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 <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#define _repargs(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _repargs(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define inf 2000000007
#define mp make_pair
#define pb push_back
typedef long long ll;
using namespace std;
template <typename T>
inline void output(T a, int p = 0) {
    if(p) cout << fixed << setprecision(p)  << a << "\n";
    else cout << a << "\n";
}
// end of template

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    // source code
    int N, M, K;
    cin >> N >> M >> K;
    vector<ll> A(N);
    vector<vector<ll>> dp(K, vector<ll>(N, 0));
    rep(i, N) {
        cin >> A[i];
    }
    dp[0] = A;
    rep(i, 1, K) {
        deque<int> dq;
        rep(j, i) {
            while(!dq.empty() && dp[i - 1][dq.back()] < dp[i - 1][j]) {
                dq.pop_back();
            }
            dq.push_back(j);
        }
        rep(j, i, N) {
            dp[i][j] = dp[i - 1][dq.front()] + A[j] * (i + 1);
            while(!dq.empty() && dp[i - 1][dq.back()] < dp[i - 1][j]) {
                dq.pop_back();
            }
            dq.push_back(j);
            while(!dq.empty() && j - dq.front() >= M) dq.pop_front();
            
        }
    }
    
    ll ans = 0;
    rep(i, N) {
        ans = max(ans, dp[K - 1][i]);
    }
    output(ans);
    return 0;
}
