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 <bits/stdc++.h>

#define mp make_pair
#define eb emplace_back
#define ff first
#define ss second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#ifdef ONPC
mt19937 rnd(228);
#else
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#endif


const int MAXN = 1;
//#define int long long
const ll INF = 1e18;

/*
task: https://atcoder.jp/contests/cf16-tournament-round3-open/tasks/asaporo_d
*/

struct sw{
    vector<ll> l, r;
    ll rmx;
    
    sw(){
        rmx = 0;
    }
    
    void push_back(ll x){
        r.push_back(x);
        rmx = max(rmx, x);
    }
    
    void pop_front(){
        if(l.empty()){
            for(ll mn = 0; !r.empty();){
                mn = max(mn, r.back());
                l.push_back(mn);
                r.pop_back();
            }
            rmx = 0;
        }
        l.pop_back();
    }
    
    ll get_max(){
        ll res = rmx;
        if(!l.empty()) res = max(res, l.back());
        return res;
    }
};

void solve2(){
  int n, m, k;
  cin >> n >> m >> k;

  vector<ll> a(n);
  for (int i =0 ; i < n; i++) cin >> a[i];

  //vector<vector<ll>> val(n, vector<ll> (k, 0));
  //vector<set<pair<ll, int>>> dp(k, set<pair<ll, int>> ());
  vector<sw> dp(k, sw());


  ll ans = 0LL;
  for (int i = 0; i < n; i++){
    //val[i][0] = a[i];
    for (int j = min(i+1, k) - 1; j > -1; j--){
      ll x = 0LL;
      if (j > 0)
        x = dp[j-1].get_max();

        //cerr << i << ' ' << j << ' ' << x<< '\n';
      //val[i][j] = x + (ll)(j + 1) * a[i];
      x += (ll)(j + 1) * a[i];
      if (j == k-1) ans = max(ans, x);
      dp[j].push_back(x);
      if (i - m - j > -1)
        dp[j].pop_front();
    }
  }

  //ll ans = 0LL;
  //for (int i = 0; i < n; i++) ans = max(ans, val[i][k-1]);

  cout << ans << '\n';
}


signed main()
{
    #ifdef FILEIO
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    #endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);

    solve2();

    return 0;
}
