<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>

from itertools import permutations
class Solution:
    '''
    def specialPerm(self, nums: List[int]) -> int:
        def check(l):
            c = True
            for i in range(len(l)-1):
                if l[i]%l[i+1]==0 or l[i+1]%l[i]==0:
                    c = True
                else:
                    c = False
                    return
            return c
        a = permutations(nums)
        ans = 0
        for i in list(a):
            
            if check(
                (i)):
                ans+=1
        return ans
        '''
    def specialPerm(self, nums: List[int]) -> int:
        n, MOD = len(nums), 10**9 + 7
        @cache
        def dfs(prev, mask):
            if mask == (1 << n) - 1: return 1
            count = 0
            for i in range(n):
                if not (mask & (1 << i)) and (nums[i] % prev == 0 or prev % nums[i] == 0):
                    count += dfs(nums[i], mask | (1 << i))
            return count % MOD
        return dfs(1, 0)

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