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: Runtime Error)

use proconio::input;

#[proconio::fastout]
pub fn main() {
    const M: i64 = 1000000007;
    input! {
    n:usize,
    k:usize,
    a:[i64;n],
    }
    let mut apos: Vec<i64> = a.iter().filter(|&x| x >= &0).cloned().collect();
    let mut aneg: Vec<i64> = a.iter().filter(|&x| x <= &0).cloned().collect();
    let ans = if apos.len() == 0 && k % 2 == 1 {
        //必ず非正。絶対値を最小化したい
        aneg.sort_by_key(|&x| std::cmp::Reverse(x));
        aneg.iter().take(k).fold(1, |pr, v| (pr * v).rem_euclid(M))
    } else {
        //必ず非負
        apos.sort();
        aneg.sort_by_key(|&x| std::cmp::Reverse(x));
        let mut tmp = 1;
        if k % 2 == 1 {
            tmp *= apos.pop().unwrap();
        }
        for _ in 0..k / 2 {
            if apos.len() <= 1 {
                tmp = (tmp * aneg.pop().unwrap()).rem_euclid(M);
                tmp = (tmp * aneg.pop().unwrap()).rem_euclid(M);
            } else if aneg.len() <= 1 {
                tmp = (tmp * apos.pop().unwrap()).rem_euclid(M);
                tmp = (tmp * apos.pop().unwrap()).rem_euclid(M);
            } else {
                if apos[apos.len() - 2] * apos[apos.len() - 1] > aneg[aneg.len() - 2] * aneg[aneg.len() - 2] {
                    tmp = (tmp * apos.pop().unwrap()).rem_euclid(M);
                    tmp = (tmp * apos.pop().unwrap()).rem_euclid(M);
                } else {
                    tmp = (tmp * aneg.pop().unwrap()).rem_euclid(M);
                    tmp = (tmp * aneg.pop().unwrap()).rem_euclid(M);
                }
            }
        }
        tmp
    };
    println!("{}", ans);
}