**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
import sys
from collections import deque, Counter, defaultdict
import heapq
from bisect import bisect_left, bisect_right, bisect
from itertools import combinations, permutations, product
from math import sin, cos, gcd
from copy import deepcopy

input = sys.stdin.readline

def I():
	return [int(i) for i in input().split()]

def F():
	return [float(i) for i in input().split()]

def II():
	return int(input())

def S():
	return input().split()

def IS():
	return input().strip('\n')

def get_s(tree, v):
	q = [v]
	p = [v] * len(tree)
	order = []
	while q:
		v = q.pop()
		order.append(v)
		for u in tree[v]:
			if p[v] != u:
				p[u] = v
				q.append(u)

	s = [1] * len(tree)
	for i in range(len(tree) - 1, 0, -1):
		s[p[order[i]]] += s[order[i]]

	return s

def normalize_p(p):
	s = list(set(p) - {-2})
	s.sort()
	s = {s[i]: i for i in range(len(s))}
	new_p = []
	for i in range(len(p)):
		if p[i] == -2:
			continue
		new_p.append(s[p[i]])

	return new_p

def step(p: list[int]):
	n = len(p)

	if n in [0, 1, 2, 3]:
		return 1
	if n == 0:
		return 0

	tree = [[] for _ in range(n)]
	for i, pi in enumerate(p):
		if i == pi:
			continue
		tree[pi].append(i)
		tree[i].append(pi)

	s = get_s(tree, 0)

	centroid_idx = 0
	while True:
		w = -1
		for u in tree[centroid_idx]:
			if s[u] > s[centroid_idx]:
				continue

			if 2 * s[u] >= n:
				w = u
				break

		if w == -1:
			break
		centroid_idx = w

	kids = tree[centroid_idx]
	s = get_s(tree, centroid_idx)
	kids.sort(key=lambda x: s[x])

	t = n - 1
	_sum = 0
	index = len(kids) - 2
	p1 = [-2] * n
	p2 = p1.copy()

	if len(kids) > 2:
		p1[centroid_idx] = centroid_idx
		q = kids[:index + 1]
		for u in q:
			p1[u] = centroid_idx

		while q:
			v = q.pop()
			for u in tree[v]:
				if p[v] != u and u != centroid_idx:
					q.append(u)
					p1[u] = v
	else:
		p1[kids[0]] = kids[0]
		q = [kids[0]]

		while q:
			v = q.pop()
			for u in tree[v]:
				if p[v] != u and u != centroid_idx:
					q.append(u)
					p1[u] = v

	p2[kids[-1]] = kids[-1]
	q = [kids[-1]]

	while q:
		v = q.pop()
		for u in tree[v]:
			if p[v] != u and u != centroid_idx:
				q.append(u)
				p2[u] = v

	# print(p1, p2)
	p1 = normalize_p(p1)
	p2 = normalize_p(p2)
	# print(p1, p2)
	return max(step(p1), step(p2)) + 1

def main():
	n = II()
	p = [0] + [i - 1 for i in I()]
	print(step(p))

if __name__ == '__main__':
	for _ in range(II()):
		main()
# main()

**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 or scenarios that might cause the code to fail. The key points to consider are:

1. **Tree Structure**: Ensure the tree structure is complex enough to test the centroid finding and tree splitting logic.
2. **Edge Cases**: Test with small trees and trees with specific structures that might cause issues in the centroid finding or normalization steps.
3. **Large Trees**: Although the output length limit restricts us to small-scale tests, we should still aim to create a tree that is not trivial.

Let's create a test input that includes a tree with a structure that might expose the bug:

- A tree with a centroid that has multiple children.
- A tree with a long chain to test the normalization and step function.
- A tree with a balanced structure to test the centroid finding.

Here is a small-scale test input that meets these criteria:

```plaintext
3
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
```

### Explanation:
1. **Test Case 1 (n = 5)**:
   - Tree: 1 -> 2, 2 -> 3, 2 -> 4, 2 -> 5
   - This tree has a single root with multiple children, which tests the centroid finding and normalization.

2. **Test Case 2 (n = 7)**:
   - Tree: 1 -> 2, 1 -> 3, 1 -> 4, 2 -> 5, 2 -> 6, 4 -> 7
   - This tree has a more complex structure with multiple levels and branches, testing the step function and normalization.

3. **Test Case 3 (n = 10)**:
   - Tree: 1 -> 2, 1 -> 3, 1 -> 4, 2 -> 5, 2 -> 6, 2 -> 7, 4 -> 8, 3 -> 9, 3 -> 10
   - This tree has a balanced structure with multiple levels and branches, testing the centroid finding and normalization thoroughly.

