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 <deque>
#include <vector>
#include <algorithm>
using namespace std;

class Slide_Max {
public:
	vector<long long>dat; long long size_, L = 0, R = -1;
	deque<pair<long long, long long>>Q;

	void init(int size__) {
		size_ = size__;
		dat.resize(size_, 0);
	}
	void update(long long p, long long x) {
		dat[p] = x;
	}
	long long query(long long l, long long r) {
		while (R < r) {
			R++;
			while (!Q.empty()) {
				pair<long long, long long>A = Q.back();
				if (A.first <= dat[R])Q.pop_back();
				else break;
			}
			Q.push_back(make_pair(dat[R], R));
		}
		while (L < l) {
			while (!Q.empty()) {
				pair<long long, long long>A = Q.front();
				if (A.second <= L)Q.pop_front();
				else break;
			}
			L++;
		}
		return Q.front().first;
	}
};

long long N, M, K, a[100009], dp[309][100009];

int main() {
	cin >> N >> M >> K;
	for (int i = 1; i <= N; i++)cin >> a[i];
	for (int i = 0; i < 309; i++) { for (int j = 0; j < 100009; j++)dp[i][j] = (-1LL << 62); }
	for (int i = 0; i < 100009; i++)dp[0][i] = 0;
	for (int i = 0; i < K; i++) {
		Slide_Max X; X.init(N + 3);
		for (int j = 0; j <= N; j++) X.update(j, dp[i][j]);
		for (int j = 1; j <= N; j++) {
			long long L = j - M, R = j - 1; if (L < 0)L = 0;
			dp[i + 1][j] = 1LL * (i + 1)*a[j] + X.query(L, R);
		}
	}
	long long maxn = 0;
	for (int i = 0; i <= N; i++)maxn = max(maxn, dp[K][i]);
	cout << maxn << endl;
	return 0;
}