<p>There are <code>n</code> piles of <code>stones</code> arranged in a row. The <code>i<sup>th</sup></code> pile has <code>stones[i]</code> stones.</p>

<p>A move consists of merging exactly <code>k</code> <strong>consecutive</strong> piles into one pile, and the cost of this move is equal to the total number of stones in these <code>k</code> piles.</p>

<p>Return <em>the minimum cost to merge all piles of stones into one pile</em>. If it is impossible, return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> stones = [3,2,4,1], k = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.
</pre>

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

<pre>
<strong>Input:</strong> stones = [3,2,4,1], k = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> After any merge operation, there are 2 piles left, and we can&#39;t merge anymore.  So the task is impossible.
</pre>

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

<pre>
<strong>Input:</strong> stones = [3,5,1,2,6], k = 3
<strong>Output:</strong> 25
<strong>Explanation:</strong> We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.
</pre>

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

<ul>
	<li><code>n == stones.length</code></li>
	<li><code>1 &lt;= n &lt;= 30</code></li>
	<li><code>1 &lt;= stones[i] &lt;= 100</code></li>
	<li><code>2 &lt;= k &lt;= 30</code></li>
</ul>

class Solution:
    def mergeStones(self, stones: List[int], k: int) -> int:
        
        n = len(stones)

        if (n-1)%(k-1): return -1
        pref = [0 for i in range(n+1)]

        for i in range(1,n+1):
            pref[i] = pref[i-1] + stones[i-1]
        

        def pref_sum(l,r):
            return pref[r+1]-pref[l]

        inf = int(2e9)

        @cache
        def solve(i,j):
    
            if j-i+1 < k:
                return 0,j-i+1
            val,sz = inf,1
            for l in range(i,j):
                left,lsz = solve(i,l)
                right,rsz = solve(l+1,j)
                if lsz+rsz<k:
                    if left+right<val:
                        val,sz = left+right,lsz+rsz
                        
                
                if lsz+rsz == k:
                    if left+right+pref_sum(i,j)<val:
                        val,sz = left+right+pref_sum(i,j),1

            return val,sz

        return solve(0,n-1)[0]
       

            
            
17300