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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <typeinfo>
#include <numeric>
#include <functional>
#include <unordered_map>
#include <bitset>
#include <stack>
#include <assert.h>
#include <unordered_set>



using namespace std;
using ll = long long;
using ull = unsigned long long;

const ll INF = 1e16;
const ll MOD = 1e9 + 7;

#define REP(i, n) for(ll i = 0; i < n; i++)




















int main(){
    ll n, m, k;
    cin >> n >> m >> k;
    vector<ll> a(n + 1);
    REP(i, n){
        cin >> a[i + 1];
    }
    
    vector<vector<ll>> dp(k + 1, vector<ll>(n + 1, -INF));
    for(ll j = 0; j <= n; j++){
        dp[0][j] = 0;
    }
    for(ll i = 1; i <= k; i++){
        deque<ll> que = {0};
        ll l = 1, r = 1;
        for(ll j = 1; j <= n; j++){
            dp[i][j] = dp[i - 1][que.front()] + i * a[j];
            while (!que.empty() && dp[i - 1][que.back()] <= dp[i - 1][r]) que.pop_back();
            que.push_back(r++);
            if(j > m){
                if (que.front() == l++) que.pop_front();
            }
        }
    }
    
    cout << *max_element(dp[k].begin() + k, dp[k].end()) << endl;
}