This test input should help expose any issues in the centroid finding, tree splitting, and normalization logic of the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
/*
   卫风·芄兰
芄兰之支,童子佩觿.
虽则佩觿,能不我知?
容兮遂兮,垂带悸兮.
芄兰之叶,童子佩韘.
虽则佩韘,能不我甲?
容兮遂兮,垂带悸兮.
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int,int>
#define tul tuple<int,int,int>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define rep_(i,a,b) for(int i=a;i>=b;--i)
#define all(x) x.begin(),x.end()
#define bp(x) __builtin_popcountll(x)
#define ctz(x) __builtin_ctzll(x)
#define cy cout<<"YES"<<endl
#define cn cout<<"NO"<<endl
#define lc (rt<<1)
#define rc (rt<<1|1)
mt19937_64 rnd(time(0));
const int N=3e5+5,yyx=1e9+7;
vector<int> to[N];
int n,m,k,p[N];
inline int mod(int x){
  return (x%yyx+yyx)%yyx;
}
inline int cmin(int &x,int y){
  return x>y?x=y,1:0;
}
inline int cmax(int &x,int y){
  return x<y?x=y,1:0;
}
int f(int x){
    if(!x||x==1) return 0;
    int j=__lg(x-1);
    return j;
}
int dfs(int now,int fa){
    priority_queue<int,vector<int>,greater<int> > q;
    for(auto c:to[now]){
        if(c==fa) continue;
        q.push(dfs(c,now));
    }
    if(q.size()==1){
        return q.top()+1;
    }
    while(q.size()>=2){
        int x=q.top();q.pop();
        x=q.top();q.pop();
        q.push(x+1);
    }
    if(!q.size()) return 0;
    return q.top();;
}
inline void solve(){
   cin>>n;
   rep(i,1,n) to[i].clear();
   rep(i,2,n){
    cin>>p[i];
    to[p[i]].push_back(i);
   }
   cout<<dfs(1,0)<<endl;
}
signed main(){
  cin.tie(0)->sync_with_stdio(0);
  //freopen("D://321//in.txt","r",stdin);
  //freopen("D://321//out.txt","w",stdout);
  int _=1;
  cin>>_;
  while(_--)
  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
1
4
1 2 2
```

**Explanation:**
This test case is designed to challenge the buggy code by creating a scenario where the tree structure is not a perfect binary tree but can be transformed into one with a minimal number of operations. The input describes a tree with 4 vertices, where vertex 1 is the root, and vertices 2 and 3 are children of vertex 2. This tree can be transformed into a perfect binary tree of depth 2 by performing an operation on the edge (1, 2). The expected output should be 2, but the buggy code might not handle this correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
//
// Created by 35395 on 2024/11/15.
//
#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N = 2e5 + 100;

int n;
vector<int> h[N];
int f[N];
void dfs(int u){
    priority_queue<int,vector<int>,greater<>> q;
    for(auto v : h[u]){
        dfs(v);
        q.push(f[v]);
    }
    while(q.size() > 1){
        int a = q.top(); q.pop();
        int b = q.top(); q.pop();
        q.push(max(a,b) + 1);
    }
    if(q.size()) f[u] = q.top() + (h[u].size() == 1);
}

void solve() {
    for(int i=1;i<=n;i++) h[i].clear(),f[i] = 1;
    cin >> n;
    for(int i=1;i<=n;i++) f[i] = 1;
    for(int i=2;i<=n;i++){
        int x;
        cin >> x;
        h[x].push_back(i);
    }
    dfs(1);
    cout << f[1] - 1<< '\n';
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(20);
    int test = 1;
    cin >> test;
    while (test--) {
        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 buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The key aspects to consider are:

1. **Single Child Nodes**: The code might handle single child nodes incorrectly, especially when the root has only one child.
2. **Multiple Levels of Single Child Nodes**: A tree where multiple levels of nodes have only one child can be a challenging case.
3. **Balanced and Unbalanced Trees**: The code should correctly handle both balanced and unbalanced trees.

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

### Test Input
```plaintext
3
5
1 1 2 2
6
1 1 1 2 2
7
1 2 2 3 3 3
```

### Explanation
1. **First Test Case (5 nodes)**
   - Tree structure: 1 -> 2, 1 -> 3, 2 -> 4, 2 -> 5
   - This tree has a root with two children, and each of those children has two children. The minimum depth of Chloe's perfect binary tree should be 2.

2. **Second Test Case (6 nodes)**
   - Tree structure: 1 -> 2, 1 -> 3, 1 -> 4, 2 -> 5, 2 -> 6
   - This tree has a root with three children, and one of those children has two children. The minimum depth of Chloe's perfect binary tree should be 2.

3. **Third Test Case (7 nodes)**
   - Tree structure: 1 -> 2, 2 -> 3, 3 -> 4, 3 -> 5, 3 -> 6
   - This tree has a root with one child, and that child has one child, which has three children. The minimum depth of Chloe's perfect binary tree should be 3.

This test input includes a mix of balanced and unbalanced trees, as well as single child nodes, which should help in exposing any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
import sys
import heapq
input = sys.stdin.readline
sys.setrecursionlimit(3*10**5)

def dfs(u, parent):
    pq = []
    for v in graph[u]:
        if v != parent:
            dfs(v, u)
            heapq.heappush(pq, depth[v])

    if not pq:
        depth[u] = 0
    elif len(pq) == 1:
        depth[u] = heapq.heappop(pq) + 1
    else:
        hold = -1
        add = False
        while pq:
            smallest = heapq.heappop(pq)
            if smallest != hold:
                if hold != -1:
                    add = True
                hold = smallest
            else:
                heapq.heappush(pq, hold + 1)
                hold = -1
        depth[u] = hold + (1 if add else 0)

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))

    graph = [[] for _ in range(n + 1)]
    for i in range(n - 1):
        graph[a[i]].append(i + 2)
        graph[i + 2].append(a[i])

    depth = [0] * (n + 1)
    dfs(1, 0)
    print(depth[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
4
1 2 2
5
1 1 2 2
6
1 2 2 3 3
```

**Explanation:**

1. **Test Case 1:**
   - Tree structure: `1 -> 2 -> 3, 4`
   - This tree is a simple linear chain with a single child at each level. The minimum depth of a perfect binary tree that can be made isomorphic to this tree is `2`.

2. **Test Case 2:**
   - Tree structure: `1 -> 2, 3 -> 4, 5`
   - This tree has a root with two children, and each child has one child. The minimum depth of a perfect binary tree that can be made isomorphic to this tree is `2`.

3. **Test Case 3:**
   - Tree structure: `1 -> 2 -> 3, 4 -> 5, 6`
   - This tree has a root with one child, which has two children, and each of those children has one child. The minimum depth of a perfect binary tree that can be made isomorphic to this tree is `3`.

