<p>Given a rectangle of size <code>n</code> x <code>m</code>, return <em>the minimum number of integer-sided squares that tile the rectangle</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_11_1592.png" style="width: 154px; height: 106px;" /></p>

<pre>
<strong>Input:</strong> n = 2, m = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> <code>3</code> squares are necessary to cover the rectangle.
<code>2</code> (squares of <code>1x1</code>)
<code>1</code> (square of <code>2x2</code>)</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_22_1592.png" style="width: 224px; height: 126px;" /></p>

<pre>
<strong>Input:</strong> n = 5, m = 8
<strong>Output:</strong> 5
</pre>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_33_1592.png" style="width: 224px; height: 189px;" /></p>

<pre>
<strong>Input:</strong> n = 11, m = 13
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= n, m &lt;= 13</code></li>
</ul>

class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        @lru_cache(None)
        def dp(n, m):
            if n == 0 or m == 0:
                return 0
            if n == m:
                return 1
            ans =  float("inf")
            for a in range(n):
                for b in range(m):
                    for c in range(a, n):
                        for d in range(b, m):
                            curr = [(c-a, d-b), (a, d), (c ,m-d), (n-c, m-b), (n-a, b)]
                            temp = 0
                            for i, j in curr:
                                if i == n and j == m:
                                    temp = float("inf")
                                    break
                                temp += dp(i, j)
                            ans = min(ans, temp)
            return ans
        return dp(n,m)
336