**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
from math import floor, ceil

output = []

t = int(input())

for _ in range(t):
    a, b, c = map(int, input().split())
    if a + 1 != c:
        output.append(-1)
        continue
    flag = ((1 + 8 * a)**0.5 - 1)/2
    min_depth, max_depth = floor(flag), ceil(flag)
    max_depth_count = 2 * (a - min_depth * (min_depth + 1) // 2)
    min_depth_count = a + 1 - max_depth_count

    q = b // c
    r = b - q
    min_added_depth, max_added_depth = q, q + int(r > 0)
    min_added_depth_count, max_added_depth_count = c - r, r

    if max_added_depth_count <= min_depth_count:
        out = max(
            max_added_depth + min_depth,
            min_added_depth + max_depth,
        )
    else:
        out = max_added_depth + max_depth

    output.append(out)

for line in output:
    print(line)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issues in the buggy code are:

1. **Incorrect condition for checking the feasibility of the tree**:
   - The condition `if a + 1 != c:` is incorrect. It should check if the total number of vertices and edges form a valid tree structure.

2. **Incorrect handling of the minimum and maximum depths**:
   - The calculation of `min_depth` and `max_depth` and their counts might not be accurate.

3. **Incorrect handling of the additional depth from vertices with one child**:
   - The way `b` (vertices with one child) is handled might not be correct.

Let's generate a test input that targets these issues:

1. **Feasibility check**:
   - Test cases where `a + 1 != c` but the tree is still valid.
   - Test cases where `a + 1 == c` but the tree is not valid.

2. **Depth calculation**:
   - Test cases with varying values of `a`, `b`, and `c` to ensure the depth calculation is correct.

3. **Handling of `b`**:
   - Test cases where `b` is significant and can affect the depth.

Here is a small-scale test input that covers these scenarios:

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

Explanation:
1. **Test Case 1: 2 1 3**
   - This is the example given in the problem statement. The tree is valid, and the height should be 2.
2. **Test Case 2: 1 2 1**
   - This is a valid tree with 1 vertex with 2 children, 2 vertices with 1 child, and 1 leaf. The height should be 2.
3. **Test Case 3: 3 0 2**
   - This is a valid tree with 3 vertices with 2 children and 2 leaves. The height should be 3.
4. **Test Case 4: 0 3 1**
   - This is a valid tree with 3 vertices with 1 child and 1 leaf. The height should be 3.
5. **Test Case 5: 4 1 2**
   - This is a valid tree with 4 vertices with 2 children, 1 vertex with 1 child, and 2 leaves. The height should be 4.

These test cases should help expose the bugs in 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: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

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

public class Main {
    static int inf = (int) 2e7;

    public static void main(String[] args) throws IOException {
        int t = f.nextInt();
        while (t > 0) {
            solve();
            t--;
        }

        w.flush();
        w.close();
        br.close();
    }

    public static void solve() {
        int a=f.nextInt();
        int b=f.nextInt();
        int c=f.nextInt();
        if (a == c-1){
            int cur = 1;
            int m = 0;
            int deep = 0;
            while (m < a){
                m += cur;
                cur *= 2;
                deep++;
            }
            if (m != a) b-= (cur/2 - (a - (m - cur / 2)));
            while (b > 0) {b -= Math.max(cur/2,1);deep++;}
            w.println(deep);
        }else w.println(-1);
    }

    static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));
    static Input f = new Input(System.in);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    static class Input {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public Input(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public String nextLine() {
            String str = null;
            try {
                str = reader.readLine();
            } catch (IOException e) {
                // TODO 自动生成的 catch 块
                e.printStackTrace();
            }
            return str;
        }

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

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

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

        public BigInteger nextBigInteger() {
            return new BigInteger(next());
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that might cause the code to fail. The code appears to handle certain cases correctly but might have issues with specific conditions or edge cases. Here are some potential scenarios to consider:

1. **Cases where the tree cannot exist**: The code should return `-1` if no such tree can be formed.
2. **Cases where the tree can exist but the height calculation is incorrect**.
3. **Cases where the values of `a`, `b`, and `c` are such that the logic in the code fails to handle them correctly**.

Let's analyze the code to identify potential issues:

- The code checks if `a == c - 1`. If this condition is true, it proceeds to calculate the height.
- If `a != c - 1`, it directly prints `-1`.

### Potential Issues:
1. **Incorrect handling of `a == c - 1`**:
   - The code might not correctly handle cases where `a` and `c` are such that the tree can be formed, but the height calculation is incorrect.
2. **Incorrect handling of `a != c - 1`**:
   - The code might not correctly identify cases where the tree cannot be formed, leading to incorrect `-1` outputs.

### Test Cases to Consider:
1. **Case where the tree cannot exist**:
   - Example: `a = 1`, `b = 0`, `c = 0` (No tree can be formed with 1 vertex having 2 children and no other vertices).
2. **Case where the tree can exist but the height calculation is incorrect**:
   - Example: `a = 2`, `b = 1`, `c = 3` (The correct height is 2, but the code might fail to handle this correctly).
3. **Case where `a == c - 1` but the height calculation is incorrect**:
   - Example: `a = 3`, `b = 1`, `c = 4` (The correct height is 3, but the code might fail to handle this correctly).

### Chosen Test Input:
```plaintext
3
1 0 0
2 1 3
3 1 4
```

### Explanation:
1. **Test Case 1**: `a = 1`, `b = 0`, `c = 0`
   - No tree can be formed with 1 vertex having 2 children and no other vertices. The expected output is `-1`.
2. **Test Case 2**: `a = 2`, `b = 1`, `c = 3`
   - The correct height is 2. The code should correctly handle this case.
3. **Test Case 3**: `a = 3`, `b = 1`, `c = 4`
   - The correct height is 3. The code should correctly handle this case, but it might fail if the logic for height calculation is incorrect.

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

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

    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    //    public static void main(String[] args) {
//        MyScanner myScanner = new MyScanner();
//        out = new PrintWriter(new BufferedOutputStream(System.out));
//        int n = myScanner.nextInt();
//        int q = myScanner.nextInt();
//        int[] a = new int[n];
//        for (int i = 0; i < n; i++) {
//            a[i] = myScanner.nextInt();
//        }
//        int[] newa = new int[a.length + 1];
//        System.arraycopy(a, 0, newa, 1, a.length);
//        for (int i = 0; i < q; i++) {
//            int l = myScanner.nextInt();
//            int r = myScanner.nextInt();
//            long res = 0;
//            long max = a[l - 1];
//            for (int j = l; j <= r; j++) {
//                max = Math.max(a[j - 1], max);
//                res += max * a[j - 1];
//            }
//            out.println(res);
//        }
//
//        out.close();
//    }
    public static void main(String[] args) {
        MyScanner myScanner = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = myScanner.nextInt();
        while (t-- != 0) {
            int a = myScanner.nextInt();
            int b = myScanner.nextInt();
            int c = myScanner.nextInt();
            if (c == 0 && (a != 0 || b != 0)) {
                out.println(-1);
                continue;
            } else if (c == 0) {
                out.println(0);
                continue;
            }
            if (c == 1) {
                if (a != 0) {
                    out.println(-1);
                } else {
                    out.println(b);
                }
                continue;
            }
            int tempc = c;
            int cnt1 = 0;
            boolean odd = false;
            int floor = 0;
            int oddfloor = -1;
            while (tempc != 1) {
                if (tempc % 2 == 1) {
                    odd = true;
                    oddfloor = floor;
                }
                cnt1 += tempc / 2;
                tempc /= 2;
                floor++;
            }
            int cansu = 0;
            if (oddfloor != -1) {
                cansu = floor - oddfloor;
            }
            if (odd) {
                cnt1++;
            }
            if (cnt1 != a) {
                out.println(-1);
                continue;
            }
//            if (c % 2 == 1) {
//                c--;
//            }
            if (b <= cansu) {
                out.println(floor + (odd ? 1 : 0));
            } else {
                b -= cansu;
                int addfloor = (int) Math.ceil((double) b / c);
                out.println(floor + addfloor + (odd ? 1 : 0));
            }

        }
        out.close();
    }

}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that challenge the logic of the algorithm. The code handles several specific conditions, but there are a few areas where it might fail:

1. **Handling of `c == 0` and `a != 0` or `b != 0`**: The code correctly returns `-1` in these cases.
2. **Handling of `c == 1` and `a != 0`**: The code correctly returns `-1` in these cases.
3. **Calculation of `cnt1` and `cansu`**: The code might have issues with the logic for calculating the number of levels and the number of nodes that can be used to balance the tree.
4. **Handling of `b` nodes**: The code might not correctly handle the distribution of `b` nodes across the levels.

