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
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)



typedef long long ll;
#define INF 1LL<<60
int N, M, K;
int A[101010];
//-----------------------------------------------------------------
ll dp[101010][300];
int main() {
	cin >> N >> M >> K;
	rep(i, 1, N + 1) scanf("%d", &A[i]);

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

	rep(j, 2, K + 1) {
		deque<int> deq;
		rep(i, 1, N + 1) {
			while (deq.size() && dp[deq.back()][j - 1] <= dp[i][j - 1]) deq.pop_back();
			deq.push_back(i);
			ll ma = dp[deq.front()][j - 1];
			if (0 < ma) dp[i + 1][j] = ma + (ll)j * A[i + 1];
			if (deq.front() == i - M + 1) deq.pop_front();
			
		}
	}

	ll ans = 0;
	rep(i, 1, N + 1) ans = max(ans, dp[i][K]);

	/*rep(j, 1, K + 1) rep(i, 1, N + 1) {
	printf("dp[%d][%d] = %lld\n", i, j, dp[i][j]);
	}*/

	cout << ans << endl;
}