<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 numberWays(vector<vector<int>>& preferences) {
        const int MOD = (int)1e9 + 7;
        int n(preferences.size()), k((1ll<<n)-1);
        map<int, vector<int>> mp;
        for (int i(0); i < n; ++i) 
            for (int &hat : preferences[i]) 
                mp[hat].emplace_back(i);
        vector<vector<int>> dp(42, vector<int>(k+1));
        for (int i(0); i < 42; ++i) 
            dp[i][k] = 1;
        for (int msk(k); msk >= 0; --msk) {
            for (int hat(40); hat > 0; --hat) {
                int ways(dp[hat+1][msk]);
                for (int i : mp[hat]) {
                    if (!((msk >> i) & 1)) 
                        ways = (ways % MOD + dp[hat+1][msk|(1<<i)] % MOD) % MOD;
                }
                dp[hat][msk] = ways;
            }
        }
        return dp[1][0];
    }
};

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