<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:
        nums = [0]*(n//2+1)
        result = 0
        while current <= n//2 +1:
            if nums[current] == 0:
                p = current
                while current*p<=n:
                    nums[current*p] = 1
                    if current*p == n:
                        n = p
                        result += current
                    else:
                        p += 1
'''
class Solution:
    def smallestValue(self, n: int) -> int:
        def sp(k):
            if k<=5: return k
            factor = 2
            s = 0
            while k - factor>0:
                if k % factor == 0:
                    s+= factor
                    k //=factor
                else:
                    factor += 1
            return s+factor
        while sp(n)!=n:
            n = sp(n)
        return sp(n)


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