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: Runtime Error)

use std::cmp::*;
use std::io::*;
const MOD: i64 = 1_000_000_007;

fn inv(mut x: i64, mut e: i64) -> i64 {
    let mut res = 1;
    while e > 0 {
        if e & 1 == 1 {
            res = res * x % MOD;
        }
        x = x % MOD * x % MOD;
        e >>= 1;
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();
    let mut a: Vec<i64> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();

    let mut b = Vec::new();
    let mut cnt = 0;
    let mut zero = 0;
    for i in 0..n {
        if a[i] == 0 {
            zero += 1;
            continue;
        }
        b.push((a[i].abs(), a[i]));
        if a[i] < 0 {
            cnt += 1;
        }
    }
    b.sort();
    b.reverse();

    if cnt == n {
        let mut ans = 1;
        if k % 2 == 0 {
            a.sort();
            for i in 0..k {
                ans *= (MOD + a[i]) % MOD;
                ans %= MOD;
            }
        } else {
            a.sort();
            a.reverse();
            for i in 0..k {
                ans *= (MOD + a[i]) % MOD;
                ans %= MOD;
            }
        }
        println!("{}", ans);
    } else {
        let mut ans = 1;
        let mut mi = 0;
        for i in 0..k {
            ans = (ans * (MOD + b[i].1) % MOD) % MOD;
            if b[i].1 < 0 {
                mi += 1;
            }
        }
        if mi % 2 == 0 {
            println!("{}", ans);
        } else {
            let mut pmin = 1 << 60;
            let mut mmax = -(1 << 60);
            for i in 0..k {
                if b[i].1 < 0 {
                    mmax = max(mmax, b[i].1);
                }
                if b[i].1 > 0 {
                    pmin = min(pmin, b[i].1);
                }
            }
            let mut pmax = 0;
            let mut mmin = 1 << 60;
            for i in k..n {
                if b[i].1 < 0 {
                    mmin = min(mmin, b[i].1);
                }
                if b[i].1 > 0 {
                    pmax = max(pmax, b[i].1);
                }
            }
            // マイナスを減らしてプラスを入れる
            let val1 = ans * inv(mmax, MOD - 2) % MOD * pmax % MOD;
            // プラスを減らしてマイナスを入れる
            let val2 = ans * inv((MOD + pmin) % MOD, MOD - 2) % MOD * (MOD + mmin) % MOD;
            println!("{}", max(val1, val2));
        }
    }
}