These test cases are designed to cover different structures and edge cases to ensure the algorithm can handle various scenarios correctly. The small scale of the test cases ensures they are complete and valid while being manageable in size.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
def dfs(cur):
    global child  # Ensure correct scope for 'child'
    t = 0
    arr = []
    for i in child[cur]:
        t2 = dfs(i)
        arr.append(t2 - 1)
        t = max(t, t2)

    x = sum(pow(2, i) for i in arr)

    while x > pow(2, t):
        t += 1

    return t + 1

for _ in range(int(input())):
    n = int(input())
    if n == 0:
        print(0)
        continue

    a = input().split()
    child = [[] for _ in range(n)]

    for i in range(n - 1):
        parent = int(a[i]) - 1
        if parent < 0 or parent >= n:
            raise ValueError("Invalid parent index")
        child[parent].append(i + 1)

    print(dfs(0) - 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
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Comparator.*;

import java.util.*;

public class E {

	public Object solve () {
		int N = sc.nextInt();
		int [] P = dec(sc.nextInts());

		int [][] E = new int [N-1][];
		for (int i : rep(N-1))
			E[i] = new int [] { P[i], i+1 };

		G = dgraph(N, E);
		int res = dfs(0);
		return res;
	}

	int [][] G;

	int dfs (int i) {
		int N = G[i].length;
		if (N == 0)
			return 0;
		else if (N == 1)
			return 1 + dfs(G[i][0]);
		else if (N == 2)
			return 1 + max(dfs(G[i][0]), dfs(G[i][1]));
		else {
			int [] H = new int [N];
			for (int j : rep(N))
				H[j] = dfs(G[i][j]);
			sort(H, reverseOrder());

			if (N == 3)
				return max(1 + H[0], 2 + H[1]);
			else {
				TreeSet<int[]> T = new TreeSet<>(by(1, 0));
				T.add(new int [] { 0, H[0] + 1 });
				T.add(new int [] { 1, H[1] + 1 });
				for (int j : rep(2, N)) {
					int [] t = T.removeFirst();
					int d = t[1] - H[t[0]] + 1;
					t[1]++; T.add(t);
					T.add(new int [] { j, H[j] + d });
				}
				return T.last()[1];
			}
		}
	}

	private static final int CONTEST_TYPE = 2;

	private static void init () {
	}

	private static Integer [] boxed (int ... A) { return stream(A).mapToObj(x -> x).toArray(Integer[]::new); }
	private static Comparator<int[]> by (int j, int ... J) {
		Comparator<int[]> res = (x, y) -> Integer.compare(x[j], y[j]);
		for (int i : J)
			res = res.thenComparing((x, y) -> Integer.compare(x[i], y[i]));
		return res;
	}
	private static int [] dec (int ... A) { for (int i = 0; i < A.length; ++i) --A[i]; return A; }
	private static int [][] dgraph (int N, int [][] E) { return dwgraph(N, E)[0]; }
	private static int [][][] dwgraph (int N, int [][] E) {
		int [] D = new int [N];
		for (int [] e : E)
			++D[e[0]];
		int [][][] res = new int [2][N][];
		for (int i = 0; i < 2; ++i)
			for (int j = 0; j < N; ++j)
				res[i][j] = new int [D[j]];
		D = new int [N];
		for (int [] e : E) {
			int x = e[0], y = e[1], z = e.length > 2 ? e[2] : 1;
			res[0][x][D[x]] = y;
			res[1][x][D[x]] = z;
			++D[x];
		}
		return res;
	}
	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	private static int [] sort (int [] A, Comparator<Integer> C) { Integer [] AA = boxed(A); Arrays.sort(AA, C); return unboxed(AA, A); }
	private static int [] unboxed (Integer [] A) { return stream(A).mapToInt(x -> x).toArray(); }
	private static int [] unboxed (Integer [] A, int [] a) { System.arraycopy(unboxed(A), 0, a, 0, A.length); return a; }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static class IOUtils {
		public static class MyScanner {
			public String next () { newLine(); return line[index++]; }
			public int nextInt () { return Integer.parseInt(next()); }
			public String nextLine () { line = null; return readLine(); }
			public String [] nextStrings () { return split(nextLine()); }
			public int [] nextInts () {
				String [] S = nextStrings(); int N = S.length;
				int [] res = new int [N];
				for (int i = 0; i < N; ++i)
					res[i] = Integer.parseInt(S[i]);
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol () { return index == line.length; }
			private String readLine () {
				try {
					int b;
					StringBuilder res = new StringBuilder();
					while ((b = System.in.read()) < ' ')
						continue;
					res.append((char)b);
					while ((b = System.in.read()) >= ' ')
						res.append((char)b);
					return res.toString();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private MyScanner () {
				try {
					while (System.in.available() == 0)
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine () {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim (String delim, Object o, Object ... A) {
			java.util.List<String> str = new java.util.ArrayList<>();
			append(str, o);
			for (Object p : A)
				append(str, p);
			return String.join(delim, str.toArray(new String [0]));
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (java.util.List<String> str, Object o) {
			append(x -> append(str, x), x -> str.add(x.toString()), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void flush () { pw.flush(); System.out.flush(); }
		private static void print (Object o, Object ... A) {
			String res = build(o, A);
			if (DEBUG == 0) {
				String out = res.split(System.lineSeparator())[0];
				if (out.length() > 80)
					out = out.substring(0, 80);
				if (out.length() < res.length())
					out += " ...";
				err(out, '(', time(), ')');
			}
			if (DEBUG == 2)
				err(res, '(', time(), ')');
			if (res.length() > 0)
				pw.println(res);
			if (DEBUG == 1)
				flush();
		}
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static int DEBUG = -1;
		private static void exit () {
			String end = "------------------" + System.lineSeparator() + time();
			switch(DEBUG) {
			case 1: print(end); break;
			case 0:
			case 2: err(end); break;
			}
			IOUtils.pw.close();
			System.out.flush();
			System.exit(0);
		}
		private static long t;
		private static long millis () { return System.currentTimeMillis(); }
		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
		private static void run (int N) {
			try { DEBUG = Integer.parseInt(System.getProperty("DEBUG")); }
			catch (Throwable t) {}

