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;

typedef long long LL;

#define MAX_N 100005
#define MAX_M 100005
#define MAX_K 305

int N, M, K;
LL A[MAX_N];

void input()
{
  cin >> N >> M >> K;
  for (size_t i = 0; i < N; i++) {
    scanf("%lld", &A[i]);
  }
}

LL dp[MAX_N][MAX_K];
deque<pair<LL, int>> numList;

void init()
{
  numList.clear();
}

void pushList(LL val, int num)
{
  pair<LL, int> a = make_pair(val, num);
  while(numList.size()){
    if(numList[numList.size()-1] < a){
      numList.pop_back();
    }
    else{
      break;
    }
  }
  numList.push_back(a);
}

LL getList(int minNum){
  while(numList.size()){
    if(numList[0].second < minNum){
      numList.pop_front();
    }
    else{
      return numList[0].first;
    }
  }
  return -1;
}

LL solve()
{
  memset(dp, -1, sizeof(dp));
  for (size_t i = 0; i < N; i++) {
    dp[i][1] = A[i];
  }

  for (int k = 1; k < K; k++) {
    init();
    for (int n = 0; n < N; n++) {
      if(getList(max(0, n-M)) >= 0){
        LL tmp = getList(max(0, n-M)) + (LL)(k+1)*A[n];
        dp[n][k+1] = max(dp[n][k+1], tmp);
      }
      pushList(dp[n][k], n);
    }
  }

  LL ret = 0;
  for (size_t i = 0; i < N; i++) {
    ret = max(ret, dp[i][K]);
  }

  return ret;
}

signed main()
{
  input();

  cout << solve() << endl;

  return 0;
}