<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:
        is_allowed_order = {n:set() for n in nums}
        for i in range(len(nums)-1):
            for j in range(i+1, len(nums)):
                if (nums[i] % nums[j] == 0):
                    is_allowed_order[nums[i]].add(nums[j])
                if (nums[j] % nums[i] == 0):
                    is_allowed_order[nums[j]].add(nums[i])

        @cache
        def helper(last=None, used=tuple()):
            #print(last, used)
            if (len(used) == len(nums)):
                return 1
            count = 0
            for n in nums:
                if (n in used):
                    continue
                if ((not n in is_allowed_order[last]) and (not last in is_allowed_order[n])):
                    continue
                used_copy = list(used)
                bisect.insort_left(used_copy, n)
                count += helper(n, tuple(used_copy))
            return count

        count = 0
        for n in nums:
            count += helper(n, tuple([n]))
        return (count % (10 ** 9 + 7))
3964