<p>Suppose you have <code>n</code> integers labeled <code>1</code> through <code>n</code>. A permutation of those <code>n</code> integers <code>perm</code> (<strong>1-indexed</strong>) is considered a <strong>beautiful arrangement</strong> if for every <code>i</code> (<code>1 &lt;= i &lt;= n</code>), <strong>either</strong> of the following is true:</p>

<ul>
	<li><code>perm[i]</code> is divisible by <code>i</code>.</li>
	<li><code>i</code> is divisible by <code>perm[i]</code>.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the <strong>number</strong> of the <strong>beautiful arrangements</strong> that you can construct</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 2
<b>Explanation:</b> 
The first beautiful arrangement is [1,2]:
    - perm[1] = 1 is divisible by i = 1
    - perm[2] = 2 is divisible by i = 2
The second beautiful arrangement is [2,1]:
    - perm[1] = 2 is divisible by i = 1
    - i = 2 is divisible by perm[2] = 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
</ul>

class Solution {
public:
    int rec(int level,set<int>&ch,set<int>&pos){
        if(level==0){ 
            return 1;
        }
        int ans=0;
        set<int>temp=ch;
        map<int,vector<int>>opt;
        for(auto v:pos){
            for(auto x:ch){
                if(!(v%x)||!(x%v)){
                    opt[v].push_back(x);
                }
            }
        }
        int minop=*pos.begin();
        for(auto v:pos){
            if(opt[v].size()<opt[minop].size())minop=v;
        }
        pos.erase(minop);
        for(auto c:opt[minop])
        {
            ch.erase(c);
            ans+=rec(level-1,ch,pos);
            ch.insert(c);
        }
        pos.insert(minop);
        return ans;
    }
    int countArrangement(int n) {
        set<int>st;
        set<int>pos;
        for(int i=1;i<=n;i++){
            st.insert(i);
            pos.insert(i);
        }
        return rec(n,st,pos);
    }
};
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