<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:
    void fun(int n,int ind,vector<int>&nums,int& cnt,vector<int>&v,vector<vector<int>>&a){
        if(ind==n){
            a.push_back(v);
            return ;
        }
        for(int i=ind;i<n;i++){
            
            if(nums[i]%(ind+1)==0 || (ind+1)%nums[i]==0){
                 swap(nums[i],nums[ind]);
                fun(n,ind+1,nums,cnt,v,a);
                 swap(nums[i],nums[ind]);
            }
          
        }
    }
    int countArrangement(int n) {
        vector<int>nums;
        for(int i=1;i<=n;i++){
            nums.push_back(i);
        }
        vector<int>v;int cnt=0;
        vector<vector<int>>a;
        fun(n,0,nums,cnt,v,a);
        return a.size();
    }
};
11187