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 <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#define INF 1e18
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n, m, k;
  cin>>n>>m>>k;
  ll a[100001];
  for(int i=1; i<=n; i++){
    cin>>a[i];
  }
  ll dp[100001], mx[100001];
  for(int i=1; i<=n; i++) dp[i]=a[i];
  for(int i=2; i<=k; i++){
    deque<int> deq;
    deq.push_back(1);
    mx[1]=-INF;
    for(int j=2; j<=n; j++){
      while(1){
        int s=deq.front();
        if(s>=j-m) break;
        deq.pop_front();
      }
      mx[j]=dp[deq.front()];
      while(!deq.empty()){
        int t=deq.back();
        if(dp[t]>dp[j]) break;
        deq.pop_back();
      }
      deq.push_back(j);
    }
    for(int j=1; j<=n; j++){
      dp[j]=mx[j]+((ll)i)*a[j];
    }
  }
  ll ans=0;
  for(int i=1; i<=n; i++) ans=max(ans, dp[i]);
  cout<<ans<<endl;
	return 0;
}