**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cstdint>

using i64 = std::int64_t;

int dfs(int node, std::vector<int> const &parent, std::vector<int> &next) {
    int next_node = node + 1;
    while (parent[next_node] == node) {
        next_node = dfs(next_node, parent, next);
    }
    return next[node] = next_node;
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    std::vector<int> parent(2e5 + 1);
    std::vector<int> next(2e5 + 1);
    std::vector<char> num_edges(2e5 + 10);
    int num_tests;
    for (std::cin >> num_tests; num_tests; --num_tests) {
        int n;
        i64 total_weight;
        std::cin >> n >> total_weight;
        std::fill(num_edges.begin(), num_edges.begin() + n + 2, 0);
        for (int vertex = 2; vertex <= n; ++vertex) {
            std::cin >> parent[vertex];
        }
        dfs(1, parent, next);
        for (int vertex = 2; vertex <= n; ++vertex) {
            num_edges[vertex - 1]++;
            num_edges[next[vertex] - 1]++;
        }
        int num_with_free_edges = n;
        i64 rem = total_weight;
        i64 sum = 0;
        for (int i = 0; i < n - 1; ++i) {
            int vertex;
            i64 weight;
            std::cin >> vertex >> weight;
            if (--num_edges[vertex - 1] <= 0) {
                --num_with_free_edges;
            }
            if (--num_edges[next[vertex] - 1] <= 0) {
                --num_with_free_edges;
            }
            sum += 2 * weight;
            rem -= weight;
            i64 ans = sum + num_with_free_edges * rem;
            std::cout << ans << ' ';
        }
        std::cout << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The main issues in the buggy code seem to be related to how it handles the calculation of the maximum possible distances and the distribution of weights.

### Steps to Generate the Test Input:
1. **Small Test Case**: Since the output length is limited, we will create a small-scale test case with a few vertices.
2. **Edge Cases**:
   - **Single Edge**: Test a tree with only two vertices to check the basic functionality.
   - **Linear Tree**: Test a linear tree where each vertex has exactly one child to ensure the depth-first search (DFS) numbering is correct.
   - **Balanced Tree**: Test a balanced tree to ensure the algorithm handles more complex structures.
3. **Weight Distribution**:
   - **All Weight on One Edge**: Test a scenario where all the weight is assigned to a single edge.
   - **Uniform Distribution**: Test a scenario where the weight is uniformly distributed among the edges.

### Chosen Test Case:
We will create a test case with a balanced tree of 4 vertices, where the total weight is 10. This will allow us to test the distribution of weights and the calculation of distances.

### Test Input:
```plaintext
1
4 10
1 1 2
2 5
3 3
4 2
```

### Explanation:
- **Test Case 1**:
  - **Number of Vertices (n)**: 4
  - **Total Weight (w)**: 10
  - **Parent Array**: [1, 1, 2] (indicating the edges: (1, 2), (1, 3), (2, 4))
  - **Events**:
    - **Event 1**: Set \( t_2 = 5 \)
    - **Event 2**: Set \( t_3 = 3 \)
    - **Event 3**: Set \( t_4 = 2 \)

### Expected Output:
- **After Event 1**: The tree has the edge weights \( t_2 = 5 \). The remaining weight is 5.
  - \(\operatorname{dist}(1, 2) = 5\)
  - \(\operatorname{dist}(2, 3) = 5 + 5 = 10\) (if \( t_3 = 5 \))
  - \(\operatorname{dist}(3, 4) = 5 + 5 = 10\) (if \( t_3 = 5 \))
  - \(\operatorname{dist}(4, 1) = 5 + 5 = 10\) (if \( t_3 = 5 \))
  - Sum: \(5 + 10 + 10 + 10 = 35\)

- **After Event 2**: The tree has the edge weights \( t_2 = 5 \), \( t_3 = 3 \). The remaining weight is 2.
  - \(\operatorname{dist}(1, 2) = 5\)
  - \(\operatorname{dist}(2, 3) = 5 + 3 = 8\)
  - \(\operatorname{dist}(3, 4) = 3 + 2 = 5\)
  - \(\operatorname{dist}(4, 1) = 2 + 5 = 7\)
  - Sum: \(5 + 8 + 5 + 7 = 25\)

- **After Event 3**: The tree has the edge weights \( t_2 = 5 \), \( t_3 = 3 \), \( t_4 = 2 \). The remaining weight is 0.
  - \(\operatorname{dist}(1, 2) = 5\)
  - \(\operatorname{dist}(2, 3) = 5 + 3 = 8\)
  - \(\operatorname{dist}(3, 4) = 3 + 2 = 5\)
  - \(\operatorname{dist}(4, 1) = 2 + 5 = 7\)
  - Sum: \(5 + 8 + 5 + 7 = 25\)

This test case should expose any issues in the distribution of weights and the calculation of distances in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

Buggy Code:
import sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)

lol = lambda func = int : list(map(func,sys.stdin.readline().split()))

def solve():
    t = lol()[0]

    def dfs(u,g,cnt,dp):
        mx = u
        for v in g[u]:
            dfs(v,g,cnt,dp)
            mx = max(dp[v],mx)

        if u != 0:
            cnt[u]+=1
            cnt[(mx+1)%n]+=1

        dp[u] = mx

    for _ in range(t):

        n,w = lol()

        p = lol()

        g = [[] for i in range(n)]

        for i in range(n-1):
            u = p[i]
            u-=1
            g[u].append(i+1)

        dp = [i for i in range(n)]
        cnt = [0]*n
        dfs(0,g,cnt,dp)
        cur = 0
        mxed = 0
        ans = []
        for _ in range(n-1):
            i,t = lol()
            cur+=t
            i-=1

            cnt[i]-=1
            cnt[(dp[i]+1)%n]-=1

            if cnt[i] == 0:
                mxed+=1
            if cnt[(dp[i]+1)%n] == 0:
                mxed+=1

            ans.append(2 * cur + (n - mxed) * (w - cur))

        print(' '.join(map(str,ans)))

threading.Thread(target=solve).start()

**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 consider scenarios where the code might fail. The key points to consider are:

1. **Edge Cases**: Test with small and large values of `n` and `w`.
2. **Special Tree Structures**: Test with trees that have specific structures, such as a linear chain or a star graph.
3. **Events that Change the Structure**: Test with events that significantly change the structure of the tree or the weights of the edges.

