<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 {
public:

    bool issqr(int a){
        int sqr = (int)sqrtl(a);
        return sqr*sqr==a;
    }

    int numSquarefulPerms(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> dp(1<<n, vector<int> (n));
        for(int i=0; i<(1<<n); i++) {
            if(__popcount(i)==1){
                for(int j=0; j<n; j++){
                    if (i&(1<<j)) dp[i][j]=1;
                }
                continue;
            }

            for(int j=0; j<n; j++){
                if(( i&(1<<j) ) == 0 ) continue;
                int s = i^(1<<j);
                for(int k =0; k<n; k++){
                    if(( s&(1<<k) ) == 0 ) continue;
                    dp[i][j] += issqr(nums[j]+nums[k])*dp[s][k];
                }
            }

        }
        for(auto p : dp){
            for(auto q: p) cout << q<< " "; cout << endl;
        }
        vector<int> fact(13);
        fact[0]=1;
        for(int i=1; i<13; i++) fact[i]=fact[i-1]*i;
        map<int,int> m;
        for(auto p : nums) m[p]++;
        int ans = 0;
        for(int i=0; i<n; i++) ans+=dp.back()[i];
        for(auto [x,y] : m){
            ans/=fact[y];
        }
        return ans;
    }
};
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