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;

#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))
#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))
#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))

typedef pair<int, int> P;
#define INF 1e9
#define INF_LL 1LL<<60
#define ll long long


/* dp[i+1][j] = (i+2)*aj + max{dp[i, k] | 1 <= j - k <= M} */
ll dp[305][100005];    /* dp[i][j]: i(0-indexed)番目にj(0-indexed)を選んだ時の得点の最大値 */

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<ll> a(n);
    REP(i, n) {
        cin >> a[i];
        dp[0][i] = a[i];
    }
    REP(i, k-1) {
        deque<ll> deq;
        /* スライド最大値でmaxを更新 */
        REPN(j, i, n) {
            int top_idx = max(0, j + 1 - m);
            /* 範囲外を除去 */
            if(!deq.empty() && deq.front() < top_idx) {
                deq.pop_front();
            }
            /* 後ろから見ていって対称より小さいものをpop */
            while(!deq.empty() && dp[i][deq.back()] < dp[i][j]) {
                deq.pop_back();
            }
            deq.push_back(j);
            dp[i+1][j+1] = dp[i][deq.front()] + a[j+1] * (i+2);
        }
    }

    ll ans = 0;
    REP(i, n) {
        ans = max(ans, dp[k-1][i]);
    }

    cout << ans << endl;


    return 0;
}