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 std::io;
use std::io::prelude::*;
use std::io::BufReader;
use std::str::FromStr;

fn read_vec<T: FromStr>(s: String) -> Result<Vec<T>, T::Err> {
    s.split(' ').map(|x| x.parse()).collect()
}

fn main() {
    const M: i64 = 1000_000_000 + 7;
    let mut lines = BufReader::new(io::stdin()).lines().map(|l| l.unwrap());
    let nk: Vec<usize> = read_vec(lines.next().unwrap()).unwrap();
    let mut v: Vec<i64> = read_vec(lines.next().unwrap()).unwrap();
    let mut P: Vec<_> = v.iter().filter(|&&x| x > 0).collect();
    let mut N: Vec<_> = v.iter().filter(|&&x| x < 0).map(|x| -x).collect();
    P.sort_unstable();
    N.sort_unstable();
    // kをなるべくPで埋めて、あまりを2の倍数に切り上げて、それがNより少ない時
    let can_positive = if P.len() == 0 {
        nk[1] & 1 == 0 && nk[1] <= N.len() // Pがゼロ個なら、kが偶数で、Nはkより多い必要がある
    } else {
        (nk[1] - P.len() + 1) & !1 <= N.len() // Pが一つ以上あれば、k-Pを2の倍数に切り上げた数より、Nが多い必要がある
    };
    if !can_positive {
        if v.contains(&0) {
            // ゼロにできる時
            println!("0");
        } else {
            // 負の数になる時
            v.sort_unstable_by_key(|x| {
                let y = x >> 31;
                x ^ y - y
            });
            let ans = v
                .iter()
                .take(nk[1]) // 小さい方から集めて
                .fold(1, |acc, x| acc * (x % M) % M);
            println!("{}", (ans + M) % M);
        }
        return;
    }
    let mut ans = 1;
    let mut k = nk[1];
    if k & 1 == 1 {
        ans *= P.pop().unwrap();
        k -= 1;
    }
    let a = P.pop();
    let b = P.pop();
    let mut ab = a.and_then(|a| b.map(|b| a * b));
    let c = N.pop();
    let d = N.pop();
    let mut cd = c.and_then(|c| d.map(|d| c * d));
    loop {
        if k == 0 {
            break;
        }
        let pn = match ab {
            Some(ab) => match cd {
                Some(cd) => ab > cd,
                _ => true,
            },
            None => false,
        };
        if pn {
            ans *= ab.unwrap() % M;
            ans %= M;
            let a = P.pop();
            let b = P.pop();
            ab = a.and_then(|a| b.map(|b| a * b));
        } else {
            ans *= cd.unwrap() % M;
            ans %= M;
            let c = N.pop();
            let d = N.pop();
            cd = c.and_then(|c| d.map(|d| c * d));
        }
        k -= 2;
    }
    println!("{}", (ans + M) % M);
}