<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 m;
    int dp[(1<<15)+1][17];
    int func(int i, int marks, unordered_map<int, set<int>>& mp){
        if(marks==((1<<m)-1)) return 1;
        if(i==(m+1)) return 0;

        if(dp[marks][i]!=-1) return dp[marks][i];

        int ans=0;
        ans=func(i+1, marks, mp);

        for(auto j : mp[i]){
            if((marks & (1<<(j-1)))==0){
                ans+=func(i+1, (marks | (1<<(j-1))), mp);
            }
        }

        return dp[marks][i] = ans;
    }

    int countArrangement(int n) {
        m=n;
        unordered_map<int, set<int>> mp;
        for(int i=1; i<=n; i++){
            for(int k=i; k<=n; k+=i){
                mp[i].insert(k);
                mp[k].insert(i);
            }
        }

        memset(dp, -1, sizeof(dp));

        return func(1, 0, mp);
    }
};
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