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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
#define mod 1000000007
#define FOR(x,to) for(int x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define long long long
inline int rei(){int x;cin>>x;return x;}
inline long rel(){long x;cin>>x;return x;}
inline string res(){string x;cin>>x;return x;}
//------------------------------------------------------- 
deque<pair<int,long>> DP[301];
long A[100000];
void Calc(){
	int N = rei();
	int M = rei();
	int K = rei();
	for(int i=0;i<N;i++){
		A[i] = rel();
	}
	long ans = 0;
	for(int i=N-1;i>=0;i--){
		for(int j=K;j>1;j--){
			while(!DP[j-1].empty()){
				if(DP[j-1].front().first > i+M){
					DP[j-1].pop_front();
				}
				else{
					break;
				}
			}
			if(!DP[j-1].empty()){
				while(!DP[j].empty() && DP[j].back().second <= DP[j-1].front().second+(K-j+1)*A[i]){
					DP[j].pop_back();
				}
				DP[j].push_back({i,DP[j-1].front().second+(K-j+1)*A[i]});
			}
		}
		{
			while(!DP[1].empty() && DP[1].back().second <= K*A[i]){
				DP[1].pop_back();
			}
			DP[1].push_back({i,K*A[i]});
		}
		if(!DP[K].empty()){
			ans = max(ans,DP[K].front().second);
		}
	}
	cout << ans << endl;
}
int main(int argc,char** argv){
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0); Calc(); return 0;
}