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;
using lint = long long int;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T> void assign(V<T>& v, int n, const T& a = T()) { v.assign(n, a); }
template<class T, class... U> void assign(V<T>& v, int n, const U&... u) { v.resize(n); for (auto&& i : v) assign(i, u...); }

int main() {
  cin.tie(NULL); ios::sync_with_stdio(false);
  int n, m, k; cin >> n >> m >> k;
  V<lint> a(n); for (int i = 0; i < n; i++) cin >> a[i];
  VV<lint> dp; assign(dp, 2, n + 1);
  for (int i = 1; i < k + 1; i++) {
    deque<int> dq;
    for (int j = max(i - m, 0); j < i; j++) {
      while(!dq.empty() and dp[i - 1 & 1][dq.back()] <= dp[i - 1 & 1][j]) dq.pop_back();
      dq.push_back(j);
    }
    for (int j = i; j < n + 1; j++) {
      dp[i & 1][j] = i * a[j - 1] + dp[i - 1 & 1][dq.front()];
      if (dq.front() == j - m) dq.pop_front();
      while (!dq.empty() and dp[i - 1 & 1][dq.back()] <= dp[i - 1 & 1][j]) dq.pop_back();
      dq.push_back(j);
    }
  }
  cout << *max_element(dp[k & 1].begin(), dp[k & 1].end()) << '\n';
}