<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:
    long long mod=1e9+7;
    int numberWays(vector<vector<int>>& hats) {
        int n=hats.size();
        vector<vector<long long > >dp((1<<n),vector<long long >(41));
        dp[0][0]=1;
        unordered_set<int>bfs;
        bfs.insert(0);
        while(bfs.size()){
            unordered_set<int>b;
            for(int m:bfs){
                for(int i=0;i<n;++i){
                    if( ((1<<i)&m) ==0){
                        int mask=(m|(1<<i));
                        for(int val:hats[i]){
                            for(int j=0;j<val;++j){
                                dp[mask][val]=(dp[mask][val]+dp[m][j])%mod;
                                if(dp[m][j]>0){
                                    b.insert(mask);
                                }
                            }
                        }
                    }
                }
            }
            swap(bfs,b);
        }
        long long ans=0;
        for(int i=1;i<=40;++i){
            ans=(ans+dp[(1<<n)-1][i])%mod;
        }
        return ans;
    }
};
17500