Problem: There are N panels arranged in a row in Takahashi's house, numbered 1 through N. The i-th panel has a number A_i written on it. Takahashi is playing by throwing balls at these panels.

Takahashi threw a ball K times. Let the panel hit by a boll in the i-th throw be panel p_i. He set the score for the i-th throw as i \times A_{p_i}.

He was about to calculate the total score for his throws, when he realized that he forgot the panels hit by balls, p_1,p_2,...,p_K. The only fact he remembers is that for every i (1 ≦ i ≦ K-1), 1 ≦ p_{i+1}-p_i ≦ M holds. Based on this fact, find the maximum possible total score for his throws.

Constraints

* 1 ≦ M ≦ N ≦ 100,000
* 1 ≦ K ≦ min(300,N)
* 1 ≦ A_i ≦ 10^{9}

Input

The input is given from Standard Input in the following format:


N M K
A_1 A_2 … A_N


Output

Print the maximum possible total score for Takahashi's throws.

Examples

Input

5 2 3
10 2 8 10 2


Output

56


Input

5 5 2
5 2 10 5 9


Output

28


Input

10 3 5
3 7 2 6 9 4 8 5 1 1000000000


Output

5000000078

Solution (in CPP):

Solution:
 #include<bits/stdc++.h>
 using namespace std;
#define INF 1000000000
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
typedef long long LL;
LL M,N;
LL K;
LL A[100001];
LL dp[302][100001];
LL deq[100001];
int main(){
    REP(i,302){
        REP(j,100001){
            dp[i][j]=0;
        }
    }
    cin>>N>>M>>K;
    REP(i,N){
        cin>>A[i];
        dp[0][i]=A[i];
    }
    for(int i=1;i<K;i++){
        int bg=0;
        int ed=0;
        deq[0]=0;
        ed++;

        REP(j,N){
            if(j==0)continue;
            if(dp[i-1][deq[bg]]!=0){
               dp[i][j]=dp[i-1][deq[bg]]+(i+1)*A[j];
            }else{
               dp[i][j]=0;
            }
            while(ed>bg){
               if(dp[i-1][j]<dp[i-1][deq[ed-1]])break;
               ed--;
            }
            deq[ed]=j;
            ed++;
            if(j>=M){
               if(deq[bg]==j-M)bg++;
            }
        }
        /*REP(j,N){
            cout<<dp[i][j]<<" ";
        }
        cout<<endl;
        */
    }
    LL ans=0;
    REP(i,N){
        ans=max(ans,dp[K-1][i]);
    }
    cout<<ans<<endl;
    return 0;
}