<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 {
    int n,kref;
    int dp[31][31][31];
    int ps[31];
    long long rec(int l,int r,int k,vector<int>& stones){
        #define int long long
        if(l==r && k==1) return stones[l];
        if(l==r) return 1e9;
        if(dp[l][r][k]!=-1) return dp[l][r][k];
        int ans =1e9;
        if(k==1){
            int ref = (l==0)?l:ps[l-1];
            ans = rec(l,r,kref,stones) + ps[r]-ref;
        }else{
            for(int i=l;i<r;i++){
                ans = min(ans, rec(l,i,1,stones)+rec(i+1,r,k-1,stones));
            }
        }
        return dp[l][r][k]=ans;
        #undef int
    }
public:
    int mergeStones(vector<int>& stones, int K) {
        // if(K>stones.size()) return -1;
        n = stones.size();
        if(n==1) return 0;
        kref =K;
        memset(dp,-1,sizeof(dp));
        ps[0]=stones[0];
        for(int i=1;i<n;i++) ps[i]=stones[i]+ps[i-1];
        long long ans= rec(0,n-1,K,stones);
        return ans>=1e9 ? -1 :ans;
    }
};
9800