<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 - k)%(k - 1) != 0)return -1;
        vector<int> pre(n);
        vector<vector<int>> dp(n + 1, vector<int> (n + 1, -1));
        for(int i = 0; i < n; i++){
            if(i == 0)pre[i] = stones[i];
            else pre[i] += pre[i - 1] + stones[i];
        }
        function<int(int i, int j)> helper;
        helper = [&](int i, int j){
            if(j - i + 1 < k)return 0;
            if(dp[i][j] != -1)return dp[i][j];
            int ans = INT_MAX;
            for(int s = i; s < j; s += k - 1){
                int temp = helper(i, s) + helper(s + 1, j);
                ans = min(ans, temp);
            }
            if((j - i)%(k - 1) == 0)ans += pre[j] - pre[i] + stones[i];
            return dp[i][j] = ans;
        };
        return helper(0, n - 1);
    }
};
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