<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 setIsPrime(){
        for(int i=2;i<1001;i++){
            if(isPrime[i]){
                int k=i+i;
                while(k<1001){
                    isPrime[k]=0;
                    k+=i;
                }
            }
        }
    }
    int solve(int n){
        if(isPrime[n])return n;
        int ans=INT_MAX;
        for(int i=2;i<=sqrt(n);i++){
            if(n%i==0){
                int n2=n/i;
                int ret1=i+solve(n2);
                int ret2=n2+solve(i);
                ans=min({ans,ret1,ret2});
            }
        }
        return ans;
    }
    int minSteps(int n) {
        isPrime.resize(100000,1);
        setIsPrime();
        if(n==1)return 0;
        return solve(n);
    }
};
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