**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$q=0$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$q=0$$$) — the number of members, the number of sections and the number of updates.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases do not exceed $$$2 \cdot 10^5$$$ respectively.

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 0
1 2 3 4
1 1
3 6 0
1 2 3
1 1 2 3 3 2
4 6 0
3 1 4 2
3 1 1 2 3 4
output:
YA
YA
TIDAK

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line.

For the second test case, the following is a possible way to move members so that the presentation is good:

1. $$$[1,2,3]$$$, do not move member $$$1$$$.
2. $$$[1,2,3]$$$, move member $$$1$$$ after member $$$3$$$.
3. $$$[2,3,1]$$$, move member $$$2$$$ after member $$$3$$$.
4. $$$[3,2,1]$$$, do not move member $$$3$$$.
5. $$$[3,2,1]$$$, move member $$$3$$$ after member $$$1$$$.
6. $$$[2,1,3]$$$, do not move member $$$2$$$.

Buggy Code:
/*
ziad-abdou1
    ┌──────────────────────────────────────────────────┐
    │                                                  │
    │      ________                   __               │
    │     /\_____  \  __             /\ \              │
    │     \/____//'/'/\_\     __     \_\ \             │
    │          //'/' \/\ \  /'__`\   /'_` \            │
    │         //'/'___\ \ \/\ \L\.\_/\ \L\ \           │
    │         /\_______\ \_\ \__/.\_\ \___,_\          │
    │         \/_______/\/_/\/__/\/_/\/__,_ /          │
    │                                                  │
    └──────────────────────────────────────────────────┘
*/

import java.io.*;
import java.util.*;

public class Main {
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter pw = new PrintWriter(System.out);

    static int[] dx = {1, 0, -1, 0};
    static int[] dy = {0, 1, 0, -1};
    static final int MOD = 1000000007;
    static final long INF = (long) 1e18;

    static final int N = 200200;

    static int n, x;

    public static void main(String[] args) throws IOException {

        int test = 1;
        test = sc.nextInt();
        while (test-- > 0) {

            int n = sc.nextInt(); // Number of members
            int m = sc.nextInt(); // Number of slides
            int q = sc.nextInt(); // Number of updates (always 0 in this problem)

            int[] a = new int[n];
            int[] b = new int[m];
            Map<Integer, Integer> mp = new HashMap<>();
            HashSet<Integer> hs = new HashSet<>();
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                mp.put(a[i], i);
            }

            for (int i = 0; i < m; i++) {
                b[i] = sc.nextInt();
            }

            boolean ok = true;
            int maxInd = -1;

            for (int i = 0; i < m; i++) {
                if ( hs.contains(b[i])){
                    continue;
                }
                int cur = mp.get(b[i]);

                if (cur < maxInd) {
                    ok = false;
                    break;
                }
                hs.add(b[i]);
                maxInd = Math.max(maxInd, cur);
            }

            if (ok) {
                System.out.println("YA");
            } else {
                System.out.println("TIDAK");
            }
        }
        pw.flush();

    }

    static ArrayList<Long> factors(int n) {  // O(sqrt(N)
        ArrayList<Long> fac = new ArrayList<>();
        for (long i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                fac.add(i);
                if (n / i != i) fac.add(n / i);
            }
        }
        return fac;
    }

    static long fastpow(long n, long ti) {
        if (ti == 0) return 1l;
        if (ti % 2 == 0) {
            long y = fastpow(n, ti / 2);
            long k = y * y;
            k %= MOD;
            return k;
        } else {
            long y = fastpow(n, ti / 2);
            long k = ((n * y) % MOD) * y;
            k %= MOD;
            return k;
        }
    }

    static long modInverse(long x) {
        return fastpow(x, MOD - 2);
    }

    static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static class Edge implements Comparable<Edge> {
        int node, cost;

        Edge(int a, int b) {
            node = a;
            cost = b;
        }

        public int compareTo(Edge e) {
            return cost - e.cost;
        }
    }

    static class Pair implements Comparable<Pair> {
        int a, b, c;

        Pair(int vv, int dis, int h) {
            this.a = vv;
            this.b = dis;
            c = h;
        }

        @Override
        public int compareTo(Pair o) {
            return a - o.a;
        }
    }

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec) f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongarr(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongarr(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntarr(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerarr(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the algorithm might fail. The key aspect to consider is the condition where the code incorrectly determines if the slide show is good or not.

### Analysis of the Buggy Code
The buggy code checks if the current member presenting the slide is already in a valid position (i.e., its index in the initial lineup is greater than or equal to the maximum index seen so far). If it is not, it sets `ok` to `false` and breaks out of the loop.

### Edge Case to Consider
1. **Repeating Members in `b`**: The code skips over members that are already in the `HashSet`. This can lead to incorrect results if the same member is required to present multiple slides, but their initial positions are not in the correct order.
2. **Non-Sequential Members**: The code might fail if the members in `b` are not in a strictly increasing order of their initial positions in `a`.

### Test Input
Let's create a test case where the same member is required to present multiple slides, but their initial positions are not in the correct order. This will expose the bug in the code.

```plaintext
1
4 5 0
3 1 4 2
3 1 3 2 4
```

### Explanation
- **`n = 4`**: There are 4 members.
- **`m = 5`**: There are 5 slides.
- **`q = 0`**: No updates.
- **`a = [3, 1, 4, 2]`**: Initial order of members.
- **`b = [3, 1, 3, 2, 4]`**: Members who should present each slide.

### Expected Output
- **`TIDAK`**: The slide show is not good because the member 3 is required to present slides 1 and 3, but the initial order of members does not allow this without violating the constraints.

### Why This Test Case is Challenging
- The member 3 is required to present slides 1 and 3, but in the initial order, member 3 is at position 0, which is not greater than the maximum index seen so far (which is 0 after the first slide).
- The code will incorrectly skip over the second occurrence of member 3 and might incorrectly determine that the slide show is good.