Given these points, a small-scale test input that can expose potential bugs is:

```plaintext
3
1 0 2
1 1 2
2 1 3
```

### Explanation:
1. **Test Case 1: `1 0 2`**
   - This case has 1 vertex with 2 children and 2 leaves.
   - The expected output is `1` because the tree can be formed with a single level.

2. **Test Case 2: `1 1 2`**
   - This case has 1 vertex with 2 children, 1 vertex with 1 child, and 2 leaves.
   - The expected output is `2` because the tree can be formed with 2 levels.

3. **Test Case 3: `2 1 3`**
   - This case has 2 vertices with 2 children, 1 vertex with 1 child, and 3 leaves.
   - The expected output is `2` because the tree can be formed with 2 levels.

These test cases cover different combinations of `a`, `b`, and `c` and should help identify any issues in the logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#pragma GCC optimize("unroll-loops,O3")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define cok cout << (ok ? "YES\n" : "NO\n");
#define dbg(x) cout << (#x) << ": " << x << endl;
#define dbga(x,l,r) cout << (#x) << ": "; for (int ii=l;ii<r;ii++) cout << x[ii] << " "; cout << endl;
#define int long long
#define pi pair<int, int>
const int N = 1e6+9, INF = 2e9;
int n;
vector<int> vec[N];
int tin[N], t, tout[N], atin[N];
int p[N], h[N];
vector<int> L;
void dfs(int v, int p = -1) {
    tin[v] = t++;
    if (vec[v].size() == 1 && v) L.pb(v);
    for (int u: vec[v]) {
        if (u == p) continue;
        ::p[u] = v;
        h[u] = h[v] + 1;
        dfs(u, v);
    }
    tout[v] = t;
}
pi d[N];
int pd[N], kol;
set<pi> st;
pi find_deepest() {
    pi res = {0, -1};
    for (int l: L) {
        int cnt = 0;
        for (int v = l; v; v = p[v]) {
            cnt += !st.count({p[v], v});
        }
        res = max(res, pi{cnt, l});
    }
    return res;
}
void add(int lx, int rx, int v, int l, int r, int x) {
    if (l >= rx || lx >= r) return;
    if (l <= lx && rx <= r) {
        d[v].fi += x;
        pd[v] += x;
        return;
    }
    int m = lx + rx >> 1;
    add(lx, m, 2 * v + 1, l, r, x);
    add(m, rx, 2 * v + 2, l, r, x);
    d[v] = max(d[2 * v + 1], d[2 * v + 2]);
    d[v].fi += pd[v];
}
int perform_operation() {
    auto [val, v] = d[0];
    v = atin[v];
    val += pd[0];
    if (val == 0) return 0;
    int cnt = 0;
    for (int u = v; u; u = p[u]) {
        if (st.count({p[u], u})) break;
        // dbg(u + 1 << " " << p[u] + 1);
        cnt++;
    }
    // dbg(val);
    // dbg(cnt);
    // dbg(v);
    int cur = 0;
    for (int u = v; u; u = p[u]) {
        cur++;
        if (st.count({p[u], u})) break;
        st.insert({p[u], u});
        // do subtree pu
        add(0, kol, 0, tin[p[u]], tout[p[u]], cur - cnt);
        add(0, kol, 0, tin[u], tout[u], cnt - cur);
    }
    add(0, kol, 0, tin[v], tin[v] + 1, -INF);
    return cnt;
}
void solve() {
    cin >> n;
    t = 0;
    // memset(tin, 0, sizeof(tin));
    // memset(tout, 0, sizeof(tout));
    memset(atin, 0, sizeof(atin));
    // memset(tin, 0, sizeof(tin));
    // memset(d, 0, sizeof(d));
    // memset(tin, 0, sizeof(d));
    st.clear();
    L.clear();
    for (int i = 0; i < n; i++) vec[i].clear(), h[i] = 0;
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        vec[a].pb(b);
        vec[b].pb(a);
    }
    dfs(0);
    for (int i = 0; i < n; i++) atin[tin[i]] = i;
    kol = 1;
    while (kol < n) kol *= 2;
    for (int i = kol - 1; i < 2 * kol - 1; i++) d[i] = {-INF, i - kol + 1}, pd[i] = 0;
    for (int l: L) d[kol - 1 + tin[l]].fi = h[l];
    for (int i = kol - 2; i >= 0; i--) d[i] = max(d[2 * i + 1], d[2 * i + 2]), pd[i] = 0;
    // dbg(d[0].fi);
    int res = n - 1;
    for (int k = 1; k <= n - 1; k++) {
        res -= perform_operation();
        if (k != 1) {
            res -= perform_operation();
        }
/*        auto [depth, v] = find_deepest();
        res -= depth;
        for (int u = v; u; u = p[u]) st.insert({p[u], u});
        if (k != 1) {
            auto [depth, v] = find_deepest();
            res -= depth;
            for (int u = v; u; u = p[u]) st.insert({p[u], u});
        }*/
        cout << res << " ";
    }
    for (int i = 0; i < n; i++) atin[i] = 0;
}
signed main() {
    cin.tie(0); ios_base::sync_with_stdio(0);
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
        cout << "\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:
```plaintext
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
//Crimson flames tied through my ears
//Rollin' high and mighty traps
//Pounced with fire on flaming roads
//Using ideas as my maps
//"We'll meet on edges, soon," said I
//Proud 'neath heated brow
//Ah, but I was so much older then
//I'm younger than that now.
//
//Half-cracked prejudice leaped forth
//"Rip down all hate," I screamed
//Lies that life is black and white
//Spoke from my skull, I dreamed
//Romantic facts of musketeers
//Foundationed deep, somehow
//Ah, but I was so much older then
//I'm younger than that now.
//
//Girls' faces formed the forward path
//From phony jealousy
//To memorizing politics
//Of ancient history
//Flung down by corpse evangelists
//Unthought of, thought, somehow
//Ah, but I was so much older then
//I'm younger than that now.
//A self-ordained professor's tongue
//Too serious to fool
//Spouted out that liberty
//Is just equality in school
//"Equality," I spoke the word
//As if a wedding vow
//Ah, but I was so much older then
//I'm younger than that now.
//
//In a soldier's stance, I aimed my hand
//At the mongrel dogs who teach
//Fearing not that I'd become my enemy
//In the instant that I preach
//My existence led by confusion boats
//Mutiny from stern to bow
//Ah, but I was so much older then
//I'm younger than that now.
//
//Yes, my guard stood hard when abstract threats
//Too noble to neglect
//Deceived me into thinking
//I had something to protect
//Good and bad, I define these terms
//Quite clear, no doubt, somehow
//Ah, but I was so much older then
//I'm younger than that now.

#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
using namespace std;
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
/*=====================================================================*/
#define ll long long
#define int ll
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pii>
#define pdd pair<double,double>
#define ull unsigned long long
#define pb push_back
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define all(s) (s).begin(),(s).end()
#define repd(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define forr(i,a,b,c) for(int i=(int)(a);i<=(int)(b);i+=(int)(c))
#define forn(i,p,n) for(int i=(int)(p);i<=(int)(n);++i)
#define ford(i,p,n) for(int i=(int)(n);i>=(int)(p);--i)
#define foreach(i,c) for(__typeof(c.begin())i=(c.begin());i!=(c).end();++i)
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define modcg(x) if(x>=mod)x-=mod;
#define modcl(x) if(x<0)x+=mod;
#define lowbit(x) ((x)&(-(x)))
#define ckmax(x,y) x=max(x,y)
#define ckmin(x,y) x=min(x,y)
/*=====================================================================*/
mt19937 GeN(chrono::system_clock::now().time_since_epoch().count());
int Rand(int l,int r)
{
	uniform_int_distribution<>RAND1(l,r);
	return RAND1(GeN);
}
struct Fastmod
{
	int mod,b;
	typedef __int128 lll;
	void init(int m)
	{
		mod=m;
		b=((lll)1<<64)/mod;
	}
	int operator()(ull a)
	{
		int q=((lll)a*b)>>64,r=a-q*mod;
		modcg(r);
		return r;
	}
}MOD;
int mul(int a,int b)
{
	return MOD(a*b);
}
/*======================================================================*/
//think twice,code once
//think once,debug forever
const int MAXN=3e5+10;
int n;
vi v[MAXN];
int dep[MAXN],son[MAXN];
int fa[MAXN];
void dfs1(int now)
{
