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 <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <queue>
#include <fstream>
#include <cstring>

using namespace std;
typedef long long LL;

#define MAXN 100010
int n, m, k;
int a[MAXN];
LL dp[2][MAXN];
deque<pair<LL, int> > q;

int main() {
	//freopen("struck_out.in", "r", stdin);
	cin>>n>>m>>k;
	for (int i=0;i<n;i++) scanf("%d", &a[i]);
	memset(dp, 0, sizeof(dp));
	for (int i=0;i<n;i++) {
		dp[(k-1)&1][i] = 1LL*k*a[i];
		//cout<<k-1<<" "<<i<<" = "<<dp[k-1][i]<<endl;
	}
	for (int i=k-2;i>=0;i--) {
		q.clear();
		for (int j=n-2;j>=0;j--) {
			LL val=dp[(i+1)&1][j+1];
			while (q.size()>0 && q.front().first<=val) q.pop_front();
			q.push_front(make_pair(val, j+1));
			while (q.size()>0 && q.back().second-j-1>=m) q.pop_back();
			dp[i&1][j]=q.back().first+1LL*(i+1)*a[j];
			//cout<<i<<" "<<j<<" = "<<dp[i][j]<<endl;
		}
	}
	LL ret=0;
	for (int i=0;i<n;i++) ret=max(ret, dp[0][i]);
	cout<<ret<<endl;
}