Given the constraints and the need for a small-scale test input, let's create a test case with a small tree and specific events that could expose the bug.

### Test Case Explanation

- **Tree Structure**: A simple linear chain.
- **Events**: Events that change the weights in a way that could cause issues in the DFS traversal and distance calculation.

### Test Input

```plaintext
1
4 10
1 2 3
2 1
3 3
4 6
```

### Explanation

1. **Test Case Count**: 1
2. **Tree Description**:
   - `n = 4`, `w = 10`
   - `p = [1, 2, 3]` (This means the tree is a linear chain: 1 -> 2 -> 3 -> 4)
3. **Events**:
   - `2 1`: Set `t_2 = 1`
   - `3 3`: Set `t_3 = 3`
   - `4 6`: Set `t_4 = 6`

### Expected Output

For this test case, the expected output should be:

```plaintext
10 14 14
```

### Reasoning

- **Initial State**: The tree is a linear chain with unknown weights.
- **After First Event (t_2 = 1)**:
  - `t_2 = 1`, `t_3` and `t_4` are unknown.
  - Maximum possible distances:
    - `dist(1, 2) = 1`
    - `dist(2, 3) = 1 + 3 = 4` (if `t_3 = 3`)
    - `dist(3, 4) = 3 + 6 = 9` (if `t_4 = 6`)
    - `dist(4, 1) = 6`
  - Sum: `1 + 4 + 9 + 6 = 20`

- **After Second Event (t_3 = 3)**:
  - `t_2 = 1`, `t_3 = 3`, `t_4` is unknown.
  - Maximum possible distances:
    - `dist(1, 2) = 1`
    - `dist(2, 3) = 1 + 3 = 4`
    - `dist(3, 4) = 3 + 6 = 9`
    - `dist(4, 1) = 6`
  - Sum: `1 + 4 + 9 + 6 = 20` (unchanged from the previous event)

- **After Third Event (t_4 = 6)**:
  - `t_2 = 1`, `t_3 = 3`, `t_4 = 6`.
  - Maximum possible distances:
    - `dist(1, 2) = 1`
    - `dist(2, 3) = 1 + 3 = 4`
    - `dist(3, 4) = 3 + 6 = 9`
    - `dist(4, 1) = 6`
  - Sum: `1 + 4 + 9 + 6 = 20` (unchanged from the previous event)

This test case should help expose any issues in the DFS traversal and distance calculation in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

Buggy Code:
import sys,threading
sys.setrecursionlimit(10**7)
threading.stack_size(10**8)

lol = lambda func = int : list(map(func,sys.stdin.readline().split()))

t = lol()[0]

def dfs(u,g,cnt,dp):
    mx = u
    for v in g[u]:
        dfs(v,g,cnt,dp)
        mx = max(dp[v],mx)

    if u != 0:
        cnt[u]+=1
        cnt[(mx+1)%n]+=1

    dp[u] = mx

for _ in range(t):

    n,w = lol()

    p = lol()

    g = [[] for i in range(n)]

    for i in range(n-1):
        u = p[i]
        u-=1
        g[u].append(i+1)

    dp = [i for i in range(n)]
    cnt = [0]*n
    dfs(0,g,cnt,dp)
    cur = 0
    mxed = 0
    ans = []
    for _ in range(n-1):
        i,t = lol()
        cur+=t
        i-=1

        cnt[i]-=1
        cnt[(dp[i]+1)%n]-=1

        if cnt[i] == 0:
            mxed+=1
        if cnt[(dp[i]+1)%n] == 0:
            mxed+=1

        ans.append(2 * cur + (n - mxed) * (w - cur))

    print(' '.join(map(str,ans)))

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

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

Think step by step.

