<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:
        # prime number? supare
        #  2*2 = 4 , 7*7 = 49 prim square 121 169...
        # 1 4 9 16X 
        # 
        # left , right 7 :x  after sqrt(x), is prime
        def isP(n):
            if n == 0 or n == 1 :
                return 0
            for i in range(2,int(sqrt(n)+1)):
                if n % i == 0 :
                    return 0
            return 1
        # primes = [isP(i) for i in range(int(sqrt(r))+1) ]
        # primes[0] = 0
        # primes[1] = 0

        special = 0
        # primes count
        for i in range((int(sqrt(r))+1)):
            if isP(i) == 1 and l <= i*i <= r   :
                special +=1
        return r+1-l - special

        return 


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