<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>

from math import ceil
M = 10**5+1
is_prime = [True for _ in range(M)]
for i in range(2, ceil(M**0.5)):
    if is_prime[i]==True:
        for j in range(i**2, M, i):

            is_prime[j] = False

def get_prime_sum(x):
    ans = 0
    while(x%2==0):
        x//=2
        ans+=2
    for j in range(3, ceil(x**0.5)+1, 2):
        while(x>1 and x%j==0):
            # print(x, j, ans)
            x//=j
            ans+=j
    if x>2:
        ans+=x
    # print(x, ans)
    return ans

class Solution:
    def smallestValue(self, n: int, pr=None) -> int:
        # print(is_prime[n], n, get_prime_sum(n),)
        if is_prime[n]:
            return n
        if n==pr:
            return n
        return self.smallestValue(get_prime_sum(n), n)
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