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)]

use std::cmp::Reverse;
use proconio::{input, fastout};
use proconio::marker::{Chars, Usize1};

const M: i64 = 1000000007;

#[fastout]
fn main() {
    input! {
        n: usize,
        k: usize,
        xs: [i64; n],
    }

    let mut negxs: Vec<i64> = xs.iter().cloned().filter(|&x| x < 0).collect();

    let nneg = negxs.len();
    let npos = xs.iter().filter(|&&x| x > 0).count();

    if npos == 0 {
        if k % 2 != 0 {
            negxs.sort_unstable_by_key(|&x| Reverse(x));
            let ans = negxs.iter().take(k).fold(1, |a, &x| {
                (a * x) % M
            });
            println!("{}", M + ans);
            return;
        }
    }

    let mut ys: Vec<(i64, usize)> = xs
        .iter()
        .filter(|&&x| x > 0)
        .map(|&x| (x, 1))
        .collect();

    negxs.sort_unstable();
    for i in 0..(negxs.len() / 2) {
        ys.push((negxs[i * 2] * negxs[i * 2 + 1], 2))
    }

    ys.sort_unstable_by_key(|&(x, _)| Reverse(x));

    let mut ans = 1;
    let mut ntaken = 0;
    for i in 0..ys.len() {
        if ntaken == k {
            break;
        }
        let (x, w) = ys[i];
        if ntaken + w <= k {
            ntaken += w;
            ans = (ans * x) % M;
        }
    }

    println!("{}", ans);
}