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 <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
using vi = vector<i64>;
using vvi = vector<vi>;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n, m, k;
    cin >> n >> m >> k;
    vi as(n);
    for (int i = 0; i < n; i++) {
        cin >> as[i];
    }

    vi dp(as);
    using ii = pair<i64, int>;
    for (int i = 0; i < k - 1; i++) {
        deque<ii> deq;
        vi dp2(n);
        for (int j = 0; j < n; j++) {
            // for (int d = 0; d < deq.size(); d++) {
            //     cout << "DEBUG " << deq[d].first << " " << deq[d].second << endl;
            // }

            if (j >= i + 1) {
                dp2[j] = (i + 2) * as[j] + deq.front().first;
            }

            if (deq.size() && deq.front().second <= j - m) {
                deq.pop_front();
            }
            while (deq.size()) {
                if (deq.back().first <= dp[j]) deq.pop_back();
                else break;
            }
            deq.push_back(ii(dp[j], j));
        }
        dp = vi(dp2);
    }
    i64 nax = -1;
    for (int i = 0; i < n; i++) {
        nax = max(nax, dp[i]);
    }
    cout << nax << endl;
}