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
/**
 *    author:  tourist
 *    created: 26.11.2019 09:39:34       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> dp(n);
  for (int i = 0; i < n; i++) {
    dp[i] = a[i];
  }
  vector<long long> new_dp(n);
  vector<int> st(n);
  for (int it = 1; it < k; it++) {
    new_dp[0] = (long long) -1e18;
    int ptr = 0;
    int end = 0;
    for (int i = 0; i < n - 1; i++) {
      while (ptr < end && st[ptr] <= i - m) {
        ++ptr;
      }
      while (ptr < end && dp[st[end - 1]] <= dp[i]) {
        --end;
      }
      st[end++] = i;
      new_dp[i + 1] = dp[st[ptr]] + (it + 1LL) * a[i + 1];
    }
    swap(dp, new_dp);
  }
  cout << *max_element(dp.begin(), dp.end()) << '\n';
  return 0;
}