This test case should expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$q=0$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$q=0$$$) — the number of members, the number of sections and the number of updates.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

It is guaranteed that the sum of $$$n$$$ and the sum of $$$m$$$ over all test cases do not exceed $$$2 \cdot 10^5$$$ respectively.

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 0
1 2 3 4
1 1
3 6 0
1 2 3
1 1 2 3 3 2
4 6 0
3 1 4 2
3 1 1 2 3 4
output:
YA
YA
TIDAK

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line.

For the second test case, the following is a possible way to move members so that the presentation is good:

1. $$$[1,2,3]$$$, do not move member $$$1$$$.
2. $$$[1,2,3]$$$, move member $$$1$$$ after member $$$3$$$.
3. $$$[2,3,1]$$$, move member $$$2$$$ after member $$$3$$$.
4. $$$[3,2,1]$$$, do not move member $$$3$$$.
5. $$$[3,2,1]$$$, move member $$$3$$$ after member $$$1$$$.
6. $$$[2,1,3]$$$, do not move member $$$2$$$.

Buggy Code:
import java.io.*;
import java.util.*;
import java.lang.*;

public class AdjustThePresentationEasy {
    public static int MOD = (int) (1e9 + 7);

    public static void main(java.lang.String[] args) throws java.io.IOException {
        try {
            FastReader x = new FastReader();
            FastWriter out = new FastWriter();
            int testcases = x.nextInt();
            test:
            while (testcases-- > 0) {
                int n = x.nextInt(), m = x.nextInt(), q = x.nextInt();
                int[] a = new int[n];
                for(int i=0;i<n;i++)
                    a[i] = x.nextInt();
                HashSet<Integer> set = new HashSet<>();
                int[] b = new int[m];
                for(int i=0;i<m;i++)
                    b[i] = x.nextInt();
                int i = 0, j = 0;
                while(j<m && i<n) {
                    if(set.contains(b[j])) j++;
                    if(j<m && a[i] != b[j]) break;
                    set.add(a[i]);
                    while(j<m && a[i] == b[j]) j++;
                    i++;
                }
                while(j<m) {
                    if(set.contains(b[j])) j++;
                    else break;
                }
                if(j < m) out.println("TIDAK");
                else out.println("YA");
            }
            out.close();
        } catch (java.lang.Exception e) {
            e.printStackTrace();
        }
    }

    // gcd function
    public static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    // lcm function
    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    // function to check prime
    public static boolean isprime(long a) {
        if (a == 2) {
            return true;
        }
        for (int i = 2; i < Math.sqrt(a); i++) {
            if (a % i == 0) {
                return false;
            }
        }
        return true;
    }

    // power function
    public static long power(long a, long b) {
        if (b == 0) {
            return 1;
        }
        long res = power(a, b / 2);
        if (b % 2 == 0) {
            return res * res;
        } else {
            return res * res * a;
        }
    }

    public static String reverseString(String str) {
        return (new StringBuilder(str)).reverse().toString();
    }

    public static String convert(long number, int base) {
        return Long.toString(number, base);
    }

    // fast input
    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
            if (System.getProperty(
                    "ONLINE_JUDGE") == null) {
                try {
                    br = new BufferedReader(
                            new InputStreamReader(
                                    new FileInputStream(
                                            "input.txt")));
                } catch (Exception ignored) {
                }
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    // fast output
    public static class FastWriter {
        BufferedWriter bwr;

        public FastWriter() {
            bwr = new BufferedWriter(new OutputStreamWriter(System.out));

            if (System.getProperty(
                    "ONLINE_JUDGE") == null) {
                try {
                    bwr = new BufferedWriter(
                            new OutputStreamWriter(
                                    new FileOutputStream(
                                            "output.txt")));
                } catch (Exception ignored) {
                }
            }
        }

        public void print(Object object) throws IOException {
            bwr.append("").append(String.valueOf(object));
        }

        public void println(Object object) throws IOException {
            print(object);
            bwr.append("\n");
        }

        public void close() throws IOException {
            bwr.close();
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4 2 0
1 2 3 4
1 4
3 3 0
1 2 3
1 2 3
4 4 0
3 1 4 2
4 3 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

typedef long long           ll;
typedef unsigned long long  ull;
typedef double              db;
typedef long double         ldb;
typedef float               fl;

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fast_io ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pb                                  push_back
#define pf                                  push_front
#define pob                                 pop_back
#define pof                                 pop_front
#define eb                                  emplace_back
#define ef                                  emplace_front
#define endl                                '\n'
#define mkp                                 make_pair
#define mkt                                 make_tuple
#define fi                                  first
#define se                                  second
#define lb                                  lower_bound
#define ub                                  upper_bound
#define sz(x)                               (x).size()
#define fastio \
   ios_base::sync_with_stdio(false); \
   cin.tie(NULL); \
   cout.tie(NULL);

void countReachable(vector<int>& arr, vector<int>& blacks, vector<bitset<200000>>& reachable, string& s, vector<bool>& visited, int i) {

    int dest = arr[i] - 1;
    if ( dest == i ) {
        blacks[i] = reachable[i].count();
        return;
    }

    if ( blacks[dest] != -1 ) {
        reachable[i] |= reachable[dest];
        blacks[i] = reachable[i].count();
        return;
    }

    visited[i] = true;
    if ( ! visited[dest] )
        countReachable(arr, blacks, reachable, s, visited, dest);
    else
        reachable[dest] |= reachable[i];
    visited[i] = false;

    reachable[i] |= reachable[dest];

    blacks[i] = reachable[i].count();
}

void solve() {
    int n;
    string s;
    cin >> n;
    vector<int> arr(n);
    for (ll i = 0; i < n; i++) {
        cin >> arr[i];
    }
    cin >> s;

    vector<int> blacks(n, -1);
    vector<bitset<200000>> reachable(n);

    vector<bool> visited(n, false);

    for ( int i = 0; i < n; i++ ) {
        if ( s[i] == '0' )
            reachable[i].set(i);
    }
    for ( int i = 0; i < n; i++ ) {
        countReachable(arr, blacks, reachable, s, visited, i);
    }

    for (ll i = 0; i < n; i++) {
        cout << blacks[i] << ((i < n - 1) ? ' ' : '\n');
    }
}

int main() {
    fast_io;
    ull t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases that might cause the algorithm to fail. The code seems to handle the basic cases correctly, but there are a few potential issues:

1. **Cycles in the Permutation**: The code might not handle cycles correctly, especially if the cycle includes both black and white elements.
2. **Self-loops**: Elements that map to themselves.
3. **Long Chains**: Long chains of elements that might cause the recursion to go deep and potentially miss some reachable elements.

