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 <string>
#include <vector>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<functional>
#include<list>
#include<deque>
#include<bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<cstring>
#include<sstream>
#include<complex>
#include<iomanip>
#include<numeric>
#include<cassert>
#define X first
#define Y second
#define pb push_back
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define repe(X,Y) for ((X) = 0;(X) < (Y);++(X))
#define peat(X,Y) for (;(X) < (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define eb emplace_back
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<class T> using vv=vector<vector<T>>;
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"{"; rep(i,t.size()) {os<<t[i]<<",";} os<<"}"<<endl; return os;}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}
const ll MOD=1e9+7;
const ll INF=1e15;
ll dp[333][112345];

int main(){
  ios_base::sync_with_stdio(false);
  cout<<fixed<<setprecision(0);
  fill(dp[1],dp[1]+332*112345,-INF);
  int n,m,t;
  cin>>n>>m>>t;
  vector<ll> a(n);
  rep(i,n) cin>>a[i];
  deque<pll> que[333];
  rep(i,n)rep(j,t){
    while(que[j].size() && que[j].back().X<=dp[j][i]) que[j].pop_back();
    que[j].eb(dp[j][i],i);
    while(i-que[j].front().Y>=m) que[j].pop_front();
    assert(que[j].size());
    dp[j+1][i+1]=que[j].front().X+(j+1ll)*a[i];
  }
  //rep(i,t+1){rep(j,n+1)cout<<dp[i][j]<<",";cout<<endl;}
  cout<<*max_element(dp[t],dp[t]+n+1)<<endl;
  return 0;
}