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)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;

#[fastout]
fn main() {
    input! {
    n: usize,
    k: usize,
    mut a: [i64; n]
    }
    let p = 1_000_000_007;
    a.sort();
    let n_a = a.iter().filter(|&&x| x < 0).copied().collect::<Vec<_>>();
    let p_a = a.iter().filter(|&&x| x >= 0).rev().copied().collect::<Vec<_>>();
    if k % 2 == 1 && p_a.is_empty() {
        let ans = (p + n_a[n_a.len() - k..].iter().fold(1, |acc, x| acc * x % p)) % p;
        println!("{}", ans);
        return;
    }
    let mut s = vec![];
    let (mut ni, mut pi) = (0, 0);
    while s.len() < k {
        let nv = if ni < n_a.len() {
            n_a[ni] * if let Some(x) = n_a.get(ni + 1) { *x } else { 1 }
        } else {
            std::i64::MIN
        };
        let pv = if pi < p_a.len() {
            if s.len() < k - 1 {
                p_a[pi] * if let Some(x) = p_a.get(pi + 1) { *x } else { 1 }
            } else {
                std::i64::MAX
            }
        } else {
            std::i64::MIN
        };
        if nv <= pv {
            s.push(p_a[pi]);
            pi += 1;
        } else {
            s.push(n_a[ni]);
            s.push(n_a[ni + 1]);
            ni += 2;
        }
    }
    let ans = (p + s.iter().fold(1, |acc, x| acc * x % p)) % p;
    println!("{}", ans);
}