<p>There is a pizza with <code>3n</code> slices of varying size, you and your friends will take slices of pizza as follows:</p>

<ul>
	<li>You will pick <strong>any</strong> pizza slice.</li>
	<li>Your friend Alice will pick the next slice in the anti-clockwise direction of your pick.</li>
	<li>Your friend Bob will pick the next slice in the clockwise direction of your pick.</li>
	<li>Repeat until there are no more slices of pizzas.</li>
</ul>

<p>Given an integer array <code>slices</code> that represent the sizes of the pizza slices in a clockwise direction, return <em>the maximum possible sum of slice sizes that you can pick</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/18/sample_3_1723.png" style="width: 500px; height: 266px;" />
<pre>
<strong>Input:</strong> slices = [1,2,3,4,5,6]
<strong>Output:</strong> 10
<strong>Explanation:</strong> Pick pizza slice of size 4, Alice and Bob will pick slices with size 3 and 5 respectively. Then Pick slices with size 6, finally Alice and Bob will pick slice of size 2 and 1 respectively. Total = 4 + 6.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/18/sample_4_1723.png" style="width: 500px; height: 299px;" />
<pre>
<strong>Input:</strong> slices = [8,9,8,6,1,1]
<strong>Output:</strong> 16
<strong>Explanation:</strong> Pick pizza slice of size 8 in each turn. If you pick slice with size 9 your partners will pick slices of size 8.
</pre>

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

<ul>
	<li><code>3 * n == slices.length</code></li>
	<li><code>1 &lt;= slices.length &lt;= 500</code></li>
	<li><code>1 &lt;= slices[i] &lt;= 1000</code></li>
</ul>

class Solution:
    def maxSizeSlices(self, slices: List[int]) -> int:
        n = len(slices)
        k = n//3
        m=n-1
        @cache
        def dfs(i,j):
            if i>=m: return 0
            res1 = slices[i]+dfs(i+2,j-1) if j>=1 else 0
            res2 = dfs(i+1,j)
            return max(res1,res2)
        r1 = dfs(0,k)
        slices = slices[1:]
        dfs.cache_clear()
        r2 = dfs(0,k)
    
        
        return max(r1,r2)
        
50000