	dep[now]=0;
	for(int u:v[now])
	{
		if(u==fa[now])
		{
			continue;
		}
		fa[u]=now;
		dfs1(u);
		if(dep[u]+1>dep[now])
		{
			dep[now]=dep[u]+1;
			son[now]=u;
		}
	}
//	dep[now]++;
}
int top[MAXN];
vi g[MAXN];
void dfs2(int now)
{
	if(!son[now])
	{
		return;
	}
	top[son[now]]=top[now];
	dfs2(son[now]);
	for(int u:v[now])
	{
		if(u==fa[now]||u==son[now])
		{
			continue;
		}
		dfs2(top[u]=u);
	}
}
int mx[MAXN];
priority_queue<pair<pii,int>> pq;
void add(int x)
{
	for(int u:g[x])
	{
		pq.push({{dep[u],mx[u]},u});
	}
}
void solve()
{
	cin>>n;
	forn(i,1,n)
	{
		v[i].clear();
		g[i].clear();
		mx[i]=top[i]=dep[i]=son[i]=0;
	}
	forn(i,1,n-1)
	{
		int x,y;
		cin>>x>>y;
		v[x].pb(y);
		v[y].pb(x);
	}
	dfs1(1);
//	forn(i,1,n)
//	{
//		cout<<son[i]<<" ";
//	}
//	cout<<endl;
	top[1]=1;
	dfs2(1);
	forn(i,2,n)
	{
		if(i==top[i])
		{
			g[top[fa[i]]].pb(i);
			mx[top[fa[i]]]=max(mx[top[fa[i]]],dep[i]);
		}
	}
//	cout<<"FFF"<<endl;
//	forn(i,1,n)
//	{
//		for(int u:g[i])
//		{
//			cout<<i<<" "<<u<<endl;
//		}
//	}
//	forn(i,1,n)
//	{
//		cout<<dep[i]<<" ";
//	}
//	cout<<endl;
	int ans=n-1;ans-=dep[1];
	add(1);
	cout<<ans<<" ";
	forn(i,1,n-2)
	{
		if(pq.empty())
		{
			cout<<ans<<" ";
			continue;
		}
		int x=0,y=0;
		ans-=pq.top().first.first+1;x=pq.top().second;pq.pop();
		if(pq.size())
		{
			ans-=pq.top().first.first+1;y=pq.top().second;pq.pop();
		}
		add(x);add(y);
		cout<<ans<<" ";
	}
	cout<<endl;
}
/*======================================================================*/
signed main()
{
    cin.tie(0);
    cout.tie(0);
	std::ios::sync_with_stdio(false);
//#define Hank2007
#ifdef Hank2007
	freopen("input.txt","r",stdin);
	freopen("output1.txt","w",stdout);
#endif
  	/*====================================================================*/
  	int TEST_CASE=1;
	cin>>TEST_CASE;
  	while(TEST_CASE--)
  	{
  		solve();
  	}
  	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to consider:

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the DFS and the priority queue operations.
2. **Root with One Child**: Ensure the root (vertex 1) has exactly one child.
3. **Bridges and Subtrees**: Create a tree where the addition of edges can significantly alter the number of bridges, especially in the subtrees of the lower vertices of bridges.
4. **Small Scale**: The test input should be small enough to be manageable but complex enough to expose bugs.

Given these considerations, let's construct a test case:

### Test Case
- **Tree Structure**: A linear chain of vertices with a single branch.
- **Root with One Child**: Vertex 1 has exactly one child (vertex 2).
- **Bridges and Subtrees**: The tree will have a structure where adding edges can significantly reduce the number of bridges.

### Example Input
```plaintext
1
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
```

### Explanation
- **Tree Structure**: The tree is a linear chain from vertex 1 to vertex 10.
- **Root with One Child**: Vertex 1 has exactly one child (vertex 2).
- **Bridges and Subtrees**: Adding edges in this structure can significantly reduce the number of bridges, especially when edges are added between vertices that are far apart.

This test case is small and simple but can expose issues in the priority queue and DFS operations, particularly in how the code handles the addition of edges and the calculation of the number of bridges.

### Expected Output
The expected output for this test case should be:
```plaintext
8 6 4 2 0 0 0 0 0
```

This output reflects the smallest number of bridges left after adding 1 to 9 edges to the tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
int t,n,u,v,m,son[400000],hi[400000],f[400000],tot,T;
struct node{
	int u;
}p1,p2;
vector<int>e[400000];
bool operator<(node x,node y){
	return hi[x.u]<hi[y.u];
}
priority_queue<node>q;
void Dfs(int u,int fa){
	f[u]=fa;
	hi[u]=1;
	for(int v:e[u]){
		if(v==fa)continue;
		Dfs(v,u);
		if(hi[v]>hi[son[u]])son[u]=v;
		hi[u]=max(hi[u],hi[v]+1);
	}
}
void dfs(int u){
	for(int v:e[u]){
		if(v==f[u]||v==son[u])continue;
		q.push(node{v});
	}
	if(son[u])dfs(son[u]);
}
int main(){
	scanf("%d",&t);
	while(t--){
		T++;
		while(!q.empty())q.pop();
		scanf("%d",&n);
		for(int i=1;i<=n;i++)e[i].clear();
		for(int i=1;i<n;i++){
			scanf("%d%d",&u,&v);
			if(T==932)cout<<u;
			e[u].push_back(v);
			e[v].push_back(u);
		}
		if(T==932)cout<<endl;
		for(int i=1;i<=n;i++)son[i]=hi[i]=f[i]=0;
		//cout<<'a';
		Dfs(1,0);
		/*for(int i=1;i<=n;i++)cout<<son[i]<<' ';
		cout<<endl;
		for(int i=1;i<=n;i++)cout<<hi[i]<<' ';
		cout<<endl;*/
		m=n;
		q.push(node{1});
		for(int k=1;k<n;k++){
			if(q.empty()){
				tot++;
				if(tot==4795)printf("%d ",T);
				else printf("%d ",0);
				continue;
			}
			//cout<<k<<": ";
			if(q.size()==1){
				p1=q.top();
				q.pop();
				//cout<<p1.u<<endl;
				m-=hi[p1.u];
				dfs(p1.u);
			}
			else{
				p1=q.top();
				q.pop();
				p2=q.top();
				q.pop();
				//cout<<p1.u<<' '<<p2.u<<endl;
				m-=hi[p1.u]+hi[p2.u];
				dfs(p1.u);
				dfs(p2.u);
			}
			tot++;
			if(tot==4795)printf("%d ",T);
			else printf("%d ",m);
		}
		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
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
// LUOGU_RID: 161997849
#include <bits/stdc++.h>
using namespace std;
int T,n,p1,p2,pans[300005],tot,ans,fa[300005],dis[300005],vis[300005];
priority_queue<pair<int,int> >pq;
pair<int,int>f[300005];
vector<int>v[300005];
void dfs(int x,int l){
	fa[x]=l;f[x]={1,x};
	for(auto i:v[x]){
		if(i==l)continue;
		dfs(i,x);f[x]=max(f[x],{f[i].first+1,f[i].second});
	}
}
int gtop(){
	while(!pq.empty())
		if(dis[pq.top().second]!=pq.top().first)pq.pop();
		else break;
	if(pq.empty())return 0;
	return pq.top().second;
}
void del(int x){
	dis[x]=-1;
	while(!vis[x]){
		if(x==0)break;
		for(auto i:v[x]){
			if(i!=fa[x]&&(!vis[i])){
				dis[f[i].second]=f[i].first;
				pq.push(f[i]);
			}
		}
		vis[x]=1;
		x=fa[x];ans++;
	}
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		cin>>n;ans=tot=0;
		for(int i=1;i<=n;i++)dis[i]=-1,v[i].clear(),vis[i]=0;
		for(int i=1;i<n;i++){
			cin>>p1>>p2;
			v[p1].push_back(p2);v[p2].push_back(p1);
		}
		dfs(1,0);
		int a=f[1].second;
		del(a);
		pans[++tot]=ans;
		while(!pq.empty()){
			int x=gtop();if(x)pq.pop();int y=gtop();
			if(!x)break;
			if(x)del(x);if(y)del(y);
			pans[++tot]=ans;
		}
		while(tot<n-1)pans[++tot]=ans;
		for(int i=1;i<n;i++)cout<<n-pans[i]<<' ';
		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:
```plaintext
1
6
1 2
2 3
3 4
4 5
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#pragma GCC optimize("Ofast,no-stack-protector")
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define pii pair<int,int>
#define piii pair<int,pii>
#define mp make_pair
#define mt make_tuple
#define x first
#define y second
#define fi first
#define se second
#define era erase
#define ins insert
#define it iterator
#define lb lower_bound
#define ub upper_bound
#define exc(exp) if(exp)continue;
#define ret(exp) if(exp)return;
#define stop(exp) if(exp)break;
#define quit(sth) {sth;return;}
#define let(var...) int var;tie(var)
#define siz(vec) ((int)((vec).size()))
#define all(vec) (vec).begin(),(vec).end()
#define unq(vec) sort(all(vec)),(vec).erase(unique(all(vec)),(vec).end())
#define deb(var) cerr<<#var<<'='<<(var)<<"; "
#define debl(var) cerr<<#var<<'='<<(var)<<";\n"
#define int long long
int n,len[300010],pos[300010];
vector<int> g[300010],h[300010];
void dfs(int u,int fa){
    len[u]=0;
    pos[u]=0,h[u].clear();
    for(auto v:g[u]){
        exc(v==fa);
        dfs(v,u);
        h[u].pb(v);
        if(len[v]+1>len[u]){
            len[u]=len[v]+1;
        }
    }
    sort(all(h[u]),
        [&](int u,int v){return len[u]>len[v];});
}

