Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<deque>
#include<algorithm>

using namespace std;

typedef pair<long long, int> P;

const long long inf = 1e18;

int M, N;
int K;
int A[100100];

long long dp[100100][330];

long long solve(){
	for(int i = 0; i < N; ++i){
		dp[i][1] = A[i];
	}
	for(int j = 2; j <= K; ++j){
		deque<P> deq;
		if(dp[0][j - 1] != -inf) deq.push_front(P(dp[0][j - 1], 0));
		dp[0][j] = -inf;
		for(int i = 1; i < N; ++i){
			if(!deq.empty()){
			if(deq.front().first == -inf){
				dp[i][j] = -inf;
			}else{
				dp[i][j] = deq.front().first + (long long)A[i] * j;
			}
			while(!deq.empty()){
				if(deq.back().first < dp[i][j - 1]){
					deq.pop_back();
				}else break;
			}
			deq.push_back(P(dp[i][j - 1], i));
			if(deq.front().second == i - M){
				deq.pop_front();
			}
			}else{
				dp[i][j] = -inf;
				if(dp[i][j - 1] != -inf){
					deq.push_front(P(dp[i][j - 1], i));
				}
			}
		}
	}
	long long ans = -1;
	for(int i = 0; i < N; ++i){
	/*	for(int j = 0; j <= K; ++j){
			printf("%lld ", dp[i][j]);
		}
		printf("\n");*/
		ans = max(ans, dp[i][K]);
	}
	return ans;
}

void input(){
	scanf("%d%d%d", &N, &M, &K);
	for(int i = 0; i < N; ++i){
		scanf("%d", A + i);
	}
}

int main(){
	input();
	long long ans = solve();
	printf("%lld\n", ans);
	return 0;
}