<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();
        if ((N-1)%(K-1)) return -1;

        auto dp = vector<vector<vector<int>>>(N, vector<vector<int>>(N, vector<int>(K+1, INT_MAX)));

        vector<int>sum(N+1, 0);
        for (int i = 0; i < N; i++)
            sum[i+1] = sum[i] + stones[i];

        for (int i = 0; i < N; i++) dp[i][i][1] = 0;

        for (int len = 2; len <= N; len++)
            for (int i = 0; i <= N-len; i++) {
                int j = i+len-1;
                for (int k = 2; k <= K; k++) {
                    if (k > len) continue;
                    for (int m = i; m < j; m++) {
                        if (dp[i][m][1] == INT_MAX || dp[m+1][j][k-1] == INT_MAX) continue;
                        dp[i][j][k] = min(dp[i][j][k], dp[i][m][1]+dp[m+1][j][k-1]);
                    }
                }

                if (dp[i][j][K] != INT_MAX)
                    dp[i][j][1] = dp[i][j][K] + sum[j+1]-sum[i];
            }
            if (dp[0][N-1][1] == INT_MAX) return -1;
            else return dp[0][N-1][1];
    }
};
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