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)

use proconio::{input, fastout};

const MOD: i64 = 1_000_000_007;

#[fastout]
// TODO: Don't use xxx.iter().take(k).sum(). it may be overflow.
fn main() {
    input! {
        n: usize,
        k: usize,
        a: [i64; n],
    };

    let mut a = a;
    let mut ans = 1;
    a.sort_by(|a, b| a.abs().partial_cmp(&b.abs()).unwrap());
    let a: Vec<i64> = a.into_iter().rev().collect();

    // let mut ans = a.iter().take(k).sum::<i64>() % MOD;
    for (i, &v) in a.iter().enumerate() {
        if i == k {
            break;
        }
        ans *= v;
        ans %= MOD;
    }

    if n == k {
        println!("{}", ans);
        return;
    }

    if ans < 0 {
        let mut is_all_neg = true;
        let mut has_zero = false;
        for &i in a.iter() {
            if i > 0 {
                is_all_neg = false;
            } else if i == 0 {
                has_zero = true;
            }
        }

        if !is_all_neg && k%2 == 1 {
            let do_switch_to_positive: bool;

            if a.iter().nth(k).unwrap() > &0 {
                do_switch_to_positive = true;
            } else {
                do_switch_to_positive = false;
            }

            for &i in a.iter().take(k).rev() {
                if i < 0 && do_switch_to_positive {
                    ans /= i;
                    break;
                } else if i > 0 && !do_switch_to_positive {
                    ans /= i;
                    break;
                }
            }

            for &j in a.iter().skip(k) {
                if j > 0 && do_switch_to_positive || j < 0 && !do_switch_to_positive {
                    ans *= j;
                    ans %= MOD;
                    break;
                }
            }
        } else if !is_all_neg && k%2 == 0 {
            for &v in a.iter().take(k) {
                if v > 0 {
                    ans /= v;
                    break;
                }
            }

            for &j in a.iter().skip(k) {
                if j < 0 {
                    ans *= j;
                    ans %= MOD;
                    break;
                }
            }

        } else {
            ans = 1;
            for &v in a.iter().rev().take(k) {
                ans *= v;
                ans %= MOD;
            }
        }

        if is_all_neg && has_zero {
            ans = 0;
        }
    }

    if ans < 0 {
        ans += MOD;
    }
    println!("{:?}", ans);
}