Let's create a small-scale test input that includes these edge cases:

1. A permutation with a cycle that includes both black and white elements.
2. A permutation with a self-loop.
3. A permutation with a long chain of elements.

Here is the test input:

```plaintext
3
6
1 2 3 4 5 6
100110
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
000000
```

### Explanation:
1. **First Test Case**:
   - Permutation: `[1, 2, 3, 4, 5, 6]`
   - Colors: `100110`
   - This is a simple permutation where each element maps to itself (self-loops). The function should return the number of black elements reachable from each position, which is 0 for all positions except the first, which is 1.

2. **Second Test Case**:
   - Permutation: `[3, 5, 6, 1, 2, 4]`
   - Colors: `010000`
   - This permutation has a cycle: `3 -> 6 -> 4 -> 1 -> 3`. The function should correctly count the black elements in this cycle.

3. **Third Test Case**:
   - Permutation: `[1, 2, 3, 4, 5, 6]`
   - Colors: `000000`
   - This is a simple permutation where each element maps to itself (self-loops). All elements are black, so the function should return 1 for each position.

This test input should help expose any issues in the code's handling of cycles, self-loops, and long chains.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
void solve() {
	int n;
	cin >> n;
	vector<int>a(n + 1);
	vector<int>b(n + 1);
	vector<int>f(n + 1);
	vector<int>c;
	vector<int>d;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	string s;
	cin >> s;//0是黑色，1白色
	for (int i = 0; i < n; i++) {
		b[i + 1] = s[i] - '0';
		if (b[i + 1] == 0)d.push_back(a[i + 1]);
	}
	for (int i = 1; i <= n; i++) {
		ll cnt = 0;
		ll x = i;
		if (f[x] != 0) {
			cnt += f[x];
			cout << cnt << " ";
			continue;
		}
		else {
			if (b[x] == 0)cnt++;
			c.push_back(x);
			x = a[x];
		}
		while (x!=i) {
			if (f[x]!=0) {
				cnt += f[x];
				break;
			}
			else {
				if (b[x] == 0)cnt++;
				c.push_back(x);
				x = a[x];
			}
		}
		cout << cnt << " ";
		f[i] = cnt;
		for (int i = 1; i < c.size(); i++) {
			if (binary_search(d.begin(), d.end(), c[i])) {
				f[c[i]] = cnt;
				cnt--;
				continue;
			}
			if (cnt <= 0)cnt = 0;
			f[c[i]] = cnt;
		}
		c.clear();
	}
	cout << endl;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's consider the following scenarios:

1. **Cycles and Loops**: The code should handle cycles and loops in the permutation correctly. If there are cycles, the code should not get stuck in an infinite loop.
2. **Reachability**: The code should correctly count the number of black integers that are reachable from each starting integer.
3. **Edge Cases**: Small and specific test cases that can easily expose bugs, such as a permutation with all elements being black or all elements being white.

Let's construct a test input that includes a permutation with a cycle and a mix of black and white elements:

```plaintext
1
5
1 2 3 4 5
01010
```

### Explanation:
- **Permutation**: `[1, 2, 3, 4, 5]`
- **Colors**: `01010` (Black, White, Black, White, Black)
- **Reachability**:
  - `F(1)`: 1 (1 is black and reachable only from itself)
  - `F(2)`: 1 (2 is white and reachable only from itself)
  - `F(3)`: 1 (3 is black and reachable only from itself)
  - `F(4)`: 1 (4 is white and reachable only from itself)
  - `F(5)`: 1 (5 is black and reachable only from itself)

This test input is designed to be small and straightforward, but it ensures that the code must correctly handle the reachability of each element in a simple permutation with a mix of black and white elements. This should help expose any issues in the code's logic for counting reachable black integers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define endl '\n'
using namespace std;
/* bitset<64>(n).to_string().substr(bitset<64>(n).to_string().find('1'));*/
/* Trust za process */

ll gcd(ll a, ll b) {
    if (b == 0)
        return a;
    return __gcd(b, a % b);
}

void Kim0z(){
    ll n;cin>>n;
    vector<ll> a(n) , ans(n , -1);
    string b;
    for(auto &i:a)cin>>i;
    cin>>b;
    ll maxc = 0;
    for(int i=0;i<n;i++){
        if(a[i] != (i+1) && a[a[i]-1] != (i+1)){
            if(b[i] == '0')maxc++ , b[i] = '1';
            if(b[a[i]-1] == '0')maxc++ , b[a[i]-1] = '1';
        }
        else if(a[i] == (i+1)){
            if(b[i] == '1')ans[i] = 0;
            else ans[i] = 1;
        }
        else{
            ll kk=0;
            if(b[i] == '0')kk++;
            if(b[a[i]-1] == '0')kk++;
            ans[i] = kk;
            ans[a[i]-1] = kk;
        }
    }
    for(int i=0;i<n;i++)if(ans[i] == -1)ans[i] = maxc;
    for(auto i:ans)cout<<i<<' ';
    cout<<'\n';
}

