<p>Given an integer <code>num</code>, find the closest two integers in absolute difference whose product equals&nbsp;<code>num + 1</code>&nbsp;or <code>num + 2</code>.</p>

<p>Return the two integers in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = 8
<strong>Output:</strong> [3,3]
<strong>Explanation:</strong> For num + 1 = 9, the closest divisors are 3 &amp; 3, for num + 2 = 10, the closest divisors are 2 &amp; 5, hence 3 &amp; 3 is chosen.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = 123
<strong>Output:</strong> [5,25]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = 999
<strong>Output:</strong> [40,25]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num &lt;= 10^9</code></li>
</ul>

from math import sqrt
class Solution:
    def closestDivisors(self, num):
        n1, n2 = num+1, num+2
        n1b, n2b = int(sqrt(n1)), int(sqrt(n2))
        for i in range(0, n1 if n1 < n2 else n2):
            n1d, n1m = divmod(n1, n1b-i)
            n2d, n2m = divmod(n2, n2b-i)
            if n1m == 0 and n2m == 0:return [n1d, n1b-i] if abs(n1d - n1b+i) < abs(n2d - n2b+i) else [n2d, n2b-i]
            if n1m == 0:return [n1d, n1b-i]
            if n2m == 0:return [n2d, n2b-i]
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