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
#include <bits/stdc++.h>
using namespace std;

int main(){
    long long i, j, k;
    long long N, M, K;
    cin >> N >> M >> K;
    vector<long long> A(N);
    for(i=0; i<N; i++) cin >> A[i];

    long long ans = 0;
    vector<long long> dp(N, 0);
    for(j=1; j<=K; j++){
        deque<long long> deq = {};
        vector<long long> ndp(N);
        for(i=0; i<N; i++){
            if(i>=j-1){
                long long m = deq.empty() ? 0 : dp[deq.front()];
                ndp[i] = m + j * A[i];
            }else{
                ndp[i] = 0;
            }
            if(!deq.empty() && deq.front() == i-M) deq.pop_front();
            while(!deq.empty() && dp[deq.back()] < dp[i]) deq.pop_back();
            deq.push_back(i);
        }
        dp = ndp;
    }
    cout << *max_element(dp.begin(), dp.end()) << endl;
    return 0;
}
