<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:
        M = 10**9 + 7
        def conjugate(x, y):
            out = {0,1,2}
            if x>=0: out.remove(x)
            if y>=0: out.discard(y)
            return out

        @lru_cache(None)
        def genPerm(col):
            ans = []
            def backtrack(curr, i):
                if i == m: 
                    ans.append(curr)
                    return
                for j in conjugate(col[i], curr[-1] if curr else -1): 
                    backtrack(curr + (j,), i + 1)
            backtrack(tuple(), 0)
            return ans
    
        @lru_cache(None)
        def dp(col,i):
            if i==n: return 1
            return sum(dp(c, i + 1) for c in genPerm(col)) % M
        return dp(tuple([-1] * m),0)
35200