<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:
        if n==0 or n==1:
            return 0
        return self.rec(0,n)
    
    def rec(self,oldnum,newnum):
        if oldnum==newnum:
            return oldnum
        s = 0
        arr = [True]*(newnum+1)
        for i in range(2,newnum+1):
            if arr[i]:
                if newnum%i==0:
                    x = newnum
                    while(x%i==0):
                        s+=i
                        x=x//i
                for j in range(i*i,newnum+1,i):
                    arr[j] = False
        return self.rec(newnum,s)
18000