<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:
        def count_spc(prev, used):
            if (used == pow(2, len(nums))-1):
                return 1
                
            res = 0

            for n in nums:
                if (used & pow(2, idxs[n])):
                    continue
                if (prev%n != 0 and n%prev != 0):
                    continue
                if ((used, n) in dp):
                    res += dp[(used, n)]
                else:
                    new_used = used | pow(2, idxs[n])
                    res += count_spc(n, new_used)
            
            used ^= pow(2, idxs[prev])
            dp[(used, prev)] = res

            return res
        

                
        MOD = pow(10, 9) + 7
        idxs = {}

        for i, n in enumerate(nums):
            idxs[n] = i


        dp = {}
        res = 0

        for n in nums:
            res += count_spc(n, pow(2, idxs[n]))
        
        return res % MOD



        
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