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;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=1<<18;
const ll INF=1LL<<60;

int main(){
    
    int N,M,K;cin>>N>>M>>K;
    
    vector<ll> P(N);
    
    for(int j=0;j<N;j++){
        cin>>P[j];
    }
    vector<ll> A(N+M,0),B(N+M-M+1,0);//B[i]はi番目(1-indexed)以前の最大値
    for(int i=0;i<M;i++) A[i]=-INF;
    /*for(int i=M;i<N+M;i++){
        A[i]=P[i-M];
    }*/
    deque<int> deq(N+M);
    /*int s=0,t=0;
    for(int i=0;i<N+M;i++){
        while(s<t&&A[deq[t-1]]<=A[i]) t--;
        deq[t]=i;
        t++;
        
        if(i-M+1>=0){
            B[i-M+1]=A[deq[s]];
            if(deq[s]==i-M+1) s++;
        }
    }*/
    
    ll ans=0;
    
    for(int i=1;i<=K;i++){
        vector<ll> S(N+1,-INF);
        S[0]=0;
        for(int j=1;j<=N;j++){
            S[j]=P[j-1]*i;
        }
        for(int j=1;j<=N;j++){
            ll a=B[j-1];
            S[j]+=a;
        }
        for(int j=1;j<=N;j++){
            A[j+M-1]=S[j];
        }
        //deq.clear();
        int s=0,t=0;
        for(int j=0;j<N+M;j++){
            while(s<t&&A[deq[t-1]]<=A[j]) t--;
            deq[t]=j;
            t++;
            
            if(j-M+1>=0){
                B[j-M+1]=A[deq[s]];
                if(deq[s]==j-M+1) s++;
            }
        }
        if(i==K){
            for(int j=1;j<=N;j++) ans=max(ans,S[j]);
        }
        /*for(int j=1;j<=N;j++) cout<<S[j]<<" ";
        cout<<endl;*/
    }
    
    cout<<ans<<endl;
}