int main(){
    ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ll t=1;
    cin>>t;
    while(t--){
        Kim0z();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
1 3 4 2
0000
4
2 3 4 1
0000
4
4 3 2 1
0000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
import sys
import threading
threading.stack_size(10**6)
sys.setrecursionlimit(200005)
t=int(input())
def count(arr,idxnode,visited,str,first,head,snt):
    if idxnode not in visited:
        head[idxnode]=first
        visited.add(idxnode)
        if str[idxnode]=='0':
           return count(arr,arr[idxnode],visited,str,first,head,snt+1)
        else:
            return count(arr,arr[idxnode],visited,str,first,head,snt)
    return snt

while t!=0:
    n=int(input())
    fake=list(map(int, input().split()))
    fakestr=input()
    str=['s' for _ in range(n+1)]
    arr=[0 for _ in range(n+1)]
    for i in range(n):
        arr[i+1]=fake[i]
        str[i+1]=fakestr[i]
    cnt=[0 for _ in range(n+1)]
    head=[0 for _ in range(n+1)]
    visited=set()
    for i in range(1,n+1):
        if i not in visited:
            cnt[i]=count(arr,i,visited,str,i,head,0)
    for i in range(1,n+1):
        print(cnt[head[i]],end=' ')
    print()
    t-=1
thread = threading.Thread(target=count)
thread.start()
thread.join()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
6
6 1 2 3 4 5
100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
cases=int(input())
i=1
black=[]
a=[]
while(i<=cases):
    list1=[]
    x=[int(input())]
    y=[int(y) for y in input().split()]
    z=input()
    j=0
    for letter in z:
        if(letter=='0'):
            list1.append(y[j])
        j+=1
    if(len(list1)>0):
        black.append(list1)
    i+=1
    a.append(y)
def f(list,black,i,check,new):
    while(1):
        if i in check:
            w=list[i-1]
            check.remove(i)
            if w in black:
                new.append(w)
            f(list,black,w,check,new)
        else:
            break
    return(new)
for i in range(0,cases):
    for j in range(0,len(a[i])):
        check = a[i][:]
        new=[]
        print(len(f(a[i],black[i],j+1,check,new)),end=' ')
    print()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following aspects:

1. **Edge Cases**: These include small permutations, permutations with all elements of the same color, and permutations with specific patterns.
2. **Complex Permutations**: These are permutations that have multiple cycles or complex reachability patterns.
3. **Coloring Patterns**: We need to test with different coloring patterns, including all black, all white, and mixed colors.

Given the constraints and the need for a small-scale test input, let's construct a test case that can expose the bug effectively.

### Test Input:
```plaintext
3
6
1 2 3 4 5 6
101010
4
4 1 2 3
0000
5
3 5 1 4 2
10101
```

### Explanation:
1. **Test Case 1**:
   - **Permutation**: `[1, 2, 3, 4, 5, 6]`
   - **Colors**: `101010` (Black: 1, 3, 5; White: 2, 4, 6)
   - **Reachability**: Each number is only reachable from itself because the permutation is the identity permutation.
   - **Expected Output**: `1 1 1 1 1 1`

2. **Test Case 2**:
   - **Permutation**: `[4, 1, 2, 3]`
   - **Colors**: `0000` (All black)
   - **Reachability**: This permutation forms a single cycle.
   - **Expected Output**: `4 4 4 4`

3. **Test Case 3**:
   - **Permutation**: `[3, 5, 1, 4, 2]`
   - **Colors**: `10101` (Black: 3, 1, 5; White: 2, 4)
   - **Reachability**: This permutation has a more complex reachability pattern.
   - **Expected Output**: `3 2 3 1 2`

### Why This Test Input is Challenging:
- **Test Case 1**: Tests the identity permutation where each number is only reachable from itself. This checks if the code correctly handles the simplest case.
- **Test Case 2**: Tests a permutation where all elements are black and form a single cycle. This checks if the code can handle cycles and all black elements.
- **Test Case 3**: Tests a more complex permutation with mixed colors and multiple reachability paths. This checks if the code can handle more complex permutations and mixed colors.

This test input should help expose any issues in the buggy code, especially in how it handles reachability and cycles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    p = list(map(int, input().split()))
    s = input()
    ans = [-1] * n
    for i in range(n):
        if ans[i] != -1:
            ans[i] = ans[ans[i]]
            continue
        g = p[i]
        check = True
        while p[g - 1] != p[i] or check:
            if p[g - 1] != p[i]:
                ans[g - 1] = i
            if s[g - 1] == "0":
                ans[i] += 1
            g = p[g - 1]
            if p[g - 1] == p[i] and check:
                break
            if check:
                check = False
        else:
            if s[i] == "0":
                ans[i] += 1
        ans[i] += 1
    print(*ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. The buggy code has issues with how it handles the cycle detection and the counting of black integers. Here are the steps to create a suitable test input:

1. **Cycle Detection**: The code might fail if there are multiple cycles in the permutation, and it doesn't correctly handle the cycle detection and counting.
2. **Black Integer Counting**: The code might incorrectly count the black integers if it doesn't properly handle the cycle structure.

Given these points, we can create a test input with a permutation that has multiple cycles and a mix of black and white integers.

### Test Input
```plaintext
1
6
1 2 3 4 5 6
001110
```

### Explanation
- **Permutation**: `[1, 2, 3, 4, 5, 6]`
  - This permutation has multiple cycles: `[1]`, `[2]`, `[3]`, `[4]`, `[5]`, `[6]`.
- **Coloring**: `001110`
  - The integers 1, 2, and 6 are black.
  - The integers 3, 4, and 5 are white.

### Expected Output
```plaintext
1 1 0 0 0 1
```

### Reasoning
- For `i = 1`: The only reachable integer is 1, which is black.
- For `i = 2`: The only reachable integer is 2, which is black.
- For `i = 3`: The only reachable integer is 3, which is white.
- For `i = 4`: The only reachable integer is 4, which is white.
- For `i = 5`: The only reachable integer is 5, which is white.
- For `i = 6`: The only reachable integer is 6, which is black.

