<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 __init__(self):
        self.prefix_sum = []
        self.k = 0
    
    def mergeStones(self, stones: List[int], k: int) -> int:
        n = len(stones)
        
        # Check if merging is possible
        if (n - 1) % (k - 1) != 0:
            return -1
        
        self.k = k
        self.prefix_sum = [0] * (n + 1)
        
        # Compute prefix sums
        for i in range(n):
            self.prefix_sum[i + 1] = self.prefix_sum[i] + stones[i]

        # Use LRU cache for memoization
        @lru_cache(None)
        def dp(l: int, r: int, piles: int) -> int:
            if r - l < piles:
                return inf
            if r - l == piles:
                return 0
            if piles == 1:
                return dp(l, r, self.k) + self.prefix_sum[r] - self.prefix_sum[l]
            return min(dp(l, m, i) + dp(m, r, piles - i) 
                       for i in range(1, piles) 
                       for m in range(l + i, r - (piles - i) + 1))
        
        return dp(0, n, 1)


18000