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;
#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};
#[allow(unused_imports)]
use std::cmp::{min, max};

fn main() {
	input! {
		n: usize,
		k: usize,
		mut a: [isize; n],
	}

	a.sort_by_key(|&x| std::cmp::Reverse(x.abs()));
	let mut count: usize = 0;
	let mut pmax: usize = 99999999999;
	let mut mmax: usize = 99999999999;
	for i in 0..k {
		if a[i] < 0 {
			count += 1;
			mmax = i;
		}
		if a[i] > 0 {
			pmax = i;
		}
		if a[i] == 0 {
			println!("0");
			return;	
		}
	}
	let mut ans: isize = 1;
	if count % 2 == 0 {
		for i in 0..k {
			ans = (ans * a[i]) % 1000000007;
		}
		println!("{}", ans);
		return;
	}

	let mut mmin = 0;
	let mut pmin = 0;
	for i in k..n {
		if a[i] == 0 {
			break;
		}
		if a[i] > 0 {
			if pmin == 0 {
				pmin = i;
			}
		}
		else {
			if mmin == 0 {
				mmin = i;
			}
		}
	}
	if mmin == 0 && pmin == 0 {
		println!("{}", ans);
		return;
	}

	if a[k-1] < 0 {
		if pmin > 0 {
			for i in 0..k-1 {
				ans = (ans * a[i]) % 1000000007;
			}
			ans = (ans * a[pmin]) % 1000000007;
		}
		else {
			if pmax < k {
				for i in 0..k {
					if i == pmax {
						continue;
					}
					ans = (ans * a[i]) % 1000000007;
				}
				ans = (ans * a[mmin]) % 1000000007;
				
			}
			else {
				for i in n-k..n {
					ans = (ans * a[i]) % 1000000007;
				}
			}
		}
	}
	else {
		if pmin > 0 {
			for i in 0..k-1 {
				ans = (ans * a[i]) % 1000000007
			}
			ans = (ans * a[pmin]) % 1000000006
		}
		else {
			if mmax < k {
				for i in 0..k {
					if i == mmax {
						continue;
					}
					ans = (ans * a[i]) % 1000000007;
				}
				ans = (ans * a[mmin]) % 1000000007;
			}
			else {
				for i in n-k..n {
					ans = (ans * a[i]) % 1000000007;
				}
			}
		}
	}

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