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 <iostream>
#include <queue>
#include <cstring>
using namespace std;

using int64 = long long;

int main() {
    int N, M, K;
    cin >> N >> M >> K;

    const int MAXN = 200005;

    static int64 A[MAXN];
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }

    static int64 dp[2][MAXN];
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 0; i < N; i++) dp[1][i] = A[i];

    int64 ans = 0;
    // スライド最大値
    for (int i = 1; i < K; i++) {
        deque<int> dq;
        for (int j = 0; j < N; j++) {
            if (dq.empty() or dp[i & 1][dq.front()] < 0) dp[(i + 1) & 1][j] = -1;
            else {
                dp[(i + 1) & 1][j] = dp[i & 1][dq.front()] + int64(i + 1) * A[j];
            }
            while (!dq.empty() and dp[i & 1][dq.back()] <= dp[i & 1][j]) dq.pop_back();
            dq.push_back(j);

            if (dq.front() == j - M) dq.pop_front();

            if (i == K - 1) ans = max(ans, dp[K & 1][j]);
        }
    }
    cout << ans << endl;

    return 0;
}