Response:
```plaintext
1
4 9
1 1 1
2 2
4 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

Buggy Code:
import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class SolutionCodeforces2006B {
    static final long MOD = 1_000_000_007;

    static ArrayList<ArrayList<Integer>> a;
    static ArrayList<Set<Integer>> path;
    static ArrayList<ArrayList<Integer>> back;
    static int current;

    static void dfs(int v) {
        current = v;
        for (int u : a.get(v)) {
            path.get(current).add(u);
            back.get(u).add(current);
            dfs(u);
            path.get(current).add(u);
            back.get(u).add(current);
        }
    }

    public static void main(String[] args) throws FileNotFoundException {
        // InputStream is = new FileInputStream("input.txt");
        InputStream is = System.in;

        // Scanner in = new Scanner(is);
        FastScanner in = new FastScanner(is);
        PrintWriter out = new PrintWriter(System.out);
        // PrintWriter out = new PrintWriter(new FileOutputStream("output.txt"));

        int tn = in.nextInt();
        for (int tc = 0; tc < tn; tc++) {
            int n = in.nextInt();
            long w = in.nextLong();
            a = Stream.generate(ArrayList<Integer>::new).limit(n).collect(Collectors.toCollection(ArrayList::new));
            for (int i = 1; i < n; i++) {
                int p = in.nextInt() - 1;
                a.get(p).add(i);
            }
            path = Stream.generate(HashSet<Integer>::new).limit(n).collect(Collectors.toCollection(ArrayList::new));
            back = Stream.generate(ArrayList<Integer>::new).limit(n).collect(Collectors.toCollection(ArrayList::new));
            long[] sums = new long[n];
            dfs(0);
            long fullSum = 0;
            long nonFullSum = w * n;
            long decrement = 0;
            int fullCount = 0;
            for (int k = 1; k < n; k++) {
                int x = in.nextInt() - 1;
                long y = in.nextLong();
                if (n == 2 || k < n - 1) {
                    nonFullSum -= (n - 2 - fullCount) * y;
                    int p1 = back.get(x).get(0);
                    sums[p1] += y;
                    path.get(p1).remove(x);
                    if (path.get(p1).isEmpty()) {
                        nonFullSum -= w - decrement + sums[p1] - y;
                        fullSum += sums[p1];
                        fullCount++;
                    }
                    int p2 = back.get(x).get(1);
                    sums[p2] += y;
                    path.get(p2).remove(x);
                    if (path.get(p2).isEmpty()) {
                        nonFullSum -= w - decrement + sums[p2] - y;
                        fullSum += sums[p2];
                        fullCount++;
                    }
                    decrement += y;
                }
                out.print((fullSum + nonFullSum) + " ");
            }

            out.println();
        }

        out.close();
    }

    static class FastScanner {
        private final BufferedReader br;
        private StringTokenizer st = new StringTokenizer("");

        public FastScanner(InputStream is) throws FileNotFoundException {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException ignored) {
                }
            return st.nextToken();
        }

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

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

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. The code seems to have issues with how it handles the removal of vertices and the calculation of the maximum possible distances.

Here are the steps to create a challenging test input:

1. **Small Tree with Minimal Vertices**: Start with a small tree to ensure the input is manageable and easy to verify.
2. **Edge Cases for Vertex Removal**: Ensure the tree structure and events lead to vertex removals that can expose the bug.
3. **Specific Edge Weights**: Use specific edge weights that can highlight incorrect calculations or logic errors.

Let's create a test input with a small tree and specific events:

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

### Explanation:
- **Test Case Count**: 1
- **Tree Size and Total Weight**: 4 vertices, total weight 10
- **Parent Array**: [1, 1, 1] (indicating edges 2->1, 3->1, 4->1)
- **Events**:
  - `2 3`: Set the weight of the edge from vertex 2 to 3
  - `3 2`: Set the weight of the edge from vertex 3 to 2
  - `4 5`: Set the weight of the edge from vertex 4 to 5

### Why This Test Input:
- **Small Tree**: The tree has only 4 vertices, making it easy to verify the correctness of the solution.
- **Specific Edge Weights**: The edge weights are chosen to ensure that the sum of all weights equals 10, which is the total weight given.
- **Vertex Removal**: The structure of the tree and the events ensure that vertices are removed in a way that can expose the bug in the code's logic for handling vertex removal and distance calculations.

This test input should help in identifying any issues with the code's handling of vertex removal and the calculation of maximum possible distances.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    static long power(long a, long b, long m) {
        a %= m;
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = (res * a) % m;
            a = (a * a) % m;
            b >>= 1;
        }
        return res;
    }
    static long gcd(long a, long b)  {
        return b == 0? a: gcd(b, a % b);
    }
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }
    static long modinv(long a) {
        return a <= 1 ? a : mod - (mod / a) * modinv(mod % a) % mod;
    }
    static boolean isPrime(int n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i * i <= n; i = i + 6) {
            if (n % i == 0 || n % (i + 2) == 0) return false;
        }
        return true;
    }
    static FastReader sc = new FastReader();
    static int mod = 1000000007;

    static HashMap<Integer, HashSet<Integer>> map;
    static int init;
    static HashSet<Integer> set;
    public static void dfs(int u, ArrayList<ArrayList<Integer>> arr, int par) {

        for(int v : arr.get(u)) {
            if(v == par) continue;
            set.add(v);
            if(v == init) {
                map.put(init, set);
                set = new HashSet<>();
                init++;
            }
            dfs(v, arr, u);
            set.add(v);
        }
    }
    public static void main(String[] args) {
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            long w = sc.nextLong();
            ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
            for(int i=0; i<n; i++) arr.add(new ArrayList<>());

            for(int i=1; i<n; i++) {
                int p = sc.nextInt()-1;
                arr.get(p).add(i);
                arr.get(i).add(p);
            }

            init = 1;
            map = new HashMap<>();
            set = new HashSet<>();
            dfs(0, arr, -1);
            map.put(0, set);

            // for(int v : map.keySet()) {
            //     System.out.println(v+" "+map.get(v));
            // }

            HashMap<Integer, HashSet<Integer>> map2 = new HashMap<>();

            for(int dist : map.keySet()) {
                for(int t : map.get(dist)) {
                    if(!map2.containsKey(t)) map2.put(t, new HashSet<>());
                    map2.get(t).add(dist);
                }
            }

            // for(int v : map2.keySet()) {
            //     System.out.println(v+" "+map2.get(v));
            // }

            int k = n;
            long sum_t = 0;

            for(int i=0; i<n-1; i++) {
                int t = sc.nextInt()-1;
                long val = sc.nextLong();
                sum_t += val;

                for(int d : map2.get(t)) {
                    map.get(d).remove(t);
                    if(map.get(d).size() == 0) map.remove(d);
                }

                int x = map.size();
                long ans = x*(w - sum_t);
                ans += 2*sum_t;

                System.out.print(ans+" ");
            } System.out.println();
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Iris and the Tree
time_limit_ms: 3000
memory_limit_mb: 256
Description: Given a rooted tree with the root at vertex $1$. For any vertex $i$ ($1 < i \leq n$) in the tree, there is an edge connecting vertices $i$ and $p_i$ ($1 \leq p_i < i$), with a weight equal to $t_i$.

Iris does not know the values of $t_i$, but she knows that $\displaystyle\sum_{i=2}^n t_i = w$ and each of the $t_i$ is a non- negative integer.

