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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

// practice with kaiboy
import java.io.*;
import java.util.*;

public class Main extends PrintWriter {
	Main() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		Main o = new Main(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int c = sc.nextInt();
		byte[] aa = sc.next().getBytes();
		int[] pp = new int[k];
		int[] qq = new int[k];
		for (int h = 0, i = 0, o = i - c - 1; i < n; i++)
			if (aa[i] == 'o' && i - o > c) {
				pp[h] = o = i;
				if (++h == k)
					break;
			}
		for (int h = 0, i = n - 1, o = i + c + 1; i >= 0; i--)
			if (aa[i] == 'o' && o - i > c) {
				qq[k - 1 - h] = o = i;
				if (++h == k)
					break;
			}
		for (int h = 0; h < k; h++)
			if (pp[h] == qq[h])
				println(pp[h] + 1);
	}
}