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
#include <deque>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int main() {
    int n, m, k; cin >> n >> m >> k;
    long long a[n]; for (int i = 0; i < n; i++) cin >> a[i];
    long long dp[2][n];
    int cur = 0, prev = 1;
    for (int i = 0; i < n; i++) dp[prev][i] = a[i];
    for (int kk = 1; kk < k; kk++) {
        deque<pair<int, long long> > deq;
        for (int i = kk; i < n; i++) {
            while (!deq.empty() && deq.front().first < i-m) deq.pop_front();
            while (!deq.empty() && deq.back().second <= dp[prev][i-1]) deq.pop_back();
            deq.push_back({i-1, dp[prev][i-1]});
            dp[cur][i] = deq.front().second + a[i]*(kk+1);
        }
        swap(prev, cur);
    }
    cout << *max_element(dp[prev]+k-1, dp[prev+1]) << endl;
}
