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<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
VI stk;
in lf,rt;
void clr(){
  stk.clear();
  lf=0;
  rt=0;
}
void ad(in a){
  while(rt>lf && stk[rt-1]<a)
    --rt;
  if(rt==sz(stk))
    stk.PB(0);
  stk[rt]=a;
  ++rt;
}
void rm(in a){
  if(lf<rt && stk[lf]==a)
    ++lf;
}
in bst(){
  if(lf==rt)
    return -1;
  return stk[lf];
}
VI a;
VI bs[2];
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  in n,m,k;
  cin>>n>>m>>k;
  a.resize(n);
  forn(i,n)
    cin>>a[i];
  forn(z,2)
    bs[z].resize(n);
  in sw=1;
  in nw=0;
  bs[nw]=a;
  forn(i,n)
    bs[nw][i]*=k;
  for(in z=1;z<k;++z){
    swap(nw,sw);
    clr();
    for(in i=n-1;i>=0;--i){
      bs[nw][i]=bst();
      if(bs[nw][i]!=-1)
	bs[nw][i]+=(k-z)*a[i];
      ad(bs[sw][i]);
      if(i+m<n)
	rm(bs[sw][i+m]);
    }
  }
  in r=0;
  forn(i,n)
    r=max(r,bs[nw][i]);
  cout<<r<<endl;
  return 0;
}