<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>

class Solution {
public:
    vector<int> isPrime;
    
    void sieve() {
        isPrime.assign(1001, 0); // 0 indicates prime, 1 indicates not prime.

        isPrime[0] = 1;
        isPrime[1] = 1;

        for(int i = 2; i * i <= 1000; i++) {
            if(isPrime[i] == 0) {
                for(int j = i * i; j <= 1000; j += i) {
                    isPrime[j] = 1;
                }
            }
        }
    }

    vector<int> fact;
    
    void factors() {
        fact.assign(1001, 1);
        for(int i = 2; i <= 1000; i++) {

            for(int j = 2; j * j <= i; j++) {

                if(i % j == 0) {
                    fact[i] = max(j, i / j);
                    break;
                }
            }
        }
    }

    int rec(int n) {
        if(n == 1) {
            return 0;
        }
        if(isPrime[n] == 0) {
            return n;
        }
        return (n / fact[n]) + rec(fact[n]);
    }

    int minSteps(int n) {
        sieve();
        factors();
        return rec(n);
    }
};

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