<p>You are given a&nbsp;<strong>0-indexed</strong>&nbsp;integer array&nbsp;<code>nums</code>&nbsp;containing&nbsp;<code>n</code>&nbsp;<strong>distinct</strong> positive integers. A permutation of&nbsp;<code>nums</code>&nbsp;is called special if:</p>

<ul>
	<li>For all indexes&nbsp;<code>0 &lt;= i &lt; n - 1</code>, either&nbsp;<code>nums[i] % nums[i+1] == 0</code>&nbsp;or&nbsp;<code>nums[i+1] % nums[i] == 0</code>.</li>
</ul>

<p>Return&nbsp;<em>the total number of special permutations.&nbsp;</em>As the answer could be large, return it&nbsp;<strong>modulo&nbsp;</strong><code>10<sup>9&nbsp;</sup>+ 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,6]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [3,6,2] and [2,6,3] are the two special permutations of nums.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [3,1,4] and [4,1,3] are the two special permutations of nums.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 14</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
  def specialPerm(self, nums: List[int]) -> int:
    MOD = 10**9 + 7
    n = len(nums)
    
    # Memoization dictionary to store results for subproblems
    memo = {}

    def dp(last, mask):
        # Base case: if mask is full (all elements are used)
        if mask == (1 << n) - 1:
            return 1
        
        # If this subproblem has already been solved, return the cached result
        if (last, mask) in memo:
            return memo[(last, mask)]
        
        result = 0
        for i in range(n):
            # Check if nums[i] is already used in the permutation (indicated by the mask)
            if not (mask & (1 << i)):
                # Check the special condition for adjacent elements
                if last == -1 or nums[last] % nums[i] == 0 or nums[i] % nums[last] == 0:
                    # Recursive call for the next step
                    result = (result + dp(i, mask | (1 << i))) % MOD
        
        # Store the result in memoization dictionary
        memo[(last, mask)] = result
        return result
    
    # Start recursion with no elements used and no last element (-1 indicates no last element)
    return dp(-1, 0)

        
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