.toLong() }
    val positiveA = A.filter { it >= 0 }.sortedDescending()
    val negativeA = A.filter { it < 0 }.sorted()

    if (N == K) {
        println(A.reduce { acc, l -> mul(acc, l) })
        return
    }
    if (positiveA.isEmpty() && K % 2 == 1) {
        println(negativeA.takeLast(K).reduce { acc, l -> mul(acc, l) })
        return
    }

    var positiveIdx = 0
    var negativeIdx = 0
    var ans = 1L
    var remK = K

    while (remK > 0) {
        if (remK == 1) {
            ans = mul(ans, positiveA[positiveIdx])
            break
        }

        val positiveX = if (positiveA.size - positiveIdx >= 2) {
            positiveA[positiveIdx] * positiveA[positiveIdx + 1]
        } else {
            0L
        }
        val negativeX = if (negativeA.size - negativeIdx >= 2) {
            negativeA[negativeIdx] * negativeA[negativeIdx + 1]
        } else {
            0L
        }

        if (positiveX >= negativeX) {
            ans = mul(ans, positiveA[positiveIdx])
            positiveIdx++
            remK--
        } else {
            ans = mul(ans, negativeA[negativeIdx])
            ans = mul(ans, negativeA[negativeIdx + 1])
            negativeIdx += 2
            remK -= 2
        }
    }

    println(ans)
}

fun mul(x: Long, y: Long): Long {
    val mod = 1000000007L
    return ((x * y) + mod * mod) % mod
}