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;

typedef long long ll;

template <typename T>
class SlideMaximum
{
    // (位置, 値)
    std::deque<std::pair<int, T>> deq;
    int pos, window;

  public:
    SlideMaximum(int _w) : pos(0), window(_w) {}
    T query(T val)
    {
        while (!deq.empty() && deq.front().first <= pos - window)
            deq.pop_front();
        while (!deq.empty() && deq.back().second <= val)
            deq.pop_back();
        deq.push_back(std::make_pair(pos++, val));
        return deq.front().second;
    }
};

ll N, M, K;
vector<ll> A(100010);

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N >> M >> K;
    vector<vector<ll>> dp(K+1,vector<ll>(N+1,0));
    for (int i = 1; i <= N; i++)
    {
        cin >> A[i];
        dp[1][i] = A[i];
    }
    for (ll i = 2; i <= K; i++)
    {
        SlideMaximum<ll> sm(M);
        ll ma = 0;
        for(int j=max(1LL, i-M);j<=i-2;j++)
        {
            ma = sm.query(dp[i-1][j]);
        }
        for (ll j = i; j <= N; j++)
        {
            ma = sm.query(dp[i-1][j-1]);
            ll v = i * A[j] + ma;
            dp[i][j] = v;
        }
    }
    ll ret = 0;
    for(int i=K;i<=N;i++) ret = max(ret, dp[K][i]);
    cout << ret << endl;
}