<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 helper(vector<int>& curr, int& res, int n, vector<int>& num, set<int>& visited){
        if(curr.size() == n){
            res++;
            return;
        }
        for(int i = 0; i < num.size(); i++){
            if(!visited.contains(num[i]) && ((curr.size()+1) % num[i] == 0 || num[i] % (curr.size()+1) == 0)){
                curr.push_back(num[i]);
                visited.insert(num[i]);
                helper(curr, res, n, num, visited);
                visited.erase(num[i]);
                curr.pop_back();
            }
        }
    }
    int countArrangement(int n) {
        int res = 0;
        vector<int> v;
        for(int i = 1; i <= n; i++){
            v.push_back(i);
        }
        vector<int> t;
        set<int> s;
        helper(t, res, n, v, s);
        return res;
    }
};
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