The vertices of the tree are numbered in a special way: the numbers of the vertices in each subtree are consecutive integers. In other words, the vertices of the tree are numbered in the order of a depth-first search.

![](CDN_BASE_URL/274244c032854fe172d47861e2eb9c02) The tree in this picture satisfies the condition. For example, in the subtree of vertex $2$, the vertex numbers are $2, 3, 4, 5$, which are consecutive integers. ![](CDN_BASE_URL/83174231191d329be697a6e3f67b5eb3) The tree in this picture does not satisfy the condition, as in the subtree of vertex $2$, the vertex numbers $2$ and $4$ are not consecutive integers.

We define $\operatorname{dist}(u, v)$ as the length of the simple path between vertices $u$ and $v$ in the tree.

Next, there will be $n - 1$ events:

  * Iris is given integers $x$ and $y$, indicating that $t_x = y$.

After each event, Iris wants to know the maximum possible value of $\operatorname{dist}(i, i \bmod n + 1)$ independently for each $i$ ($1\le i\le n$). She only needs to know the sum of these $n$ values. Please help Iris quickly get the answers.

Note that when calculating the maximum possible values of $\operatorname{dist}(i, i \bmod n + 1)$ and $\operatorname{dist}(j, j \bmod n + 1)$ for $i \ne j$, the unknown edge weights may be different.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $w$ ($2 \le n \le 2 \cdot 10^5$, $0 \leq w \leq 10^{12}$) — the number of vertices in the tree and the sum of the edge weights.

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

Then follow $n-1$ lines indicating the events. Each line contains two integers $x$ and $y$ ($2 \leq x \leq n$, $0 \leq y \leq w$), indicating that $t_x = y$.

It is guaranteed that all $x$ in the events are distinct. It is also guaranteed that the sum of all $y$ equals $w$.

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

Output: For each test case, output one line containing $n-1$ integers, each representing the answer after each event.

Examples:
input:
4
2 1000000000000
1
2 1000000000000
4 9
1 1 1
2 2
4 4
3 3
6 100
1 2 3 2 1
6 17
3 32
2 4
4 26
5 21
10 511
1 2 2 4 2 1 1 8 8
3 2
6 16
10 256
9 128
2 1
5 8
8 64
4 4
7 32
output:
2000000000000
25 18 18
449 302 247 200 200
4585 4473 2681 1567 1454 1322 1094 1022 1022

Note: In the first test case, $\operatorname{dist}(1, 2) = \operatorname{dist}(2, 1) = t_2 = w = 10^{12}$, so $\operatorname{dist}(1, 2) + \operatorname{dist}(2, 1) = 2 \cdot 10^{12}$.

In the second test case, the tree after Iris found out all $t_x$ is shown below:

![](CDN_BASE_URL/2f461ea7db7bb0cda4536a2a78190c2f)

$\operatorname{dist}(1, 2) = t_2$, $\operatorname{dist}(2, 3) = t_2 + t_3$, $\operatorname{dist}(3, 4) = t_3 + t_4$, $\operatorname{dist}(4, 1) = t_4$. After the first event, she found out that $t_2 = 2$, so $\operatorname{dist}(1, 2) = 2$. At the same time:

  * $\operatorname{dist}(2, 3)$ is maximized if $t_3 = 7$, $t_4 = 0$. Then $\operatorname{dist}(2, 3) = 9$.    * $\operatorname{dist}(3, 4)$ and $\operatorname{dist}(4, 1)$ are maximized if $t_3 = 0$, $t_4 = 7$. Then $\operatorname{dist}(3, 4) = \operatorname{dist}(4, 1) = 7$.

Thus, the answer is $2 + 9 + 7 + 7 = 25$.

After the second event, she found out that $t_4 = 4$, then $t_3 = w - t_2 - t_4 = 4$. $\operatorname{dist}(1, 2) = 2$, $\operatorname{dist}(2, 3) = 2 + 3 = 5$, $\operatorname{dist}(3, 4) = 3 + 4 = 7$, $\operatorname{dist}(4, 1) = 4$. Thus, the answer is $2 + 5 + 7 + 4 = 18$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cstdint>

using i64 = std::int64_t;

int dfs(int node, std::vector<int> const &parent, std::vector<i64> const &weight,
        std::vector<int> &next, i64 &sum) {
    int next_node = node + 1;
    while (parent[next_node] == node) {
        sum += weight[next_node];
        next_node = dfs(next_node, parent, weight, next, sum);
    }
    next[node] = next_node;
    sum += weight[node];
    return next_node;
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    std::vector<int> parent(2e5 + 1);
    std::vector<int> next(2e5 + 1);
    std::vector<i64> weight(2e5 + 1);
    std::vector<int> vertex(2e5 + 1);
    std::vector<char> has_free_edge(2e5 + 10);
    std::vector<i64> ans(2e5);
    int num_tests;
    for (std::cin >> num_tests; num_tests; --num_tests) {
        int n;
        i64 total_weight;
        std::cin >> n >> total_weight;
        std::fill(has_free_edge.begin(), has_free_edge.begin() + n + 2, 0);
        for (int vertex = 2; vertex <= n; ++vertex) {
            std::cin >> parent[vertex];
        }
        for (int i = 0; i < n - 1; ++i) {
            i64 len;
            std::cin >> vertex[i] >> len;
            weight[vertex[i]] = len;
        }
        int num_with_free_edges = 0;
        i64 rem = 0;
        i64 sum = 0;
        dfs(1, parent, weight, next, sum);
        for (int i = n - 2; i >= 0; --i) {
            ans[i] = sum + num_with_free_edges * rem;
            rem += weight[vertex[i]];
            if (vertex[i] > 1) {
                sum -= weight[vertex[i]];
                if (!has_free_edge[vertex[i]]) {
                    has_free_edge[vertex[i]] = true;
                    ++num_with_free_edges;
                }
            }
            sum -= weight[vertex[i]];
            if (!has_free_edge[next[vertex[i]]]) {
                has_free_edge[next[vertex[i]]] = true;
                ++num_with_free_edges;
            }
        }
        for (int i = 0; i < n - 1; ++i) {
            std::cout << ans[i] << ' ';
        }
        std::cout << '\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
1
3 5
1 1
2 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <deque>
#include <unordered_map>
#include <map>
using namespace std;

int lg2;

void dfs(int nod,int pr,vector <int>& d1,vector <int>& d2,vector<int>& lv,vector <vector<int>>& v)
{
    lv[nod]=lv[pr]+1;
    for(auto& i : v[nod])
        if(i!=pr)
        {
            dfs(i,nod,d1,d2,lv,v);
            if(d2[nod] < d1[i] + 1)
            {
                d2[nod] = d1[i]+1;
            }
            if(d2[nod] > d1[nod])
                swap(d2[nod],d1[nod]);
        }
}
void uplift(int nod,int pr,vector <int>& d1,vector <int>& d2,vector<int>& lv,vector <vector<int>>& v,vector <vector<pair<int,int>>>& best)
{
    best[0][nod]= {d1[pr],pr};
    if(d1[pr] == d1[nod]+1)
        best[0][nod]= {d2[pr],pr};
    best[0][nod].first-=lv[pr];
    for(int i=1; i<=lg2; i++)
    {
        best[i][nod].second=best[i-1][best[i-1][nod].second].second;
        best[i][nod].first=max(best[i-1][nod].first,best[i-1][best[i-1][nod].second].first);
    }
    for(auto & i : v[nod])
    {
        if(i==pr) continue;
        uplift(i,nod,d1,d2,lv,v,best);
    }

}

int get_sol(int nod,int stam,vector<vector<pair<int,int>>>& best,vector<int>& d1,vector<int>& d2,vector<int>& lv)
{
    int dst = min(stam,lv[nod]);
    int sol = d1[nod]-lv[nod];
    int h1 = lv[nod];
    for(int i = lg2; i>=0; i--)
    {
        if((1<<i)<=dst)
        {
            sol=max(sol,best[i][nod].first);
            nod=best[i][nod].second;
            dst-=(1<<i);
        }
    }
    return sol + h1;

}

void solve()
{
    int n;
    cin>>n;
    vector <vector<int>> v(n+3);
    lg2 = 0;
    for(int i=3; i<=20; i++)
        if((1<<i) >=n)
        {
            lg2 = i;
            break;
        }
    vector <vector<pair<int,int>>> best(lg2+1,vector<pair<int,int>>(n+3));
    vector <int> d1(n+3);
    vector <int> d2(n+3);
    vector <int> lv(n+3);
    for(int i=1; i<n; i++)
    {
        int x,y;
        cin>>x>>y;
        v[x].push_back(y);
        v[y].push_back(x);
    }

    dfs(1,0,d1,d2,lv,v);
    uplift(1,0,d1,d2,lv,v,best);
    int q;
    cin>>q;
    for(int i=1; i<=q; i++)
    {
        int nod,stam;
        cin>>nod>>stam;
        cout<<get_sol(nod,stam,best,d1,d2,lv)<<' ';
    }
    cout<<'\n';
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    for(int i=1; i<=t; i++)
        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 generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are some considerations:

1. **Tree Structure**: Create a tree that has a deep and complex structure, which can challenge the algorithm's ability to correctly calculate distances and stamina usage.
2. **Queries**: Include queries with low stamina and high stamina to test both edge cases and typical cases.
3. **Ancestors and Descendants**: Ensure that the tree has multiple levels and branches to test the algorithm's ability to handle ancestor-descendant relationships and stamina consumption accurately.

Given these considerations, let's create a small-scale test input:

