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)

fn main() {
    proconio::input! {
        n: usize,
        k: usize,
        mut a_s: [i64; n],
    }
    const M: i64 = 1_000_000_007;
    a_s.sort();
    a_s.reverse();
    let negs = a_s
        .iter()
        .filter(|a| **a < 0)
        .map(|a| *a)
        .collect::<Vec<_>>();
    let poss = a_s
        .iter()
        .filter(|a| **a > 0)
        .map(|a| *a)
        .collect::<Vec<_>>();
    let mut ans = 1;
    if n == k {
        for a in a_s {
            ans *= a;
            ans %= M;
        }
    } else if n == negs.len() && k % 2 == 1 {
        for neg in negs.iter().take(k) {
            ans *= neg;
            ans %= M;
        }
        ans += M;
        ans %= M;
    } else if poss.len() == 0 && k % 2 == 1 {
        ans = 0;
    } else if poss.len() + negs.len() / 2 * 2 < k {
        ans = 0;
    } else {
        let mut negs = negs;
        negs.reverse();
        let mut inegs = 0;
        let mut iposs = 0;
        if k % 2 == 1 {
            ans *= poss[iposs];
            ans %= M;
            iposs += 1;
        }
        while iposs + inegs != k {
            let neg2 = if inegs + 1 < negs.len() {
                negs[inegs] * negs[inegs + 1]
            } else {
                0
            };
            let pos2 = if iposs + 1 < poss.len() {
                poss[iposs] * poss[iposs + 1]
            } else {
                0
            };
            if neg2 >= pos2 {
                assert!(neg2 != 0);
                ans *= negs[inegs];
                ans %= M;
                ans *= negs[inegs + 1];
                ans %= M;
                inegs += 2;
            } else {
                assert!(pos2 != 0);
                ans *= poss[iposs];
                ans %= M;
                ans *= poss[iposs + 1];
                ans %= M;
                iposs += 2;
            }
        }
    }
    println!("{}", ans);
}