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;

constexpr ll inf = 1e18;

int main() {
    int n, m, k; cin >> n >> m >> k;
    vector<ll> a(n);
    for(int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    vector<vector<ll>> dp(n + 1, vector<ll>(k + 1, -inf));
    dp[0][0] = 0;
    vector<deque<pair<ll, int>>> deq(k + 1);
    deq[0].emplace_back(0, n + 1);
    for(int i = 0; i < n; ++i) {
        for(int j = k - 1; j >= 0; --j) {
            if(deq[j].empty()) continue;
            while(m < i - deq[j].front().second) deq[j].pop_front();
            dp[i + 1][j + 1] = max(dp[i + 1][j + 1], deq[j].front().first + (j + 1) * a[i]);
            while(!deq[j + 1].empty() && deq[j + 1].back().first <= dp[i + 1][j + 1]) {
                deq[j + 1].pop_back();
            }
            if(j + 1 != k) {
                deq[j + 1].emplace_back(dp[i + 1][j + 1], i);
            }
        }
        dp[i + 1][k] = max(dp[i + 1][k], dp[i][k]);
    }

    cout << dp[n][k] << endl;
}