```plaintext
1
7
1 2
1 3
2 4
2 5
3 6
3 7
4
1 0
4 2
5 3
7 1
```

### Explanation:
- **Tree Structure**:
  - Root at vertex 1.
  - Vertex 1 has children 2 and 3.
  - Vertex 2 has children 4 and 5.
  - Vertex 3 has children 6 and 7.
- **Queries**:
  - Query 1: Starting at vertex 1 with 0 stamina. Expected output: 0 (since no moves are possible).
  - Query 2: Starting at vertex 4 with 2 stamina. Expected output: 3 (can move to 2, 1, and 3).
  - Query 3: Starting at vertex 5 with 3 stamina. Expected output: 4 (can move to 2, 1, 3, and 6 or 7).
  - Query 4: Starting at vertex 7 with 1 stamina. Expected output: 2 (can move to 3 and 6).

This test input is designed to challenge the algorithm by testing various scenarios including low stamina, high stamina, and different levels of the tree structure.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

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

using namespace std;
using ll = long long;
const int MAX_N = 2e5 + 1;
const int MAX_LG = 26;
int jumps[MAX_LG][MAX_N];
int step[MAX_LG];

int raise(int v, int k) {
    int cur = MAX_LG - 1;
    while (cur >= 0) {
        if (k >= step[cur]) {
            k -= step[cur];
            v = jumps[cur][v];
        }
        cur--;
    }
    return v;
}

vector<vector<int>> g;
vector<int> dist, parent, sets, cur_dist, cur_max;
vector<pair<int, int>> block;
vector<set<int>> a;
vector<map<int, int>> all;

void dfs1(int v, int pr) {
    parent[v] = pr;
    dist[v] = 0;
    int mx1 = -1, mx2 = -1;
    for (int u: g[v]) {
        if (u != pr) {
            dfs1(u, v);
            dist[v] = max(dist[v], dist[u] + 1);
            if (mx1 == -1 || dist[mx1] < dist[u]) {
                mx2 = mx1;
                mx1 = u;
            } else if (mx2 == -1 || dist[mx2] < dist[u]) {
                mx2 = u;
            }
        }
    }
    block[v].first = mx1;
    block[v].second = mx2;
}

void dfs(int v, int pr) {
    int mx = -1;
    vector<pair<int, int>> to;
    for (int u: g[v]) {
        if (u != pr) {
            dfs(u, v);
            to.emplace_back(sets[u], u);
            if (mx == -1 || int(a[mx].size()) < int(a[sets[u]].size())) {
                mx = sets[u];
            }
        }
    }
    if (mx != -1) {
        a[mx].insert(v);
        for (auto [s, u]: to) {
            for (int e: a[s]) {
                int val = block[v].first;
                int per = max(cur_dist[e], cur_max[e]);
                if (val != -1) {
                    if (val != u) {
                        per = max(per, cur_dist[e] + dist[block[v].first] + 1);
                    } else if (block[v].second != -1) {
                        per = max(per, cur_dist[e] + dist[block[v].second] + 1);
                    }
                }
                all[e][v] = per;
                cur_max[e] = max(cur_max[e], per);
            }
            if (s != mx) {
                for (int e: a[s]) {
                    a[mx].insert(e);
                }
                a[s].clear();
            }
        }
        sets[v] = mx;
        for (int e: a[mx]) {
            cur_dist[e]++;
        }
    } else {
        a[v].insert(v);
        cur_dist[v]++;
    }
}

void prime_antony(int n) {
    g.assign(n, vector<int>());
    dist.assign(n, 0);
    parent.assign(n, 0);
    cur_max.assign(n, 0);
    cur_dist.assign(n, 0);
    cur_max.assign(n, 0);
    block.assign(n, {-1, -1});
    sets.assign(n, 0);
    all.assign(n, map<int, int>());
    a.assign(n, set<int>());
}

void nicholas_djekson() {
    int n;
    cin >> n;
    prime_antony(n);
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        g[u - 1].push_back(v - 1);
        g[v - 1].push_back(u - 1);
    }
    dfs1(0, -1);
    parent[0] = 0;
    for (int i = 0; i < n; i++) {
        sets[i] = i;
        jumps[0][i] = parent[i];
    }
    for (int lg = 1; lg < MAX_LG; lg++) {
        for (int i = 0; i < n; i++) {
            jumps[lg][i] = jumps[lg - 1][jumps[lg - 1][i]];
        }
    }
    dfs(0, -1);
    int q;
    cin >> q;
    while (q--) {
        int v, k;
        cin >> v >> k;
        int pr = raise(v - 1, k);
        cout << max(dist[v - 1], all[v - 1][pr]) << " ";
    }
    cout << "\n";
}

signed main() {
    step[0] = 1;
    for (int i = 1; i < MAX_LG; i++) {
        step[i] = step[i - 1] * 2;
    }
    int _;
    cin >> _;
    while (_--) {
        nicholas_djekson();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
//AshrafSustS19

#include<bits/stdc++.h>
#include <random>
#include <chrono>

using namespace std;
using ll = long long;
using ull = unsigned long long;

#ifndef ONLINE_JUDGE
#include "H:\New folder\AshrafSustS19\app_debug.cpp"
#else
#define debug(...)
#define debugArr(arr, n)
#endif

#define int long long

int lgn;
int n, q;
vector<vector<ll>> lp;
vector<vector<ll>> adj;
vector<array<ll, 2>> MEM;
vector<ll> ldpt;

ll getLCA(int a, int b){
    if (ldpt[a] < ldpt[b]){
        swap(a, b);
    }
    int diff = ldpt[a] - ldpt[b];
    for (int i = 0; i < lgn; i++){
        if ((1ll << i) & diff){
            a = lp[a][i];
        }
    }
    if (a == b){
        return a;
    }
    for (int i = lgn - 1; i >= 0; i--){
        if (lp[a][i] != lp[b][i]){
            a = lp[a][i];
            b = lp[b][i];
        }
    }
    return lp[a][0];
}

ll dis(int a, int b){
    int p = getLCA(a, b);
    return ldpt[a] + ldpt[b] - ldpt[p] * 2;
}

void dp(int u, int p){
    lp[u][0] = p;
    for (int i = 1; i < lgn && lp[u][i - 1] != -1; i++){
        lp[u][i] = lp[lp[u][i - 1]][i - 1];
    }
    MEM[u] = {u, u};
    if (u != 0 && adj[u].size() == 1){
        return;
