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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
#define N 100010
ll dp[310][N],a[N],n,m,k;
deque<pair<ll,ll> >Q;
int main(){
    cin>>n>>m>>k;
    lol(i,n)cin>>a[i+1];
    lol(i,k+1)lol(j,n+1)dp[i][j]=-1;
    dp[0][0]=0;
    for(ll i=1;i<=k;i++){
	while(!Q.empty())Q.pop_front();
	for(ll j=0;j<=n;j++){
	    if(j>=i)dp[i][j]=a[j]*i+Q.front().first;
	    while(!Q.empty()){
		if(Q.back().first<=dp[i-1][j]){
		    Q.pop_back();
		}
		else break;
	    }
	    Q.push_back(make_pair(dp[i-1][j],j));
	    while(!Q.empty()){
		if(Q.front().second<j-m)Q.pop_front();
		else break;
	    }
	}
    }
    ll ans=0;
    lol(i,n+1)ans=max(ans,dp[k][i]);
    cout<<ans<<endl;
    return 0;
}