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: Runtime Error)

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();
        char[] c = S.toCharArray();
        int[][] left = new int[N+1][2]; // left[0][:] : not worked, left[1][:] : worked
        for (int i=0;i<N;i++) {
            if (c[i]=='o' && i>=C) {
                left[i+1][0] = Math.max(left[i][0], left[i-C][1]);

                left[i+1][1] = Math.max(left[i][0]+1, left[i][1]);
                left[i+1][1] = Math.max(left[i+1][1], left[i-C][1]+1);
                left[i+1][1] = Math.max(left[i+1][1], left[i-C][0]+1);
            } else if (c[i]=='o' && i<C) {
                left[i+1][0] = left[i][0];
                left[i+1][1] = Math.max(left[i][0]+1, left[i][1]);
            } else { // c[i]=='x'
                left[i+1][0] = left[i][0];
                left[i+1][1] = left[i][1];
            }
        }

        int[][] right = new int[N+1][2]; // right[0][:] : not worked, right[1][:] : worked
        for (int i=N;i>=1;i--) {
            if (c[i-1]=='o' && i<=N-C) {
                right[i-1][0] = Math.max(right[i][0], right[i+C][1]);

                right[i-1][1] = Math.max(right[i][0]+1, right[i][1]);
                right[i-1][1] = Math.max(right[i-1][1], right[i+C][1]+1);
                right[i-1][1] = Math.max(right[i-1][1], right[i+C][0]+1);
            } else if (c[i-1]=='o' && i>N-C) {
                right[i-1][0] = right[i][0];
                right[i-1][1] = Math.max(right[i][0]+1, right[i][1]);
            } else { // c[i]=='x'
                right[i-1][0] = right[i][0];
                right[i-1][1] = right[i][1];
            }
        }

        int[][] new_left = new int[N][2];
        for (int i=0;i<N;i++) {
            new_left[i][0] = left[i+1][0];
            // new_left[i][1] = left[i+1][1];
            new_left[i][1] = Math.max(left[i+1][0], left[i+1][1]);
        }

        ////////////////////// ??
        for (int i=0;i<N+1;i++) {
            right[i][1] = Math.max(right[i][0], right[i][1]);
        }


        long[] imos = new long[N+1];
        for (int i=0;i<N-C-1;i++) {
            if (new_left[i][1]+right[i+C+1][1]<K) {
                imos[i+1]++;
                imos[i+C+1]--;
            }
        }

        for (int i=0;i<C;i++) {
            if (right[i+1][1]<K) {
                imos[0]++;
                imos[i+1]--;
            }
        }

        for (int i=N-1;i>=N-C;i--) {
            if (new_left[i-1][1]<K) {
                imos[i]++;
                imos[N]--;
            }
        }


        // for (int i=0;i<N;i++) {
        //     if (new_left[i][1]>=K) {
        //         imos[i]++;
        //         imos[N]--;
        //     }
        // }

        // for (int i=N-1;i>=0;i--) {
        //     if (right[i][1]>=K) {
        //         imos[0]++;
        //         imos[i+1]--;
        //     }
        // }

        for (int i=0;i<N;i++) {
            imos[i+1] = imos[i]+imos[i+1];
        }

        // System.out.println(Arrays.deepToString(new_left));
        // System.out.println(Arrays.deepToString(right));
        // System.out.println(Arrays.toString(imos));
        if (C==0) {
            System.out.println("");
        } else {
            for (int i=0;i<N;i++) {
                if (imos[i]==C) System.out.println(i+1);
            }
        }
    }
}