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)

#include <bits/stdc++.h>
using namespace std;
     
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef pair<LL, LL> PLL;
     
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())
     
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
     
#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
     
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;
     
     
LL dp[310][100010];
LL mx[310][100010];

// 幅Wの窓で値pを追加した時の最大値
void add(deque<PLL>& dq, PLL p, int W){
  while(!dq.empty() && dq.back().FF < p.FF)
	dq.pop_front();
  dq.push_back(p);
  while(!dq.empty() && p.SS - dq.front().SS + 1 > W)
	dq.pop_front();
}

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);
     
  LL N, M, K; cin >> N >> M >> K;
  vector<LL> xs(N);
  REP(i,N) cin >> xs[i];
  deque<PLL> dq;
  REP(i,N){
	dp[1][i] = xs[i];
	add(dq, MP(dp[1][i],i), K);
	mx[1][i] = dq.front().FF;
  }

  for(LL i=2;i<=K;++i){
	dq.clear();
	for(int j=1;j<N;++j){
	  LL nxt = mx[i-1][j-1] + i * xs[j];
	  dp[i][j] = max(dp[i][j], nxt);

	  add(dq, MP(dp[i][j],j), K);
	  mx[i][j] = dq.front().FF;
	}
  }
     
  LL ans = 0;
  REP(i,N) ans = max(ans, dp[K][i]);
  cout << ans << endl;
      
  return 0;
}