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<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];
    lol(i,k+1)lol(j,n+1)dp[i][j]=-1;
    lol(i,n)dp[1][i+1]=a[i];
    for(ll i=2;i<=k;i++){
	while(!Q.empty())Q.pop_front();
	for(ll j=0;j<=n;j++){
	    if(!Q.empty()&&Q.front().first!=-1){
		dp[i][j]=a[j-1]*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;
}
