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;
using ld = long double;
using D = double;
using uint = unsigned int;

#ifdef WIN32
    #define LLD "%I64d"
#else
    #define LLD "%lld"
#endif

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

const int maxn = 100005;
const int maxk = 305;

deque<pair<ll, int>> curmax[maxk];

ll del[2 * maxn][maxk];
int a[maxn];
int n, m, k;

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < k; i++) curmax[i].push_back({-1e18 * (i > 0), -1});
    ll answer = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = k - 1; j >= 0; j--)
        {
//             cout << curmax[j].size() << endl;
            ll curans;
            if (j > 0) {
                while (curmax[j].front().se < i - m) curmax[j].pop_front();
                curans = curmax[j].front().fi + (ll)(j + 1) * a[i];
            } else curans = a[i];
            if (j + 1 == k) answer = max(answer, curans);
            while (!curmax[j + 1].empty() && curmax[j + 1].back().fi <= curans) curmax[j + 1].pop_back();
            curmax[j + 1].push_back({curans, i});
        }
    }
    cout << answer << endl;
    return 0;
}