<p>You have a <code>grid</code> of size <code>n x 3</code> and you want to paint each cell of the grid with exactly one of the three colors: <strong>Red</strong>, <strong>Yellow,</strong> or <strong>Green</strong> while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).</p>

<p>Given <code>n</code> the number of rows of the grid, return <em>the number of ways</em> you can paint this <code>grid</code>. As the answer may grow large, the answer <strong>must be</strong> computed modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/26/e1.png" style="width: 400px; height: 257px;" />
<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 12
<strong>Explanation:</strong> There are 12 possible way to paint the grid as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5000
<strong>Output:</strong> 30228214
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

class Solution {
public:
    int m=1e9+7;
    vector<vector<long long>>dp;
    vector<int>v;
long long fn(int n,int bm){
    if(n==1)      return 1;

    if(dp[n][bm]!=-1)
        return dp[n][bm];
int r=bm%3,b=(bm-r)/9,g=(bm-9*b-r)/3;
    long long int ans=0;
 for(int j=0;j<12;j++){
int rmi=v[j]%3,bmi=(v[j]-rmi)/9, gmi=(v[j]-9*bmi-rmi)/3;
    
if(r!=rmi && gmi!=g && bmi!=b)
 ans=(ans+fn(n-1,v[j]))%m; }
    return dp[n][bm]=ans;         
  }
int numOfWays(int n) {      
long long int ans=0;    dp.resize(5001,vector<long long>(27,-1));
 for(int i=0;i<3;i++){
   for(int j=0;j<3;j++){
     for(int k=0;k<3;k++){
       if(i!=j && j!=k)
 v.push_back(i+3*j+9*k);}      }}

  for(int j=0;j<12;j++)
    ans=(ans+fn(n,v[j]))%m;
   return ans;
    }
};
72905