int T=0;
void work(){
    cin>>n;
    for(int i=1;i<=n;i++)g[i].clear();
    for(int i=1,u,v;i<n;i++)
        cin>>u>>v,g[u].pb(v),g[v].pb(u);
    dfs(1,0);
    priority_queue<pii> Q;
    Q.push({len[1],1});
    int ans=n;
    for(int i=1;i<n;i++){
        if(Q.empty()){
            ++T;
            assert(ans==0);
            cout<<ans<<' ';continue;
        }
        int u=Q.top().y;Q.pop();
        int v=Q.empty()?0:Q.top().y;if(!Q.empty())Q.pop();

        // deb(u),debl(pos[u]);
        // deb(v),debl(pos[v]);
        for(int w=u;;w=h[w][pos[w]-1]){
            if(!pos[w]){
                ans--;
                for(int i=1;i<siz(h[w]);i++)
                    Q.push({len[h[w][i]],w});
            }
            ++pos[w];
            stop(pos[w]-1>=siz(h[w]));
        }
        if(!v){
            ++T;
            cout<<ans<<' ';continue;
        }
        for(int w=v;;w=h[w][pos[w]-1]){
            if(!pos[w]){
                ans--;
                for(int i=1;i<siz(h[w]);i++)
                    Q.push({len[h[w][i]],w});
            }
            ++pos[w];
            stop(pos[w]-1>=siz(h[w]));
        }               cout<<ans<<' ';
    }                                       cout<<'\n';
}
signed main(){
	ios::sync_with_stdio(0),
	cin.tie(0),cout.tie(0);
	int T=1;cin>>T;while(T--)work();
}
/*
 * CONTINUE, NON-STOPPING, FOR THE FAITH
 * START TYPING IF YOU DON'T KNOW WHAT TO DO
 * STOP TYPING IF YOU DON'T KNOW WHAT YOU'RE DOING
 */

**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
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

#ifdef LOCAL
#include "../prettyprint.hpp"
#define cerr cout
#define debug(x) cerr << "\033[36m"<< __LINE__ << ":\033[39m "  << (#x) << " = " << (x) << '\n'
#else
#define cerr if(false) cout
#define debug(x) 42
#endif

using namespace std;
using ll = long long;

const int N = 300300;

int n;
vector<int> g[N];
vector<pair<int, int>> d[N];
int par[N];

void dfs(int s, int p) {
    par[s] = p;
    for (int to : g[s]) if (to != p) {
        dfs(to, s);
        if (d[to].empty())
            d[s].push_back({1, to});
        else
            d[s].push_back({d[to].back().first + 1, to});
    }
    sort(d[s].begin(), d[s].end());
}

bool used[N] = {};
priority_queue<pair<int, int>> pq;

void del(int v) {
    if (d[v].empty()) return;
    del(d[v].back().second);
    d[v].pop_back();
    if (!used[v]) {
        used[v] = true;
        for (auto c : d[v])
            pq.push(c);
    }
    return;
};

void test_case() {
    for (int i = 1; i <= n; i++) {
        g[i].clear();
        d[i].clear();
    }
    cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    for (int i = 1; i <= n; i++)
        used[i] = false;

    dfs(1, 1);

    used[1] = true;
    for (auto c : d[1])
        pq.push(c);

    int res[n];
    int cur = 0, p = 1;
    while (!pq.empty()) {
        int x[2] = {};

        x[0] = par[pq.top().second];
        cur += pq.top().first;
        pq.pop();

        if (!pq.empty()) {
            x[1] = par[pq.top().second];
            cur += pq.top().first;
            pq.pop();
        }

        del(x[0]);
        if (x[1])
            del(x[1]);

        res[p++] = cur;
    }
    while (p < n) res[p++] = cur;

    for (int i = 1; i < n; i++)
        cout << n - 1 - res[i] << ' ';
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);

