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;

template <typename T> class SlideMin {
public:
  vector<int> deq;
  vector<T> res;
  int n, k;
  SlideMin(vector<T> Array, int sz) {
    n = (int) Array.size(), k = sz;
    deq.resize(n), res.resize(n);
    int s = 0, t = 0;
    for (int i = 0; i < n; i++) {
      while (s < t && Array.at(deq.at(t - 1)) <= Array.at(i)) {
        t--;
      }
      deq.at(t++) = i;
      res.at(i) = Array.at(deq.at(s));
      if (deq.at(s) == i - k + 1) {
        s++;
      }
    }
  }
};

int main() {
  int N, M, K;
  cin >> N >> M >> K;
  vector<long> A(N);
  for (int i = 0; i < N; i++) cin >> A.at(i);
  SlideMin<long> SM(A, M);
  for (int i = 0; i + 1 < K; i++) {
    vector<long> tmp(N, -1e18);
    for (int j = 1; j < N; j++) {
      tmp.at(j) = max(tmp.at(j), SM.res.at(j - 1) + (i + 2) * A.at(j));
    }
    SM = SlideMin<long>(tmp, M);
  }
  cout << *max_element(SM.res.begin(), SM.res.end()) << "\n";
}