<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 getBitPos(self, n):
        rtn = []
        sft = 0 
        while n:
            if (n>>sft & 1)==1:
                rtn.append(sft)
                n^=1<<sft
            sft+=1       
        return rtn           
    def specialPerm(self, nums: List[int]) -> int:
        mod = 10**9+7
        dp = [[0]*len(nums) for  _ in range(2**(len(nums)))]
        for i in range(len(nums)): dp[1<<i][i] = 1 
        for seen in range(1, 2**len(nums)):
            bits = self.getBitPos(seen)
            for prev in bits:
                #curr is anything but the prev 
                for curr in bits:
                    if curr!=prev:
                        if nums[prev]%nums[curr]==0 or nums[curr]%nums[prev]==0:
                            dp[seen][curr]+=dp[seen^(1<<curr)][prev]
                            dp[seen][curr]%=mod
        return sum(dp[-1]) % mod
4114