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;

#define int long long

template< typename T >
vector< T > slide_max(const vector< T > &v, int k)
{
  deque< int > deq;
  vector< T > ret;
  for(int i = 0; i < v.size(); i++) {
    while(!deq.empty() && v[deq.back()] <= v[i]) {
      deq.pop_back();
    }
    deq.push_back(i);
    if(i - k + 1 >= 0) {
      ret.emplace_back(v[deq.front()]);
      if(deq.front() == i - k + 1) deq.pop_front();
    }
  }
  return ret;
}

int dp[310][100010] = {};


signed main(){
    int n, m, k;

    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> n >> m >> k;

    int a[100010];
    for(int i = 0;i < n;i++){
        cin >> a[i];
    }

    for(int i = 0;i < n;i++){
        dp[k][i] = a[i] * k;
    }

    for(int i = k-1;i >= 1;i--){

        vector<int> vec;

        for(int j = 0;j < n;j++){
            vec.push_back(dp[i+1][j]);
        }

        for(int j = 0;j < m;j++){
            vec.push_back(LLONG_MIN);
        }

        auto p = slide_max(vec, m);


        for(int j = 0;j < n;j++){
            dp[i][j] = a[j] * i + p[j+1];
        }
    }

    int ans = LLONG_MIN;
    for(int i = 0;i < n;i++){
        ans = max(ans, dp[1][i]);
    }

    cout << ans << endl;
    return 0;
}