<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:
bool prime(int n)
{
    if(n==2)
        return true;
    if(n%2==0)
        return false;
    for(int i=3;i<=sqrt(n);i=i+2)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

int factor(int n,vector<int> dp)
{
    int t = 0;
    for(int i=2;i<=n/2;i++)
    {
        if(n%i==0)
        {
            t+=dp[i];
            n = n/i;
            break;
        }
    }
    t+=dp[n];
    return t;
}

    int minSteps(int n) {
        if(n==1)
        return 0;
    
    vector<int> dp(n+1,0);
    for(int i=2;i<=n;i++)
    {
        if(prime(i))
            dp[i] = i;
        else
        {
            dp[i] = factor(i,dp);
        }
    }
    return dp[n];        
    }
};
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