<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>

using ll = long long;

class Solution {
public:
    int numberWays(vector<vector<int>>& hats) {
        vector<vector<int>> d(41);
        int n = hats.size();
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            for (int& h : hats[i]) {
                d[h].push_back(i);
                mx = max(mx, h);
            }
        }
        vector<vector<ll>> dp(mx + 1, vector<ll>(1 << n));
        dp[0][0] = 1;
        int mod = 1e9 + 7;
        for (int i = 1; i <= mx; ++i) {
            for (int mask = 0; mask < 1 << n; ++mask) {
                dp[i][mask] = dp[i - 1][mask];
                for (int& j : d[i]) {
                    if ((mask >> j) & 1) {
                        dp[i][mask] = (dp[i][mask] + dp[i - 1][mask ^ (1 << j)]) % mod;
                    }
                }
            }
        }
        return dp[mx][(1 << n) - 1];
    }
};
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