<p>You are given 2 <strong>positive</strong> integers <code>l</code> and <code>r</code>. For any number <code>x</code>, all positive divisors of <code>x</code> <em>except</em> <code>x</code> are called the <strong>proper divisors</strong> of <code>x</code>.</p>

<p>A number is called <strong>special</strong> if it has exactly 2 <strong>proper divisors</strong>. For example:</p>

<ul>
	<li>The number 4 is <em>special</em> because it has proper divisors 1 and 2.</li>
	<li>The number 6 is <em>not special</em> because it has proper divisors 1, 2, and 3.</li>
</ul>

<p>Return the count of numbers in the range <code>[l, r]</code> that are <strong>not</strong> <strong>special</strong>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">l = 5, r = 7</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>There are no special numbers in the range <code>[5, 7]</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">l = 4, r = 16</span></p>

<p><strong>Output:</strong> <span class="example-io">11</span></p>

<p><strong>Explanation:</strong></p>

<p>The special numbers in the range <code>[4, 16]</code> are 4 and 9.</p>
</div>

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

<ul>
	<li><code>1 &lt;= l &lt;= r &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def nonSpecialCount(self, l: int, r: int) -> int:
        def is_prime(num):
            if num == 1:
                return False
            if num == 2:
                return True
            for i in range(2, int(sqrt(num))+2):
                if num % i == 0:
                    return False
            return True
        
        cnt = 0
        start = int(sqrt(l))
        end = int(sqrt(r))
        for n in range(start, end+1):
            if n**2 >= l and n**2 <= r:
                if is_prime(n):
                    cnt += 1
        return r-l+1 - cnt
16500