    }
    for (int i = 0; i < adj[u].size(); i++){
        int v = adj[u][i];
        if (v == p) continue;
        ldpt[v] = ldpt[u] + 1;
        dp(v, u);
        if (ldpt[MEM[v][0]] > ldpt[MEM[u][0]]){
            MEM[u][0] = MEM[v][0];
        }
    }
    for (int i = 0; i < adj[u].size(); i++){
        int ind = MEM[u][0];
        int v = adj[u][i];
        if (v == p) continue;
        for (int l = 0; l < 2; l++){
            int cind = MEM[u][1];
            int ci = MEM[v][l];
            if (dis(ind, ci) > dis(ind , cind)){
                MEM[u][1] = ci;
            }
        }
    }
    if (ldpt[MEM[u][0] < ldpt[MEM[u][1]]]){
        swap(MEM[u][0], MEM[u][1]);
    }
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--){
        cin >> n;
        lgn = (__lg(n) + 1);
        lp.clear();
        lp.resize(n, vector<ll> (lgn, -1));
        adj.clear();
        adj.resize(n);
        MEM.clear();
        MEM.resize(n);
        ldpt.clear();
        ldpt.resize(n, 0);
        for (int i = 0; i < n - 1; i++){
            int a, b;
            cin >> a >> b;
            adj[--a].push_back(--b);
            adj[b].push_back(a);
        }
        dp(0, -1);
        cin >> q;

        while (q--){
            ll u, k;
            cin >> u >> k;
            u--;
            k = min(k, ldpt[u]);
            int v = u;
            for (int i = 0; i < lgn; i++){
                if ((1ll << i) & k){
                    v = lp[v][i];
                }
            }
            int j = MEM[v][0], l = MEM[v][1];
            ll res = max(dis(u, j), dis(u, l));
            cout << res << " ";
        }
        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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios that can maximize the failure probability. The key points to consider are:

1. **Tree Structure**: The tree should have a structure that can cause the LCA (Lowest Common Ancestor) and distance calculations to be complex and error-prone.
2. **Queries**: The queries should test the limits of the stamina and the ability to move through the tree.

Let's construct a small-scale test input that can expose potential bugs in the code:

1. **Tree Structure**: We can create a tree with a deep path and some branches to test the LCA and distance calculations.
2. **Queries**: We will include queries with different starting points and stamina values to test various scenarios.

Here is the test input:

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

### Explanation:
- **Tree Structure**:
  - The tree has 5 vertices.
  - The edges form a deep path from 1 to 4 with a branch from 3 to 5.
- **Queries**:
  - Query 1: Starting at vertex 4 with 1 stamina. This tests the ability to move up the tree with limited stamina.
  - Query 2: Starting at vertex 3 with 2 stamina. This tests the ability to move to both children and the parent with sufficient stamina.
  - Query 3: Starting at vertex 2 with 0 stamina. This tests the ability to stay at the current vertex without moving.

This test input should expose any issues with the LCA and distance calculations in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
import sys
from collections import defaultdict,deque,Counter
from heapq import heappush,heappop,heapify
from math import inf,comb,ceil,log,lcm,gcd
from functools import cache
from itertools import accumulate,combinations,permutations
from bisect import bisect_left,bisect_right,insort_left
from random import randint
class LazySegmentTree:
    def __init__(self, data, default=inf, func=min):
        """initialize the lazy segment tree with data"""
        self._default = default
        self._func = func

        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()
        self._lazy = [0] * (2 * _size)

        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(range(_size)):
            self.data[i] = func(self.data[i<<1], self.data[i<<1|1])

    def __len__(self):
        return self._len

    def _push(self, idx):
        """push query on idx to its children"""
        # Let the children know of the queries
        q, self._lazy[idx] = self._lazy[idx], 0

        self._lazy[idx<<1] += q
        self._lazy[idx<<1|1] += q
        self.data[idx<<1] += q
        self.data[idx<<1|1] += q

    def _update(self, idx):
        """updates the node idx to know of all queries applied to it via its ancestors"""
        for i in reversed(range(1, idx.bit_length())):
            self._push(idx >> i)

    def _build(self, idx):
        """make the changes to idx be known to its ancestors"""
        idx >>= 1
        while idx:
            self.data[idx] = self._func(self.data[idx<<1], self.data[idx<<1|1]) + self._lazy[idx]
            idx >>= 1

    def add(self, start, stop, value):
        """lazily add value to [start, stop)"""
        start = start + self._size
        stop = stop + self._size
        cr=cl=False
        while start < stop:
            if cl:
                sp=start-1
                self.data[sp]=self._func(self.data[sp<<1],self.data[sp<<1|1])+self._lazy[sp]
            if cr:
                self.data[stop]=self._func(self.data[stop<<1],self.data[stop<<1|1])+self._lazy[stop]
            if start & 1:
                self._lazy[start] += value
                self.data[start] += value
                start += 1
                cl=True
            if stop & 1:
                stop -= 1
                self._lazy[stop] += value
                self.data[stop] += value
                cr=True
            start >>= 1
            stop >>= 1

        # Tell all nodes above of the updated area of the updates
