<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:

unordered_map<int, vector<int>> mp;

int dp[41][1<<10];

int mod= 1e9+7;

int help( int i, int n, int mask ){

    if(mask==((1<<n)-1 )) return 1;

    if(i>40) return 0;

    if(dp[i][mask]!=-1) return dp[i][mask];

      int a=0,b=0;

      for(auto t: mp[i]){
          
          if(mask & (1<<t)) continue;

          a= (a%mod+ (help(i+1,n, mask|(1<<t)))%mod)%mod;

      }

      b= help(i+1,n,mask);

      return dp[i][mask]= (a%mod+b%mod)%mod;;
}


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

          int n= hats.size();
          mp.clear();

          for(int i=0; i<n; i++){

            for(auto t: hats[i]){

                mp[t].push_back(i);
            }
          }

          memset(dp,-1,sizeof(dp));

          return help(1,n,0)%mod;


           
        
    }
};
13000