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 <deque>
#define llint long long
#define inf 1000000000000000000

using namespace std;

llint M, N, K;
llint A[100005];
llint dp[100005][305];
deque<llint> deq[305];

int main(void)
{
	cin >> N >> M >> K;
	for(int i = 1; i <= N; i++) cin >> A[i];
	
	for(int i = 0; i <= K; i++){
		dp[0][i] = -inf;
		deq[i].push_back(0);
	}
	dp[0][0] = 0;
	
	for(int i = 1; i <= N; i++){
		for(int j = 1; j <= K; j++){
			dp[i][j] = dp[deq[j-1].back()][j-1] + j * A[i];
		}
		for(int j = 1; j <= K; j++){
			if(deq[j].back() == i-M) deq[j].pop_back();
			while(deq[j].size() && dp[deq[j].front()][j] <= dp[i][j]) deq[j].pop_front();
			deq[j].push_front(i);
		}
	}
	
	llint ans = 0;
	for(int i = 0; i <= N; i++) ans = max(ans, dp[i][K]);
	
	cout << ans << endl;
	return 0;
}