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 <algorithm>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;

class SlideRangeQuery {
 public:
  using Compare = function<bool(int, int)>;

 private:
  Compare comp;
  deque<int> deq;
  int x = 0;

 public:
  SlideRangeQuery(const Compare& comp_) : comp(comp_) {}

  // [a, b)
  int query(int a, int b) {
    while (!deq.empty() && deq.front() < a) {
      deq.pop_front();
    }
    for (x = max(a, x); x < b; x++) {
      while (!deq.empty() && not comp(deq.back(), x)) {
        deq.pop_back();
      }
      deq.push_back(x);
    }
    return deq.empty() ? -1 : deq.front();
  }
};

int main() {
  int M, N, K;
  while (cin >> N >> M >> K) {
    vector<ll> A(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    vector<ll> dp = A;
    for (int i = 2; i <= K; i++) {
      vector<ll> ndp(N);
      SlideRangeQuery sq([&](int a, int b) { return dp[a] > dp[b]; });
      for (int j = i - 1; j < N; j++) {
        ndp[j] = i * A[j] + dp[sq.query(j - M, j)];
      }
      dp = ndp;
    }
    ll res = *max_element(dp.begin(), dp.end());
    cout << res << endl;
  }
  return 0;
}