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: Time Limit Exceeded)

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();
        String S = sc.next();
        boolean[] sche = new boolean[N];
        for (int i = 0; i < N; i++) {
            if(S.toCharArray()[i] == 'o'){
                sche[i] = true;
            }
        }

        int[] left = new int[N];
        int[] right= new int[N];

        int count = 0;

        int lcount =0;
        for (int i = 0; i < N; i++) {
            if(sche[i]){
                left[i] = ++lcount;
                i+=C;
            }
            if(lcount==K) break;
        }
        int rcount =0;
        for (int i = N-1; i >= 0; i--) {
            if(sche[i]){
                right[i] = K-rcount;
                ++rcount;
                i-=C;
            }
            if(rcount==K) break;
        }


        for (int i = 0; i < N ; i++) {
            if(left[i]==right[i] && right[i]!=0){
                System.out.println(i+1);
            }
            
        }

    }

}