<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>

MOD = 10 ** 9 + 7

class Solution:
    def colorTheGrid(self, m: int, n: int) -> int:

        def other(x, y):
            for i in range(1, 4):
                if i != x and i != y:
                    return i

        def possibilities(col1, col2):
            if col1 == col2 == -1:
                return [1, 2, 3]
            if (col1 == -1 or col2 == -1) or col1 == col2:
                return [x for x in range(1, 4) if x != col1 and x != col2]
            if col1 != col2:
                return [other(col1, col2)]
                
        def get_next(x, y):
            if y == m - 1:
                return x + 1, 0
            return x, y + 1

        @lru_cache(None)
        def dp(x, y, prev):
            if x == n:
                return 1
            prev_list = list(prev)
            nx, ny = get_next(x, y)
            col1, col2 = prev_list[y], prev_list[y + 1]
            cols = possibilities(col1, col2)
            res = 0
            for col in cols:
                res += dp(nx, ny, tuple(prev_list[:y+1] + [col] + prev_list[y+2:]))
                res %= MOD
            return res

        return dp(0, 0, tuple([-1] * (m + 1)))
        
2401