			Runnable task = new Runnable() {
				@Override
				public void run() {
					for (int n = 1; n <= N; ++n) {
						Object res = new E().solve();
						if (res != null) {
							@SuppressWarnings("all")
							Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
							print(o);
						}
					}
				}
			};
			Thread t = new Thread(null, task, "", 1L << 32);
			t.start();
			try { t.join(); } catch (Exception e) { throw new Error(e); }
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main (String[] args) {
		@SuppressWarnings("all")
		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
		init();
		IOUtils.run(N);
	}
}

**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, we need to consider scenarios that are likely to cause the algorithm to fail. The key aspects to focus on are:

1. **Tree Structure**: The tree should be structured in a way that requires careful handling of the operations to transform it into a perfect binary tree.
2. **Edge Cases**: Consider trees with few vertices, trees with a single child, and trees with multiple children.
3. **Depth Handling**: Ensure that the tree's depth is such that it requires the algorithm to handle multiple levels of recursion and comparisons.

Given the constraints and the nature of the problem, a good test case would be a tree that is nearly a perfect binary tree but with a slight deviation. This will test the algorithm's ability to handle the transformation correctly.

### Test Case
Let's create a tree with 10 vertices where the tree is nearly a perfect binary tree but with one vertex missing at a deeper level.

```plaintext
1
10
1 1 2 2 3 3 4 4 5
```

### Explanation
- **Number of Test Cases**: 1
- **Number of Vertices**: 10
- **Parent Array**: [1, 1, 2, 2, 3, 3, 4, 4, 5]

This tree structure is as follows:
```
1
├── 2
│   ├── 4
│   │   ├── 8
│   │   └── 9
│   └── 5
│       └── 10
└── 3
    ├── 6
    └── 7
```

This tree is almost a perfect binary tree of depth 3, but it is missing vertex 9's child. This should require the algorithm to handle the transformation correctly and output the minimum depth of 3.

### Output
For this test case, the expected output is:
```
3
```

This test case should challenge the algorithm to correctly handle the transformation and determine the minimum depth required.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Comparator.*;

import java.util.*;

public class E {

	public Object solve () {
		int N = sc.nextInt();
		int [] P = dec(sc.nextInts());

		int [][] E = new int [N-1][];
		for (int i : rep(N-1))
			E[i] = new int [] { P[i], i+1 };

		G = dgraph(N, E);
		int res = dfs(0);
		return res;
	}

	int [][] G;

	int dfs (int i) {
		int N = G[i].length;
		if (N == 0)
			return 0;
		else if (N == 1)
			return 1 + dfs(G[i][0]);
		else if (N == 2)
			return 1 + max(dfs(G[i][0]), dfs(G[i][1]));
		else {
			int [] H = new int [N];
			for (int j : rep(N))
				H[j] = dfs(G[i][j]);
			sort(H, reverseOrder());

			if (N == 3)
				return max(1 + H[0], 2 + H[1]);
			else {
				TreeSet<int[]> T = new TreeSet<>(by(1, 0));
				T.add(new int [] { 0, H[0] + 1 });
				T.add(new int [] { 1, H[1] + 1 });
				for (int j : rep(2, N)) {
					int [] t = T.removeFirst();
					int d = t[1] - H[t[0]] + 1;
					t[1]++; T.add(t);
					T.add(new int [] { j, H[j] + d });
				}
				return T.last()[1];
			}
		}
	}

	private static final int CONTEST_TYPE = 2;

	private static void init () {
	}

	private static Integer [] boxed (int ... A) { return stream(A).mapToObj(x -> x).toArray(Integer[]::new); }
	private static Comparator<int[]> by (int j, int ... J) {
		Comparator<int[]> res = (x, y) -> Integer.compare(x[j], y[j]);
		for (int i : J)
			res = res.thenComparing((x, y) -> Integer.compare(x[i], y[i]));
		return res;
	}
	private static int [] dec (int ... A) { for (int i = 0; i < A.length; ++i) --A[i]; return A; }
	private static int [][] dgraph (int N, int [][] E) { return dwgraph(N, E)[0]; }
	private static int [][][] dwgraph (int N, int [][] E) {
		int [] D = new int [N];
		for (int [] e : E)
			++D[e[0]];
		int [][][] res = new int [2][N][];
		for (int i = 0; i < 2; ++i)
			for (int j = 0; j < N; ++j)
				res[i][j] = new int [D[j]];
		D = new int [N];
		for (int [] e : E) {
			int x = e[0], y = e[1], z = e.length > 2 ? e[2] : 1;
			res[0][x][D[x]] = y;
			res[1][x][D[x]] = z;
			++D[x];
		}
		return res;
	}
	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	private static int [] sort (int [] A, Comparator<Integer> C) { Integer [] AA = boxed(A); Arrays.sort(AA, C); return unboxed(AA, A); }
	private static int [] unboxed (Integer [] A) { return stream(A).mapToInt(x -> x).toArray(); }
	private static int [] unboxed (Integer [] A, int [] a) { System.arraycopy(unboxed(A), 0, a, 0, A.length); return a; }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static class IOUtils {
		public static class MyScanner {
			public String next () { newLine(); return line[index++]; }
			public int nextInt () { return Integer.parseInt(next()); }
			public String nextLine () { line = null; return readLine(); }
			public String [] nextStrings () { return split(nextLine()); }
			public int [] nextInts () {
				String [] S = nextStrings(); int N = S.length;
				int [] res = new int [N];
				for (int i = 0; i < N; ++i)
					res[i] = Integer.parseInt(S[i]);
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol () { return index == line.length; }
			private String readLine () {
				try {
					int b;
					StringBuilder res = new StringBuilder();
					while ((b = System.in.read()) < ' ')
						continue;
					res.append((char)b);
					while ((b = System.in.read()) >= ' ')
						res.append((char)b);
					return res.toString();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private MyScanner () {
				try {
					while (System.in.available() == 0)
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine () {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim (String delim, Object o, Object ... A) {
			java.util.List<String> str = new java.util.ArrayList<>();
			append(str, o);
			for (Object p : A)
				append(str, p);
			return String.join(delim, str.toArray(new String [0]));
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (java.util.List<String> str, Object o) {
			append(x -> append(str, x), x -> str.add(x.toString()), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void flush () { pw.flush(); System.out.flush(); }
		private static void print (Object o, Object ... A) {
			String res = build(o, A);
			if (DEBUG == 0) {
				String out = res.split(System.lineSeparator())[0];
				if (out.length() > 80)
					out = out.substring(0, 80);
				if (out.length() < res.length())
					out += " ...";
				err(out, '(', time(), ')');
			}
			if (DEBUG == 2)
				err(res, '(', time(), ')');
			if (res.length() > 0)
				pw.println(res);
			if (DEBUG == 1)
				flush();
		}
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static int DEBUG = -1;
		private static void exit () {
			String end = "------------------" + System.lineSeparator() + time();
			switch(DEBUG) {
			case 1: print(end); break;
			case 0:
			case 2: err(end); break;
			}
			IOUtils.pw.close();
			System.out.flush();
			System.exit(0);
		}
		private static long t;
		private static long millis () { return System.currentTimeMillis(); }
		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
		private static void run (int N) {
			try { DEBUG = Integer.parseInt(System.getProperty("DEBUG")); }
			catch (Throwable t) {}

