<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 {
public:
    int n,m;
    vector<int>Hats[41]; // Hats[i] = list of the people who prefer i-th hat

    // Reverseing the dp style.
    int dp[41][(1<<11)]; // dp[i][mask] = count of additional ways to to use i-th hat to m-th hat to fill unused persons.


    int mod = 1000000007;



    int f(int pos,int mask)
    {
        if(pos == m+1)
        {
            return (mask == (1<<n)-1); // all person got hat
        }

        int &rat = dp[pos][mask];

        if(rat != -1)
        {
            return rat;
        }

        int ans = f(pos+1,mask); // nobody took this hat
        ans %= mod;


        for(int i = 0; i<Hats[pos].size(); i++)
        {
            int person = Hats[pos][i];
            //cout<<" hat "<<pos<<" person "<<person<<endl;
            if( (1<<person) & mask)
            {
                // in this combination, this person already got a hat
            }
            else 
            {
                ans += f(pos+1, (1<<person) | mask);
                ans %= mod;
            }
        }


        return rat = ans;

        
    }

    int numberWays(vector<vector<int>>& hats) {
        n = hats.size();
        for(int i=0;i<hats.size();i++)
        {
            for(int h : hats[i])
            {
                m = max(h,m);
                Hats[h].push_back(i); // h-th hat is prefered by these persons.
            }
        }

        memset(dp, -1, sizeof(dp));


        return f(0,0);
    }
};
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