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)

#![allow(unused_imports)]
use itertools::Itertools;
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
use superslice::*;

const MOD: i64 = 1_000_000_007;
const MAX: i64 = 1_000_000_001;

fn main() {
    input! {
        n: usize,
        k: usize,
        mut a: [i64; n],
    }

    a.sort();

    let mut negatives: Vec<i64> = Vec::new();
    let mut positives: Vec<i64> = Vec::new();

    for v in a {
        if v < 0 {
            negatives.push(v);
        } else {
            positives.push(v);
        }
    }

    positives.reverse();

    let mut ans = 1;
    let mut cur = 0;
    let mut neg_cur = 0;
    let mut pos_cur = 0;

    if positives.len() == 0 && k % 2 == 1 || k == n {
        negatives.reverse();

        while cur < k {
            let neg_v = negatives.get(neg_cur).unwrap_or(&MAX);
            let pos_v = positives.get(pos_cur).unwrap_or(&MAX);

            if neg_v.abs() < pos_v.abs() {
                ans = ans * neg_v % MOD;
                neg_cur += 1;
            } else {
                ans = ans * pos_v % MOD;
                pos_cur += 1;
            }

            cur += 1;
        }
    } else {
        while cur < k {
            let neg_v1 = negatives.get(neg_cur);
            let neg_v2 = negatives.get(neg_cur + 1);
            let pos_v1 = positives.get(pos_cur);
            let pos_v2 = positives.get(pos_cur + 1);

            let neg_v = if neg_v1.and(neg_v2).is_some() {
                neg_v1.unwrap() * neg_v2.unwrap() % MOD
            } else {
                -MAX * MAX
            };

            let pos_v = if pos_v1.and(pos_v2).is_some() {
                pos_v1.unwrap() * pos_v2.unwrap() % MOD
            // } else if pos_v1.is_some() {
            //     *pos_v1.unwrap()
            } else {
                -MAX * MAX
            };

            if neg_v < pos_v || cur == k - 1 {
                ans = ans * pos_v1.unwrap() % MOD;
                pos_cur += 1;
                cur += 1;
            } else {
                ans = ans * neg_v % MOD;
                neg_cur += 2;
                cur += 2;
            }
        }
    }

    if ans < 0 {
        println!("{}", ans + MOD);
    } else {
        println!("{}", ans);
    }
}