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,s,e) for(int i=(s); i<(e);i++)
#define rep(i,n) REP(i,0,n)
#define rep1(i,n) REP(i,1,n)
#define repe(i,n) for(auto &&i:n)
#define all(v) (v).begin(),(v).end()
#define decimal fixed<<setprecision(20)
#define fastcin() cin.tie(0);ios::sync_with_stdio(false)
using namespace std;
using LL = long long;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = 1e9;
const LL LLINF = 1e16;

vector<LL> slide_max(vector<LL> &a, int k){
    deque<LL> deq;
    vector<LL> b;
    rep(i, a.size()){
        while(!deq.empty() && a[deq.back()] <= a[i]) deq.pop_back();
        deq.push_back(i);
        b.push_back(a[deq.front()]);
        if(i-k+1 >= 0 && deq.front() == i-k+1) deq.pop_front();
    }
    return b;
}

int main(){
    int n,m,k;
    cin >> n >> m >> k;
    vector<LL> a(n);
    rep(i, n) cin >> a[i];

    vector<LL> dp = a;
    REP(j, 2, k+1){
        vector<LL> mx = slide_max(dp, m);
        REP(i, j-1, n){
            dp[i] = j*a[i] + mx[i-1];
        }
        // repe(i, dp) cout << i << " "; cout << endl;
    }
    cout << *max_element(dp.begin()+k-1, dp.end()) << endl;


}