<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 it's impossible to merge into one pile, return -1
        if (n - 1) % (k - 1) != 0:
            return -1
        
        # Prefix sum array to quickly calculate sum of any subarray
        prefix_sum = [0] * (n + 1)
        for i in range(n):
            prefix_sum[i + 1] = prefix_sum[i] + stones[i]
        
        # dp[i][j][m]: Minimum cost to merge subarray stones[i:j+1] into m piles
        dp = [[[float('inf')] * (k + 1) for _ in range(n)] for _ in range(n)]
        
        # Base case: cost of merging one pile to one pile is 0
        for i in range(n):
            dp[i][i][1] = 0
        
        # Length of the subarray
        for length in range(2, n + 1):  # subarray lengths from 2 to n
            for i in range(n - length + 1):  # starting index of subarray
                j = i + length - 1  # ending index of subarray
                
                # Merge into different pile counts
                for m in range(2, k + 1):  # number of piles from 2 to k
                    # Partition the subarray into two parts
                    for mid in range(i, j, k - 1):
                        dp[i][j][m] = min(dp[i][j][m], dp[i][mid][1] + dp[mid + 1][j][m - 1])
                
                # Merge k piles into 1 pile
                if dp[i][j][k] < float('inf'):
                    dp[i][j][1] = dp[i][j][k] + prefix_sum[j + 1] - prefix_sum[i]
        
        return dp[0][n - 1][1] if dp[0][n - 1][1] < float('inf') else -1

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