This test input should expose the bug in the code by challenging its ability to handle multiple cycles and correctly count the black integers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
import java.util.Scanner;
import java.util.HashSet;

public class Solution {

    public static int[] helper(int n, int[] arr, String str) {
        int[] ans = new int[n + 1];
        char[] ch = str.toCharArray();
        boolean[] visited = new boolean[n + 1];
        for (int i = 1; i <= n && !visited[i]; i++) {
            int idx = i;
            int count = 0;
            HashSet<Integer> currentPath = new HashSet<>();
            if (arr[idx] == i) {
                currentPath.add(idx);
                if (ch[idx - 1] == '0') {
                    count++;
                }
            }
            while (arr[idx] != i) {
                currentPath.add(idx);
                if (ch[idx - 1] == '0') {
                    count++;
                }
                idx = arr[idx];
                if (arr[idx] == i) {
                    currentPath.add(idx);
                    if (ch[idx - 1] == '0') {
                        count++;
                    }
                }
            }
            for (int index : currentPath) {
                visited[index] = true;
                ans[index] = count;
            }

        }
        return ans;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] arr = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                arr[i] = sc.nextInt();
            }
            sc.nextLine();
            String str = sc.nextLine();
            int ans[] = helper(n, arr, str);
            for (int i = 1; i <= n; i++) {
                System.out.print(ans[i] + " ");
            }
            System.out.println();
        }
    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
4
2 3 4 1
0000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main {
    public static int mod = (int)1e9+7;

    public static void main(String[] args) throws Exception {
        fastscanner sc = new fastscanner();
        PrintWriter pr = new PrintWriter(System.out);
        int t = sc.nextInt();

        while (t-- != 0) {

            int n=sc.nextInt();
            int a[]=new int[n+1];
            for(int i=1;i<=n;i++) a[i]=sc.nextInt();
            char b[]=sc.next().toCharArray();
            boolean[] flag=new boolean[n+1];
            int ans[]=new int[n+1];
            for(int i=1;i<=n;i++) flag[i]=false;
            for(int i=1;i<=n&&(!flag[i]);i++){
                ArrayList<Integer> l=new ArrayList<>();
                flag[i]=true;
                l.add(i);
                int cnt=0;
                int tmp=a[i];
                if(b[i-1]=='0') cnt++;
                while(a[tmp]!=a[i]){
                    if(b[tmp-1]=='0') cnt++;
                    flag[tmp]=true;
                    l.add(tmp);
                    tmp=a[tmp];
                }

                flag[tmp]=true;

                for (Integer integer : l) {
                    ans[integer] = cnt;
                }

            }
            for(int i=1;i<=n;i++) pr.print(ans[i]+" ");
            if(t!=53) pr.println();
            if(t==53){
                pr.print('x');
                pr.print(Arrays.toString(a));
                pr.print(Arrays.toString(b));
            }

        }

            pr.flush();
    }

    static boolean primeCheck(int a){
        if(a==2||a==3) return true;
        if(a==1||a%2==0||a%3==0) return false;
        for(int i=5;i*i<=a;i+=6){
            if(a%i==0||a%(i+2)==0) return false;
        }
        return true;
    }

    static long divisorCount(long a){
        long cnt=0;
        for(long i=1;i*i<=a;i++){
            if(a%i==0){
                if(i*i==a) cnt++;
                else cnt+=2;
                cnt%=mod;
            }
        }
        return cnt;
    }

    static long nc2(int n){
        if(n<2) return 0;
        return ((long) n *(n-1))/2;
    }
    static long nCr(long n,long r){
        if(n<r) return 0;
        if(r==0) return 1;
        return ((fact((int)n)*modInv( fact((int)r)))%mod*modInv( fact((int)(n-r))))%mod;
    }

    public static long ncr(long n,long r){
        if(n-r<r)
            return ncr(n,n-r);
        long res=1;
        for(long i=n-r+1;i<=n;++i){
            res*=i;
            res%=mod;
        }
        for(long i=2;i<=r;++i){
            res*=power(i,mod-2);
            res%=mod;
        }
        return res;
    }

    static long modInv(long n){
        return power(n,mod-2)%mod;
    }

    static long power(long a,long b){
        a=a%mod;
        long res=1;
        while(b>0){
            if((b&1)!=0) res=(res*a)%mod;
            b>>=1;
            a=(a*a)%mod;
        }
        return res%mod;
    }

    static long binExp(int a ,int b){
        long res=1;
        while(b>0){
            if((b&1)!=0) res*=a;
            a*=a;
            b>>=1;
        }
        return res%mod;
    }

    public static int fact(int n)
    {
        int res = 1, i;
        for (i = 2; i <= n; i++) res =(res*i)%mod;
        return res%mod;
    }

    public static int[] sieve(int n){
        boolean []isPrime=new boolean[n+1];
        for(int i=2;i<=n;i++) isPrime[i]=true;
        isPrime[0]=isPrime[1]=false;
        for(int i=2;i<=n;i++){
            if(isPrime[i]&&((long) i *i)<=n){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        int []primes=new int[n];
        primes[0]=2;
        int cnt=1;
        for(int i=3;i<=n;i+=2){
            if(isPrime[i]){
                primes[cnt]=i;cnt++;
            }
        }
        return primes;
    }

    public static int gcd(int a,int b){
        while(b>0){
            int tmp=b;
            b=a%b;
            a=tmp;
        }
        return a;
    }

    public static int gcdArray(int a[]){
        int gcd=a[0];
        for(int i=1;i<a.length;i++){
            gcd=gcd(gcd,a[i]);
        }
        return gcd;
    }

}

class fastscanner
{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    String next()
    {
        while (st == null || !st.hasMoreTokens())
        {
            try {

                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e) {}
        }
        return st.nextToken();
    }
    int nextInt()
    {
        return Integer.parseInt(next());
    }
    float nextFloat()
    {
        return Float.parseFloat(next());
    }
    long nextLong()
    {
        return Long.parseLong(next());
    }
    double nextDouble()
    {
        return Double.parseDouble(next());
    }
    String nextLine()
    {
        String s = "";
        try {
            s = br.readLine();
        }
        catch (IOException e) {}
        return s;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's consider the following aspects:

1. **Cycle Detection**: The code uses a while loop to detect cycles in the permutation. However, it might not correctly handle all cases, especially when there are multiple cycles or when the cycle includes the starting point itself.

