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;

typedef long long ll;
typedef pair<ll, ll> pll;

#define F first
#define S second
#define pb push_back
#define all(x) (x).begin(), (x).end()

const ll inf = 1e18;
const int N = 100100;

ll n, m, k, a[N], dp[N][305], ans;

int main() {
	cin >> n >> m >> k;
	for (ll i = 1; i <= n; i++) {
		cin >> a[i];
		dp[i][1] = a[i];
	}
	for (ll i = 2; i <= k; i++) {
		deque<ll> d;
		ll j;
		for (j = max(1LL, i - m); j < i; j++) {
			while (!d.empty() && dp[j][i - 1] >= dp[d.back()][i - 1])
				d.pop_back();
			d.push_back(j);
		}
		for ( ; j <= n; j++) {
			ll mx = dp[d.front()][i - 1];
			dp[j][i] = max(dp[j][i], mx + i * a[j]);
			while (!d.empty() && d.front() <= j - m)
				d.pop_front();
			while (!d.empty() && dp[j][i - 1] >= dp[d.back()][i - 1])
				d.pop_back();
			d.push_back(j);
		}
	}
	for (ll i = 1; i <= n; i++)
		ans = max(ans, dp[i][k]);
	cout << ans;
}
