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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const double pi = acos(-1.0);
const double EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

int main(){
    int N, M, K;
    cin >> N >> M >> K;
    ll A[N];
    rep(i, N) cin >> A[i];
    ll dp[K+1][N+1];
    fill(dp[0], dp[K+1], 0);
    rep(i, K){
        deque<int> que;
        rep(j, N){
            if(!que.empty() && que.front() <= j-M) que.pop_front();
            while(!que.empty() && dp[i][que.back()] <= dp[i][j]) que.pop_back();
            que.push_back(j);
            if(j >= i) dp[i+1][j+1] = dp[i][que.front()]+A[j]*(i+1);
        }
    }
    ll ans = 0;
    rep(i, N+1) chmax(ans, dp[K][i]);
    cout << ans << endl;
}