**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

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

using namespace std;

vector<vector<long long>> gr;
vector<long long> de;
vector<long long> ma;
vector<bool> used;
vector<long long> roof;
vector<vector<long long>> vert;
long long s;
long long n;
long long zn = 0;

void dfs(long long h, long long pr, long long d) {
    roof[h] = d;
    for (auto v : gr[h]) {
        if (v != pr) {
            dfs(v, h, d + 1);
            ma[h] = max(ma[h], ma[v]);
        }
    }
    if (h != 0 && gr[h].size() == 1) {
        vert[d].push_back(h);
    }
    ma[h]++;
    de[d]++;
}

void dfs2(long long h, long long w) {
    if (ma[h] > w) {
        return;
    }
    used[h] = true;
    zn++;
    for (auto v : gr[h]) {
        if (roof[h] > roof[v]) {
            dfs2(v, w);
        }
    }
}

int main() {
    long long t;
    cin >> t;
    for (long long test = 0; test < t; test++) {
        cin >> n;
        gr.assign(n, {});
        de.assign(n, 0);
        ma.assign(n, 0);
        used.assign(n, false);
        vert.assign(n, {});
        roof.assign(n, 0);
        s = 0;
        for (long long j = 0; j < n - 1; j++) {
            long long x, y;
            cin >> x >> y;
            x--;
            y--;
            gr[x].push_back(y);
            gr[y].push_back(x);
        }
        dfs(0, -1, 0);
        long long s = n - 1;
        long long ans = s;
        for (long long j = 1; j < n; j++) {
            zn = 0;
            for (auto v : vert[j - 1]) {
                if (!used[v]) {
                    dfs2(v, j - 1);
                }
            }
            s += zn;
            s -= de[j];
            ans = min(ans, s);
        }
        cout << ans << '\n';
    }
}

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

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

Think step by step.

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

1. **Tree Structure**: The tree should be such that the algorithm might not handle certain edge cases correctly. For example, trees with very long chains, trees with many leaves at different levels, or trees with a complex branching structure.
2. **Multiple Test Cases**: The input should include multiple test cases to ensure the algorithm is robust.
3. **Edge Cases**: Test cases should include edge cases like a tree with a single path, a star-shaped tree, and a balanced tree.

Given these considerations, let's construct a test input that includes a tree with a long chain and a tree with a star-like structure:

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

### Explanation:
1. **First Test Case (Long Chain)**
   - The tree is a simple chain: `1-2-3-4-5`.
   - The goal is to make all leaves at the same distance from the root. In this case, removing the edges `(1, 2)` and `(4, 5)` will make the leaves `3` and `4` at the same distance from the root.
   - Expected output: `2`.

2. **Second Test Case (Star-shaped Tree)**
   - The tree is a star with the root at node `1` and leaves at nodes `2, 3, 4, 5, 6, 7`.
   - The goal is to make all leaves at the same distance from the root. In this case, no operations are needed because all leaves are already at the same distance from the root.
   - Expected output: `0`.

This test input should help expose any issues in the buggy code, especially in handling long chains and star-shaped trees.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
// Source: https://usaco.guide/general/io

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

#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define ii pair<int, int>
#define vii vector<ii>
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define int ll
#define PB push_back

struct Ftree{
    vi f;
    int sz;

    Ftree(int n){
        f.assign(n, 0);
        sz = n;
    }
    void upd(int i, int v){
        while(i < sz){
            f[i] += v;
            i = (i | (i + 1));
        }
    }
    int get(int l){
        int res = 0;
        while(l >= 0){
            res += f[l];
            l = ((l+1)&l) - 1;
        }
        return res;
    }
    int range(int l, int r){
        return get(r) - (l ? get(l-1) : 0);
    }
};

void solve(){
    int n;
    cin >> n;
    vi g[n+1];
    FOR(i, 1, n){
        int u, v;
        cin >> u >> v;
        g[u].PB(v);
        g[v].PB(u);
    }
    vi h1(n+1, 0), h2(n+1, 0);
    h1[0] = -1;
    function<void(int, int)>dfs = [&](int u, int p){
        for(int v : g[u]){
            if(v == p)continue;
            h1[v] = h1[u] + 1;
            dfs(v, u);
            h2[u] = max(h2[u], h2[v] + 1);
        }
        h2[u] += h1[u];
    };
    dfs(1, 0);
    // h1[i] ...
    int ans = 0;
    vi order(n + 1);
    iota(order.begin(), order.end(), 0);
    sort(order.begin(), order.end(), [&](int i, int j){
        if(h1[i] != h1[j])return h1[i] < h1[j];
        else return false;
    });
    vii v;
    FOR(i, 1, n+1){
        v.PB({h1[i], 0});
        v.PB({h2[i], 1});
    }
    sort(v.begin(), v.end());
    int curr = 0;
    for(auto [x, y] : v){
        if(y)curr--;
        else curr++;
        ans = max(ans, curr);
    }
    cout << n - ans << '\n';
}

