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 <bits/stdc++.h>
#define rep(i,x,y) for(ll i=x;i<y;i++)
#define Rrep(i,y,x) for(ll i=y;i>x;i--)
#define SIZE(a) ll(a.size())
#define vll vector<ll>
#define MEMSET(a, n, m) for(ll i=0;i<=n;i++) a[i] = m
#define BIT(n) (ll(1)<<n)
#define UNIQUE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define UNIQUE_ARRAY(a,n) n = unique(a + 1, a + x + 1) - a - 1
#define SORT(a,n) sort(a+1,a+n+1)
#define SORT_O(a,n,order) sort(a+1,a+n+1,order)
#define print(x) cout << (x) << endl
#define printvec(x) for(ll i=0;i<x.size();i++) cout << (x.at(i)) <<' ' 
typedef long long ll;
using namespace std;

vector<ll> f(vector<ll> A, ll N, ll M){
    deque<ll> H;
    vector<ll> L(N);
    H.push_back(0);
    rep(i,1,min(M+1,N)){
        L.at(i)=H.at(0);
        while ( H.size() >0 && A.at(H.at(H.size()-1)) <= A.at(i)){
            H.pop_back();
        }
        H.push_back(i);
    }
    rep(i,M+1,N){
        if (H.at(0)<i-M){
            H.pop_front();
        }
        L.at(i)=H.at(0);
        while ( H.size() >0 && A.at(H.at(H.size()-1)) <= A.at(i)){
            H.pop_back();
        }
        H.push_back(i);
    }
    vector<ll> ans(N);
    rep(i,1,N){
        ans.at(i)=A.at(L.at(i));
    }
    return ans;
}

int main(){
    ll N,M,K;
    cin>>N>>M>>K;
    vector<ll> A(N);
    rep(i,0,N){
        cin>>A.at(i);
    }
    vector<ll> dp(N+1),ndp(N+1);
    ll ans=0;
    vector<ll> B(N+1);
    rep (k,1,K+1){
        B = f(dp,N+1,M);
        rep(i,k,N+1){
            ndp.at(i)=B.at(i)+ k*A.at(i-1);
        }
        dp=ndp;
    }
    rep(i,0,N+1){
        ans=max(ans,dp.at(i));
    }
    print(ans);


}