<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(vector<vector<int>>& dp, vector<bool>& vis, int i, int& n, int bit){
        
        if(dp[bit][i]!=-1){
            //cout<<dp[bit][i]<<" ";
            return dp[bit][i];
        }
        int sm = 0;
        for(int u=1;u<=n;u++){
            if(!vis[u] && (u%i==0 || i%u==0)){
                if(i<n){
                    vis[u] = true;
                    sm += rec(dp, vis, i+1, n, bit+pow(2, u));
                    vis[u] = false;
                }
                else{
                    sm += 1;
                }
            }
        }
        return dp[bit][i] = sm;
    }
    int countArrangement(int n) {
        vector<vector<int>> dp(pow(2, 16), vector<int>(n+1, -1));
        vector<bool> vis(n+1, false);
        return rec(dp, vis, 1, n, 0);
    }
};
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