signed main() {
    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 generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause the code to fail. The main focus should be on:

1. **Trees with a large number of nodes** but within the constraints.
2. **Trees with complex structures** that might cause the depth calculation to go wrong.
3. **Trees with multiple levels of depth** to ensure the pruning logic is thoroughly tested.

Given the constraints and the need for a small-scale test input, let's consider a tree with a specific structure that might expose the bug:

- A tree with a root node and several levels of children, where the leaves are not all at the same depth.
- A tree with a branch that has a different depth from the others.

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

```plaintext
1
13
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
6 12
6 13
```

Explanation:
- The tree has 13 nodes.
- Node 1 is the root.
- Node 1 has two children: Node 2 and Node 3.
- Node 2 has two children: Node 4 and Node 5.
- Node 3 has two children: Node 6 and Node 7.
- Node 4 has two children: Node 8 and Node 9.
- Node 5 has two children: Node 10 and Node 11.
- Node 6 has two children: Node 12 and Node 13.

This structure ensures that the leaves are not all at the same depth from the root, and it will challenge the depth calculation and pruning logic in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import sys
from collections import defaultdict

input = sys.stdin.readline
N = int(input())

graph = defaultdict(list)
nodetolevel = defaultdict(int)
levels = defaultdict(list)
tree = defaultdict(list)

def createTree(val, tree, visited,level):

    visited.add(val)
    levels[level].append(val)
    nodetolevel[val] = level

    for nei in graph[val]:
        if nei not in visited:
            tree[val].append(nei)
            createTree(nei, tree, visited,level+1)

def numofChildren(node,tree):
    if node not in tree:
        return 0
    res = 0
    for child in tree[node]:
        res += (1+numofChildren(child,tree))

    return res

def computelca(node1,node2):

    return

def find_path(root, node, path):

    path.append(root)

    if root == node:
        return True

    for child in tree[root]:
        if find_path(child, node, path):
            return True

    path.pop()
    return False

def lca_nary_tree(root, node1, node2):
    path1 = []
    path2 = []

    if not find_path(root, node1, path1) or not find_path(root, node2, path2):
        return None

    i = 0
    while i < len(path1) and i < len(path2) and path1[i] == path2[i]:
        i += 1

    return path1[i-1]

def helper(graph):
    visited = set()
    createTree(1, tree, visited,0)

    leaves = []
    for i in range(1,M+1):
        if i not in tree:
            leaves.append(i)

    cand = 0

    res = float('inf')
    for level, nodes in levels.items():
        cand = 0
        for node in nodes:
            cand += numofChildren(node,tree)

        for leaf in leaves:
            if leaf != node:
                dist = float('inf')
                for node in nodes:
                    if leaf == node:
                        continue
                    if nodetolevel[leaf] < nodetolevel[node]:
                        #print(level, leaf, node)
                        ancestor = lca_nary_tree(1, leaf, node)
                        dist = min(nodetolevel[leaf]- nodetolevel[ancestor], dist)
                if dist != float('inf'):
                    cand += dist
        res = min(cand, res)
        #print(level, cand)

    return res

for i in range(N):
    M = int(input())
    for j in range(M-1):
        u,v = list(map(int,input().split()))

        graph[u].append(v)
        graph[v].append(u)

    print(helper(graph))
    graph.clear()
    nodetolevel.clear()
    levels.clear()
    tree.clear()

