<p>You are given a positive integer <code>n</code>.</p>

<p>Continuously replace <code>n</code> with the sum of its <strong>prime factors</strong>.</p>

<ul>
	<li>Note that if a prime factor divides <code>n</code> multiple times, it should be included in the sum as many times as it divides <code>n</code>.</li>
</ul>

<p>Return <em>the smallest value </em><code>n</code><em> will take on.</em></p>

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

<pre>
<strong>Input:</strong> n = 15
<strong>Output:</strong> 5
<strong>Explanation:</strong> Initially, n = 15.
15 = 3 * 5, so replace n with 3 + 5 = 8.
8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6.
6 = 2 * 3, so replace n with 2 + 3 = 5.
5 is the smallest value n will take on.
</pre>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, n = 3.
3 is the smallest value n will take on.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def smallestValue(self, n: int) -> int:

        # def isPrime(v):
        #     if v == 2:
        #         return True
        #     if v%2 == 0:
        #         return False
        #     upper = math.floor(math.sqrt(v)) + 1
        #     for i in range(3, upper, 2):
        #         if v%i == 0:
        #             return False
        #         return True
        def primeFactors(v):
            factors = []
            while v%2 == 0:
                v = v//2
                factors.append(2)

            i = 3
        
            while i <= v:
                while v%i == 0:
                    factors.append(i)
                    v //= i
                i+=2
            return factors
        
        

        while True:
            f = primeFactors(n)
            nf = sum(f)
            if nf == n:
                return n
            n = nf

            
        
51