<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:
        @cache
        def f(i,j):
            if i == len(pizza)-1 and j == len(pizza[0])-1:
                return 1 if pizza[i][j] == 'A' else 0

            if pizza[i][j] == 'A':
                addl = 1
            else:
                addl = 0
            
            if i == len(pizza)-1:
                return addl + f(i,j+1)
            elif j == len(pizza[0]) - 1:
                return addl + f(i+1,j)
            else:
                return addl + f(i+1,j) + f(i,j+1) - f(i+1,j+1)

        q_apples = [[f(i,j) for j in range(len(pizza[0]))] for i in range(len(pizza))]
        
        @cache
        def h(i,j,k_rem):
            if k_rem ==1:
                if q_apples[i][j] > 0:
                    return 1
                else:
                    return 0
            if k_rem == 2:
                total = 0
                for new_i in range(i,len(pizza)):
                    if q_apples[new_i][j] > 0 and q_apples[i][j] - q_apples[new_i][j] != 0:
                        total+=1
                for new_j in range(j,len(pizza[0])):
                    if q_apples[i][new_j] > 0 and q_apples[i][j] - q_apples[i][new_j] != 0:
                        total+=1
                return total%(10**9+7)

            total = 0
            for new_i in range(i,len(pizza)):
                if q_apples[new_i][j] > 0 and q_apples[i][j] - q_apples[new_i][j] != 0 and q_apples[new_i][j] >= k_rem-1:
                    total += h(new_i,j,k_rem-1)
            for new_j in range(j,len(pizza[0])):
                if q_apples[i][new_j] > 0 and q_apples[i][j] - q_apples[i][new_j] != 0 and q_apples[i][new_j] >= k_rem-1:
                    total += h(i,new_j,k_rem-1)
            return total%(10**9+7)
            
        return h(0,0,k)

        
19100