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 all(x) (x).begin(), (x).end()

typedef long long ll;

const ll INF = 1e18;

const int N = 1e5 + 10, LOG = 18;
ll a[N], dp[N], ndp[N];
int n, m, K;

void solve() {
  cin >> n >> m >> K;
  for (int i = 1; i <= n; i++) {
    cin >> a[i]; dp[i] = a[i];
  }
  for (int k = 2; k <= K; k++) {
    ndp[1] = -INF;
    deque<ll> q; q.push_back(dp[1]);
    for (int i = 2; i <= n; i++) {
      int l = max(1, i - m), r = i - 1;
      ll upd = q.front();
      ndp[i] = upd + k * 1LL * a[i];
      while (!q.empty() && q.back() < dp[i])
        q.pop_back();
      q.push_back(dp[i]);
      if (1 <= i - m && !q.empty() && q.front() == dp[i - m])
        q.pop_front();
    }
    for (int i = 1; i <= n; i++) {
      dp[i] = ndp[i];
    }
  }
  cout << *max_element(dp + 1, dp + 1 + n);
}

// CHECK LIMITS (n <= 10^5)
// CHECK CORNER CASES (n == 1)

int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  //cout << 1.0 * clock() / CLOCKS_PER_SEC << endl;
  solve();
}
