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;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::f64::consts::*;

#[allow(unused)]
const INF: usize = std::usize::MAX / 4;
#[allow(unused)]
const M: isize = 1000000007;

fn main() {
    input! {
        n: usize,
        k: usize,
        mut a: [isize; n],
    }

    let mut a_plus = vec![];
    let mut a_minus = vec![];
    for &ai in &a {
        if ai > 0 {
            a_plus.push((ai, 1));
        } else if ai < 0 {
            a_minus.push(ai);
        }
    }

    let np = a_plus.len();
    let nm = a_minus.len();

    if np + nm < k {
        println!("0");
        return;
    }

    a_minus.sort();

    for i in (0..nm).step_by(2) {
        if i + 1 < nm {
            a_plus.push((a_minus[i] * a_minus[i + 1], 2));
        }
    }

    a_plus.sort();
    a_plus.reverse();

    let mut count = 0;
    let mut result = 1;
    for &(ai, c) in &a_plus {
        if c == 2 {
            if count + c <= k {
                result = result * ai % M;
                count += c;
            }
        } else {
            result = result * ai % M;
            count += c;
        }
        if count == k {
            break;
        }
    }

    if count < k {
        a.sort_by_key(|&ai| ai.abs());
        result = 1;
        for i in 0..k {
            result = result * a[i].abs() % M;
        }
        result = M - result;
    }

    println!("{}", result);
}