			for (int n = 1; n <= N; ++n) {
				Object res = new E().solve();
				if (res != null) {
					@SuppressWarnings("all")
					Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
					print(o);
				}
			}
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main (String[] args) {
		@SuppressWarnings("all")
		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
		init();
		IOUtils.run(N);
	}
}

**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 that can cause the algorithm to fail. Here are the steps to think through the process:

1. **Identify Edge Cases:**
   - Trees with a single child node.
   - Trees with multiple child nodes.
   - Trees with a deep structure.
   - Trees with a balanced structure.
   - Trees with unbalanced structures.

2. **Consider the Buggy Code:**
   - The code uses a depth-first search (DFS) to determine the minimum depth of Chloe's perfect binary tree.
   - It handles different cases based on the number of children a node has.
   - The logic for handling more than three children nodes seems complex and could be a source of bugs.

3. **Generate a Small-Scale Test Input:**
   - We need a test case that can trigger the complex logic in the DFS function.
   - A tree with a structure that forces the code to handle multiple children and requires careful balancing of depths.

Given these considerations, a small-scale test input that can expose the bug could be a tree with a structure that forces the code to handle multiple children and requires careful balancing of depths. Here is a test input that fits this criteria:

```plaintext
5
4
1 1 1
```

### Explanation:
- **Test Case:**
  - **Number of test cases:** 5
  - **Tree structure:** A tree with 4 nodes where node 1 is the root and has three children (nodes 2, 3, and 4).

- **Expected Output:**
  - The minimum depth of Chloe's perfect binary tree should be 2. This is because the tree can be transformed into a perfect binary tree of depth 2 by performing the operations.

This test case is designed to trigger the logic in the DFS function that handles nodes with multiple children, which is a complex part of the code and a potential source of bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
#include <stdio.h>
#include <string.h>

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        getchar();
        int a[501]={};
        int m=0;
        for (int i = 0; i < n; i++) {
            char s[5];
            scanf("%s", s);
            for (int j = 0; j < 4; j++) {
                if (s[j] == '#') {
                  a[m]=j+1;
                }
            }
            m++;
            }
        for(int i=3;i>=0;i--){
            if(a[i]!=0){
            printf("%d ",a[i]);
            }
        }

        printf("\n");
    }
    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
4
#...
.#..
..#.
...#
2
...#
...#
1
#...
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
#include<stdio.h>

int main()
{
    int t,m;
    char c;
    scanf("%d",&t);
    for(int i=0;i<t;i++){

        scanf("%d",&m);
        int a[m-1];

        for(int i1=0;i1<m;i1++){
        	char C[5];
            //for(int i2=1;i2<=4;i2++){

            scanf("%s",C);
            //}
            for(int i4=0;i4<4;i4++){
				if(C[i4]=='#'){
					a[i1]=i4+1;
					break;
				}
			}

        }
		for(int i3=m-1;i3>=0;i3--){
        printf("%d ",a[i3]);}
    }
    printf("\n");

    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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Let's analyze the code and identify potential issues:

1. **Array Indexing Issue**: The array `a` is declared with a size of `m-1`, but it should be `m` to store all `m` notes.
2. **Output Order**: The code outputs the notes in reverse order, which is incorrect according to the problem statement.

Given these issues, we can create a test case that will highlight these problems:

- **Test Case 1**: A single row with a note in the first column.
- **Test Case 2**: A single row with a note in the last column.
- **Test Case 3**: Multiple rows with notes in different columns, including the first and last columns.

Here is a small-scale test input that should expose the bugs:

```plaintext
3
1
#...
1
...#
2
#...
...#
```

Explanation:
- **Test Case 1**: A single row with a note in the first column. The correct output should be `1`.
- **Test Case 2**: A single row with a note in the last column. The correct output should be `4`.
- **Test Case 3**: Two rows with notes in the first and last columns. The correct output should be `1 4`.

This test input will help identify the issues with the array size and the output order in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
#include <iostream>
using namespace std;

int main()
{
    int t;cin>>t;
    while(t--){
      int a;
      cin>>a;
      string arr[4];
      for(int i=0;i<a;i++){
          cin>>arr[i];
      }
      int count=1;
       for(int i=a;i>0;i--){
         count=1;
         for(int j=0;j<4;j++){
           if(arr[i-1][j]=='#'){
            cout<<count<<" ";
          }
          count++;
         }

      }
      cout<<endl;

    }

}

**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 test input that exposes the bug in the given code, let's analyze the code and identify the potential issues:

### Code Analysis:
1. **Input Reading:**
   - The code reads the number of test cases and the number of rows for each test case.
   - It then reads the rows of the beatmap.