        start-=1
        while stop>0:
            if cl:
                self.data[start]=self._func(self.data[start<<1],self.data[start<<1|1])+self._lazy[start]
            if cr and (not cl or start!=stop):
                self.data[stop]=self._func(self.data[stop<<1],self.data[stop<<1|1])+self._lazy[stop]

            start>>=1
            stop>>=1
    def query(self, start, stop):
        """func of data[start, stop)"""
        start += self._size
        stop += self._size

        # Apply all the lazily stored queries
        self._update(start)
        self._update(stop - 1)

        res = self._default
        while start < stop:
            if start & 1:
                res = self._func(res, self.data[start])
                start += 1
            if stop & 1:
                stop -= 1
                res = self._func(res, self.data[stop])
            start >>= 1
            stop >>= 1
        return res

    def __repr__(self):
        return "LazySegmentTree({0})".format(self.data)
input=lambda:sys.stdin.readline().rstrip()
inp=lambda:map(int,input().split())
t=int(input())
xor=randint(1,2**32-1)
for _ in range(t):
    n=int(input())
    g=[[]for _ in range(n)]
    par=[0]*n
    for _ in range(n-1):
        u,v=inp()
        u-=1
        v-=1
        g[u].append(v)
        g[v].append(u)
    q=[(0,0,0)]
    depth=[0]*n
    depth[0]=0
    euler=[]
    while q:
        x,lx,d=q.pop()
        euler+=[x]
        if d==0:
            if x!=0:depth[x]=1+depth[lx]
            for v in g[x]:
                if v!=lx:
                    q.append((x,lx,1))
                    q.append((v,x,0))
    st=LazySegmentTree(data=[depth[e]for e in euler])
    last=[0]*n
    for i,e in enumerate(euler):
        last[e]=i
    leaves=[last[e]for e in range(1,n)if len(g[e])==1]
    leaves.sort()
    st2=LazySegmentTree(data=[depth[euler[e]]for e in leaves],func=max,default=-inf)
    Q=int(input())
    o=[0]*Q
    for i in range(Q):
        v,k=inp()
        v-=1
        o[i]=min(depth[v],k)
        u=last[v]
        ind=bisect_left(leaves,u)
        c1=c2=None

        lo=ind-1
        hi=len(leaves)
        while hi-lo>1:
            mid=(lo+hi)//2
            if st.query(u,leaves[mid]+1)>=depth[v]-k:
                lo=mid
            else:
                hi=mid
        if lo!=ind-1:
            olo=lo
            V=st.query(u,leaves[lo]+1)
            lo=ind-1
            hi=len(leaves)
            while hi-lo>1:
                mid=(lo+hi)//2
                if st.query(u,leaves[mid]+1)>V:
                    lo=mid
                else:
                    hi=mid
            n1=st2.query(hi,olo+1)
            o[i]=max(o[i],depth[v]+n1-2*V)

        lo=-1
        hi=ind
        while hi-lo>1:
            mid=(lo+hi)//2
            if st.query(leaves[mid],u+1)<depth[v]-k:
                lo=mid
            else:
                hi=mid
        if hi!=ind:
            ohi=hi
            V=st.query(leaves[hi],u+1)
            lo=-1
            hi=ind
            while hi-lo>1:
                mid=(lo+hi)//2
                if st.query(leaves[mid],u+1)<=V:
                    lo=mid
                else:
                    hi=mid
            n2=st2.query(ohi,lo+1)
            o[i]=max(o[i],depth[v]+n2-2*V)
    print(*o)

**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
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
9
8 1
1 7
1 4
7 3
4 9
3 2
1 5
3 6
7
6 0
2 3
6 2
8 2
2 4
9 2
6 3
6
2 1
2 5
2 4
5 6
4 3
3
3 1
1 3
6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
from sys import stdin
input = lambda: stdin.buffer.readline().decode().strip()

for _ in range(int(input())):
    n = int(input())
    adj = [[] for _ in range(n+1)]
    logn = n.bit_length()

    for _ in range(n-1):
        u, v = map(int, input().split())
        adj[u] += [v]
        adj[v] += [u]

    bfs = [1]
    par = [[0]*(n+1) for _ in range(logn + 1)]
    depth = [0]*(n+1)

    for u in bfs:
        for v in adj[u]:
            if v != par[0][u]:
                bfs += [v]
                par[0][v] = u
                depth[v] = depth[u] + 1

        # assign ancestors
        for i in range(1, logn + 1):
            par[i][u] = par[i-1][par[i-1][u]]

    # assign height
    height = [[0]*(n+1) for _ in range(logn + 1)]
    st = [[(0, -1)] for _ in range(n + 1)]

    for u in bfs[::-1]:
        p = par[0][u]
        st[u].sort(reverse=True)

        for v in adj[u]:
            if v != p:
                height[1][v] = st[u][0][0] if st[u][0][1] != v else st[u][1][0]

        height[0][p] = max(height[0][p], height[0][u] + 1)
        st[p] += (height[0][u] + 1, u),

    for u in bfs:
        for i in range(1, logn):
            height[i + 1][u] = max(height[i][u] - 1, height[i][par[i-1][u]])

    q = int(input())
    ans = [0]*q

    for i in range(q):
        u, k = map(int, input().split())
        ans[i] = height[0][u]
        k = min(k, depth[u])

        up = 0
        for j in range(logn):
            if k >> j & 1:
                up |= 1 << j
                ans[i] = max(ans[i], height[j+1][u] + up)
                u = par[j][u]

    print(*ans)

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

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

Think step by step.

Response:
```plaintext
1
6
1 2
2 3
2 4
4 5
4 6
3
3 2
4 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Chefir
time_limit_ms: 4000
memory_limit_mb: 256
Description: Given a tree with $n$ vertices rooted at vertex $1$. While walking through it with her cat Chefir, Sakurako got distracted, and Chefir ran away.

