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 int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define f first
#define s second
#define Time (double)clock()/CLOCKS_PER_SEC
#define debug(x) std::cout << #x << ": " << x << '\n';

const int N = 1e5+7, K = 307, INF = 1e18;

int a[N];
int dp[K][N];

struct Queue_Min {
//pair (i, x)
deque <ii> a;
void add(ii e) {
    while (a.size() && a.back().s >= e.s)   
        a.pop_back();
    a.push_back(e);
}
void del(ii e) {
    if (a.size() && a.front() == e)
        a.pop_front();
}   
int get() {
    if (a.empty())
        return INF;
    else
        return a.front().s;
}   
};  

signed main() {
    #ifdef HOME
    freopen("input.txt", "r", stdin);
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cout.setf(ios::fixed); cout.precision(20); 
    #endif

    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        dp[1][i] = a[i];
    }   

    for (int i = 2; i <= k; ++i) {
        Queue_Min q;
        for (int j = 1; j <= n; ++j) {
            if (j - m - 1 > 0) {
                q.del(mp(j - m - 1, -dp[i - 1][j - m - 1]));
            }   
            if (j > 1) {
                q.add(mp(j - 1, -dp[i - 1][j - 1]));
            }   
            dp[i][j] = i * a[j] - q.get();
        }   
    }   

    int ans = 0;
    for (int i = 1; i <= n; ++i)
        ans = max(ans, dp[k][i]);
    cout << ans << endl;
}