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 <deque>
#include <iostream>
using namespace std;
using ll = long long;

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

    ll A[N + 1];
    for (ll i = 1; i <= N; ++i) {
        cin >> A[i];
    }

    ll dp[N + 1];
    fill(dp, dp + N + 1, 0);
    // dp[n] = A_nをk個目として選んだときの最大スコア
    // kについて逐次更新

    for (ll k = 1; k <= K; ++k) {
        ll ndp[N + 1];

        deque<ll> que;
        // スライド最大値
        // [i - M + 1, i]の最大値を求める

        for (ll i = 0; i < N; ++i) {
            // 古い上に値の小さいindexを破棄
            while (!que.empty() && dp[que.back()] <= dp[i]) {
                que.pop_back();
            }

            que.push_back(i);

            // 先頭が範囲内か確認
            while (que.front() <= i - M) {
                que.pop_front();
            }

            ndp[i + 1] = dp[que.front()] + A[i + 1] * k;
        }

        // k未満を更新するとWA
        // A_1がやたら大きいときにA_1 * Kを答えにしてしまう
        for (ll i = k; i <= N; ++i) {
            dp[i] = ndp[i];
        }
    }

    ll ans = 0;
    for (ll i = 1; i <= N; ++i) {
        ans = max(ans, dp[i]);
    }

    cout << ans << endl;
    return 0;
}