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(x) push_back(x)
#define all(x) x.begin(), x.end()
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int inf = 1e9;
const ll INF = 1e18;
const ld EPS = 1e-10;

int main(){
    int N, M, K;
    cin >> N >> M >> K;
    ll A[N];
    rep(i, N) cin >> A[i];
    //dp[i][j]:=A[0]〜A[i-1]まででj個まで取るときの最大値(A[i-1]は必ず取る)
    ll dp[N+1][K+1];
    fill(dp[0], dp[N+1], 0);
    //deq[i]の要素x,yについてx<yならばdp[x][j]>dp[y][j]
    deque<ll> deq[K+1];
    rep2(i, 1, N){
        rep(j, K+1){
            while(!deq[j].empty() && deq[j].front() < i-M) deq[j].pop_front();
        }
        dp[i][0] = 0;
        rep2(j, 1, K){
            //j個目をA[i-1]に取る
            if(j > i) dp[i][j] = dp[i][j-1];
            elif(deq[j-1].empty()) dp[i][j] = j*A[i-1];
            else dp[i][j] = dp[deq[j-1].front()][j-1] + j*A[i-1];
        }
        rep(j, K+1){
            while(!deq[j].empty() && dp[deq[j].back()][j] <= dp[i][j]) deq[j].pop_back();
            deq[j].push_back(i);
        }
    }
    ll ans = 0;
    rep(i, N+1){
        //rep(j, K+1) cout << dp[i][j] << ' ';
        //cout << endl; 
        ans = max(ans, dp[i][K]);
    }
    cout << ans << endl;
}