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 <string>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <cstring>
#include <numeric>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define F0(i,n) for (int i = 0; i < n; i++)
#define F1(i,n) for (int i = 1; i <= n; i++)
#define CL(a,x) memset(x, a, sizeof(x));
#define SZ(x) ((int)x.size())
const double eps = 1e-10;
const int inf = 1000000009;
int i, j, k, m, n, l;
int insd;
const int N = 100001;
ll a[N], dp[2][N];

ll s1[N], smax[N], c1;
ll s2[N], s2max, c2;

void reset() {
	c1 = c2 = s2max = 0;
}

void push(ll x) {
	s2[c2++] = x;
	if (x > s2max) s2max = x;

	while (c1 + c2 > m) {
		if (c1) c1--;
		else {
			for (int i = c2 - 1; i >= 0; i--) {
				smax[c1] = s1[c1] = s2[i];
				if (c1 > 0 && smax[c1 - 1] > smax[c1]) smax[c1] = smax[c1 - 1];
				c1++;
			}
			c2 = 0;
			s2max = 0;
		}
	}
}

ll get_max() {
	if (c1 == 0) return s2max;
	return max(s2max, smax[c1 - 1]);
}

int main() {
	//freopen("x.in", "r", stdin);
	cin >> n >> m >> k;
	F1(i, n) cin >> a[i];

	F1(i, n) dp[0][i] = a[i];

	for (int kk = 2; kk <= k; kk++) {
		reset();
		int i1 = kk % 2;
		int i2 = 1 - i1;

		insd = kk - 1;
		for (int i = kk; i <= n; i++) {
			while (insd < i) {
				push(dp[i1][insd]);
				insd++;
			}
			dp[i2][i] = kk * a[i] + get_max();
		}

		if (kk == k) {
			ll ans = 0;
			for (int i = k; i <= n; i++) {
				ans = max(ans, dp[i2][i]);
			}
			cout << ans << endl;
		}
	}

	return 0;
}