2. **Processing:**
   - The code processes the rows in reverse order (from bottom to top).
   - For each row, it iterates over the columns to find the note and prints the column number.

### Potential Issues:
1. **Resetting the `count` variable:**
   - The `count` variable is reset to 1 at the start of each row, which is correct.
   - However, the `count` variable is also reset at the start of each test case, which is unnecessary and redundant but does not cause a bug.

2. **Output Format:**
   - The code prints the column numbers for each row in reverse order, which is correct according to the problem statement.

### Edge Cases to Consider:
1. **Single Note in Each Row:**
   - Ensure that each row contains exactly one note, as specified in the problem statement.
2. **Multiple Test Cases:**
   - Test with multiple test cases to ensure the code handles them correctly.
3. **Rows with Notes in Different Columns:**
   - Test with rows where notes are in different columns to ensure the correct column numbers are printed.

### Test Input:
To create a challenging test input, we can use a small number of test cases with a mix of rows where notes are in different columns. This will ensure that the code correctly processes the rows in reverse order and prints the correct column numbers.

```plaintext
3
3
#...
..#.
...#
2
.#..
..#.
1
...#
```

### Explanation:
1. **Test Case 1:**
   - 3 rows with notes in columns 1, 3, and 4.
   - Expected output: `4 3 1`

2. **Test Case 2:**
   - 2 rows with notes in columns 2 and 3.
   - Expected output: `3 2`

3. **Test Case 3:**
   - 1 row with a note in column 4.
   - Expected output: `4`

This test input covers a variety of scenarios and should help expose any issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
def main(n):
    final_list=[]
    for _ in range(n):
        a=int(input())
        c=[]
        for __ in range(a):
            b=input()
            for i in range(len(b)):
                if b[i]=="#":
                    c.append(i+1)
        d=sorted(c,reverse=True)
        final_list.append(d)
    for i in final_list:
        for j in i:
            print(j,end=" ")
        print()
main(int(input()))

**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 order of processing the notes is critical. The code sorts the columns in reverse order, which is incorrect. We need to ensure that the notes are processed from the bottom row to the top row without altering their order.

### Step-by-Step Reasoning:
1. **Understanding the Bug**: The code sorts the columns in reverse order, which means it will output the columns in descending order instead of the order in which they appear from bottom to top.
2. **Edge Cases**:
   - A single row with a note in any column.
   - Multiple rows with notes in different columns.
   - Multiple rows with notes in the same column.
3. **Small-Scale Test Input**:
   - We can create a test case with a few rows to clearly demonstrate the issue.
   - For example, a test case with 3 rows where the notes are in columns 1, 2, and 3 from the bottom to the top.

### Test Input:
```plaintext
1
3
...#
..#.
.#..
```

### Explanation:
- The first line is `1`, indicating there is one test case.
- The second line is `3`, indicating there are 3 rows in the beatmap.
- The following 3 lines represent the rows from the top to the bottom:
  - `...#` (note in column 4)
  - `..#. (note in column 3)
  - `.#..` (note in column 2)

### Expected Output:
The correct output should be:
```
4 3 2
```

### Why This Test Input Exposes the Bug:
- The code will process the rows from the top to the bottom, but it will sort the columns in reverse order.
- The output will be `2 3 4` instead of `4 3 2`, clearly showing the bug.

### Conclusion:
This test input is small, complete, and valid, and it effectively demonstrates the issue with the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
t=int(input())
lst=[]
for _ in range(t):
    lst1=[]
    lst2=[]
    n=int(input())
    for i in range(n):
        k=str(input())
        lst1.append(k)
    for i in lst1:
        dup=list(i)
        if '#' in i:
            lst2.append(dup.index('#')+1)
    lst.append(lst2)
for i in lst:
    i.sort(reverse=True)
    for j in i:
        print(j,'',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 test input that exposes the bug in the provided buggy code, we need to focus on the logic that processes the notes in the beatmap. The key issue in the buggy code is that it sorts the columns in reverse order, which is not required by the problem statement. The correct output should be the columns in the order they appear from the bottom to the top.

### Step-by-Step Analysis:
1. **Identify the Bug:**
   - The buggy code sorts the columns in reverse order before printing, which is incorrect. The problem requires the columns to be printed in the order they appear from the bottom to the top.

2. **Edge Cases to Consider:**
   - **Single Note in Each Row:** Ensure that the test case includes notes in different columns to check if the sorting is affecting the output.
   - **Multiple Test Cases:** Include multiple test cases to ensure the sorting is not affecting the output across different test cases.
   - **Notes in All Columns:** Ensure that notes appear in all columns to test the sorting logic.

3. **Construct the Test Input:**
   - **Test Case 1:** A small test case with notes in all columns.
   - **Test Case 2:** A test case with notes in the same column to check if the sorting logic is affecting the output.

### Test Input:
```plaintext
2
4
#...
.#..
..#.
...#
4
...#
..#.
.#..
#...
```

### Explanation:
- **Test Case 1:**
  - 4 rows with notes in columns 1, 2, 3, and 4 respectively.
  - The correct output should be `4 3 2 1`.

- **Test Case 2:**
  - 4 rows with notes in columns 4, 3, 2, and 1 respectively.
  - The correct output should be `1 2 3 4`.

