<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:
        rows = []
        M = int(1e9 + 7)
        def search(x, cur):
            if x == m:
                rows.append(tuple(cur))
            else:
                cur.append((cur[-1] + 1) % 3)
                search(x + 1, cur)
                cur.pop()
                cur.append((cur[-1] + 2) % 3)
                search(x + 1, cur)
                cur.pop()
        search(1, [0])
        search(1, [1])
        search(1, [2])
        
        dp = {r : 1 for r in rows}
        for c in range(n-1):
            ndp = {}
            for cr in dp:
                for pr in dp:
                    if any([cr[i] == pr[i] for i in range(m)]):
                        continue
                    ndp[cr] = (ndp.get(cr, 0) + dp[pr]) % M
            dp = ndp
        ans = 0
        for cnt in dp.values():
            ans = (ans + cnt) % M
        return ans


                
            

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