    int test_cases;
    cin >> test_cases;
    for (int _ = 1; _ <= test_cases; _++) {
        cerr << "\n++++++++++++++++++++ " << _ << " TEST STARTS ++++++++++++++++++++\n";
        test_case();
        cerr << "\n+++++++++++++++++++++++++++++++++++++++++++++++++++++++\n";
        cout << '\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:
```plaintext
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
/*
           _                   _         _       __  __  _____
     /\   | |            /\   | |       | |     |  \/  |/ ____|
    /  \  | |__   ___   /  \  | |__   __| | ___ | \  / | |
   / /\ \ | '_ \ / _ \ / /\ \ | '_ \ / _` |/ _ \| |\/| | |
  / ____ \| |_) | (_) / ____ \| |_) | (_| | (_) | |  | | |____
 /_/    \_\_.__/ \___/_/    \_\_.__/ \__,_|\___/|_|  |_|\_____|
*/

// build command:
// g++ -std=gnu++17 -O3 -DDEBUG -g -fsanitize=signed-integer-overflow -fsanitize=bounds-strict -fsanitize=address -fsanitize=integer-divide-by-zero -fsanitize=float-divide-by-zero -fsanitize=pointer-overflow -fsanitize=shift-exponent -fsplit-stack -Wshadow -Wall -fconcepts -Wno-unused-result

// REMEMBER:
// BS, Offline, Persistent SegTree, SQRT, Treap, MaxFlow, FFT, Matrices

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

#pragma GCC optimize("O3,unroll-loops")

// DEBUG STUFF
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef DEBUG
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

#define F first
#define S second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

// RANDOM NUMBER GENERATOR
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define SHUF(v) shuffle(all(v), RNG);
// Use mt19937_64 for 64 bit random numbers.

int getrand(int l,int r) {
	return uniform_int_distribution<int>(l, r)(RNG);
}

const ld eps = 1e-9;
const int mod = 1e9 + 7;
const int oo = 1e9 + 7;
const ll lloo = 1e18 + 7;
const int N = 1e6 + 7;

void solve(int tc) {
	int n;
	cin >> n;
	vector<vector<int>> g(n+1);
	for(int i = 0 ; i < n-1 ; i++) {
		int u,v;
		cin >> u >> v;
		g[u].pb(v);
		g[v].pb(u);
	}

	vector<int> dpth(n+1),mxleaf(n+1),par(n+1);

	function<void(int,int)> dfs = [&](int u,int p){
		mxleaf[u] = u;
		for(int v:g[u]) {
			if (v == p) continue;
			par[v] = u;
			dpth[v] = dpth[u]+1;
			dfs(v,u);
			if (dpth[mxleaf[v]] > dpth[mxleaf[u]]) {
				mxleaf[u] = mxleaf[v];
			}
		}
	};

	dfs(1,-1);

	vector<int> ans(n);

	ans[1] = dpth[mxleaf[1]];
	multiset<pii> st; //{dpth,leaf}
	vector<bool> vis(n+1);
	function<void(int)> goup = [&](int u) {
		if (vis[u]) return;
		vis[u] = 1;
		if (u == 1) return;
		for(auto v:g[u]) {
			if (v == par[u] || mxleaf[v] == mxleaf[u]) continue;
			st.insert({dpth[mxleaf[v]]-dpth[u],mxleaf[v]});
		}
		goup(par[u]);
	};
	goup(mxleaf[1]);

	for(int k = 2 ; k < n ; k++) {
		ans[k] = ans[k-1];
		if (sz(st) >= 2) {
			auto it = prev(st.end());
			pii x = *it;
			st.erase(it);
			it = prev(st.end());
			pii y = *it;
			st.erase(it);
			ans[k] += x.F+y.F;
			goup(x.S);
			goup(y.S);
		} else if (sz(st) == 1) {
			auto it = st.begin();
			pii x = *it;
			st.erase(it);
			ans[k] += x.F;
			goup(x.S);
		}
	}
	for(int i = 1 ; i < n ; i++) cout << n-1-ans[i] << ' ';
	cout << '\n';
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	// freopen("in","r",stdin);
	// freopen("out","w",stdout);
	int T = 1;
	cin >> T;
	for(int i = 0 ; i < T ; i++) solve(i+1);
	return 0;
}

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

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

Think step by step.

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

1. **Tree Structure**: The tree should have a structure that can lead to a complex set of bridges, especially when adding edges.
2. **Root with One Child**: The root (vertex 1) has exactly one child, which is a given constraint.
3. **Small Scale**: The test input should be small enough to be manageable but complex enough to reveal the bug.

### Step-by-Step Analysis

1. **Tree with a Long Path**: A tree with a long path from the root to a leaf can create a scenario where adding edges can significantly reduce the number of bridges.
2. **Multiple Branches**: Adding multiple branches at different levels can create a more complex scenario for the algorithm to handle.
3. **Balanced Tree**: A balanced tree with a root having one child can also create interesting scenarios.

### Test Input

Let's create a tree with a long path and a few branches to test the algorithm's ability to handle such structures.

```plaintext
1
7
1 2
2 3
3 4
4 5
5 6
6 7
```

This tree has a long path from the root to the leaf, which can help in testing the algorithm's ability to handle long paths and reduce the number of bridges effectively.

### Explanation

- **Test Case 1**: The tree has 7 vertices.
- **Edges**:
  - 1 -> 2
  - 2 -> 3
  - 3 -> 4
  - 4 -> 5
  - 5 -> 6
  - 6 -> 7

This structure ensures that the root (vertex 1) has exactly one child (vertex 2), and the path from the root to the leaf (vertex 7) is long. Adding edges in this structure can significantly reduce the number of bridges, making it a good test case to expose potential bugs in the algorithm.

### Expected Output

For this test input, the expected output should be:

```plaintext
5 3 1 0 0 0
```

This output represents the smallest number of bridges left after adding 1 to 6 edges, respectively.

### Final Test Input

```plaintext
1
7
1 2
2 3
3 4
4 5
5 6
6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define PII std::pair<int,int>
struct HLD {
	int n;
	std::vector<int> siz, top, dep, parent, in, out, seq;
	std::vector<std::vector<int>> adj;
	int cur;

	HLD() {}
	HLD(int n) {
		init(n);
	}
	void init(int n) {
		this->n = n;
		siz.resize(n);
		top.resize(n);
		dep.resize(n);
		parent.resize(n);
		in.resize(n);
		out.resize(n);
		seq.resize(n);
		cur = 0;
		adj.assign(n, {});
	}
	void addEdge(int u, int v) {
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	void work(int root = 0) {
		top[root] = root;
		dep[root] = 0;
		parent[root] = -1;
		dfs1(root);
		dfs2(root);
	}
	void dfs1(int u) {
		if (parent[u] != -1) {
			adj[u].erase(std::find(adj[u].begin(), adj[u].end(), parent[u]));
		}

		siz[u] = 0;
		for (auto &v : adj[u]) {
			parent[v] = u;
			dep[v] = dep[u] + 1;
			dfs1(v);
			siz[u] = std::max(siz[u], siz[v] + 1);
			if (siz[v] > siz[adj[u][0]]) {
				std::swap(v, adj[u][0]);
			}
		}
	}
	void dfs2(int u) {
		in[u] = cur++;
		seq[in[u]] = u;
		for (auto v : adj[u]) {
			top[v] = v == adj[u][0] ? top[u] : v;
			dfs2(v);
		}
		out[u] = cur;
	}
	int lca(int u, int v) {
		while (top[u] != top[v]) {
			if (dep[top[u]] > dep[top[v]]) {
				u = parent[top[u]];
			} else {
				v = parent[top[v]];
			}
		}
		return dep[u] < dep[v] ? u : v;
	}

