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>
#define show(x) std::cerr << #x << " = " << (x) << std::endl
using ll = long long;
template <typename T>
constexpr T INF = std::numeric_limits<T>::max() / 10;
template <typename T>
std::vector<T> slideMaximum(const std::vector<T>& a, const int k)  // 連続するk個の要素の最小値
{
    const int n = a.size();
    assert(n >= k);
    std::deque<T> deq;
    std::vector<T> b(n);
    for (int i = 0; i < n + k - 1; i++) {
        if (i < n) {
            while (not deq.empty() and a[deq.back()] < a[i]) { deq.pop_back(); }
            deq.push_back(i);
        }
        if (i >= k - 1) {
            b[i - k + 1] = a[deq.front()];
            if (deq.front() == i - k + 1) { deq.pop_front(); }
        }
    }
    return b;
}
int main()
{
    int N, M, K;
    std::cin >> N >> M >> K;
    std::vector<ll> A(N);
    for (int i = 0; i < N; i++) { std::cin >> A[i]; }
    std::vector<ll> sl(N + M, 0);
    for (ll i = K; i >= 1; i--) {
        std::vector<ll> tmp(N + M, -INF<ll>);
        for (int j = 0; j < N + i - K; j++) { tmp[j] = sl[j + 1] + A[j] * i; }
        sl = slideMaximum(tmp, M);
    }
    std::cout << slideMaximum(sl, sl.size())[0] << std::endl;
    return 0;
}
