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 <bits/stdc++.h>
using namespace std;
using Int = long long;
struct SlideMax {
    deque<Int> A;
    Int value() {
        return A.empty() ? -1e18 : A[0];
    }
    void push(Int x) {
        while (!A.empty() && A.back() < x) A.pop_back();
        A.push_back(x);
    }
    void pop(Int x) {
        if (x == A.front()) A.pop_front();
    }
};
int main() {
    int N, M, K; cin >> N >> M >> K;
    vector<Int> A(N); for (auto &a : A) cin >> a;
    vector<Int> B(A);
    for (int i = 1; i < K; i++) {
        SlideMax sm;
        for (int j = 0; j < M; j++) sm.push(B[N-1-j]);
        for (int j = N-1; j >= 0; j--) {
            sm.pop(B[j]);
            if (j-M >= 0) sm.push(B[j-M]);
            B[j] = sm.value() + (i+1) * A[j];
        }
    }
    cout << *max_element(begin(B), end(B)) << '\n';
    return 0;
}