<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>

typedef long long ll;

using vi = vector<ll>;
using vvi = vector<vi>;
class Solution {
public:
    ll MOD = 1e9 + 7;
    // ll dp[41][(1LL<<10)] = {0};
    vvi dp;

    vvi adj;

    int n;

    int solve(int hat, int state){
        if(state == ((1<<n) - 1)) return 1;

        if(hat > 40) return 0;

        if(dp[hat][state] != -1) return dp[hat][state] % MOD;

        ll ways = solve(hat+1, state) % MOD;

        for(auto p: adj[hat]){
            if(! (state & (1<<p) )){
                ways += solve(hat+1, (state | (1<<p) ) ) % MOD;
            }
        }

        return dp[hat][state] = ways % MOD;
    }


    int numberWays(vector<vector<int>>& hats) {

        dp.assign(41, vi((1LL<<10), -1));
        
        n = hats.size();
        adj.resize(42);

        for(int i=0; i<n; i++){
            for(auto e: hats[i]) adj[e].push_back(i);
        }

        int ans = solve(1, 0);

        return ans;

    }
};
41000