	int dist(int u, int v) {
		return dep[u] + dep[v] - 2 * dep[lca(u, v)];
	}

	int jump(int u, int k) {
		if (dep[u] < k) {
			return -1;
		}

		int d = dep[u] - k;

		while (dep[top[u]] > d) {
			u = parent[top[u]];
		}

		return seq[in[u] - dep[u] + d];
	}

	bool isAncester(int u, int v) {
		return in[u] <= in[v] && in[v] < out[u];
	}

	int rootedParent(int u, int v) {
		std::swap(u, v);
		if (u == v) {
			return u;
		}
		if (!isAncester(u, v)) {
			return parent[u];
		}
		auto it = std::upper_bound(adj[u].begin(), adj[u].end(), v, [&](int x, int y) {
			return in[x] < in[y];
		}) - 1;
		return *it;
	}

	int rootedSize(int u, int v) {
		if (u == v) {
			return n;
		}
		if (!isAncester(v, u)) {
			return siz[v];
		}
		return n - siz[rootedParent(u, v)];
	}

	int rootedLca(int a, int b, int c) {
		return lca(a, b) ^ lca(b, c) ^ lca(c, a);
	}
};

void solve() {
	int n;
	std::cin >> n;
	HLD h(n);
	std::priority_queue<PII> q;
	for (int i = 1; i < n; i++) {
		int u, v;
		std::cin >> u >> v;
		u--, v--;
		h.addEdge(u, v);
	}
	h.work();
	std::vector<int> ans;

	q.push({h.siz[0], 0});

	int tot = n - 1;
	auto dfs = [&](auto self, int x)->void{
		for (auto j : h.adj[x]) {
			if (j == h.adj[x][0]) continue;
			q.push({h.siz[j] + 1, j});
		}
		if (!h.adj[x].empty()) self(self, h.adj[x][0]);

	};
	while (!q.empty()) {
		auto p1 = q.top();
		q.pop();
		PII p2 = { -1, -1};
		if (!q.empty()) {
			p2 = q.top();
			q.pop();
		}
		tot -= p1.first;
		dfs(dfs, p1.second);
		if (p2.second != -1) {
			tot -= p2.first;
			dfs(dfs, p2.second);

			if (!h.adj[p1.second].empty()) {
				int t = h.adj[p1.second][0];
				if (h.lca(t, p2.second) == t) {
					while (1);
				}
			}
			if (!h.adj[p2.second].empty()) {
				int t = h.adj[p2.second][0];
				if (h.lca(t, p1.second) == t) {
					while (1);
				}
			}
		}
		ans.push_back(tot);
	}
	while (ans.size() < n - 1) {
		ans.push_back(0);
	}
	for (auto j : ans) {
		std::cout << j << " ";
	}
	std::cout << '\n';
}
signed main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0), std::cout.tie(0);
	int t = 1;
	std::cin >> t;
	while (t--) {
		solve();
	}
}

**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 for the given buggy code, we need to focus on edge cases and scenarios that can expose potential issues in the algorithm. Here are the key considerations:

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the bridge removal process.
2. **Edge Cases**: Consider trees with very few nodes, very deep trees, and trees with a lot of branching.
3. **Constraints**: Ensure the test case adheres to the constraints: \(2 \le n \le 3 \cdot 10^5\), and the root (vertex 1) has exactly one child.

Given these considerations, a small-scale test input that can be challenging for the algorithm is a tree with a structure that creates a complex hierarchy and forces the algorithm to handle multiple bridges and subtrees efficiently.

### Test Input
```plaintext
1
7
1 2
2 3
3 4
4 5
5 6
6 7
```

### Explanation
- **Number of Testcases**: 1
- **Number of Vertices**: 7
- **Edges**:
  - 1 -> 2
  - 2 -> 3
  - 3 -> 4
  - 4 -> 5
  - 5 -> 6
  - 6 -> 7

This tree is a simple linear chain where each vertex has exactly one child, except the root (vertex 1). This structure ensures that the tree is deep and forces the algorithm to handle a sequence of bridges efficiently. The linear structure also ensures that the root (vertex 1) has exactly one child, adhering to the problem constraints.

This test case is designed to be small enough to be manageable but complex enough to potentially expose issues in the algorithm's handling of bridges and subtrees.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

using namespace std;

#define int long long
#define inf 0x3F3F3F3F3F3F3F3F

const int MXN = 5e5 + 5;

struct SegTree
{
	int sz;
	vector<array<int, 2>> st;
	vector<int> lz;
	void build(int l, int r, int x)
	{
		if (l == r)
		{
			st[x][1] = l;
			return;
		}
		int mid = (l + r) >> 1;
		build(l, mid, 2*x), build(mid + 1, r, 2*x + 1);
		st[x] = max(st[2*x], st[2*x + 1]);
	}
	void init(int n)
	{
		sz = n;
		st.assign((n + 1) << 2, {0, 0});
		lz.assign((n + 1) << 2, 0);
	}
	void relax(int l, int r, int x)
	{
		if (!lz[x]) return;
		st[x][0] += lz[x];
		if (l == r)
		{
			lz[x] = 0;
			return;
		}
		lz[2*x] += lz[x], lz[2*x + 1] += lz[x];
		lz[x] = 0;
	}
	void add(int l, int r, int x, int lx, int rx, int val)
	{
		if (lx > rx) return;
		relax(l, r, x);
		if (l > rx || r < lx) return;
		if (l >= lx && r <= rx)
		{
			lz[x] += val;
			relax(l, r, x);
			return;
		}
		int mid = (l + r) >> 1;
		add(l, mid, 2*x, lx, rx, val);
		add(mid + 1, r, 2*x + 1, lx, rx, val);
		st[x] = max(st[2*x], st[2*x + 1]);
	}
	array<int, 2> get(int l, int r, int x, int lx, int rx)
	{
		if (lx > rx) return {-inf, -1};
		if (l > rx || r < lx) return {-inf, -1};
		relax(l, r, x);
		if (l >= lx && r <= rx) return st[x];
		int mid = (l + r) >> 1;
		return max(get(l, mid, 2*x, lx, rx), get(mid + 1, r, 2*x + 1, lx, rx));
	}
};

int n;
vector<int> adj[MXN];
int in[MXN], out[MXN], tim = 0;
int par[MXN], used[MXN], rev[MXN];
SegTree st;

void dfs(int a, int p)
{
	par[a] = p;
	in[a] = ++tim;
	rev[tim] = a;
	for (int &v : adj[a])
	{
		if (v == p) continue;
		dfs(v, a);
	}
	out[a] = tim;
}

priority_queue<array<int, 2>> pq;

void upd(int a)
{
	while (!used[a])
	{
		st.add(1, n, 1, in[a], out[a], -1);
		for (int &v : adj[a])
		{
			if (v == par[a]) continue;
			pq.push({st.get(1, n, 1, in[v], out[v])});
		}
		used[a] = 1;
		a = par[a];
	}
}
int ok(array<int, 2> x, int f)
{
	return !used[rev[x[1]]] && x[1] != f && st.get(1, n, 1, x[1], x[1])[0] == x[0];
}
void _()
{
	while (!pq.empty()) pq.pop();
	tim = 0;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		adj[i].clear();
		used[i] = 0;
	}
	for (int i = 1; i < n; i++)
	{
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	st.init(n);
	st.build(1, n, 1);
	dfs(1, 1);
	for (int i = 2; i <= n; i++) st.add(1, n, 1, in[i], out[i], 1);
	used[0] = used[1] = 1;
	int ans = n - 1;
	pq.push(st.get(1, n, 1, 1, n));
	for (int i = 1; i <= n - 1; i++)
	{
		while (!pq.empty() && pq.top()[0] > 0 && !ok(pq.top(), -123))
		{
			int y = pq.top()[1];
			pq.pop();
			if (y != -123 && !used[rev[y]]) pq.push(st.get(1, n, 1, y, y));
		}
		if (pq.empty() || !pq.top()[0])
		{
			cout << ans << ' ';
			continue;
		}
		array<int, 2> x = pq.top();
		pq.pop();
		while (!pq.empty() && pq.top()[0] > 0 && !ok(pq.top(), x[1]))
		{
			int y = pq.top()[1];
			pq.pop();
			if (y != x[1] && !used[rev[y]]) pq.push(st.get(1, n, 1, y, y));
		}
		if (pq.empty())
		{
			ans -= x[0];
			upd(rev[x[1]]);
		}
		else
		{
			array<int, 2> y = pq.top();
			pq.pop();
			ans -= x[0] + y[0];
			upd(rev[x[1]]), upd(rev[y[1]]);
		}
		cout << ans << ' ';
	}
	cout << '\n';
}

signed main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	for (int tt = 1; tt <= t; tt++)
	{
		_();
	}
}

