Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square.

Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].

Example 1:

Input: [1,17,8]
Output: 2
Explanation: 
[1,8,17] and [17,8,1] are the valid permutations.

Example 2:

Input: [2,2,2]
Output: 1

Note:

1. 1 <= A.length <= 12
2. 0 <= A[i] <= 1e9

from collections import Counter
class Solution(object):
    def numSquarefulPerms(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        d=Counter(A)
        next={i:set([j for j in d if int((i+j)**0.5)**2==(i+j)]) for i in d}
        self.res = 0
        
        def dfs(x, cnt):
            d[x]-=1
            if cnt==len(A): self.res+=1
            for n in next[x]:
                if d[n]>0: dfs(n, cnt+1)
            d[x]+=1
            
        for x in d: dfs(x, 1)
        return self.res