**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
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import sys
import random
from math import *
from string import ascii_lowercase,ascii_uppercase
from collections import Counter, defaultdict, deque
from itertools import accumulate, combinations, permutations
from heapq import heappushpop, heapify, heappop, heappush
from bisect import bisect_left,bisect_right
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
inp = lambda: sys.stdin.buffer.readline().decode().strip()
out=sys.stdout.write
def S():
    return inp()
def I():
    return int(inp())
def MI():
    return map(int, inp().split())
def MF():
    return map(float, inp().split())
def MS():
    return inp().split()
def LS():
    return list(inp().split())
def LI():
    return list(map(int,inp().split()))
def print1(x):
    return out(str(x)+"\n")
def print2(x,y):
    return out(str(x)+" "+str(y)+"\n")
def print3(x,y,z):
    return out(str(x)+" "+str(y)+" "+str(z)+"\n")
def Query(i,j):
    print3('?', i, j)
    sys.stdout.flush()
    qi=int(inp())
    return qi
def print_arr(arr):
    for num in arr:
        out(str(num)+" ")
    out(str("\n"))
for _ in range(I()):
    n=I()
    graph,ans,curr,invalid,child=[[] for _ in range(n)],n,0,[0]*n,[-1]*n
    ff=[]
    for _ in range(n-1):
        u,v=MI()
        u-=1; v-=1;
        graph[u].append(v)
        graph[v].append(u)
        ff.append(u); ff.append(v);
    dq=deque()
    for i in range(1,n):
        if len(graph[i])==1: dq.append((i,-1))
    while dq:
        node,par=dq.popleft()
        if par!=-1: child[par]=node
        for chi in graph[node]:
            if chi==par or chi==0: continue
            dq.append((chi,node))
    dq=deque()
    dq.append((0,-1))
    def rollback(x):
        global curr
        node=x
        while node>0 and len(graph[node])-invalid[node]==1:
            curr-=1
            node=child[node]
            invalid[node]+=1
    while dq:
        t,gg=len(dq),[]
        for _ in range(t):
            node,par=dq.popleft()
            curr+=1
            for nei in graph[node]:
                if nei==par: continue
                dq.append((nei,node))
            if node>0 and len(graph[node])==1: gg.append(node)
        ans=min(ans,n-curr)
        for c in gg: rollback(c)
    print1(ans)

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given code, we need to consider edge cases that can cause the algorithm to fail. Here are some potential scenarios:

1. **Tree with a single path from the root to a leaf**: This can help test if the algorithm correctly handles linear trees.
2. **Tree with multiple leaves at different distances from the root**: This can help test if the algorithm correctly identifies the minimum number of operations needed to make all leaves at the same distance from the root.
3. **Tree with a complex structure**: This can help test if the algorithm correctly handles more intricate tree structures.

Given the constraints and the need for a small-scale test input, let's create a test case that combines these scenarios.

### Test Case:
- **Tree with a single path from the root to a leaf**:
  - This will help ensure the algorithm can handle linear trees.
- **Tree with multiple leaves at different distances from the root**:
  - This will help ensure the algorithm can correctly identify the minimum number of operations needed.

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

### Explanation:
1. **First Test Case (n=5)**:
   - The tree is a linear path from the root (1) to the leaf (5).
   - The correct answer should be 3, as we need to remove the edges (1, 2), (2, 3), and (3, 4) to make all leaves at the same distance from the root.

2. **Second Test Case (n=7)**:
   - The tree has multiple leaves at different distances from the root.
   - The correct answer should be 2, as we need to remove the edges (2, 5) and (3, 7) to make all leaves at the same distance from the root.

3. **Third Test Case (n=10)**:
   - The tree has a more complex structure with multiple leaves at different distances from the root.
   - The correct answer should be 5, as we need to remove the edges (1, 2), (2, 3), (4, 6), (5, 7), and (5, 8) to make all leaves at the same distance from the root.

