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 itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{fastout, input, marker::*};
use std::cmp::*;
use std::collections::*;
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: usize, k: usize,
        mut a: [i64; n],
    }
    let m = 1000000007i64;

    if k % 2 == 1 && a.iter().all(|v| *v <= 0) {
        a.sort();
        a.reverse();
        let ans = a.iter().take(k).fold(1, |a, v| (a * v).rem_euclid(m));
        println!("{}", ans);
        return;
    }

    a.sort_by_key(|v| Reverse((v.abs(), *v)));
    let sign = a.iter().cloned().map(i64::signum).collect_vec();
    let cum_sign = sign
        .iter()
        .scan(1, |a, &v| {
            *a *= v;
            Some(*a)
        })
        .collect_vec();

    let mut remove_plus = n;
    let mut add_minus = n;
    let mut remove_minus = n;
    let mut add_plus = n;
    a.push(0);
    for i in (0..=k - 1).rev() {
        if a[i] > 0 {
            remove_plus = i;
            break;
        }
    }
    for i in (0..=k - 1).rev() {
        if a[i] < 0 {
            remove_minus = i;
            break;
        }
    }
    for i in k..n {
        if a[i] > 0 {
            add_plus = i;
            break;
        }
    }
    for i in k..n {
        if a[i] < 0 {
            add_minus = i;
            break;
        }
    }

    let (mut remove, mut add) = if a[add_minus] * a[remove_minus] > a[add_plus] * a[remove_plus] {
        (remove_plus, add_minus)
    } else {
        (remove_minus, add_plus)
    };
    if cum_sign[k - 1] != -1 {
        remove = n;
        add = n;
    }

    let mut ans = 1i64;
    for i in 0..k {
        if i == remove {
            continue;
        }
        ans *= a[i].abs();
        ans %= m;
    }
    for i in k..n {
        if i == add {
            ans *= a[i].abs();
            ans %= m;
            break;
        }
    }
    println!("{}", ans);
}