This test input will expose the bug in the buggy code because the sorting in reverse order will produce incorrect results for both test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
t=int(input())
for i in range(t):
    a=int(input())
    l1=[]
    for i in range(a):
        s=input()
        l=list(s)
        ans=l.index('#')
        l1.append(ans+1)
    l1.sort(reverse=True)
    print(' '.join(map(str,l1)))

**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
#...
#...
1
#...
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

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

public class CodeForces {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.next();
            }
            ArrayList<Integer> res = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < 4; j++) {
                    if (arr[i].charAt(j) == '#') {
                        res.add(j + 1);
                    }
                }
            }
            Collections.sort(res, Collections.reverseOrder());
            for(int i : res){
                System.out.print(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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the issue in the code. The main problem in the buggy code is that it sorts the results in reverse order, which is incorrect for the problem's requirements. The correct output should reflect the order of the notes from the bottom to the top of the beatmap.

To expose this bug, we can create a test case where the notes are in a specific order that, when sorted in reverse, will produce an incorrect result. Here's a step-by-step approach to create such a test case:

1. **Single Test Case with Notes in Ascending Column Order:**
   - If the notes are in columns 1, 2, 3, and 4 from the bottom to the top, the correct output should be `1 2 3 4`.
   - The buggy code will sort these in reverse order, resulting in `4 3 2 1`, which is incorrect.

2. **Multiple Test Cases:**
   - We can include multiple test cases to ensure the bug is consistently exposed.
   - For example, one test case with notes in ascending column order and another with notes in descending column order.

Here is the test input:

```plaintext
2
4
...#
..#.
.#..
#...
4
#...
.#..
..#.
...#
```

Explanation:
- The first test case has notes in columns 4, 3, 2, and 1 from the bottom to the top. The correct output should be `4 3 2 1`.
- The second test case has notes in columns 1, 2, 3, and 4 from the bottom to the top. The correct output should be `1 2 3 4`.

The buggy code will sort the results in reverse order for both test cases, producing incorrect outputs. This will clearly expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int t = in.nextInt();
        int xx[][]=new int[t][4];
        for (int i = 0; i <t ; i++) {
            int x=in.nextInt();
            int n=0;
            for (int j = 0; j <x ; j++) {
                String s=in.next();
                int nextInt = s.indexOf('#');
                xx[i][n]=nextInt+1;
                n++;
            }
        }
        for (int i = 0; i <t ; i++) {
            for (int j = 3; j>=0 ; j--) {
                if (xx[i][j]!=0)
                    System.out.print(xx[i][j]+" ");
            }
            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
3
4
#...
.#..
..#.
...#
2
.#..
...#
1
...#
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template<typename T>
using ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define int long long
#define el '\n'

const int N = 1e3 + 5, mod = 1e9 + 7;

int dx[] = {0, 0, -1, 1, -1, 1, -1, 1};
int dy[] = {-1, 1, 0, 0, 1, -1, -1, 1};

char a[N][4];
void setAnswer(int Case)
{
     int n;
     cin >> n;
     vector<int> ans;
     for (int i = 0; i < n; i++)
     {
           for (int j = 0; j < 4; j++)
           {
               cin >> a[i][j];
           }

     }
      for (int i = 4; i >= 0; i--)
      {
           for (int j = n - 1; j >= 0; j--)
           {
               if(a[j][i] == '#') ans.push_back(i + 1);
           }

      }

    //  reverse(ans.begin(),ans.end());
     for(auto i : ans) cout << i << " ";
     cout << el;

}

signed main()
{

     ios_base::sync_with_stdio(0);
     cin.tie(0);
     cout.tie(0);

	 #ifndef ONLINE_JUDGE
      freopen("i.txt", "r", stdin);
      freopen("o.txt", "w", stdout);
     #endif
     int testCases = 1;
     cin >> testCases;
     for (int i = 1; i <= testCases; i++)
        setAnswer(i);

    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 provided buggy code, we need to focus on the logic that processes the notes. The main issue in the code is the nested loop that iterates over the rows and columns in the wrong order. Specifically, the outer loop should iterate over the columns, and the inner loop should iterate over the rows from bottom to top. However, the current code has the outer loop iterating over the columns incorrectly, and the inner loop is not correctly processing the rows from bottom to top.

Here’s a step-by-step approach to create a test input that exposes this bug:

1. **Single Test Case with a Single Row:**
   - This should be straightforward and should not expose the bug, but it's a good starting point to ensure basic functionality.

2. **Single Test Case with Multiple Rows:**
   - Ensure that the notes are placed in different columns to see if the code correctly identifies the column for each note.

3. **Multiple Test Cases with Varying Row Counts:**
   - This will help ensure that the code handles multiple test cases correctly.

4. **Edge Case: All Notes in the Same Column:**
   - This can help identify if the code correctly processes notes in the same column but in different rows.

5. **Edge Case: Notes in Reverse Order:**
   - Place notes in the reverse order to ensure the code processes them from bottom to top correctly.

Given these considerations, here is a small-scale test input that should expose the bug:

```plaintext
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
```

This test input includes:
- A test case with 4 rows, each with a note in a different column.
- A test case with 2 rows, both with notes in the same column.
- A test case with 1 row, with a note in the last column.

This input should help identify the issue with the nested loops in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
#include <bits/stdc++.h>
// #include <vector>

using namespace std;

int main() {
  int t; cin>>t;
  while(t--){
    int n; cin>>n; vector<int> kq(4);
    vector<vector<char>> m(4,vector<char>(n));
    for(int i=0;i<n;i++){
        cin>>m[i][0]>>m[i][1]>>m[i][2]>>m[i][3];
        for(int j=0;j<4;j++){
          if(m[i][j]=='#') kq[i]=j+1;

    }

  } reverse(kq.begin(),kq.end());
  for(auto x:kq){ if(x!=0){cout<<x<<" ";}}
    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:
To expose the bug in the given buggy code, we need to focus on the following issues:

1. **Incorrect Storage of Results**: The code uses `kq` as a vector of size 4, but it should be a vector of size `n` to store the results for each row.
2. **Incorrect Input Reading**: The code reads the input incorrectly by using `cin>>m[i][0]>>m[i][1]>>m[i][2]>>m[i][3]`, which will not read the input as a string and store it in the matrix correctly.
3. **Reversing the Results**: The code reverses the results, but it should not reverse them since the notes are processed from bottom to top.

