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
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

VL slide_max(VL a, int k){
    VL ret;
    deque<int> dq;
    REP(i,a.size()){
        while (!dq.empty() && a[dq.back()] <= a[i]) dq.pop_back();
        dq.push_back(i);
        ret.push_back(a[dq.front()]);
        if (i - k + 1 >= 0){
            if (dq.front() == i - k + 1) dq.pop_front();
        }
    }
    return ret;
}

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    VL a(n);
    REP(i,n) scanf("%lld", &a[i]);

    VL dp(n);
    REP(i,n) dp[i] = a[i];

    REP(x,k-1){
        VL ma = slide_max(dp, m);

        dp[0] = -1e18;
        REP(i,n-1) dp[i+1] = ma[i] + a[i+1] * (x + 2);

        // REP(i,n) cout << dp[i] << " ";
        // cout << endl;
    }

    ll ma = 0;
    REP(i,n) ma = max(ma, dp[i]);
    cout << ma << endl;
    
    return 0;
}
