<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:
    def minSteps(self, n: int) -> int:
        dp=[[] for i in range(n+1)]
        dp[1].append((1,0))
        if n==1: return 0
        dp[2].append((2,1))
        for i in range(3,n+1):
            start=i//2 if i%2==0 else i//2+1
            for u in range(start,i):
                if i%u==0:
                    for op,number in dp[u]:
                        dp[i].append((op+2,u))
                else:
                    for op,number in dp[u]:
                        if u+number==i:
                            dp[i].append((op+1,number))
            dp[i].sort()
            # index,find=0,False
            # for u in range(1,len(dp[i])):
            #     if dp[i][u][0]>dp[i][u-1][0]:
            #         index=u
            #         find=True
            #         break
            # if find:
            #     dp[i]=dp[i][:index]
            
        print(dp)
        return dp[n][0][0]
        


            


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