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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

const input = require("fs").readFileSync("/dev/stdin", "utf8")

const mod = 1000000007
function mul (a, b) {
    return ((a >> 16) * b % mod * 65536 + (a & 65535) * b) % mod
}

;(input => {

    const lines = input.split('\n')
    const [N, K] = lines[0].split(" ").map(Number)
    const A = lines[1].split(" ").map(Number)

    A.sort((a, b) => Math.abs(b) - Math.abs(a))

    let sign = 0
    let last = 0
    let lastP = 0
    let p = 1
    for (let i = 0; i < K; i++) {
        if (A[i] < 0) {
            sign++
            last = i
        } else {
            lastP = i
        }
        p = mul(p, A[i])
    }

  if (N === K && sign % 2) {
        p = 1
        for (let i = N - 1; i >= N - K; i--) {
            p = mul(p, A[i])
            p = (p*mod<0)*mod+p%mod
        }
    } else if (p < 0) {
        if (A[K] >= 0) {
            p /= A[last]
            p = mul(p, A[K])
        }
        else {
            p /= A[lastP]
            p = mul(p, A[K])
        }
    }

    console.log(p % mod)

})(input)