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 <cstdlib>
#include <algorithm>
using namespace std;

typedef long long ll;
const ll INF = (ll)1e18;
//const int N = 12;
const int N = 100100;
int n, m, k;
ll a[N];
ll b[N];
ll c[N];

struct Queue
{
	int sz1, sz2;
	ll a[N][2], b[N][2];

	Queue() : sz1(0), sz2(0), a(), b() {}

	void push(ll x)
	{
		a[sz1][0] = x;
		if (sz1 > 0)
			a[sz1][1] = max(a[sz1 - 1][1], x);
		else
			a[sz1][1] = x;
		sz1++;
	}
	void pop()
	{
		if (sz2 == 0)
		{
			if (sz1 == 0) throw;
			while(sz1 > 0)
			{
				sz1--;
				b[sz2][0] = a[sz1][0];
				if (sz2 > 0)
					b[sz2][1] = max(b[sz2 - 1][1], a[sz1][0]);
				else
					b[sz2][1] = b[sz2][0];
				sz2++;
			}
		}
		sz2--;
	}
	ll getMax()
	{
		if (sz1 == 0 && sz2 == 0) return -INF;
		if (sz2 == 0) return a[sz1 - 1][1];
		if (sz1 == 0) return b[sz2 - 1][1];
		return max(a[sz1 - 1][1], b[sz2 - 1][1]);
	}
};

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	for (int i = 0; i < n; i++)
		b[i] = a[i];
	for (int it = 1; it < k; it++)
	{
//		cerr << "it = " << it << endl;
		Queue q;
		for (int i = 0; i < m; i++)
		{
//			cerr << i << endl;
			c[i] = (it + 1) * a[i] + q.getMax();
			q.push(b[i]);
		}
		for (int i = m; i < n; i++)
		{
//			cerr << i << endl;
			c[i] = (it + 1) * a[i] + q.getMax();
			q.push(b[i]);
			q.pop();
		}
		for (int i = 0; i < n; i++)
			b[i] = c[i];
	}
	ll ans = 0;
	for (int i = 0; i < n; i++)
		ans = max(ans, b[i]);
	printf("%lld\n", ans);

	return 0;
}