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, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {

         n:usize,k:usize,
    mut      a:[i64;n],
     }
    let mut native_a = a.clone();
    native_a.sort();
    native_a.reverse();
    a.sort_by_key(|&x| -x.abs() as i64);
    // dbg!(&a);

    let mut neg = Vec::new();
    let mut pos = Vec::new();
    let mut zero = 0;
    for &x in a.iter() {
        if x <= 0 {
            neg.push(x);
        } else if x > 0 {
            pos.push(x);
        }
        if x == 0 {
            zero += 1;
        }
    }
    let nl = neg.len();
    let pl = pos.len();
    // // 負になる
    // if k > pl && (k - pl) % 2 == 1 {
    //     if zero > 0 {
    //         println!("{}", 0);
    //         return;
    //     } else {
    //         let mut ans = 1i64;
    //         for i in 0..k {
    //             ans *= a[n - i - 1];
    //             ans = (ans + 5 * MOD) % MOD;
    //         }
    //         println!("{}", ans);
    //         return;
    //     }
    // }
    // else
    if pl == 0 {
        let mut ans = 1i64;
        for i in 0..k {
            ans *= (neg[nl - i - 1]);
            ans = (ans + 111 * MOD) % MOD;
        }
        println!("{}", ans);
    } else if nl == 0 {
        let mut ans = 1i64;
        for i in 0..k {
            ans *= pos[i];
            ans = ans % MOD;
        }
        println!("{}", ans);
    } else if n == k {
        let mut ans = 1i64;
        for i in 0..n {
            ans = (ans * a[i] + MOD) % MOD;
        }
        println!("{}", ans);
    } else {
        let mut indn = 0;
        let mut indp = 0;
        let mut ans = 1i64;
        let mut k = k;
        if k == 1 {
            println!("{}", pos[0]);
            return;
        }

        while k >= 2 {
            if if indn + 1 < nl {
                neg[indn] * neg[indn + 1]
            } else {
                -INF
            } >= if indp + 1 < pl {
                pos[indp] * pos[indp + 1]
            } else {
                -INF
            } {
                ans *= neg[indn] * neg[indn + 1];
                indn += 2;
            } else {
                ans *= pos[indp] * pos[indp + 1];
                indp += 2;
            }
            ans %= MOD;
            ans = (ans + 3333 * MOD) % MOD;
            k -= 2;
        }
        // // k = 3 or k = 2 になる
        // if neg[indn] * neg[indn + 1] >= pos[indp] * pos[indp + 1] {
        //     ans *= neg[indn] * neg[indn + 1];
        //     ans %= MOD;
        // } else {
        //     ans *= pos[indp] * pos[indp + 1];
        //     ans %= MOD;
        // }
        // k -= 2;
        if k == 0 {
            println!("{}", (ans + 333 * MOD) % MOD);
            return;
        } else {
            if indp < pl {
                println!("{}", (ans * pos[indp] % MOD + 5555 * MOD) % MOD);
                return;
            } else {
                ans = 1;
                for i in 0..k {
                    ans *= a[i];
                    ans = (ans + 5555 * MOD) % MOD;
                }
                println!("{}", ans);
                return;
            }
        }
    }

    // let mut dp = vec![vec![0; n + 1]; 2];

    // for i in 0..n {
    //     if a[i] <= 0 {
    //         dp[i][1] = dp[i - 1][0] * a[i];
    //     }
    // }
}
const MOD: i64 = 1000000007;

const INF: i64 = 1 << 62;