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: Time Limit Exceeded)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef long long ll;

int n, m, k;
ll a[111111];
ll dp[333][33];
ll dp2[2][333];

// dp[111111][333];
ll dp3[111111][333];
priority_queue<ll> q[333], dq[333];

int main(void) {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", a+i);
  }
  if (n == m && false) {
    ll res = 0;
    memset(dp2, -1, sizeof(dp2));
    dp2[1][0] = 0;
    for (int i = 0; i < n; i++) {
      dp2[i%2][0] = 0;
      for (int j = 1; j <= k; j++) {
        if (dp2[i%2][j-1] >= 0) {
          dp2[(i+1)%2][j] = max({
                dp2[(i+1)%2][j],
                    dp2[i%2][j-1] + a[i]*j,
                    dp2[i%2][j]
                  });
        }
      }
      for (int j = 0; j <= k; j++) {
        res = max(res, dp2[(i+1)%2][j]);
        if (dp2[i%2][j] >= 0) {
          dp2[i%2][j] = 0;
        }
        // printf("%5lld", dp2[(i+1)%2][j]);
      }
      // puts("");
    }
    printf("%lld\n", res);
  } else if (n <= 300 && k <= 30 && false) {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i <= n; i++) {
      dp[i][0] = 0;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < k; j++) {
        for (int l = 1; l <= m && i-l >= 0; l++) {
          if (dp[i-l][j] >= 0) {
            dp[i][j+1] = max(dp[i][j+1],
                             dp[i-l][j]+a[i-1]*(j+1));
          }
        }
      }
    }
    ll res = 0;
    for (int i = 0; i <= n+1; i++) {
      for (int j = 0; j <= k; j++) {
        // printf("%5lld", dp[i][j]);
        res = max(res, dp[i][j]);
      }
      // puts("");
    }
    printf("%lld\n", res);
  } else {
    q[0].push(0);
    memset(dp3, -1, sizeof(dp3));
    for (int i = 1; i <= n; i++) {
      for (int j = k-1; j >= 0; j--) {
        if (q[j].size() == 0) continue;
        while (!dq[j].empty() && q[j].top() == dq[j].top()) {
          q[j].pop();
          dq[j].pop();
        }
        dp3[i][j+1] = q[j].top()+a[i-1]*(j+1);
        q[j+1].push(q[j].top()+a[i-1]*(j+1));
        /*
        dp[i][j+1];
        for (int l = 1; l <= m && i-l >= 0; l++) {
          if (dp[i-l][j] >= 0) {
            dp[i][j+1] = max(dp[i][j+1],
                             dp[i-l][j]+a[i-1]*(j+1));
          }
        }
        */
      }
      if (i-m >= 0) {
        for (int j = 0; j < k; j++) {
          dq[j].push(dp3[i-m][j]);
        }
      }
    }
    printf("%lld\n", q[k].top());
  }
  return 0;
}