<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:
    int md = 1e9 + 7;
    int key = 71;
    long long p[100];
    set<long long> dp[(1 << 13)][13];
    bool can[50][50];
    set<long long> res;
    int a[2000000];
    int cnt;
    int l;

    int f(int x){
        int k = (int)sqrt(x);
        if(k * k == x)return 1;
        return 0;
    }

    int numSquarefulPerms(vector<int>& nums) {

        int n = (int)nums.size();

        p[0] = key;
        for(int i = 1; i <= n; ++i){
            p[i] = p[i - 1] * 1ll * key % md;
        }

        for(int i = 0; i < n; ++i){
            for(int j = i + 1; j < n; ++j){
                if(f(nums[i] + nums[j])) can[i][j] = 1, can[j][i] = 1;
            }
        }

        int cntMasks = (1 << n);

        for(int i = 0; i < n; ++i){
            dp[1 << i][i].insert(nums[i] * p[0] % md);
        }

        for(int mask = 1; mask < cntMasks; ++mask){
            for(int i = 0; i < n; ++i){
                if(dp[mask][i].size() > 0){
                    cnt = 0;
                    for(auto hash : dp[mask][i]){
                        a[++cnt] = hash;
                    }
                    l = __builtin_popcount(mask);
                    for(int j = 0; j < n; ++j){
                        if((mask & (1 << j)) == 0 && can[i][j]){
                            for(int k = 1; k <= cnt; ++k){
                                dp[mask | (1 << j)][j].insert((a[k] + p[l] * nums[j]) % md);
                            }
                        }
                    }
                }
            }
        }

        for(int i = 0; i < n; ++i){
            for(auto hash : dp[(1 << n) - 1][i]){
                res.insert(hash);
            }
        }
        return res.size();
    }
};
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