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>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
//#define ll __int128
//#define int ll
//#dLefine char ll
//#define double ld
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<vvc> vvvc;
typedef pair<int, int> pii;
typedef pair<pii, pii> piii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";


template<class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template<class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
    return in >> a.first >> a.second;
}

template<class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template<class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " << get<3>(a) << "}";
}

template<class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void smain();


signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef MAX_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
#ifdef HK_CUP
    freopen("finput.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    cout << setprecision(12) << fixed;
    smain();
#ifdef MAX_HOME
#ifndef HK_CUP
    cout << "\n\nTOTAL EXECUTION TIME: " << float(clock() - start) / CLOCKS_PER_SEC << endl;
#endif
#endif
    return 0;
}

const int N = 1e5 + 10;
const ll oo = 1e15;
int n, m, k;

ll dp[N];
int a[N];
const int LOG = 17;
ll st[N][LOG];
int pw2[N];

inline ll getmin(int l, int r) {
    return max(st[r][pw2[r - l + 1]], st[l + (1 << pw2[r - l + 1]) - 1][pw2[r - l + 1]]);
}

void smain() {
    pw2[1] = 0;
    int cur = 0;
    for (int i = 2; i < N; ++i) {
        while ((1 << cur) <= i)
            cur++;
        cur--;
        pw2[i] = cur;
    }
    cin >> n >> m >> k;
    fori (i, n) {
        cin >> a[i];
        dp[i] = a[i];
    }
    for (int i = 2; i <= k; ++i) {
        for (int j = 0; j < n; ++j) {
            ll cur = (ll) i * a[j];
            if (j + 1 >= i) {
                cur += getmin(max(0, j - m), j - 1);
            } else {
                cur = -oo;
            }
            st[j][0] = dp[j];
            for (int h = 1; h < LOG; ++h) {
                st[j][h] = max(st[j][h - 1], st[max(0, j - (1 << (h - 1)))][h - 1]);
            }
            dp[j] = cur;
        }
    }
    cout << *max_element(dp, dp + n);
}