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.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		int C = sc.nextInt();
		char[] s = sc.next().toCharArray();
		int[] L = new int[N], R = new int[N];
		int wait = C, sum = 0;
		for(int i = 0; i < N; i++) {
			if(wait >= C) {
				if(s[i] == 'o') {
					wait = -1;
					L[sum] = i;
					sum++;
				}
			}
			wait++;
			if(sum == K)
				break;
		}
		boolean l = false;
		wait = C; sum = K;
		for(int i = N-1; i >= 0; i--) {
			if(s[i] == 'o')
				l = true;
			if(!l)
				continue;
			if(wait >= C) {
				if(s[i] == 'o') {
					wait = -1;
					sum--;
					R[sum] = i;
				}
			}
			wait++;
			if(sum == 0)
				break;
		}
		for(int i = 0; i < K; i++)
			if(L[i] == R[i])
				System.out.println(L[i]+1);
	}
}