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 ll = long long;

const int N = 1e5 + 1, K = 301;

ll dp[K][N];
int a[N];

int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n, m, k; cin >> n >> m >> k;
  for(int i = 0; i < n; i++) cin >> a[i + 1];
  for(int i = 1; i <= n + 1 - k; i++) dp[1][i] = 1 * a[i];

  // slide minimum
  // 貰うDP
  for(int i = 2; i <= k; i++) {
    deque<int> deq;
    for(int j = max(1, i - m); j <= i - 1; j++) {
      // cerr << " j1 = " << j << endl;
      while(deq.size() && dp[i-1][deq.back()] <= dp[i-1][j]) deq.pop_back();
      deq.emplace_back(j);
    }
    cerr << " i = " << i << endl;
    for(int j = i; j <= n + i - k; j++) {
      if(deq.front() < j - m) deq.pop_front();
      dp[i][j] = dp[i-1][deq.front()] + (ll) i * a[j];
      //
      // cerr << "dp i j  = " << dp[i][j] << endl;
      while(deq.size() && dp[i-1][deq.back()] <= dp[i-1][j]) deq.pop_back();
      deq.emplace_back(j);
    }
  }
  ll ans = 0;
  for(int i = k; i <= n; i++) ans = max(ans, dp[k][i]);
  cout << ans << endl;
}