<p>Given a rectangular pizza represented as a <code>rows x cols</code>&nbsp;matrix containing the following characters: <code>&#39;A&#39;</code> (an apple) and <code>&#39;.&#39;</code> (empty cell) and given the integer <code>k</code>. You have to cut the pizza into <code>k</code> pieces using <code>k-1</code> cuts.&nbsp;</p>

<p>For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.</p>

<p><em>Return the number of ways of cutting the pizza such that each piece contains <strong>at least</strong> one apple.&nbsp;</em>Since the answer can be a huge number, return this modulo 10^9 + 7.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/ways_to_cut_apple_1.png" style="width: 500px; height: 378px;" /></strong></p>

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AAA&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 3 
<strong>Explanation:</strong> The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.
</pre>

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

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AA.&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;A..&quot;,&quot;...&quot;], k = 1
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= rows, cols &lt;= 50</code></li>
	<li><code>rows ==&nbsp;pizza.length</code></li>
	<li><code>cols ==&nbsp;pizza[i].length</code></li>
	<li><code>1 &lt;= k &lt;= 10</code></li>
	<li><code>pizza</code> consists of characters <code>&#39;A&#39;</code>&nbsp;and <code>&#39;.&#39;</code> only.</li>
</ul>

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        mod = 1000000000 + 7
        rows = len(pizza)
        cols = len(pizza[0])
        dp = [[[0 for _ in range(11)] for _ in range(55)] for _ in range(55)]
        cnt = [[0 for _ in range(55)] for _ in range(55)]

        for i in range(1, rows+1):
            for j in range(1, cols+1):
                cnt[i][j] = cnt[i-1][j] + cnt[i][j-1] - cnt[i-1][j-1] + (pizza[rows-i][cols-j] == 'A')

        for i in range(rows+1):
            dp[i][0][0] = 1

        for i in range(rows+1):
            for j in range(cols+1):
                for p in range(k):
                    if dp[i][j][p] != 0:
                        for ii in range(i+1, rows+1):
                            if cnt[ii][j] - cnt[i][j] > 0:
                                dp[ii][j][p+1] = (dp[ii][j][p+1] + dp[i][j][p]) % mod
                        for jj in range(j+1, cols+1):
                            if cnt[i][jj] - cnt[i][j] > 0:
                                dp[i][jj][p+1] = (dp[i][jj][p+1] + dp[i][j][p]) % mod

        return dp[rows][cols][k]
17900