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: Time Limit Exceeded)

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

int main(void) {
  int n, k, c, i, tmp, cou=0, j;
  char ox;
  scanf(" %d %d %d", &n, &k, &c);
  int s[n];
  for(i=0; i < n; i++){
    scanf(" %c", &ox);
    if(ox == 'o') s[i]=0;
    else s[i]=1;
  }

  //for(i=0; i < n; i++)
   // printf("%d", s[i]);
  //printf("\n");

  for(i=0; i < n; i++){
    if(s[i] == 1) continue;
    tmp=0;

    s[i]=1;

    for(j=0; j < n; j++){
      if(s[j] == 1) continue;
      tmp=j;
      cou++;
      //printf("\n %d ", s[tmp]);
      break;
    }

    while(tmp < n){
      //printf("%d ", tmp); //////
      tmp+=c + 1;
      while(tmp < n && s[tmp] == 1)
        tmp++;
      if(tmp < n) cou++;
      if(cou >= k) break;
    }
    if(cou < k) printf("%d\n", i + 1);

    s[i]=0;
    cou=0;
  }
}