<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 {
private:
    vector<int> spf;

    void sieve(){
        spf[0] = 0;
        for(int i = 2; i <= 100001; i++){
            if(spf[i] == 1){
                for(int j = i; j <= 100001; j += i){
                    if(spf[j] == 1){
                        spf[j] = i;
                    }
                }
            }
        }
    }

    vector<int> getFactorization(int x){
        vector<int> ret;
        while(x != 1){
            ret.push_back(spf[x]);
            x /= spf[x];
        }

        return ret;
    }

public:
    Solution() : spf(100002, 1) { 
        sieve();  
    }

    int minSteps(int n) {
        vector<int> primeFactors = getFactorization(n);
        unordered_map<int, int> mp;
        int i = 2;
        int ans = 0;

        while(n != 1){
            while(n % spf[i] == 0){
                mp[spf[i]]++;
                n /= spf[i];
            }
            i++;
        }

        for(auto it = mp.begin(); it != mp.end(); it++){
            ans += (it->first * it->second);
        }

        return ans;
    }
};

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