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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
list($N, $K) = array_map(function ($n) {
    return (int)$n;
}, explode(' ', fgets(STDIN)));

$A = array_map(function ($n) {
    return (int)$n;
}, explode(' ', fgets(STDIN)));

$ans = 1;
$n = 0;
$mod = 1000000007;

if ($N === $K) {
    foreach ($A as $a) {
        $ans *= $a;
        $ans %= $mod;
    }
    echo $ans;
    return;
}

$u_zero = [];
$o_zero = [];
foreach ($A as $a) {
    if ($a < 0) {
        $u_zero[] = $a;
    } else {
        $o_zero[] = $a;
    }
}

sort($u_zero);
rsort($o_zero);

while ($n !== $K) {
    if (count($u_zero) < 2) {
        for ($i = $n; $i < $K; $i++) {
            $ans *= array_shift($o_zero);
            $ans %= $mod;
        }
        break;
    }
    if (empty($o_zero)) {
        if (($K - $n) % 2 === 0) {
            for ($i = $n; $i < $K; $i += 2) {
                $ans *= array_shift($u_zero);
                $ans %= $mod;
                $ans *= array_shift($u_zero);
                $ans %= $mod;
            }
            break;
        } else {
            for ($i = $n; $i < $K; $i++) {
                $ans *= array_pop($u_zero);
                $ans %= $mod;
            }
            break;
        }
    }
    if (count($o_zero) === 1) {
        while ($K - $n > 1) {
            $n += 2;
            $ans *= array_shift($u_zero);
            $ans %= $mod;
            $ans *= array_shift($u_zero);
            $ans %= $mod;
        }
        if ($K - $n === 1) {
            $ans *= array_shift($o_zero);
            $ans %= $mod;
        }
        break;
        /*
        if ($K - $n !== 1) {
            // 先に count($u_zero) < 2 をしているから存在する
            $n += 2;
            $ans *= array_shift($u_zero) * array_shift($u_zero);
        } else {
            $n += 1;
            $ans *= array_shift($o_zero);
        }
        */
    } elseif ($u_zero[0] * $u_zero[1] < $o_zero[0] * $o_zero[1] || $n + 1 === $K) {
        $n += 1;
        $ans *= array_shift($o_zero);
    } else {
        $n += 2;
        $ans *= array_shift($u_zero);
        $ans %= $mod;
        $ans *= array_shift($u_zero);
    }
    $ans %= $mod;
}

echo $ans >= 0 ? $ans : $mod + $ans;