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 ll = long long;

const ll INF = 2e18;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m, K;
    cin >> n >> m >> K;
    vector<ll> a(n);
    vector<ll> dp(n, -INF);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        dp[i] = a[i];
    }

    for (int i = 2; i <= K; i++) {
        vector<ll> ndp(n, -INF);
        deque<int> q;
        for (int j = 0; j < n; j++) {
            while (!q.empty()) {
                if (j - q.front() > m) q.pop_front();
                else break;
            }
            if (!q.empty()) {
                ndp[j] = dp[q.front()] + i * a[j];
            }

            while (!q.empty()) {
                if (dp[q.back()] <= dp[j]) q.pop_back();
                else break;
            }
            q.push_back(j);
        }
        dp = move(ndp);
    }
    
    ll ans = 0;
    for (int i = 0; i < n; i++) {
        ans = max(ans, dp[i]);
    }
    cout << ans << endl;
    return 0;
}