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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<lint, int> pi;
const int mod = 1e9 + 7;
const int MAXN = 100005;

int n, m, k;
lint dp[305][MAXN];
int a[MAXN];

int main(){
	scanf("%d %d %d",&n,&m,&k);
	for(int i=1; i<=n; i++){
		scanf("%d",&a[i]);
	}
	for(int i=1; i<=k; i++){
		deque<pi> dq;
		dp[i][0] = -1e18;
		for(int j=1; j<=n; j++){
			while(!dq.empty() && dq.front().second < j - m){
				dq.pop_front();
			}
			while(!dq.empty() && dq.back().first < dp[i-1][j-1]){
				dq.pop_back();
			}
			dq.push_back(pi(dp[i-1][j-1], j-1));
			dp[i][j] = dq.front().first + 1ll * a[j] * i;
		}
	}
	cout << *max_element(dp[k], dp[k] + n + 1) << endl;
}
