<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>

int dp[100004][14];
class Solution {
public:
    int help(int mask,int last,vector<int>&nums,int n)
    {
        if(mask==(1<<n)-1)return 1;
        if(dp[mask][last+1]!=-1)return dp[mask][last+1];

        int ans=0;
        for(int i=0;i<n;i++)
        {
            if(mask&(1<<i))continue;
            if(i>0 && nums[i]==nums[i-1] && (mask & (1<<(i-1))))continue;
            if(last==-1)ans=ans+help(mask|(1<<i),i,nums,n);
            else
            {
                int sum=nums[i]+nums[last];
                int tp=sqrt(sum);
                if((tp*tp)==sum){ans=ans+help(mask|(1<<i),i,nums,n);}
            }
        }
        return dp[mask][last+1]=ans;
    }
    int numSquarefulPerms(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        sort(nums.begin(),nums.end());
        return help(0,-1,nums,nums.size());
    }
};
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