<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:
    //vector<vector<int>>dp(18,vector<int>(100005,-1));
    int solve(int last , int mask , int curr,vector<vector<int>>&dp,int n){
    if(mask == (1<<n)-1)return 1;
    int &res = dp[last][mask];
    if(res!=-1)return res;
    res=0;
    for(int i=1;i<=n;i++){
        if(mask & (1<<(i-1)))continue;
        if(i%curr==0 || curr%i==0){
            res+=solve(i,mask | (1<<(i-1)),curr+1,dp,n);
        }
    }
    return res;
    }
    
    int countArrangement(int n) {
        vector<vector<int>>dp(18,vector<int>(100005,-1));
        int ans=0;
        for(int i=1;i<=n;i++){
            ans+=solve(i,(1<<(i-1)),2,dp,n);
        }
        return ans;
    }
};
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