<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:
        from functools import lru_cache
        n = len(stones)
        if n == 1:
            return 0
        if n < k or ((n - 1) % (k - 1) != 0):
            return -1
        sums = [0] * n
        sum_ = 0
        for i, s in enumerate(stones):
            sum_ += s
            sums[i] = sum_
        print(sums)
        
        @lru_cache(None)
        def helper(i, j, k_):
            print(i, j, k_)
            n = j - i + 1
            if n == k_:
                print("a")
                return 0
            # if n < k or ((n - k_) % (k - 1) != 0):
            #     print("b", n, k, (n - 1) % (k - 1))
            #     return float('inf')
            if k_ == 1:
                print("c")
                return helper(i, j, k) + sums[j] - sums[i - 1] if i > 0 else helper(i, j, k) + sums[j]
            else:
                min_cost = float('inf')
                for idx in range(i + 1, j + 1, k - 1):
                    print("d", (i, idx, 1), (idx + 1, j, k_ - 1))
                    cost = helper(i, idx - 1, 1) + helper(idx, j, k_ - 1)
                    min_cost = min(min_cost, cost)
                return min_cost
        return helper(0, n - 1, 1)
                

            
            
            
        
17400