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: Runtime Error)

#include <stdio.h>

int main(void){

  long n,k,c;
  scanf("%ld %ld %ld", &n, &k, &c);

	char s[n];
	scanf("%s", s);

  long first[k];
  long last[k];
  long work = 0;
  long day = 0;

  while (1) {
    if (s[day] == 'o') {
      first[work] = day;
      work++;
      day += c+1;
    } else {
      day++;
    }
    if (work == k) {
      break;
    }
  }

  /*for (long i = 0; i < k; i++) {
    printf("%ld ", first[i]);
  }
  printf("\n");*/

  work = 0;
  day = n-1;
  while (1) {
    if (s[day] == 'o') {
      last[k-work-1] = day;
      work++;
      day -= c+1;
    } else {
      day--;
    }
    if (work == k) {
      break;
    }
  }

  /*for (long i = 0; i < k; i++) {
    printf("%ld ", last[i]);
  }
  printf("\n");*/

  for (long i = 0; i < k; i++) {
    if (first[i] == last[i]) {
      printf("%ld\n", first[i]+1);
    }
  }

  return 0;
}