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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll MOD = 998244353;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-11;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;
typedef pair<ld, ld> LDP;
ll a[100000];
ll dp[100000];
ll maque[100000];
ll nex[100000];
int main(){
	int n,m, k; cin >> n >> m >> k;
	rep(i, n) {
		cin >> a[i];
	}
	rep(i, n) {
		dp[i] = a[i];
	}
	Rep1(j,2, k) {
		int s = 0;
		int t = 0;
		rep(i, n) {
			if (dp[i] == -1)continue;
			if (t > s) {
				nex[i] = j * a[i] + dp[maque[s]];
			}
			else {
				nex[i] = -1;
			}
			while (t - 1 >= s && dp[i] > dp[maque[t - 1]]) {
				t--;
			}
			maque[t] = i; t++;
			if (i >= m) {
				if (maque[s] == i - m) {
					s++;
				}
			}
		}
		rep(i, n) {
			dp[i] = nex[i];
		}
	}
	ll out = 0;
	rep(i, n) {
		out = max(out, dp[i]);
	}
	cout << out << endl;
	return 0;
}