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 <bitset>
#include <cmath>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <vector>

using namespace std;

#define rep(i, n) for (int64_t i = 0; i < (int64_t)(n); i++)
#define irep(i, n) for (int64_t i = 0; i <= (int64_t)(n); i++)
#define rrep(i, n) for (int64_t i = (n)-1; i >= 0; i--)
#define rirep(i, n) for (int64_t i = n; i >= 0; i--)

#define chmax(a, b) (a) = max(a, b)
#define chmin(a, b) (a) = min(a, b)

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M, K;
  cin >> N >> M >> K;

  vector<int> a(N);
  rep(i, N) { cin >> a[i]; }

  vector<vector<int64_t>> dp(
      K + 1, vector<int64_t>(N + 1, numeric_limits<int64_t>::min()));
  fill(dp[0].begin(), dp[0].end(), 0);
  rep(i, K) {
    deque<int64_t> window;
    rep(j, N) {
      while (not window.empty() and dp[i][window.back()] <= dp[i][j]) {
        window.pop_back();
      }
      window.push_back(j);

      dp[i + 1][j + 1] = dp[i][window.front()] + (int64_t)(i + 1) * a[j];

      if (window.front() <= j - (M - 1)) {
        window.pop_front();
      }
    }
  }

  int64_t result = 0;
  irep(i, N - K) { chmax(result, dp[K][K + i]); }
  cout << result << endl;

  return 0;
}