<p>An array is <strong>squareful</strong> if the sum of every pair of adjacent elements is a <strong>perfect square</strong>.</p>

<p>Given an integer array nums, return <em>the number of permutations of </em><code>nums</code><em> that are <strong>squareful</strong></em>.</p>

<p>Two permutations <code>perm1</code> and <code>perm2</code> are different if there is some index <code>i</code> such that <code>perm1[i] != perm2[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,17,8]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [1,8,17] and [17,8,1] are the valid permutations.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,2]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 12</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def numSquarefulPerms(self, nums: List[int]) -> int:
        ans = 0
        counter= Counter(nums)

        def isquare(a, b):
            root = int(math.sqrt(a+b))
            return root*root == a+b
    
        def dfs(temp):
            nonlocal ans
            if len(temp) > 1 and not isquare(temp[-1], temp[-2]):
                return
            if len(temp) == len(nums):
                ans +=1
                return
            
            for n in counter:
                if counter[n]==0:
                    continue
                temp.append(n)
                counter[n] -=1
                dfs(temp)
                counter[n] +=1
                temp.pop()
            
        dfs([])
        return ans  
16500