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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
class Main{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int c = scan.nextInt();
    String plan = scan.next();
    boolean[] can_work_days = new boolean[n];
    for(int i = 0;i < n;i++){
      if(plan.charAt(i) == 'o'){
        can_work_days[i] = true;
      }else{
        can_work_days[i] = false;
      }
    }
    
    int[] earliest_days = new int[k];
    int[] latest_days = new int[k];
    
    for(int i = 0, work_day = 0;work_day < k;i++){
      if(can_work_days[i]){
        earliest_days[work_day] = i;
        i += c;
        work_day++;
      }
    }
    for(int i = n - 1, work_day = 0;work_day < k;i--){
      if(can_work_days[i]){
        latest_days[work_day] = i;
        i -= c;
        work_day++;
      }
    }
    for(int i = 0;i < k;i++){
      if(earliest_days[i] == latest_days[k - i - 1])System.out.println(earliest_days[i] + 1);
    }
  }
}