**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 buggy code for the "Remove Bridges" problem, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code seems to have issues with handling the segment tree and the priority queue operations, especially when dealing with multiple edges and complex tree structures.

2. **Edge Cases**:
   - **Single Child Root**: The root has exactly one child, which is a given constraint.
   - **Long Chains**: Long chains of nodes can cause issues with the segment tree and priority queue operations.
   - **Multiple Levels**: Trees with multiple levels and multiple children for nodes can expose issues in the segment tree updates and queries.
   - **Self-loops and Multiple Edges**: Although not allowed by the problem constraints, ensuring that the input is strictly a tree without such edges is important.

3. **Generate a Test Input**:
   - **Small Scale**: Since the output length is limited, we should generate a small-scale test input that is complete and valid.
   - **Complex Structure**: A tree with a root and multiple levels, ensuring that the root has exactly one child.

Given these considerations, let's generate a test input:

```plaintext
1
7
1 2
2 3
3 4
4 5
5 6
6 7
```

This test input represents a tree with 7 vertices where the root (vertex 1) has exactly one child (vertex 2), and each subsequent vertex has exactly one child. This structure is simple but can expose issues in the segment tree and priority queue operations if not handled correctly.

Explanation:
- **Test Case Count**: 1
- **Number of Vertices**: 7
- **Edges**:
  - Vertex 1 is the root and has one child (vertex 2).
  - Vertex 2 has one child (vertex 3).
  - Vertex 3 has one child (vertex 4).
  - Vertex 4 has one child (vertex 5).
  - Vertex 5 has one child (vertex 6).
  - Vertex 6 has one child (vertex 7).

This test input should be sufficient to challenge the buggy code and help identify any issues in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#ifndef LAZY_SEGMENT_TREE
#define LAZY_SEGMENT_TREE

#include <vector>

/**
 * Zero-indexed segment tree
 * min and add on segment
 * Intervals is expected to be [l, r)
 * 0 <= l < r <= n
 */

template <class T>
class lazy_segment_tree_t {
   private:
    int n;

    using vt = std::vector<T>;
    vt tr;
    vt delay;

    T f(const T& x, const T& y) { return x[0] > y[0] ? x : y; }

    void build(const vt& vec, int id, int lo, int hi) {
        if (lo + 1 == hi) {
            tr[id] = vec[lo];
            return;
        }
        int mid = (lo + hi) / 2;
        build(vec, id * 2 + 1, lo, mid);
        build(vec, id * 2 + 2, mid, hi);
        tr[id] = f(tr[id * 2 + 1], tr[id * 2 + 2]);
    }

    void push(int id, int lo, int hi) {
        if (delay[id][0] == 0) {
            return;
        }
        tr[id][0] += delay[id][0];
        if (lo + 1 < hi) {
            delay[id * 2 + 1][0] += delay[id][0];
            delay[id * 2 + 2][0] += delay[id][0];
        }
        delay[id][0] = 0;
    }

    T get(int id, int lo, int hi, int l, int r) {
        push(id, lo, hi);
        if (r <= lo || hi <= l) {
            return {0, 0};
        }
        if (l <= lo && hi <= r) {
            return tr[id];
        }
        int mid = (lo + hi) / 2;
        return f(get(id * 2 + 1, lo, mid, l, r),
                 get(id * 2 + 2, mid, hi, l, r));
    }

    void upd(int id, int lo, int hi, int l, int r, const T& val) {
        push(id, lo, hi);
        if (r <= lo || hi <= l) {
            return;
        }
        if (l <= lo && hi <= r) {
            delay[id][0] += val[0];
            push(id, lo, hi);
            return;
        }
        int mid = (lo + hi) / 2;
        upd(id * 2 + 1, lo, mid, l, r, val);
        upd(id * 2 + 2, mid, hi, l, r, val);
        tr[id] = f(tr[id * 2 + 1], tr[id * 2 + 2]);
    }

   public:
    lazy_segment_tree_t(int _n) : n(_n), tr(4 * n), delay(4 * n) {}

    lazy_segment_tree_t(const vt& vec) : lazy_segment_tree_t(vec.size()) {
        build(vec, 0, 0, n);
    }

    T get(int l, int r) { return get(0, 0, n, l, r); }

    void upd(int l, int r, const T& val) { upd(0, 0, n, l, r, val); }
};

#endif /* LAZY_SEGMENT_TREE */

#include <bits/stdc++.h>

using ll = signed long long int;

#define all(x) (x).begin(), (x).end()

using pii = std::array<int, 2>;
using pll = std::array<ll, 2>;
using vi = std::vector<int>;
using vl = std::vector<ll>;

using namespace std;

using graph = vector<vi>;

int timer = 0;

void dfs(int u, int prev, const graph& g, vi& p, vi& d, vi& ord, vi& tin,
         vi& tout) {
    ord.push_back(u);
    tin[u] = timer++;
    for (int v : g[u]) {
        if (v == prev) {
            continue;
        }
        d[v] = d[u] + 1;
        p[v] = u;
        dfs(v, u, g, p, d, ord, tin, tout);
    }
    ord.push_back(u);
    tout[u] = timer++;
}

void solve() {
    int n;
    cin >> n;
    graph g(n);
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        --u, --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    timer = 0;
    vi p(n, -1), d(n, 0), ord, tin(n), tout(n);
    dfs(0, 0, g, p, d, ord, tin, tout);
    int ans = n - 1;
    vector<pii> vv;
    for (int el : ord) {
        vv.push_back({el && g[el].size() == 1 ? d[el] : 0, el});
    }
    lazy_segment_tree_t<pii> st(vv);
    multiset<pii> cand;
    for (int v : g[0]) {
        cand.insert(st.get(tin[v], tout[v] + 1));
    }
    set<int> proc;
    proc.insert(0);
    for (int i = 0; i < n - 1; ++i) {
        if (ans == 0) {
            cout << "0 ";
            continue;
        }
        vector<pii> todo;
        for (int _ = 0; _ < 2; ++_) {
            if (!cand.size()) {
                continue;
            }
            auto [x, y] = *prev(cand.end());
            cand.erase(prev(cand.end()));
            ans -= x;
            int u = y;
            vi path;
            while (!proc.count(u)) {
                proc.insert(u);
                path.push_back(u);
                u = p[u];
            }
            for (int el : path) {
                st.upd(tin[el], tout[el] + 1, {-1, 0});
            }
            for (int el : path) {
                for (int v : g[el]) {
                    if (proc.count(v)) {
                        continue;
                    }
                    pii tt = st.get(tin[v], tout[v] + 1);
                    if (tt[0] > 0) {
                        todo.push_back(tt);
                    }
                }
            }
        }
        for (pii el : todo) {
            cand.insert(el);
        }
        cout << ans << ' ';
    }
    cout << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

**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 given algorithm, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Tree Structure**: The tree should be structured in a way that the conditions for bridges and subtrees are non-trivial. This means the tree should have multiple levels and branches.

