<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]:
        x = num+1
        y = num+2
        
        i = 1
        lddx = x - i
        dx = i
        while i * i <= x: 
            if x % i == 0:
                if abs(x / i - i) < lddx:
                    lddx = abs(x / i - i)
                    dx = i
            i += 1

        i = 1
        dy = i
        lddy = y - i
        while i * i <= y:
            if y % i == 0:
                if abs(y / i - i) < lddy:
                    lddy = abs(y / i - i)
                    dy = i
            i += 1
        
        if abs(lddx) < abs(lddy):
            return [dx, x//dx]
        else:
            return [dy, y//dy]
        
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