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: Accepted)

use proconio::{fastout, input};

use std::cmp::{max, min};

fn pow(mut x: i64, mut y: i64, z: i64) -> i64 {
    let mut r = 1;
    while y > 0 {
        if y & 1 == 1 {
            r = r * x % z;
        }
        x = x * x % z;
        y >>= 1;
    }
    r
}

fn inv(x: i64, y: i64) -> i64 {
    pow((x % y + y) % y, y - 2, y)
}

#[fastout]
fn main() {
    const M: i64 = 1_000_000_007;
    input! {
        n: usize, mut k: usize,
        mut a: [i64; n],
    }
    let mut pos = 0;
    let mut zero = 0;
    let mut neg = 0;
    for &x in a.iter() {
        if x > 0 {
            pos += 1;
        } else if x < 0 {
            neg += 1;
        } else {
            zero += 1;
        }
    }
    if k > pos + neg {
        println!("0");
        return;
    }
    a.sort();
    // let mut l = 0;
    // let mut acc = 1;
    // while k > pos && k >= 2 {
    //     acc *= acc * (a[l] * a[l + 1] % M) % M;
    //     l += 2;
    //     k -= 2;
    // }
    // for i in n - k..n {
    //     acc *= acc * a[i] % M;
    // }
    // while k >= 2 && l < neg {
    //     if a[l] * a[l + 1] >= a[n - k] * a[n - k + 1] {
    //     } else {
    //         break;
    //     }
    //     acc = acc * a[l] % M * a[l + 1] % M;
    //     acc = acc * inv(a[n - k], M) % M * inv(a[n - k + 1], M) % M;
    //     l += 2;
    //     k -= 2;
    // }
    // println!("{}", acc);
    // return;

    if pos == 0 {
        if k % 2 == 0 {
            let mut acc = 1;
            for i in 0..k {
                acc = acc * a[i] % M;
            }
            acc = (acc % M + M) % M;
            if zero > 0 && acc < 0 {
                println!("0");
            } else {
                println!("{}", acc);
            }
        } else {
            let mut acc = 1;
            for i in n - k..n {
                acc = acc * a[i] % M;
            }
            acc = (acc % M + M) % M;
            if zero > 0 && acc < 0 {
                println!("0");
            } else {
                println!("{}", acc);
            }
        }
        return;
    }
    {
        let mut l = 0;
        let mut acc = 1;
        while k > pos {
            k -= 2;
            acc = acc * a[l] % M * a[l + 1] % M;
            l += 2;
        }
        for i in n - k..n {
            acc = acc * a[i] % M;
        }
        let mut ans = acc;
        for i in 0..(neg - l) / 2 {
            if n - k + i * 2 + 1 >= n {
                break;
            }
            if a[l + i * 2] * a[l + i * 2 + 1] > a[n - k + i * 2] * a[n - k + i * 2 + 1] {
            } else {
                break;
            }
            acc = acc * a[l + i * 2] % M * a[l + i * 2 + 1] % M;
            acc = acc * inv(a[n - k + i * 2], M) % M * inv(a[n - k + i * 2 + 1], M) % M;
            // ans = max(ans, acc);
        }
        acc = (acc % M + M) % M;
        println!("{}", acc);
    }
}