2. **Coloring**: The code increments the count of black integers as it traverses the cycle. If there is an issue with the cycle detection or the traversal, it might not correctly count the black integers.

3. **Edge Cases**: We should consider edge cases such as:
   - Permutations with a single element.
   - Permutations with all elements in a single cycle.
   - Permutations with multiple cycles.
   - Permutations where the starting point is part of a cycle.

4. **Input Size**: Since the input size can be up to \(2 \times 10^5\), we should create a small-scale test input to ensure it fits within the constraints.

Let's create a test input that includes multiple cycles and a mix of black and white elements to challenge the cycle detection and counting logic.

```plaintext
1
6
1 2 3 4 5 6
010101
```

This test input has:
- \(n = 6\)
- The permutation \(p = [1, 2, 3, 4, 5, 6]\), which consists of 6 cycles of length 1.
- The color string \(s = "010101"\), where each element is alternately black and white.

This input should expose any issues with the cycle detection and counting logic, as each element is its own cycle, and the code should correctly count the black elements for each position.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
//In The Name Of ALLAH

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const char nl = '\n';
const int N = 2e5 + 9;
int p[N];

void solveTest() {
    int n; cin >> n;
    for(int i=1 ; i<=n ; i++) cin >> p[i];
    string temp; cin >> temp;
    string black = '$' + temp;

    for(int i=1 ; i<=n ; i++) {
        int cnt = (black[i] == '0');
        int num = p[i];
        int it = 4;
        while(num != i) {
            //cout << num << " " << black[num] << " " << cnt << nl;
            if(black[num] == '0') {
                cnt++;
            }
            num = p[num];
        }
        cout << cnt << " ";
    }
    cout << nl;
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll tt; cin >> tt;
    while(tt--)
       solveTest();
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
6
2 3 4 5 6 1
010101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#define int long long int
#define f first
#define s second
#define endl "\n"
#define loop(i, a , b) for (ll i = a; i < b; i++)
#define countsetbits(x , count) while(x>0)  {x&= (x-1); count++;}
#define p_q priority_queue
#define maxheap p_q<int,vector<int>>
#define minheap p_q<int,vector<int>,greater<int>>

int gcd(int a , int b){
    if(a == 0)  return b;
    return gcd(b%a , a);
}

//-------------------------------------------------------------------------------
//_______________________________________________________________________________

const int mod = 1e9+7;
const int N = 2e5+10;

int precom[N];
void precompute()
{

}

void solve()
{
    int n;  cin>>n;
    vector<int> arr(n);
    loop(i,0,n) cin>>arr[i];
    string s;   cin>>s;

    vector<int> dist(n , 0);

    loop(i,0,n){
        vector<bool> currVis(n , 0);
        int k = arr[i];
        while(!currVis[i]){
            dist[i] += (s[k-1]=='0');
            currVis[k-1] = 1;
            k = arr[k-1];
        }
    }
    for(int x : dist)   cout<<x<<" ";
    cout<<endl;

}

//-------------------------------------------------------------------------------
//_______________________________________________________________________________

#undef int
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    precompute();

    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
from math import *
from heapq import *
from collections import deque
import sys,os
sys.setrecursionlimit(10**9)
'''
This code is written and submitted by adipuli_2004.
NOT to be copied or plagiarised

helper functions:

sys.stdout.write(str(n)+"\n")
sys.stdout.write( " ".join(map(str, res)) + "\n")
'''
def bfs(node,graph,a):
    if not len(graph[node]):
        return a[node]
    res=float('inf')
    for i in graph[node]:
        res=min(res,bfs(i,graph,a))

    if a[node]>res:
        a[node]=res
        return res
    else:
        return (a[node]+res)//2
def seive(num):
    prime = [True for i in range(num+1)]
    p = 2
    while (p * p <= num):
        if (prime[p] == True):
            for i in range(p * p, num+1, p):
                prime[i] = False
        p += 1
    return prime
def bins(target,arr):
    low=0
    high=len(arr)-1
    ans=len(arr)
    while low<=high:
        mid=(low+high)//2
        if arr[mid]>target:
            ans=mid
            high=mid-1
        else:
            low=mid+1
    return ans
def mex(arr):
    a=set(arr)
    for i in range(len(a)):
        if i not in a:
            return i
    return len(a)
def rotate(matrix):
    n=len(matrix)
    for i in range(n):
        for j in range(i,n):
            matrix[i][j],matrix[j][i]=matrix[j][i],matrix[i][j]
    for i in range(n):
        for j in range(n//2):
            matrix[i][j],matrix[i][-j-1]=matrix[i][-j-1],matrix[i][j]

def dfs(a1,a2,dp,vis,i,curr=0):

    if i in vis:
        return curr
    vis.add(i)

    if a2[i]=='0':
        curr+=1

    temp=dfs(a1,a2,dp,vis,a1[i]-1,curr)
    if temp>curr:
        dp[i]=temp
    else:
        dp[i]=curr
    return dp[i]

for _ in range(int(input())):
    n=int(input())
    a=list(map(int,input().split()))
    b=input()
    dp=[0]*n
    vis=set()
    for i in range(n):
        if i not in vis:
            dfs(a,b,dp,vis,i,0)
    for i in dp:
        print(i,end=" ")
    print("")

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. Let's break down the requirements and consider the potential issues:

1. **Permutation with Cycles**: The permutation might contain cycles, and the code needs to handle these cycles correctly.
2. **Coloring of Elements**: The elements can be colored either black ('0') or white ('1'), and the function `F(i)` should correctly count the number of black elements reachable from each `i`.
3. **Large Input Size**: Although the input size is limited, we can still create a small but complex test case that might reveal the bug.

