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 itertools::Itertools;
use proconio::input;

fn main() {
    input!(n: usize, k: usize, a: [i64; n]);

    let ans = solve(n, k, a)
        .iter()
        .fold(1, |acc, &a| (acc * a).rem_euclid(1_000_000_007));

    println!("{}", ans)
}

fn solve(n: usize, k: usize, a: Vec<i64>) -> Vec<i64> {
    if n == k {
        return a;
    }

    let (pos, neg): (Vec<_>, _) = a
        .iter()
        .sorted_by(|a, b| b.abs().cmp(&a.abs()))
        .partition(|&&a| a >= 0);
    let (mut pos, mut neg) = (&pos[..], &neg[..]);

    if pos.len() == 0 && k % 2 == 1 {
        return neg.iter().copied().skip(n - k).collect_vec();
    }

    let mut rest = k;
    let mut ret = Vec::new();
    while rest > 0 {
        let push_pos = if pos.len() >= 2 && neg.len() >= 2 {
            pos[0] * pos[1] >= neg[0] * neg[1]
        } else {
            neg.len() < 2
        };

        if push_pos || rest == 1 {
            ret.push(pos[0]);
            pos = &pos[1..];
            rest -= 1;
        } else {
            ret.extend(&neg[..2]);
            neg = &neg[2..];
            rest -= 2;
        }
    }

    ret
}