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: Wrong Answer)

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		// n：全日数　k：働く日数　c：働いた後働かない日数
		int n, k, c;
		String line = scanner.nextLine();
		n = Integer.parseInt(line.split(" ")[0]);
		k = Integer.parseInt(line.split(" ")[1]);
		c = Integer.parseInt(line.split(" ")[2]);
		
		char[] ch = scanner.nextLine().toCharArray();
		int i = 0;
		int num = 0;
		List<Integer> llist = new ArrayList<>();
		// 前から貪欲的に働く日を求める llist
		while(num < k) {
			if(ch[i] == 'o') {
				llist.add(i);
				i += c + 1;
				num++;
			}
			else {
				i++;
			}
		}
		
		i = n - 1;
		num = 0;
		List<Integer> rlist = new ArrayList<>();
		// 後ろから貪欲的に働く日を求める rlist
		while(num < k) {
			if(ch[i] == 'o') {
				rlist.add(i);
				i -= c + 1;
				num++;
			}
			else
				i--;
		}
		Collections.reverse(rlist);
		for(int j = 0; j < k; j++) {
			if(llist.get(j) == rlist.get(j))
				System.out.println(llist.get(j) + 1);
		}
		
		scanner.close();
	}

}