<p>You are given two integers <code>m</code> and <code>n</code>. Consider an <code>m x n</code> grid where each cell is initially white. You can paint each cell <strong>red</strong>, <strong>green</strong>, or <strong>blue</strong>. All cells <strong>must</strong> be painted.</p>

<p>Return<em> the number of ways to color the grid with <strong>no two adjacent cells having the same color</strong></em>. Since the answer can be very large, return it <strong>modulo</strong> <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/2021/06/22/colorthegrid.png" style="width: 200px; height: 50px;" />
<pre>
<strong>Input:</strong> m = 1, n = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The three possible colorings are shown in the image above.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/copy-of-colorthegrid.png" style="width: 321px; height: 121px;" />
<pre>
<strong>Input:</strong> m = 1, n = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> The six possible colorings are shown in the image above.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> m = 5, n = 5
<strong>Output:</strong> 580986
</pre>

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

<ul>
	<li><code>1 &lt;= m &lt;= 5</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution:
    def colorTheGrid(self, m: int, n: int) -> int:
        MOD = 10**9 + 7
        
        # Function to check if two columns are valid next to each other
        def is_valid(col1, col2):
            for i in range(m):
                if col1[i] == col2[i]:
                    return False
            return True
        
        # Generate all possible columns
        def generate_cols():
            cols = []
            
            def backtrack(col, i):
                if i == m:
                    cols.append(col)
                    return
                for color in range(3):
                    if i > 0 and col[i-1] == color:
                        continue
                    backtrack(col + [color], i + 1)
            
            backtrack([], 0)
            return cols
        
        cols = generate_cols()
        num_cols = len(cols)
        
        # Create a graph where each node represents a column and edges represent validity
        graph = [[] for _ in range(num_cols)]
        for i in range(num_cols):
            for j in range(num_cols):
                if is_valid(cols[i], cols[j]):
                    graph[i].append(j)
        
        # DP array to store the number of ways to color the grid
        dp = [[0] * num_cols for _ in range(n)]
        
        # Base case: The number of ways to start with each column in the first row
        for j in range(num_cols):
            dp[0][j] = 1
        
        # Fill the DP array
        for i in range(1, n):
            for j in range(num_cols):
                for k in graph[j]:
                    dp[i][j] = (dp[i][j] + dp[i-1][k]) % MOD
        
        # Sum up the number of ways for all columns in the last row
        return sum(dp[n-1]) % MOD
328