2. **Root with One Child**: Ensure that the root (vertex 1) has exactly one child, as per the problem constraints.

3. **Edge Cases**: Consider edge cases where adding edges can significantly change the number of bridges, especially when the tree has a specific structure that can easily form cycles.

4. **Small Scale**: The test input should be small enough to be manageable but large enough to expose the bug.

Given these considerations, let's construct a small tree with 7 vertices. This tree will have a root with one child, and the child will have multiple children, creating a structure that can be challenging for the algorithm.

```plaintext
1
7
1 2
2 3
2 4
2 5
3 6
3 7
```

This test input represents a tree where:
- Vertex 1 is the root and has one child (vertex 2).
- Vertex 2 has three children (vertices 3, 4, and 5).
- Vertex 3 has two children (vertices 6 and 7).

This structure ensures that adding edges can significantly change the number of bridges, especially when considering the subtrees of vertices 3, 4, and 5. This should help expose any issues in the algorithm's handling of bridge conditions and subtree constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

#define LL long long
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(),x.end()

const double eps=1e-7,PI=3.14159265358979323846;
const int N=1e6+10;

using namespace std;

int n,q,m,k,x,y,a[N],mx=-1,mn=1e9,sum;
string s,s1,s2;
map < int , int > mp;
vector < int > vec[N];

int dp[N];
int ans[N];
int par[N];

void dfs(int v,int p){
    dp[v]=0;
    par[v]=p;
    ans[v]=v;
    for (auto x:vec[v]){
        if (x==p)continue;
        dfs(x,v);
        if (1+dp[x]>dp[v]){
            dp[v]=1+dp[x];
            ans[v]=ans[x];
        }
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>q;
    while(q--){
        cin>>n;
        for (int i=1;i<n;i++){
            cin>>x>>y;
            vec[x].pb(y);
            vec[y].pb(x);
        }
        dfs(1,1);
        set < pair < int , pair < int , int > > > st;
        st.insert({dp[vec[1][0]],{vec[1][0],ans[vec[1][0]]}});
        int sum=n-1;
        a[1]=1;
        int cnt=0;
        while (st.size()){
            auto [best,tmp]=(*st.rbegin());
            st.erase(--st.end());
            if (st.size()==0){
                int p=tmp.S;
                while (!a[p]){
                    for (auto x:vec[p]){
                        if (x==par[p]||a[x])continue;
                        st.insert({dp[x],{x,ans[x]}});
                    }
                    a[p]=1;
                    p=par[p];
                    sum--;
                }
            }
            else {
                auto [best2,tmp2] = (*st.rbegin());
                st.erase(--st.end());
                int p=tmp.S;
                while (!a[p]){
                    for (auto x:vec[p]){
                        if (x==par[p]||a[x])continue;
                        st.insert({dp[x],{x,ans[x]}});
                    }
                    a[p]=1;
                    p=par[p];
                    sum--;
                }
                p=tmp2.S;
                while (!a[p]){
                    for (auto x:vec[p]){
                        if (x==par[p]||a[x])continue;
                        st.insert({dp[x],{x,ans[x]}});
                    }
                    a[p]=1;
                    p=par[p];
                    sum--;
                }
            }
            cout<<sum<<" ";
            cnt++;
        }
        for (int i=cnt;i<n-1;i++){
            cout<<0<<" ";
        }
        cout<<"\n";

        for (int i=1;i<=n;i++){
            vec[i].clear();
            a[i]=0;
        }
    }
    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
4
2
1 2
12
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

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

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
//  #6741f3c2eb1f238513f4b6f6@1732375490676
#include <bits/stdc++.h>
using namespace std;
namespace Radon{
    void Main();
}
int main(){
    // cout << "int main()\n";
    Radon::Main();
    return 0;
}

namespace Radon{
    #define N 300010

    int n;
    vector<int>g[N];
    int dep[N],chain_top[N],son[N],maxn[N],len[N];
    int point[N<<1],cnt=0;

    bool cmp(int a,int b){
        return a>b;
    }

    void dfs1(int u,int father){
        // cout << u << father;
        // cout << "dfs1:" << u << endl;
        dep[u]=dep[father]+1;
        maxn[u]=dep[u];
        for(int x=0;x<g[u].size();x++){
            int to=g[u][x];
            // cout << u << x;
            if(to==father){
                continue;
            }
            dfs1(to,u);
            if(maxn[to]>maxn[son[u]]){
                maxn[u]=maxn[to];
                son[u]=to;
            }
        }
        return;
    }
    void dfs2(int u,int top,int l,int father){
        // cout << u << top << l << father;
        // cout << "dfs2:" << u << endl;
        chain_top[u]=top;
        len[u]=l;
        if(g[u].size()==0){
            return;
        }
        dfs2(son[u],top,l+1,u);
        for(int x=0;x<g[u].size();x++){
            // cout << x;
            int to=g[u][x];
            if(to==father){
                continue;
            }
            if(to==son[u]){
                continue;
            }
            dfs2(to,to,1,u);
        }
        return;
    }

    void init(){
        memset(dep,0,sizeof(dep));
        memset(chain_top,0,sizeof(chain_top));
        memset(son,0,sizeof(son));
        memset(maxn,0,sizeof(maxn));
        memset(len,0,sizeof(len));
        memset(point,0,sizeof(0));
        cnt=0;
        for(int x=1;x<=n;x++){
            // cout << x;
            dep[x]=0;
            chain_top[x]=0;
            son[x]=0;
            maxn[x]=0;
            len[x]=0;
            g[x].clear();
        }
        return;
    }

    void Main(){
        // cout << "iueajbgiauebgiauwbfiauwbf";
        //asdasd
        // ios::sync_with_stdio(0);
        // cin.tie(0);cout.tie(0);
        int T;
        cin >> T;
        // cout << "sjehviaebgsijbnosengoaenokn: " << T << endl;
        while(T--){
            // cout << T << ':' << '\n';
            cin >> n;
            for(int x=1;x<n;x++){
                int a,b;
                cin >> a >> b;
                g[a].push_back(b);
                g[b].push_back(a);
                // cout << a << b;
            }
            dfs1(1,0);
            // cout << "dfs1 is over\n";
            // for(int x=1;x<=n;x++){
            //     printf("%d:  size:%d  dep:%d  son:%d  maxn:%d\n",x,g[x].size(),dep[x],son[x],maxn[x]);
            // }
            dfs2(1,1,0,0);
            // cout << "LEN:!!!!!!!!!\n";
            // for(int x=1;x<=n;x++){
            //     cout << x << ':' << len[x] << endl;
            // }
            // cout << "dfs2 is over\n";
            for(int x=1;x<=n;x++){
                if(g[x].size()==1){
                    // cout << x << g[x].size();
                    point[++cnt]=len[x];
                    // cout << cnt-1;
                }
            }
            sort(point+1,point+1+cnt,cmp);
            int num=n-1;
            int th=0;
            for(int x=1;x<n;x++){
                // cout << x << th << num;
                // cout << th << ' ' << point[th] << "    ";
                num=max(0,num-point[th]);
                th++;
                // cout << th;
                // cout << th << ' ' << point[th] << '\n';
                num=max(0,num-point[th]);
                th++;
                // cout << th;
                cout << num << ' ';
            }
            cout << '\n';
            init();
        }
        return;
    }
/*

4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5

*/
}

**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
2
1 2
4
1 2
2 3
3 4
6
1 2
2 3
3 4
4 5
5 6
```