<p>There are <code>n</code> people and <code>40</code> types of hats labeled from <code>1</code> to <code>40</code>.</p>

<p>Given a 2D integer array <code>hats</code>, where <code>hats[i]</code> is a list of all hats preferred by the <code>i<sup>th</sup></code> person.</p>

<p>Return <em>the number of ways that the <code>n</code> people wear different hats to each other</em>.</p>

<p>Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> hats = [[3,4],[4,5],[5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one way to choose hats given the conditions. 
First person choose hat 3, Second person choose hat 4 and last one hat 5.
</pre>

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

<pre>
<strong>Input:</strong> hats = [[3,5,1],[3,5]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 ways to choose hats:
(3,5), (5,3), (1,3) and (1,5)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> Each person can choose hats labeled from 1 to 4.
Number of Permutations of (1,2,3,4) = 24.
</pre>

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

<ul>
	<li><code>n == hats.length</code></li>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>1 &lt;= hats[i].length &lt;= 40</code></li>
	<li><code>1 &lt;= hats[i][j] &lt;= 40</code></li>
	<li><code>hats[i]</code> contains a list of <strong>unique</strong> integers.</li>
</ul>

class Solution 
{
private:
    #define mod (long long)(pow(10, 9) + 7)
    long long solve(long long i, long long &n, long long mask, unordered_map<long long, unordered_map<long long, bool>> &mp, vector<vector<long long>> &dp)
    {
        if(i == 41)
            return (mask == ((1 << n) - 1));
        if(dp[i][mask] != -1)
            return dp[i][mask];
        long long not_take = solve(i + 1, n, mask, mp, dp) % mod;
        long long take = 0;
        for(int j = 0; j < n; j++)
            if(((1 << j) & mask) == 0 and mp[j].find(i) != mp[j].end())
                take += solve(i + 1, n, mask | (1 << j), mp, dp) % mod, take %= mod;
        return dp[i][mask] = (take + not_take) % mod; 
    }
public:
    long long numberWays(vector<vector<int>>& hats) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        long long n = hats.size();
        vector<vector<long long>> dp(41, vector<long long> (1 << n, -1));
        unordered_map<long long, unordered_map<long long, bool>> mp;
        for(long long i = 0; i < n; i++)
            for(long long j = 0; j < hats[i].size(); j++)
                mp[i][hats[i][j]] = true;
        return solve(1, n, 0, mp, dp);
    }
};
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