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
#include <bits/stdc++.h>

using namespace std;

typedef long long int64;
const int64 INF = 1LL << 58;


template< class T >
std::vector< int > slideMaximum(const std::vector< T > &a, int k)
{
  int n = a.size();
  std::deque< int > deq;
  std::vector< int > res;
  for(int i = 0; i < n; i++) {
    while(deq.size() && a[deq.back()] <= a[i]) deq.pop_back();
    deq.push_back(i);
    res.push_back(deq.front());
    if(i - k + 1 >= 0) {
      if(deq.front() == i - k + 1) deq.pop_front();
    }
  }
  return res;
}


int main()
{
  int64 N, M, K, A[100000];

  cin >> N >> M >> K;
  for(int i = 0; i < N; i++) cin >> A[i];
  // dp[k][idx] := dp[k-1][j(idx-j<M)]+A_idx*k

  vector< int64 > dp1(N, 0);
  for(int i = 0; i < N; i++) dp1[i] = A[i];

  for(int i = 2; i <= K; i++) {
    vector< int64 > dp2(N, 0);
    auto dp = slideMaximum(dp1, M);
    for(int j = i - 1; j < N; j++) {
      dp2[j] = max(dp2[j], dp1[dp[j - 1]] + i * A[j]);
    }
    dp1.swap(dp2);
  }
  cout << *max_element(begin(dp1), end(dp1)) << endl;
}