<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 {
    set<vector<int>>ans;
    void recursion(int n , vector<int>&temp, int ind){
       if(ind==n-1){
          if((ind+1)%temp[ind]==0)
          ans.insert(temp);
          
          return;
       }
       if(((ind+1)%temp[ind])==0 || (temp[ind]%(ind+1)==0) ){
          recursion(n , temp , ind+1);
       }
       for(int i=ind+1;i<n;i++){
          if(((ind+1)%temp[i]) && (temp[i]%(ind+1))){continue;}
          swap(temp[ind] , temp[i]);
          recursion(n , temp , ind+1);
          swap(temp[ind] , temp[i]);
       }
    }
public:
    int countArrangement(int n) {
        vector<int>temp;
        for(int i=0;i<n;i++)temp.push_back(i+1);
        recursion(n , temp , 0);
       
         return ans.size();
    }
};
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