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
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <deque>
#define MOD 1000000007
typedef long long ll;
using namespace std;

int N,M,K;
vector<ll> A;

const ll INF=-1e18;

int main(){
  cin>>N>>M>>K;
  A.resize(N);
  for(int i=0;i<N;i++) cin>>A[i];

  vector<ll> dp(N);
  for(int i=0;i<N;i++) dp[i]=A[i];

  for(int i=2;i<=K;i++){
      vector<ll> next(N,INF);
      deque<ll> DQ;
      for(int j=0;j<N;j++){
          while(!DQ.empty()){
              if(DQ.front()<j-M) DQ.pop_front();
              else break;
          }

          if(!DQ.empty()){
              next[j]=dp[DQ.front()]+A[j]*i;
          }

          while(!DQ.empty()){
              if(dp[DQ.back()]<=dp[j]) DQ.pop_back();
              else break;
          }

          DQ.push_back(j);
      }
      dp=next;
  }

  ll ans=0;
  for(int i=0;i<N;i++) ans=max(ans,dp[i]);
  cout<<ans<<endl;

  return 0;
}
