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 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) = a
        .iter()
        .sorted_by(|a, b| b.abs().cmp(&a.abs()))
        .partition::<Vec<i64>, _>(|&&a| a >= 0);
    let (pl, nl) = (pos.len(), neg.len());

    if pl == 0 && k % 2 == 1 {
        return neg.into_iter().skip(n - k).collect_vec();
    }

    let mut pi = 0;
    let mut ni = 0;
    let mut rest = k;
    let mut ret = Vec::new();
    while rest > 0 {
        let push_p = if pi + 1 < pl && ni + 1 < nl {
            pos[pi] * pos[pi + 1] >= neg[ni] * neg[ni + 1]
        } else {
            !(ni + 1 < pos.len())
        };

        if push_p || rest == 1 {
            ret.push(pos[pi]);
            pi += 1;
            rest -= 1;
        } else {
            ret.extend(&neg[ni..=ni + 1]);
            ni += 2;
            rest -= 2;
        }
    }

    ret
}