<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:
    def numberWays(self, hats: List[List[int]]) -> int:
        N = len(hats)
        M = max(itertools.chain.from_iterable(hats))

        h2p = [0] * (M + 1)
        for i, v in enumerate(hats):
            for h in v:
                h2p[h] |= (1 << i)

        status = (1 << N) - 1
        mod = 10 ** 9 + 7

        dp = [[-1] * (M + 1) for _ in range(1 << (N + 1))]
  
        def dfs(people, i):
            # i is the ith hat
            if i == M + 1:
                return 1 if people == 0 else 0

            print(people, i)
            if dp[people][i] != -1:
                return dp[people][i]

            ans = dfs(people, i + 1)

            for j in range(N):
                if (people & (1 << j)) == 0: continue
                if (h2p[i] & (1 << j)) == 0: continue
                
                ans = (ans + dfs(people ^ (1 << j), i + 1)) % mod
            
            dp[people][i] = ans % mod
            return ans % mod

        return dfs(status, 1)

22600