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>
using namespace std;
typedef long long ll;

#define EPS (1e-7)
#define INF (1e16)
#define PI (acos(-1))
//const ll mod = 1000000007;
ll dp[100005][305];
int main() {
    //cout.precision(10);
    ll n, m, k;
    cin >> n >> m >> k;
    ll a[100005];
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        dp[i][1] = a[i];
    }
    for(int time = 2; time <= k; time++){
        dp[1][time] = -INF;
        list<int> l;
        for(ll index = 2; index <= n; index++){
            while(!l.empty()){
                if(dp[l.back()][time-1] <= dp[index - 1][time-1]){
                    l.pop_back();
                } else {
                    break;
                }
            }
            l.push_back(index - 1);
            dp[index][time] = dp[l.front()][time - 1] + time * a[index];
            if(l.front() == (index - m)){
                l.pop_front();
            }
        }
        /*cout << time << endl;
        for(int i = 1; i <= n; i++){
            cout << dp[i][time] << endl;
        }
        cout << endl;*/
    }
    ll ans = 0;
    for(int i = 1; i <= n; i++){
        ans = max(ans, dp[i][k]);
    }
    cout << ans << endl;
    return 0;
}