### Step-by-Step Analysis:

1. **Cycle in Permutation**: The permutation might have multiple cycles, and the code needs to correctly traverse these cycles without getting stuck or incorrectly counting the black elements.
2. **Complex Coloring**: The coloring should be such that it tests the code's ability to correctly count black elements in various scenarios, including cycles with different lengths and different color patterns.

### Test Input:

Let's create a test case with a permutation that has multiple cycles and a complex coloring pattern.

```plaintext
5
6
1 2 3 4 5 6
010101
6
1 3 5 2 4 6
010101
6
3 5 6 1 2 4
010000
6
4 1 2 3 6 5
000000
6
6 1 2 3 4 5
101010
```

### Explanation:

1. **First Test Case**:
   - Permutation: `[1, 2, 3, 4, 5, 6]`
   - Coloring: `010101`
   - Each element is in its own cycle. The function `F(i)` should return the color of each element itself.

2. **Second Test Case**:
   - Permutation: `[1, 3, 5, 2, 4, 6]`
   - Coloring: `010101`
   - Cycles: `[1, 3, 5]` and `[2, 4]`
   - The function `F(i)` should correctly count the black elements in each cycle.

3. **Third Test Case**:
   - Permutation: `[3, 5, 6, 1, 2, 4]`
   - Coloring: `010000`
   - Cycles: `[3, 6, 4, 1, 2, 5]`
   - The function `F(i)` should correctly count the black elements in the single cycle.

4. **Fourth Test Case**:
   - Permutation: `[4, 1, 2, 3, 6, 5]`
   - Coloring: `000000`
   - Cycles: `[4, 1, 2, 3]` and `[6, 5]`
   - All elements are black, so `F(i)` should return 4 for the first cycle and 2 for the second cycle.

5. **Fifth Test Case**:
   - Permutation: `[6, 1, 2, 3, 4, 5]`
   - Coloring: `101010`
   - Cycles: `[6, 1, 2, 3, 4, 5]`
   - The function `F(i)` should correctly count the black elements in the single cycle.

This test input should help expose any issues in the buggy code, especially in handling cycles and complex coloring patterns.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
def dfs(i,all, count):
    if i in all:
        for x in all:
            sol[x] = count
        return
    if st[i]=='0':
        count+=1
    all.add(i)
    dfs(li[i], all, count)

for _ in range(int(input())):
    n = int(input())
    li = list(map(int, input().split()))
    li.insert(0, -1)
    st = input()
    st = "a"+st
    sol=[0]*(n+1)
    v=set()
    for i in range(1, n+1):
        if i not in v :
            dfs(i, set(), 0)
    print(*sol[1:])

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
3
1 2 3
010
4
2 3 4 1
1001
5
3 5 1 4 2
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class D_Sakurako_s_Hobby {
static CustomPrintStream out;
static int fun(int i,int[] p,String str){
    boolean[] visited=new boolean[p.length];
    int count=0;
    while(!visited[i]){
        if(str.charAt(i-1)=='0'){
            count++;
        }
        visited[i]=true;
        i=p[i];
    }
    return count;
}
static void solve(FastReader cin,CustomPrintStream ps){
    int t=cin.nextInt();
    while(t-->0){
        int n=cin.nextInt();
        int[] p=new int[n+1];
        for(int i = 1; i <= n; i++){
            p[i]=cin.nextInt();
        }
        String str=cin.nextLine();
        int[] f=new int[n+1];

        for(int i=1;i<=n;i++){
            f[i]=fun(i,p,str);
        }
        for(int i=1;i<=n;i++){
            ps.print(f[i]+" ");
        }
        ps.println();
    }
}
public static void main(String[] args) throws IOException {
    out = new CustomPrintStream(System.out);
    FastReader in = new FastReader();
    solve(in, out);
}
static void cout(Object x){System.out.println(x);}
static class FastReader {
BufferedReader br; StringTokenizer st;
public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); }
String next() {
while (st == null || !st.hasMoreElements()) {
try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next()); }
String nextLine() {
String str = "";
try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); }
return str;
}}
static class CustomPrintStream extends PrintStream {
public CustomPrintStream(OutputStream out) { super(out); }
void yes(){this.println("YES");}
void no(){this.println("NO");}
}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might fail. The code provided has a potential issue in the `fun` function, where the index `i` is used without proper bounds checking. Specifically, the line `i = p[i]` can lead to an `ArrayIndexOutOfBoundsException` if `i` is out of the valid range.

To create a test case that maximizes the probability of this failure, we can construct a permutation where the cycle leads to an out-of-bounds index. However, since the permutation is guaranteed to be a valid permutation of integers from 1 to n, we need to focus on other potential issues, such as infinite loops or incorrect cycle detection.

### Step-by-Step Reasoning:
1. **Cycle Detection**: The code uses a `visited` array to detect cycles, which is generally correct for permutations.
2. **Indexing**: The code uses `i-1` to access the string `str`, which is correct since `i` ranges from 1 to n.
3. **Potential Infinite Loop**: The code might fail if there is a cycle that includes an index that is not properly handled.

### Test Case:
We can create a test case with a permutation that has a cycle that includes an index that might cause an issue. For example, a permutation with a cycle that includes all elements can be used to test the cycle detection and counting logic.

### Small-Scale Test Input:
```plaintext
1
3
2 3 1
010
```

### Explanation:
- **Permutation**: `[2, 3, 1]`
  - 1 → 2 → 3 → 1 (cycle)
- **String**: `010`
  - 0: black
  - 1: white
  - 2: black

### Expected Output:
- For `i = 1`: Reachable black integers are 1 and 3 (2 black integers).
- For `i = 2`: Reachable black integers are 3 (1 black integer).
- For `i = 3`: Reachable black integers are 1 and 3 (2 black integers).

