<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>> ans;
    
    // Recursive function to place numbers in the arrangement
    void solve(int n, vector<int>& curr, vector<bool>& used, int i) {
        if (i > n) {
            ans.push_back(curr);  // If all positions are filled, store the arrangement
            return;
        }
        
        // Try placing each number from 1 to n in position 'i'
        for (int j = 1; j <= n; j++) {
            if (!used[j] && (j % i == 0 || i % j == 0)) {
                used[j] = true;  // Mark number 'j' as used
                curr.push_back(j);
                
                // Recur for the next position
                solve(n, curr, used, i + 1);
                
                // Backtrack
                used[j] = false;
                curr.pop_back();
            }
        }
    }
    
    // Main function to count all valid arrangements
    int countArrangement(int n) {
        vector<int> curr;  // Stores the current arrangement
        vector<bool> used(n + 1, false);  // Keeps track of used numbers
        
        solve(n, curr, used, 1);  // Start solving from position 1
        
        return ans.size();  // Return the number of valid arrangements
    }
};

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