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 <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <utility>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <istream>
#include <ostream>

#include <cstdlib>
#include <cmath>
#include <cstdio>

using namespace std;

#define fi first
#define se second
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define rep(i,n) for(ll i=0; i < (n); ++i)
#define rrep(i,n) for(ll i=((n)-1); i >= 0; --i)

#define OPLT(T) bool operator<(const T & lop_, const T & rop_)
#define OPEQ(T) bool operator==(const T & lop_, const T & rop_)

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

//istream& operator>>(istream& istr, __float128& obj) { double d; istr >> d; obj = d; return istr; };
//ostream& operator<<(ostream& ostr, __float128& obj) { ostr << static_cast<double>(obj); return ostr; };

ll dp[2][100100];

int main() {
	ll N, M, K;
	ll res = 0;
	cin >> N >> M >> K;
	vector<ll> A(N);
	rep(i,N) cin >> A[i];
	rep(i,K) {
		deque<ll> mx;
		rep(j,N) {
			while(mx.size() && mx.front() < j-M)
				mx.pop_front();

			ll m;
			if(mx.size() == 0)
				m = 0;
			else
				m = dp[(i+1)%2][mx.front()];
			//cout << i << "," << j << ":" << m << endl;
			dp[i%2][j] = m + A[j] * (i+1);
			if(j<i) dp[i%2][j] = 0;
			while(mx.size() && dp[(i+1)%2][mx.back()] <= dp[(i+1)%2][j])
				mx.pop_back();
			mx.push_back(j);
		}
		/*
		cout << i << ":";
		rep(j,N) cout << dp[i%2][j] << ",";
		cout << endl;
		// */
	}
	for(int j = 0; j < N; j++) {
		res = max(res, dp[(K-1)%2][j]);
	}
	cout << res << endl;
	return 0;
}