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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"

using namespace std;

const long long int MOD = 1000000007;
//const int MOD = 998244353;

//long long int N, M, K, H, W, L, R;
int N, M, K, H, W, L, R;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N >> M >> K;
	vector<long long int>v(N + 1);
	for (int i = 1; i <= N; i++)cin >> v[i];
	vector<long long int>dp(N + 1);
	for (int i = 1; i <= K; i++) {
		deque<int>Q;
		for (int j = N - 1; j >= N - M; j--) {
			if (Q.empty())Q.push_back(j);
			else {
				while (!Q.empty() && dp[Q.back()] <= dp[j])Q.pop_back();
				Q.push_back(j);
			}
		}
		for (int j = N; j>=i; j--) {
			if (Q.front() == j)Q.pop_front();
			dp[j] = dp[Q.front()] + v[j] * i;
			if (j - M - 1 >= 0) {
				while (!Q.empty() && dp[Q.back()] <= dp[j - M - 1])Q.pop_back();
				Q.push_back(j - M - 1);
			}
		}
		//for (int j = 1; j <= N; j++)cout << i << " " << j << " " << dp[j] << endl;
	}
	cout << *max_element(dp.begin(), dp.end()) << endl;
	return 0;
}