<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 {
public:
    int mergeStones(vector<int>& stones, int k) {
        int n = stones.size();
        v = vector<vector<vector<int>>>(k, vector<vector<int>>(n, vector<int>(n, -2)));
        
        preSum = vector<int>(n + 1, 0);
        for(int i = 0; i < n; i++) preSum[i + 1] = preSum[i] + stones[i];

        dp = vector<vector<int>>(n, vector<int>(n, -1));
        for(int i = 0; i < n; i++) dp[i][i] = 0;
        for(int i = 0; i + k - 1 < n; i++) dp[i][i + k - 1] = preSum[i + k] - preSum[i];

        for(int length = k + 1; length <= n; length++)
        {
            for(int start = 0; start + length - 1 < n; start++)
            {
                int tmp = dfs(start, start + length - 1, k - 1);
                if(tmp > 0) dp[start][start + length - 1] = tmp + preSum[start + length] - preSum[start];
            }
        }

        return dp[0][n - 1];
    }

private:
    int dfs(int start, int end, int left)
    {
        if(left <= 0)
        {
            return dp[start][end];
        }

        if(v[left][start][end] != -2) return v[left][start][end];

        int res = -1;
        for(int i = start + 1; i <= end; i++)
        {
            int tmp = dfs(i, end, left - 1);
            if(tmp >= 0 && dp[start][i - 1] >= 0) 
            {   
                if(res == -1) res = tmp + dp[start][i - 1];
                else res = min(res, tmp + dp[start][i - 1]);
            }
        }

        v[left][start][end] = res;

        return res;
    }

private:
    vector<vector<int>> dp;
    vector<int> preSum;
    vector<vector<vector<int>>> v;
};
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