<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>

#define ll long long
const int INF=1e5;
class Solution {
public:
    int mergeStones(vector<int>& stones, int k) {
        int n=stones.size();
        vector<vector<vector<int>>>dp(n,vector<vector<int>>(n,vector<int>(k+1,INF)));
        for(int i=0;i<n;i++){
            dp[i][i][1]=0;
        }
        vector<int>presum(n,stones[0]);
        for(int i=1;i<n;i++){
            presum[i]=presum[i-1]+stones[i];
            // cout<<presum[i]<<" ";
        }
        // cout<<endl;
        for(int l=2;l<=n;l++){
            for(int i=0;i<=n-l;i++){
                for(int j=k;j>0;j--){
                    // cout<<dp[i][j][k]<<endl;
                    if(j==l){
                        dp[i][i+l-1][j]=0;
                        continue;
                    }
                    if(j>l)continue;
                    // cout<<"k"<<endl;
                    // int ind=i+l-1;
                    if(j==1){

                        dp[i][i+l-1][1]=dp[i][i+l-1][k]+
                        presum[i+l-1]-((i==0)?0:presum[i-1]);
                        continue;
                    }
                    // cout<<"k"<<endl;
                    for(int l1=1;l1<l;l1++){
                        if(i==0&&l==4){
                            cout<<l1<<endl;
                        }
                        dp[i][i+l-1][j]=min(dp[i][i+l-1][j],
                         dp[i][i+l1-1][j-1]+dp[i+l1][i+l-1][1]);
                    }
                }
            }
        }
        if(dp[0][n-1][1]>=INF)return -1;
        return dp[0][n-1][1];
    }
};
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