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.Scanner;

public 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 s = scan.next();
		
		int[] left = new int[n];
		int count = 0;
		for(int i = 0; i < n; i++){
			if(s.charAt(i) != 'x') {
				left[i] = i+1;
				i += c;
				count++;
				if(count == k) break;
			}
		}
		
		int[] right = new int[n];
		count = 0;
		for(int i = n-1; i >= 0; i--) {
			if(s.charAt(i) != 'x') {
				right[i] = i+1;
				i -= c;
				count++;
				if(count == k) break;
			}
		}
		
		for(int i = 0; i < n; i++) {
			if(left[i] == right[i] && left[i] == i+1)
				System.out.println(i+1);
		}
	}
}