<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<long long> primes;
    void seive(int n, vector<int>& v) {
        bool prime[n + 1];
        memset(prime, true, sizeof(prime));

        for (int p = 2; p * p <= n; p++) {
            if (prime[p] == true) {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        for (int p = 2; p <= n; p++) {
            if (prime[p]) {
                primes.push_back(p);
                v[p] = p;
            }
        }
    }

    int minSteps(int n) {

        vector<int> v(1001, -1);
        seive(1000, v);
        if (n == 1) {
            return 0;
        }

        for (int i = 2; i <= 1000; i++) {
            if (v[i] == -1) {
                for (int j = i / 2; j >= 2; j--) {
                    if (v[j] != -1 && i % j == 0) {
                        v[i] = v[j] + i / j;
                        break;
                    }
                }
            }
        }
        cout << endl;

        return v[n];
    }
};
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