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;

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

    a.sort();

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

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

    positives.reverse();

    if k > negatives.len() + positives.len() {
        println!("0");
        return;
    }

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

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

    while cur < k {
        let neg_v;
        let neg_cur_progress;
        if cur + 1 < k && neg_cur + 1 < negatives.len() {
            neg_v = negatives[neg_cur] * negatives[neg_cur + 1] % MOD;
            neg_cur_progress = 2;
        } else if neg_cur + 1 <= negatives.len() {
            neg_v = negatives[neg_cur];
            neg_cur_progress = 1;
        } else {
            neg_v = -1_000_000_001;
            neg_cur_progress = 0;
        };

        let pos_v;
        let pos_cur_progress;
        if cur + 1 < k && pos_cur + 1 < positives.len() {
            pos_v = positives[pos_cur] * positives[pos_cur + 1] % MOD;
            pos_cur_progress = 2;
        } else if pos_cur + 1 <= positives.len() {
            pos_v = positives[pos_cur];
            pos_cur_progress = 1;
        } else {
            pos_v = -1_000_000_001;
            pos_cur_progress = 0;
        };

        if neg_v >= pos_v {
            ans = ans * neg_v % MOD;
            cur += neg_cur_progress;
            neg_cur += neg_cur_progress;
        } else {
            ans = ans * pos_v % MOD;
            cur += pos_cur_progress;
            pos_cur += pos_cur_progress;
        }
    }

    if ans < 0 && zeros.len() > 0 {
        println!("0");
        return;
    } else if ans < 0 {
        println!("{}", ans + MOD);
    } else {
        println!("{}", ans);
    }
}