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: Accepted)

use proconio::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        k: usize,
        a: [i64; n],
    }
    let mut x = vec![];
    let mut y = vec![];
    let mut z = 0;
    for a in a {
        if a == 0 {
            z += 1;
        } else if a > 0 {
            x.push(a);
        } else {
            y.push(a);
        }
    }
    if k > n - z {
        println!("0");
        return;
    }
    const MOD: i64 = 1_000_000_007;
    if k == n - z {
        if y.len() % 2 == 1 && z > 0 {
            println!("0");
            return;
        }
        let mut ans = 1;
        for x in x.iter().chain(y.iter()) {
            ans = ans * (*x + MOD) % MOD;
        }
        println!("{}", ans);
        return;
    }
    x.sort();
    y.sort();
    if k % 2 == 1 && x.is_empty() {
        if z > 0 {
            println!("0");
            return;
        }
        let mut ans = 1;
        for y in y.iter().rev().take(k) {
            ans = ans * (*y + MOD) % MOD;
        }
        println!("{}", ans);
        return;
    }
    // +
    let (mul, k) = if k % 2 == 1 {
        (x.pop().unwrap(), (k - 1) / 2)
    } else {
        (1, k / 2)
    };
    x.reverse();
    let mut x = x.chunks_exact(2).map(|x| x[1] * x[0]).collect::<Vec<_>>();
    let mut y = y.chunks_exact(2).map(|y| y[1] * y[0]).collect::<Vec<_>>();
    while x.len() < k {
        x.push(0);
    }
    while y.len() < k {
        y.push(0);
    }
    let mut i = 0;
    let mut j = 0;
    let mut ans = mul;
    while i + j < k {
        if x[i] > y[j] {
            ans = ans * (x[i] % MOD) % MOD;
            i += 1;
        } else {
            ans = ans * (y[j] % MOD) % MOD;
            j += 1;
        }
    }
    println!("{}", ans);
}

fn main() {
    run();
}