Problem:
Score : 700 points

Problem Statement

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}

Partial Scores

In the test set worth 100 points, M = N.

In the test set worth another 200 points, N ≦ 300 and K ≦ 30.

In the test set worth another 300 points, K ≦ 30.

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.

Sample Input 1

5 2 3
10 2 8 10 2

Sample Output 1

56

The total score is maximized when panels 1,3 and 4 are hit, in this order.

Sample Input 2

5 5 2
5 2 10 5 9

Sample Output 2

28

This case satisfies the additional constraint M = N for a partial score.

Sample Input 3

10 3 5
3 7 2 6 9 4 8 5 1 1000000000

Sample Output 3

5000000078

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;


ll dp[2][100010];

int main(int argc, char const* argv[])
{
  int n, m, k; cin >> n >> m >> k;
  vector<ll> a(n); cin >> a;

  rep(i, n)dp[1][i] = a[i];
  FOR(i, 1, k){
    deque<pair<ll, int>> dq;
    rep(j, n){
      if(sz(dq) == 0)dp[(i+1)&1][j] = 0;
      else{
        dp[(i+1)&1][j] = dq[0].first + ((ll)i + 1LL) * a[j];
      }
      if(dq.front().second + m == j)dq.pop_front();
      while(dq.size() && dq.back().first <= dp[(i)&1][j])dq.pop_back();
      dq.push_back(mk(dp[(i)&1][j], j));
    }
  }
  ll res = 0;
  rep(i, n)res = max(res, dp[k&1][i]);
  cout << res << endl;
  return 0;
}