<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:
    int minSteps(int n) {
        if (n == 1) return 0;
        array<int, 10> rem = {1, 2, 3, 4, 5, 5, 7, 6, 6};
        if (n < 10) return rem[n - 1];

        unordered_set<int> non_primes;
        for (int i = 1; i <= 500; ++i) {
            for (int j = 1; j <= 1000 / i; ++j) {
                non_primes.insert(i * j);
            }
        }
        unordered_set<int> primes;
        for (int i = 1; i <= 1000; ++i) {
            if (!non_primes.contains(i)) {
                primes.insert(i);
                if (i == n) return n;
            }
        }

        int res = 0;

        while (n > 1) {
            if (n < 10) {
                res += rem[n - 1];
                break;
            }
            if (primes.contains(n)) {
                res += n;
                break;
            }
            for (int i = n / 2; i > 0; --i) {
                if (n % i == 0) {
                    int d = n / i;
                    res += d;
                    n = i;
                    break;
                }
            }
        }
        return res;
    }
};
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