Thus, the expected output is:
```plaintext
2 1 2
```

This test case should help expose any issues in the cycle detection and counting logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    public static void solve(FastScanner in, PrintWriter out) throws Exception {
        int n = in.nextInt();

        int[] p = new int[n + 1];
        for (int i = 1; i <= n; i++)
            p[i] = in.nextInt();

        String s = "_" + in.next();

        int[] ans = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            Set<Integer> st = new HashSet<>();

            int node = p[i];
            int count = 0;
            do {
                if (st.contains(node))
                    break;
                count += (s.charAt(node) == '0') ? 1 : 0;
                st.add(node);
                node = p[node];
            } while (true);

            for (int number : st) {
                ans[number] = count;
            }
        }
        StringBuilder ansString = new StringBuilder();
        for (int i = 1; i <= n; i++)
            ansString.append(ans[i] + " ");

        out.println(ansString);
    }

    public static void main(String[] args) {
        try {
            FastScanner in = new FastScanner("in.in");// _/ I \
            PrintWriter out = outter("outout.out"); // ______\ O /

            int t = 1;
            t = in.nextInt(); // for test-cases
            while (t-- > 0) {
                solve(in, out);
            }

            out.close();

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static PrintWriter outter(String fileName) throws Exception {
        return (new File(fileName).exists()) ? new PrintWriter(fileName)
                : new PrintWriter(System.out);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(String fileName) throws Exception {
            br = (new File(fileName).exists()) ? new BufferedReader(new FileReader(fileName))
                    : new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() throws Exception {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());

            return st.nextToken();
        }

        public int nextInt() throws Exception {
            return Integer.parseInt(next());
        }

        public long nextLong() throws Exception {
            return Long.parseLong(next());
        }

        public double nextDouble() throws Exception {
            return Double.parseDouble(next());
        }

    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the code to fail. The bug in the code is that it does not handle cycles in the permutation correctly. Specifically, it does not properly account for the fact that once a cycle is detected, all elements in that cycle should have the same count of reachable black integers.

Let's consider a permutation that forms a cycle and includes both black and white elements. A small-scale test case that can expose this bug is:

```plaintext
1
6
3 5 6 1 2 4
010000
```

### Explanation:
- **Permutation:** `p = [3, 5, 6, 1, 2, 4]`
- **Colors:** `s = "010000"`

### Cycle Analysis:
- Start from `1`:
  - `1 -> 3 -> 6 -> 4 -> 1` (forms a cycle)
  - Black integers in this cycle: `3, 6, 4`
  - Count of black integers: `3`

- Start from `2`:
  - `2 -> 5 -> 2` (forms a cycle)
  - Black integers in this cycle: `5`
  - Count of black integers: `1`

- Start from `3`, `4`, `5`, `6`:
  - These will eventually map back to the cycle starting from `1` or `2`.

### Expected Output:
- For `1`, `3`, `4`, `6`: `3` (since they are all part of the cycle `1 -> 3 -> 6 -> 4`)
- For `2`, `5`: `1` (since they are part of the cycle `2 -> 5 -> 2`)

### Correct Output:
```plaintext
3 1 3 3 1 3
```

This test case should expose the bug in the code because the buggy code will not correctly handle the cycles and will not assign the correct count to all elements in the cycle.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Hobby
time_limit_ms: 2000
memory_limit_mb: 256
Description: For a certain permutation $p$$^{\text{∗}}$ Sakurako calls an integer $j$ reachable from an integer $i$ if it is possible to make $i$ equal to $j$ by assigning $i=p_i$ a certain number of times.

If $p=[3,5,6,1,2,4]$, then, for example, $4$ is reachable from $1$, because: $i=1$ $\rightarrow$ $i=p_1=3$ $\rightarrow$ $i=p_3=6$ $\rightarrow$ $i=p_6=4$. Now $i=4$, so $4$ is reachable from $1$.

Each number in the permutation is colored either black or white.

Sakurako defines the function $F(i)$ as the number of black integers that are reachable from $i$.

Sakurako is interested in $F(i)$ for each $1\le i\le n$, but calculating all values becomes very difficult, so she asks you, as her good friend, to compute this.

$^{\text{∗}}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation (the number $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$, but the array contains $4$).

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$)  — the number of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot 10^5$)  — the number of elements in the array.

The second line of each test case contains $n$ integers $p_1, p_2, \dots, p_n$ ($1\le p_i\le n$)  — the elements of the permutation.

The third line of each test case contains a string $s$ of length $n$, consisting of '0' and '1'. If $s_i=0$, then the number $p_i$ is colored black; if $s_i=1$, then the number $p_i$ is colored white.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output $n$ integers $F(1), F(2), \dots, F(n)$.

Examples:
input:
5
1
1
0
5
1 2 4 5 3
10101
5
5 4 1 3 2
10011
6
3 5 6 1 2 4
010000
6
1 2 3 4 5 6
100110
output:
1
0 1 1 1 1
2 2 2 2 2
4 1 4 4 1 4
0 1 1 0 0 1

Note:

Buggy Code:
#include <iostream>
using namespace std;
#define MAX 0x3f3f3f3f3f
const int MAXN = 2e5 + 10;
int n, arr[MAXN], zo[MAXN], dp[MAXN] = { 0 }, ans=0,start;

string s;

void dfs(int i,int cnt) {
	if (i == start && cnt >= 1||i>n) {
		return;
	}
	if (s[i] == '0') ans++;
	dfs(arr[i],cnt+1);
}
int main() {
	int t;
	cin >> t;
	while (t--) {
		cin >> n;
		for (int i = 1; i <= n; i++) cin >> arr[i], dp[i] = 0;
		s = ' ';
		string s1;
		cin >> s1;
		s += s1;
		for (int i = 1; i <= n; i++) {
			start = i;
			ans = 0;
			dfs(i,0);
			cout << ans << " ";
		}
		cout << endl;
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
1
1
2
2 1
00
4
4 1 3 2
0000
```