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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{input, fastout};

#[fastout]
fn main() {
    input!{ n: usize, k: usize, mut a: [isize; n] };
    a.sort_by_key(|v|{ v.abs(); });
    let mut ans: isize = 1;
    const N: isize = 1000000000 + 7;
    if k % 2 != 0 && (&a).into_iter().all(|v|{ *v < 0 }) {
        for i in 0 .. k {
            ans = ((ans % N) * (a[i] % N)) % N;
        }
    } else {
        for i in 0 .. k {
            ans = ((ans % N) * (a[a.len() - i - 1] % N)) % N;
        }
        if ans < 0 {
            ans = 1;
            for i in 0 .. k - 1 {
                ans = ((ans % N) * (a[a.len() - i - 1] % N)) % N;
            }
            let b = if a[k - 1] < 0 {
                a[k-1..].into_iter().max_by_key(|v|{ -**v })
            } else {
                a[k-1..].into_iter().max()
            }.unwrap();
            ans = ((ans % N) * (b % N)) % N;
        }
    }
    if ans < 0 {
        ans = N + ans;
    }
    println!("{}", ans % N);
}