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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))

typedef struct indexVal{
  int64 val;
  int index;
} node;

typedef struct deque{
  node *a;
  int front;
  int last;
  int len;
} deque;

deque* newDeque(int len){
  deque *d=(deque *)calloc(1,sizeof(deque));
  d->a=(node *)calloc(len,sizeof(node));
  d->front=d->last=0;
  d->len=len;
  return d;
}

void push_back(deque *d,node v){
  d->a[d->last++]=v;
  d->last%=d->len;
}

node pop_front(deque *d){
  node t=d->a[d->front++];
  d->front%=d->len;
  return t;
}

node pop_back(deque *d){
  d->last=(d->last+d->len-1)%d->len;
  return d->a[d->last];
}

node get_front(deque *d){
  return d->a[d->front];
}

node get_back(deque *d){
  return d->a[(d->last+d->len-1)%d->len];
}

int isEmpty(deque *d){
  return d->front==d->last;
}

void clear(deque *d){
  d->front=d->last=0;
}

#define POS(i,j) ((i)*(k+1)+(j))

void run(void){
  int n,m,k;
  scanf("%d%d%d",&n,&m,&k);
  int *a=(int *)calloc(n+1,sizeof(int));
  int i;
  for(i=1;i<=n;i++) scanf("%d",a+i);
  int64 *dp=(int64 *)calloc(n+1,sizeof(int64));
  deque *deq=newDeque(m+1);
  for(i=1;i<=k;i++){
    clear(deq);
    push_back(deq,(node){dp[i-1],i-1});
    for(int j=i;j<=n;j++){
      node t=get_front(deq);
      while(t.index<j-m){
	pop_front(deq);
	t=get_front(deq);
      }
      int64 local=(int64)i*a[j]+t.val;
      while(!isEmpty(deq)){
	t=get_back(deq);
	if(t.val>dp[j]) break;
	pop_back(deq);
      }
      push_back(deq,(node){dp[j],j});
      dp[j]=local;
    }
  }
  int64 ans=0;
  for(i=m;i<=n;i++) ans=MAX(ans,dp[i]);
  printf("%lld\n",ans);
}

int main(void){
  run();
  return 0;
}