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;
public class Main {
	public static void main(String[] args)throws Exception{
		Scanner stdIn=new Scanner(System.in);
		int N=stdIn.nextInt();
		int K=stdIn.nextInt();
		int C=stdIn.nextInt();
		String a=stdIn.next();
		boolean work[]=new boolean[N];
		int L[]=new int[K];
		int R[]=new int[K];
		int x=0;
		for(int i=0;i<N;i++){
			if(a.charAt(i)=='o')
				work[i]=true;
			else
				work[i]=false;
		}
		for(int i=0;i<N;i++){
			if(work[i]){
				L[x]=i;x++;
				i+=C;
			}
			if(x==K)
				break;
		}x--;
		for(int i=N-1;i>=0;i--){
			if(work[i]){
				R[x]=i;x--;
				i-=C;
			}
			if(x==-1)
				break;
		}
		for(int i=0;i<K;i++){
			if(L[i]==R[i])
				System.out.println(L[i]+1);
		}
	}
}