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: Accepted)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use permutohedron::LexicalPermutation as _;
use proconio::input;
//use proconio::marker::*;
use std::collections::*;
//use std::cmp::*;
//use std::num;
//use petgraph::unionfind::UnionFind;
const MOD: i64 = 1_000_000_007;
fn main() {
    input! {
        N:usize,mut K:usize,A:[i64;N]
    }

    let mut poss = Vec::with_capacity(N);
    let mut negs = Vec::with_capacity(N);
    for a in A {
        if a > 0 {
            poss.push(a);
        } else if a < 0 {
            negs.push(a);
        }
    }
    let z = N - poss.len() - negs.len();

    // 0の利用が不可避な場合
    if poss.len() + negs.len() < K || (z > 0 && poss.len() + negs.len() == K && negs.len() % 2 != 0)
    {
        println!("0");
        return;
    }

    // 正の数がない場合。
    if poss.len() == 0 {
        if K % 2 != 0 && z > 0 {
            // negs を奇数個選んだらマイナスになってしまうので、0があるなら0の方がマシ。
            println!("0");
            return;
        }
        negs.sort();
        if K % 2 != 0 {
            // 負の数を絶対値の小さい方からK個選ぶ。
            negs.reverse();
        }
        let mut ans = 1;
        for i in 0..K {
            ans = ans * negs[i] % MOD;
            //println!("negs_i={},ans={}", negs[i], ans);
        }
        ans = (ans + MOD) % MOD;
        println!("{}", ans);
        return;
    }

    let mut i = 0;
    let mut j = 0;
    let mut ans = 1;
    if N == K {
        // 全てを選ぶしかない場合
        i = poss.len();
        j = negs.len();
    } else {
        // それ以外の場合、まず絶対値の大きい順に並べる
        // possは最低1個以上ある。
        poss.sort();
        poss.reverse();
        negs.sort();
        if K % 2 != 0 {
            // Kが奇数の場合、possを一個追加して、Kを減らす。
            K = K - 1;
            i += 1;
        }
        // 2個ずつ追加していく。
        K = K / 2;
        for k in 0..K {
            if i + 1 >= poss.len() {
                j += 2;
            } else if j + 1 >= negs.len() {
                i += 2;
            } else if poss[i] * poss[i + 1] > negs[j] * negs[j + 1] {
                i += 2;
            } else {
                j += 2;
            }
        }
    }
    for a in 0..i {
        ans = ans * poss[a] % MOD;
    }
    for a in 0..j {
        ans = ans * negs[a] % MOD;
    }
    ans = (ans + MOD) % MOD;
    println!("{}", ans);
}