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 <cstdio>
#include <vector>
#include <cmath>
#include <map>
#include <algorithm>
#include <string>
#include <utility>
#include <set>
#include <stack>
#include <deque>
#include <ctime>
#include <random>
#include <cassert>
#include <cmath>
#include <climits>
#include <queue>
#include <cstring>
#include <bitset>
#include <iomanip>
#include <chrono>


#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")



#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << x << endl;
#else 
    #define dbg(x)
#endif

#define pb push_back
#define ppb pop_back()
#define mp make_pair
#define fi(a,b) for (int i=a;i<b;i++)
#define fj(a,b) for (int j=a;j<b;j++)
#define fk(a,b) for (int k=a;k<b;k++)
#define fi1(a,b) for (int i=a-1;i>=b;i--)
#define fj1(a,b) for (int j=a-1;j>=b;j--)
#define fk1(a,b) for (int k=a-1;k>=b;k--)
#define fx(x,a) for (auto& x : a)
#define rep(i, a, b) for (int i = a; i < b; ++i)
#define rep1(i, a, b) for (int i = a - 1; i >= b; --i)
#define siz(x) (int)x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

using namespace std;

template<typename T1, typename T2>inline void mine(T1 &x, T2 y) { if (x > y) x = y; }
template<typename T1, typename T2>inline void maxe(T1 &x, T2 y) { if (x < y) x = y; }

ostream& operator << (ostream &a, const vector<int> &b) {
    for (auto k : b) cout << k << ' ';
    return a;
}


typedef long long ll;
typedef unsigned long long ull;
typedef char ch;
typedef string str;
typedef pair<int, int> pii;
typedef map<int, int> mii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;
typedef vector<ch> vch;
typedef vector<vch> vvch;
typedef vector<str> vs;



const int MOD = 1000000007;
const int INF = 1000000050;
const long long BIG = (long long)2e18 + 50;
const int MX = 100010;
const int D = 310;
const double EPS = 1e-9;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll dp[MX][D];
deque<ll> mx[D];


ll get(int i) {
    if (mx[i].empty())
        return 0;
    return mx[i].front();
}

void add(int i, ll x) {
    while (mx[i].size() && mx[i].back() < x)
        mx[i].ppb;
    mx[i].pb(x);
}

void del(int i, ll x) {
    if (mx[i].size() && mx[i].front() == x)
        mx[i].pop_front();
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
       
    int n, m, k;
    cin >> n >> m >> k;
    ll ans = 0;
    fi(0, n) {
        ll a;
        cin >> a;
        fj1(min(i + 1, k) + 1, 1) {
            if (i >= m)
                del(j, dp[i - m][j]);
            dp[i][j] = a * j + get(j - 1);
            add(j, dp[i][j]);
        }
        maxe(ans, dp[i][k]);
    }
    cout << ans << '\n';
}