<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:
        def dfs(n, m, grid, count):
            nonlocal result
            if count >= result:
                return
            if all(all(row) for row in grid):
                result = min(result, count)
                return
            for i in range(n):
                for j in range(m):
                    if grid[i][j] == 0:
                        break
                if grid[i][j] == 0:
                    break
            
            max_size = min(n - i, m - j)
            while max_size > 0:
                if all(grid[x][j:j + max_size] == [0] * max_size for x in range(i, i + max_size)):
                    for x in range(i, i + max_size):
                        grid[x][j:j + max_size] = [1] * max_size
                    dfs(n, m, grid, count + 1)
                    for x in range(i, i + max_size):
                        grid[x][j:j + max_size] = [0] * max_size
                max_size -= 1

        grid = [[0] * m for _ in range(n)]
        result = float('inf')
        dfs(n, m, grid, 0)
        return result

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