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 be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n, m, k;
    cin >> n >> m >> k;
    ll a[n];
    for(int i=0;i<n;i++){
    	cin >> a[i];
    }
    deque<pair<ll,ll> > dq[k+1];
    for(ll i=0;i<n;i++){
    	for(ll j=k-1;j>=1;j--){
    		while(dq[j-1].size() && dq[j-1].front().second < i - m)dq[j-1].pop_front();
    		if(dq[j-1].empty()) continue;
    		ll v = a[i]*(j+1) + dq[j-1].front().first;
    		while(dq[j].size() && dq[j].back().first < v) dq[j].pop_back();
    		dq[j].push_back({v, i});
    	}
    	while(dq[0].size() && a[i] > dq[0].back().first) dq[0].pop_back();
    	dq[0].push_back({a[i],i});
    }
    cout << dq[k-1].front().first << endl;
    return 0;
}
