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.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }

    static void solve(InputStream is, PrintStream os) {
        // Your code here!
        Scanner scan = new Scanner(is);
        int N = scan.nextInt();
        int K = scan.nextInt();
        int C = scan.nextInt();
        String S = scan.next();

        int l[] = new int[N];
        int r[] = new int[N];

        int day = 0;
        int lastL[] = new int[N];
        int lastPos = -1;
        int notAvailable = 0;
        for(int i = 0; i < N; i++) {
            if(S.charAt(i) == 'o' && notAvailable == 0) {
                notAvailable = C;
                l[i] = day++;
                lastL[i] = lastPos;
                lastPos = i;
            } else {
                if(notAvailable != 0)
                    notAvailable--;
                l[i] = day;
                lastL[i] = lastPos;
            }
        }

        day = 0;
        int lastR[] = new int[N];
        lastPos = -1;
        notAvailable = 0;
        for(int i = N-1; i >=0 ; i--) {
            if(S.charAt(i) == 'o' && notAvailable == 0) {
                notAvailable = C;
                r[i] = day++;
                lastR[i] = lastPos;
                lastPos = i;
            } else {
                if(notAvailable != 0)
                    notAvailable--;
                r[i] = day;
                lastR[i] = lastPos;
            }
        }

        for(int i = 0; i < N; i++) {
            if(S.charAt(i) == 'o'){
                int lrc = l[i] + r[i];
                if(lastL[i] != -1 && lastR[i] != -1 && lastR[i]-lastL[i] <= C) lrc--;
                if(lrc >= K) continue;
                os.println(i+1);
            }
        }
    }
}