<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:
vector<vector<int>>dp;
    int mergeStones(vector<int>& a, int k) {
        int n= a.size();
        dp = vector<vector<int>>(30, vector<int>(30, -1));
        vector<int>sum(n+5, 0);
        sum[0]= a[0];
        if((n-1)%(k-1)) return -1;
        for(int i=0;i<n;i++){
            sum[i+1] = sum[i]+a[i];
        }
        return sol(0, n-1, a, k, sum);
    }
    int sol(int i, int j, vector<int>& a, int k, vector<int>& sum){
        if(i>=j) return 0;
        if(dp[i][j]!=-1) return dp[i][j];
        int ans = INT_MAX;
        for(int p=i;p<j;p=p+k-1){
            ans = min(ans, sol(i, p, a, k, sum)+sol(p+1, j, a, k, sum));
        }
        if((j-i)%(k-1) ==0) ans+=sum[j+1]- sum[i];
        return dp[i][j] =  ans;
    }
};
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