Given these issues, we can create a test input that will highlight these problems. Specifically, we can create a test case where the number of rows `n` is greater than 4, and the notes are placed in such a way that the incorrect storage and input reading will be evident.

Here is a small-scale test input that should expose the bug:

```plaintext
1
5
#...
...#
..#.
.#..
...#
```

### Explanation:
- **Test Case 1**: There is one test case with 5 rows.
- **Rows**: The rows are structured such that the notes are placed in different columns to ensure the code correctly processes them from bottom to top.
- **Expected Output**: The correct output should be `4 1 3 2 1`, but the buggy code will likely produce an incorrect result due to the issues mentioned.

This test input should help in identifying the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
def find_columns(t, test):
    res = []
    for i in test:
        n = i[0]
        r = i[1:]
        c = []
        for r in reversed(r):
            c.append(r.index('#') + 1)
        res.append(" ".join(map(str, c)))
    return res
t = 3
test = [
    (4, "#...", ".#..", "..#.", "...#"),
    (2, ".#..", ".#.."),
    (1, "...#")
]

res = find_columns(t, test)
for result in res:
    print(result)

**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 that can maximize the failure probability. The buggy code processes the rows in reverse order and appends the column indices to the list. However, it does not handle the case where the number of rows is very small or where the notes are in the same column across multiple rows.

Let's consider the following edge cases:
1. A single row with a note in each column.
2. Multiple rows with notes in the same column.
3. A large number of rows with notes in a specific pattern.

Given the constraints and the need for a small-scale test input, we can create a test input with the following characteristics:
- A test case with a single row.
- A test case with multiple rows where notes are in the same column.
- A test case with a small number of rows where notes are in different columns.

Here is the test input:

```plaintext
3
1
#...
2
#...
#...
4
#...
.#..
..#.
...#
```

Explanation:
1. The first test case has a single row with a note in the first column.
2. The second test case has two rows with notes in the first column.
3. The third test case has four rows with notes in different columns, processed from bottom to top.

This test input should help expose the bug in the buggy code by ensuring that the code correctly handles notes in the same column and different columns across multiple rows.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    a = int(input())
    l = []
    for _ in range(a):
        row = input().strip()
        for i in range(len(row)):
            if row[i] == '#':
                l.append(i + 1)
                break
    l.sort()
    print(' '.join(map(str, l[::-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
4
#...
.#..
..#.
...#
3
...#
..#.
.#..
1
...#
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: osu!mania
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are playing your favorite rhythm game, osu!mania. The layout of your beatmap consists of $$$n$$$ rows and $$$4$$$ columns. Because notes at the bottom are closer, you will process the bottommost row first and the topmost row last. Each row will contain exactly one note, represented as a '#'.

For each note $$$1, 2, \dots, n$$$, in the order of processing, output the column in which the note appears.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

For each test case, the first line contains $$$n$$$ ($$$1 \leq n \leq 500$$$) — the number of rows of the beatmap.

The following $$$n$$$ lines contain $$$4$$$ characters. The $$$i$$$-th line represents the $$$i$$$-th row of the beatmap from the top. It is guaranteed that the characters are either '.' or '#', and exactly one of the characters is '#'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.

Output: For each test case, output $$$n$$$ integers on a new line, the column that the $$$i$$$-th note appears in for all $$$i$$$ from $$$1$$$ to $$$n$$$.

Examples:
input:
3
4
#...
.#..
..#.
...#
2
.#..
.#..
1
...#
output:
4 3 2 1
2 2
4

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class Solution {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      sc.nextLine();
      while (n > 0) {
        int a = sc.nextInt();
        sc.nextLine();

        if (a == 1) {
           String c = sc.nextLine();
           int len = c.length();
           int[] ans = new int[len];
           for (int i = 0; i < len; i++) {
              if (c.charAt(i) == '#') {
                ans[i] = i + 1;
              }
           }
           for (int i = len - 1; i >= 0; i--) {
              if (ans[i] != 0) {
                System.out.print(ans[i] + " ");
              }
           }
           System.out.println();
        } else{
           char [][]arr=new char[a][a];
           int []b=new int[a];
           for (int i = 0; i <a; i++) {
              String row = sc.next();
              for (int j =0; j < a; j++) {
                arr[i][j] = row.charAt(j);
              }
           }
           for (int i=0;i<a;i++)
           {
              for (int j=0;j<a;j++)
              {
                if(arr[i][j]=='#')
                {
                   b[i]=j;
                }
              }
           }
           for(int i=a-1;i>=0;i--){
              b[i]=b[i]+1;
              System.out.print(b[i]+" ");
           }
           System.out.println();
        }
        n-=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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Input Parsing Issues**:
   - The code reads the number of test cases and then processes each test case.
   - The code uses `sc.nextLine()` after reading the number of rows, which can cause issues if the input format is not as expected.

2. **Array Initialization and Usage**:
   - The code initializes a 2D array `arr` with dimensions `[a][a]` instead of `[a][4]`.
   - The code uses a 1D array `b` to store the column indices, but it incorrectly increments the values by 1 in the wrong loop.

3. **Output Format**:
   - The code prints the column indices in reverse order, which is correct according to the problem statement.

Given these issues, a test input that can expose these bugs should:
- Have a test case with a single row to check the single-row handling.
- Have a test case with multiple rows to check the multi-row handling.
- Ensure that the input format is strictly followed to avoid any parsing issues.

Here is a small-scale test input that can expose the bugs:

```plaintext
3
1
...#
4
#...
.#..
..#.
...#
2
...#
...#
```

This test input includes:
- A test case with a single row to check the single-row handling.
- A test case with multiple rows to check the multi-row handling.
- A test case with multiple rows where all notes are in the same column to check the handling of such cases.

This input should help identify issues in the buggy code, particularly with the array initialization and usage.