Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

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

int main(void)
{
  int N,K,C,i,j;
  int L[200000];
  int R[200000];
  char S[200000];
  scanf("%d %d %d",&N,&K,&C);
  scanf("%s",&S);

  j = 0;

  for(i=0;j<K;j++)
  {
    while(S[i] == 'x')
    {
      i++;
    }
    L[j] = i;
    i += C+1;
  }

  j = K-1;

  for(i=N-1;j>=0;j--)
  {
    while(S[i] == 'x')
    {
      i--;
    }
    R[j] = i;
    i -= C+1;
  }

  for(i=0;i<K;i++)
  {
    if(L[i] == R[i])
    {
      printf("%d\n",L[i]+1);
    }
  }

  return 0;
}