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)

#![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 (mut l, mut r) = (n - k, n - 1);
    let is_p = a[n - k..].iter().filter(|&&x| x < 0).count() % 2 == 0;
    loop {
        let s = (a[l % n] * a[(l + 1) % n]).abs();
        let t = a[(r + 1) % n] * a[(r + 2) % n];
        let b = if is_p {
            s < t
        } else {
            s > t
        };
        if b {
            l += 2;
            r += 2;
        } else {
            break;
        }
    }
    let mut ans = 1;
    for i in l..=r {
        ans *= a[i % n];
        ans %= p;
    }
    if ans < 0 {
        ans = p + ans;
    }
    println!("{}", ans);
}