<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:
    # TLE
    def smallestValue(self, n: int) -> int:
        if n <= 3:
            return n
        dp = [None]*(n+1)
        dp[0], dp[1], dp[2], dp[3] = 0, 1, 2, 3
        for i in range(4, n+1):
            for j in range(2, i):
                q, r = divmod(i, j)
                if r: continue
                dp[i] = dp[q]+dp[j]
                break
            if dp[i] == None:
                dp[i] = i
        curr = dp[n]
        while curr != dp[curr]:
            curr = dp[curr]
        return curr

    def smallestValue(self, n: int) -> int:
        if n <= 5: return n
        while n:
            curr, total = n, 0
            while curr > 1:
                for d in range(2, curr+1):
                    q, r = divmod(curr, d)
                    if r: continue
                    total += d
                    curr = q
                    break
            if total == n: break
            n = total
        return n
80