<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>

class Solution:
    def closestDivisors(self, num: int) -> List[int]:
        divisors1, divisors2 = set({}), set({})
        ret = [1,num+2]
        for i in range(1, int(math.sqrt(num+2))+1):
            if i not in divisors1:
                if (num+1) % i == 0:
                    tmp = (num+1)//i
                    divisors1.add(tmp)
                    divisors1.add(i)
                    if (tmp - i) == 0: 
                        return [i, tmp]
                    else:
                        if (ret[1]-ret[0]) > (tmp-i):
                            ret = [i, tmp]
            if i not in divisors2:
                if (num+2) % i == 0:
                    tmp = (num+2)//i
                    divisors2.add(tmp)
                    divisors2.add(i)
                    if (tmp - i) == 0: 
                        return [i, tmp]
                    else:
                        if (ret[1]-ret[0]) > (tmp-i):
                            ret = [i, tmp]
        return ret
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