Problem:
Score : 700 points

Problem Statement

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}

Partial Scores

In the test set worth 100 points, M = N.

In the test set worth another 200 points, N ≦ 300 and K ≦ 30.

In the test set worth another 300 points, K ≦ 30.

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.

Sample Input 1

5 2 3
10 2 8 10 2

Sample Output 1

56

The total score is maximized when panels 1,3 and 4 are hit, in this order.

Sample Input 2

5 5 2
5 2 10 5 9

Sample Output 2

28

This case satisfies the additional constraint M = N for a partial score.

Sample Input 3

10 3 5
3 7 2 6 9 4 8 5 1 1000000000

Sample Output 3

5000000078

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

ll dp[2][100010];

int M, N, K;
ll A[100010];

int main() {
	cin >> N >> M >> K;
	rep(i, N) cin >> A[i];

	int c = 0, f = 1;

	memset(dp, -1, sizeof(dp));

	rep(i, N) dp[0][i+1] = A[i];

	rep(t, K-1) {
		for (int j = 0; j <= N; ++j) {
			dp[f][j] = -1;
		}

		deque<int> deq;
		
		for (int i = 0; i < N; i++) {
			if (dp[c][i] != -1) {
				while (deq.size() && dp[c][deq.back()] <= dp[c][i]) deq.pop_back();
				deq.push_back(i);
			}

			if (deq.size()) {
				dp[f][i+1] = dp[c][deq.front()] + (ll)(t + 2) * A[i];
				while (deq.front() <= i - M) deq.pop_front();
			}
		}

		swap(c, f);
	}

	printf("%lld\n", *max_element(dp[c], dp[c] + N + 1));

	return 0;
}