<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:
        t = ['a', 'b', 'c']
        for i in range(m-1):
            new = []
            for s in t:
                if s[-1] != 'a': new.append(s+'a')
                if s[-1] != 'b': new.append(s+'b')
                if s[-1] != 'c': new.append(s+'c')
            t = new
        d = defaultdict(set)
        for s1, s2 in itertools.combinations(t, 2):
            for i in range(m):
                if s1[i] == s2[i]: break
                if i == m-1: 
                    d[s1].add(s2)
                    d[s2].add(s1)
        @cache
        def dp(s, n):
            if n == 0: return 1
            res = 0
            for ch in d[s]:
                res += dp(ch, n-1)
            return res
        return sum(dp(s, n-1) for s in t)%(10**9+7)        
55400