<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

from functools import lru_cache

class Solution:
    def __init__(self):
        self.factors = [3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
        self.primes = [1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
        self.last_prime = 47
        self.next_prime = {}
        for i in range(1, len(self.primes)):
            self.next_prime[self.primes[i - 1]] = self.primes[i]
        self.primes = set(self.primes)
        for i in range(53, 1000, 2):
            if self.isPrime(i):
                self.next_prime[self.last_prime] = i
                self.primes.add(i)
                self.last_prime = i

    @lru_cache(None)
    def isPrime(self, p: int) -> bool:
        if p < 53: return (p in self.primes)
        if p & 1 == 0: return False
        for x in self.factors:
            if x * x > p:
                break
            if p % x == 0:
                return False
        return True

    def minSteps(self, n: int) -> int:
        if n == 1: return 0
        if self.isPrime(n): return n
        count = 0
        p = 1
        while n > 1:
            p = self.next_prime[p]
            while n % p == 0:
                count += p
                n //= p
        return count
25283