To help Sakurako, Kosuke recorded his $q$ guesses. In the $i$-th guess, he assumes that Chefir got lost at vertex $v_i$ and had $k_i$ stamina.

Also, for each guess, Kosuke assumes that Chefir could move along the edges an arbitrary number of times:

  * from vertex $a$ to vertex $b$, if $a$ is an ancestor$^{\text{∗}}$ of $b$, the stamina will not change;    * from vertex $a$ to vertex $b$, if $a$ is not an ancestor of $b$, then Chefir's stamina decreases by $1$.

If Chefir's stamina is $0$, he cannot make a move of the second type.

For each assumption, your task is to find the distance to the farthest vertex that Chefir could reach from vertex $v_i$, having $k_i$ stamina.

$^{\text{∗}}$Vertex $a$ is an ancestor of vertex $b$ if the shortest path from $b$ to the root passes through $a$.

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

Each test case is described as follows:

  * The first line contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.   * The next $n-1$ lines contain the edges of the tree. It is guaranteed that the given edges form a tree.   * The next line consists of a single integer $q$ $(1\le q\le 2 \cdot 10^5)$, which denotes the number of guesses made by Kosuke.   * The next $q$ lines describe the guesses made by Kosuke, with two integers $v_i$, $k_i$ $(1\le v_i \le n, 0 \le k_i\le n)$.

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

Output: For each test case and for each guess, output the maximum distance to the farthest vertex that Chefir could reach from the starting point $v_i$ having $k_i$ stamina.

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

Note: In the first example:

  * In the first query, you can go from vertex $5$ to vertex $3$ (after which your stamina will decrease by $1$ and become $0$), and then you can go to vertex $4$;    * In the second query, from vertex $3$ with $1$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;    * In the third query, from vertex $2$ with $0$ stamina, you can only reach vertices $2$, $3$, $4$, and $5$;

Buggy Code:
from collections import defaultdict, deque

def build_tree(n, edges):
    """Build adjacency list representation of the tree."""
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)
    return graph

def find_ancestors(graph, root=1):
    """Find all ancestors for each vertex using BFS."""
    n = max(graph.keys()) + 1
    ancestors = defaultdict(set)
    parent = [-1] * n

    queue = deque([root])
    while queue:
        vertex = queue.popleft()
        # Add all ancestors of parent to current vertex
        if parent[vertex] != -1:
            ancestors[vertex] = ancestors[parent[vertex]].union({parent[vertex]})

        for neighbor in graph[vertex]:
            if parent[vertex] != neighbor:
                parent[neighbor] = vertex
                queue.append(neighbor)

    return ancestors

def find_max_distance(graph, start, stamina, ancestors):
    """Find maximum distance that can be reached with given stamina."""
    n = max(graph.keys()) + 1
    # Initialize distances to infinity
    distances = [float('inf')] * n
    distances[start] = 0

    # Use BFS to find reachable vertices
    queue = deque([(start, stamina)])
    visited = {(start, stamina)}

    while queue:
        vertex, curr_stamina = queue.popleft()

        for next_vertex in graph[vertex]:
            new_stamina = curr_stamina
            # If moving to non-ancestor vertex, decrease stamina
            if vertex not in ancestors[next_vertex]:
                new_stamina -= 1

            # Skip if no stamina left for non-ancestor moves
            if new_stamina < 0:
                continue

            # Update distance and continue BFS if state not visited
            new_state = (next_vertex, new_stamina)
            if new_state not in visited:
                visited.add(new_state)
                distances[next_vertex] = min(distances[next_vertex],
                                          distances[vertex] + 1)
                queue.append(new_state)

    # Find maximum finite distance
    max_distance = 0
    for d in distances:
        if d != float('inf'):
            max_distance = max(max_distance, d)

    return max_distance

def solve():
    """Main solution function."""
    t = int(input())  # Number of test cases

    for _ in range(t):
        # Read tree size
        n = int(input())

        # Read edges
        edges = []
        for _ in range(n - 1):
            u, v = map(int, input().split())
            edges.append((u, v))

        # Build tree and find ancestors
        graph = build_tree(n, edges)
        ancestors = find_ancestors(graph)

        # Read number of guesses
        q = int(input())

        # Process each guess
        for _ in range(q):
            vertex, stamina = map(int, input().split())
            max_dist = find_max_distance(graph, vertex, stamina, ancestors)
            print(max_dist, end=' ')
        print()  # New line after each test case

if __name__ == "__main__":
    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:
```plaintext
1
5
1 2
2 3
3 4
3 5
3
5 1
3 1
2 0
```