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;
long long INF = 100000000000000;
int main(){
  int N, M, K;
  cin >> N >> M >> K;
  vector<long long> A(N);
  for (int i = 0; i < N; i++){
    cin >> A[i];
  }
  vector<vector<long long>> dp(K, vector<long long>(N, -INF));
  for (int i = 0; i < N; i++){
    dp[0][i] = A[i];
  }
  for (int i = 1; i < K; i++){
    int L = 0, R = 0;
    deque<pair<long long, int>> dq;
    for (int j = i; j < N; j++){
      while (R < j){
        //push dp[i-1][R]
        if (!dq.empty()){
          while (dq.back().first <= dp[i - 1][R]){
            dq.pop_back();
            if (dq.empty()){
              break;
            }
          }
        }
        dq.push_back(make_pair(dp[i - 1][R], R));
        R++;
      }
      while (L <= max(j - M, 0)){
        if (!dq.empty()){
          while (dq.front().second < L){
            dq.pop_front();
            if (dq.empty()){
              break;
            }
          }
        }
        L++;
      }
      dp[i][j] = dq.front().first + A[j] * (i + 1);
    }
  }
  long long ans = 0;
  for (int i = 0; i < N; i++){
    ans = max(ans, dp[K - 1][i]);
  }
  cout << ans << endl;
}