There are n people and 40 types of hats labeled from 1 to 40.
Given a list of list of integers hats, where hats[i] is a list of all hats preferred by the i-th person.
Return the number of ways that the n people wear different hats to each other.
Since the answer may be too large, return it modulo 10^9 + 7.
Example 1:
Input: hats = [[3,4],[4,5],[5]]
Output: 1
Explanation: 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.
Example 2:
Input: hats = [[3,5,1],[3,5]]
Output: 4
Explanation: There are 4 ways to choose hats
(3,5), (5,3), (1,3) and (1,5)
Example 3:
Input: hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
Output: 24
Explanation: Each person can choose hats labeled from 1 to 4.
Number of Permutations of (1,2,3,4) = 24.
Example 4:
Input: hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]
Output: 111
Constraints:
n == hats.length
1 <= n <= 10
1 <= hats[i].length <= 40
1 <= hats[i][j] <= 40
hats[i] contains a list of unique integers.
class Solution:
    def numberWays(self, hats: List[List[int]]) -> int:
        N = len(hats)
        for i, h in enumerate(hats):
            hats[i] = set(h)
        mod = (10 ** 9) + 7
        
        @lru_cache(None)
        def rec(cur, mask):
            if cur > 41:
                return 0
            if mask == 0:
                return 1
            
            ans = 0
            for i in range(N):
                if (mask & (1<<i)) == 0:
                    continue
                if cur not in hats[i]:
                    continue
                ans += rec(cur+1, mask ^ (1<<i))
            ans += rec(cur+1, mask)
            while ans >= mod:
                ans -= mod
            return ans
        
        return rec(0, (2**N)-1)
class Solution:
    def numberWays(self, hats: List[List[int]]) -> int:
        N = len(hats)
        for i, h in enumerate(hats):
            hats[i] = set(h)
        mod = (10 ** 9) + 7
        
        @lru_cache(None)
        def rec(cur, mask):
            if cur > 41:
                return 0
            if mask == 0:
                return 1
            
            ans = 0
            for i in range(N):
                if (mask & (1<<i)) == 0:
                    continue
                if cur not in hats[i]:
                    continue
                ans += rec(cur+1, mask ^ (1<<i))
            ans += rec(cur+1, mask)
            while ans >= mod:
                ans -= mod
            return ans
        
        return rec(0, (2**N)-1)