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;
#ifdef _DEBUG
#define _GLIBCXX_DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

#define int long long
#define ll long long
#define ll1 1ll
#define ONE 1ll
#define DBG 1
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define loop(n) rep(loop, (0), (n))
#define all(c) begin(c), end(c)
const int INF =
sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
using pii = pair<int, int>;
// template<class T> ostream &operator<<(ostream &os,T &t){dump(t);return os;}
template <typename T, typename S>
istream &operator>>(istream &is, pair<T, S> &p) {
	is >> p.first >> p.second;
	return is;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, pair<T, S> &p) {
	os << p.first << " " << p.second;
	return os;
}

template <typename T> void printvv(const vector<vector<T>> &v) {
	cerr << endl;
	rep(i, 0, v.size()) rep(j, 0, v[i].size()) {
		if (typeid(v[i][j]).name() == typeid(INF).name() and v[i][j] == INF) {
			cerr << "INF";
		}
		else
			cerr << v[i][j];
		cerr << (j == v[i].size() - 1 ? '\n' : ' ');
	}
	cerr << endl;
}
/*
typedef __int128_t Int;
std::ostream &operator<<(std::ostream &dest, __int128_t value) {
std::ostream::sentry s(dest);
  if (s) {
	__uint128_t tmp = value < 0 ? -value : value;
	char buffer[128];
	char *d = std::end(buffer);
	do {
	  --d;
	  *d = "0123456789"[tmp % 10];
	  tmp /= 10;
	} while (tmp != 0);
	if (value < 0) {
	  --d;
	  *d = '-';
	}
	int len = std::end(buffer) - d;
	if (dest.rdbuf()->sputn(d, len) != len) {
	  dest.setstate(std::ios_base::badbit);
	}
  }
  return dest;
}

__int128 parse(string &s) {
  __int128 ret = 0;
  for (int i = 0; i < s.length(); i++)
	if ('0' <= s[i] && s[i] <= '9')
	  ret = 10 * ret + s[i] - '0';
  return ret;
}
*/

#ifndef _DEBUG
#define printvv(...)
#endif
void YES(bool f) { cout << (f ? "YES" : "NO") << endl; }
void Yes(bool f) { cout << (f ? "Yes" : "No") << endl; }
template <class T> bool chmax(T &a, const T &b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
template <class T> bool chmin(T &a, const T &b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}

signed main(signed argc, char *argv[]) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(12);

	int N, M; cin >> N >> M;
	int K; cin >> K;
	vector<int>A(N); rep(i, 0, N) { cin >> A[i]; }
	vector<vector<int>>dp(K + 1, vector<int>(N + 2));
	rep(k, 1, K + 1) {
		rep(i, 0, N) {
			dp[k][i] = dp[k - 1][i + 1] + (K - k + 1) * A[i];
		}
		deque<int>deq;
		rep(i, 0, N + M - 1) {
			if (i < N) {
				while (deq.size() and dp[k][deq.back()] <= dp[k][i])deq.pop_back();
				deq.push_back(i);
			}
			if (i - M + 1 >= 0) {
				dp[k][i - M + 1] = dp[k][deq.front()];
				if (deq.front() == i - M + 1)deq.pop_front();
			}
		}
	}
	cout << *max_element(all(dp[K])) << endl;

	return 0;
}