This test input is designed to cover a range of scenarios that can help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;
public class treeprun
{
    Scanner sc = new Scanner(System.in);
    static final int inf = Integer.MAX_VALUE;
    static final int mod = 1000000007;
    static final int modl = 1000000007;
    static int maxd=0;
    static int count=0;
    static int max(int a, int b)
    {
        return a>b?a:b;
    }
    static int min(int a, int b)
    {
        return a<b?a:b;
    }
    static void in(int n, int arr[], Scanner sc)
    {
        for(int i=0;i<n;i++)
        arr[i]=sc.nextInt();
    }
    static int in(Scanner sc)
    {
        return sc.nextInt();
    }
    static void println(long a)
    {
        System.out.println(a);
    }
    static void println(double a)
    {
        System.out.println(a);
    }
    static void println(String a)
    {
        System.out.println(a);
    }
    static void print(long a)
    {
        System.out.print(a+" ");
    }
    static void print(double a)
    {
        System.out.print(a+" ");
    }
    static void print(String a)
    {
        System.out.print(a+" ");
    }
    static void println()
    {
        System.out.println();
    }
    static void printarr(int arr[])
    {
        for(int i=0;i<arr.length;i++)
        {
            System.out.print(arr[i]+" ");
        }
        System.out.println();
    }
    static void sort(int a[])
    {
        Arrays.sort(a);
    }
    static void revsort(int arr[])
    {
        Arrays.sort(arr);
        for (int i = 0; i < arr.length / 2; i++)
        {
           int temp = arr[i];
           arr[i] = arr[arr.length - 1 - i];
           arr[arr.length - 1 - i] = temp;
        }
    }
    static void puts(HashMap<Integer, Integer> map, int n)
    {
        map.put(n,map.getOrDefault(n,0)+1);
    }
    static void dfs1(int current, int parent, ArrayList<ArrayList<Integer>> list, HashMap<Integer, Integer> map, int cd)
    {
        if(cd>maxd)maxd=cd;
        map.put(cd,map.getOrDefault(cd,0)+1);
        for(int x:list.get(current))
        {
            if(x==parent)continue;
            dfs1(x,current,list,map,cd+1);
        }
    }

    static void dfs2(int current, int parent, ArrayList<ArrayList<Integer>> list, int val[], int cd)
    {
        count++;
        if(val[cd]==0)val[cd]+=cd;
        else val[cd]+=count;
        for(int x:list.get(current))
        {
            if(x==parent)continue;

            dfs2(x,current,list,val,cd+1);
        }
        count=0;
    }
    static void solve(Scanner sc)
    {
        int n=in(sc);
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();
        for(int i=0;i<=n;i++)
        {
            list.add(new ArrayList<Integer>());
        }
        for(int i=0;i<n-1;i++)
        {
            int u=in(sc);
            int v=in(sc);
            list.get(u).add(v);
            list.get(v).add(u);
        }

        HashMap<Integer, Integer> map = new HashMap<>();
        dfs1(1,0,list,map,1);

        int greater[] = new int[maxd+1];
        greater[maxd]=0;
        for(int i=maxd-1;i>=1;i--)
        {
            greater[i]=greater[i+1]+map.get(i+1);
        }
        int val[] = new int[n+1];

        dfs2(1,0,list,val,1);
        count=0;
        int summin=inf;
        for(int i=1;i<=maxd;i++)
        {
            int sum=greater[i]+n-greater[i]-val[i];
            if(sum<summin)summin=sum;
        }
        println(summin);
        maxd=0;
    }
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();
        while(cases-->0)solve(sc);
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider scenarios that might cause the code to fail. Here are some potential edge cases and scenarios to consider:

1. **Tree with a single path from root to leaves**: This can test the logic for handling linear structures.
2. **Balanced tree**: This can test the logic for handling symmetric structures.
3. **Unbalanced tree with deep branches**: This can test the logic for handling deeply nested structures.
4. **Tree with multiple levels of leaves at different distances**: This can test the logic for handling complex structures with varying leaf distances.

Let's create a small-scale test input that combines some of these elements to maximize the failure probability:

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

### Explanation
1. **First Test Case (5 nodes)**:
   - The tree is a simple path from node 1 to node 5.
   - This tests the logic for handling a linear structure.
   - The expected output is 4, as all nodes except the root need to be removed to make the tree have all leaves at the same distance from the root.

2. **Second Test Case (7 nodes)**:
   - The tree is a balanced structure with the root at node 1 and two branches.
   - This tests the logic for handling symmetric structures.
   - The expected output is 2, as removing edges (1, 4) and (3, 7) will make all leaves at the same distance from the root.

3. **Third Test Case (9 nodes)**:
   - The tree has multiple levels of leaves at different distances.
   - This tests the logic for handling complex structures with varying leaf distances.
   - The expected output is 4, as removing edges (2, 4), (2, 5), (3, 7), and (6, 9) will make all leaves at the same distance from the root.

This test input should help expose any bugs in the given code, particularly in the logic for handling different tree structures and leaf distances.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;
public class TreePruning { static ArrayList<ArrayList<Integer>>list=new ArrayList<>();static int dp[][];
static boolean visit[];
    public static void main(String args[])
    {
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        while(t-->0)
        {
            int n=in.nextInt();
           list=new ArrayList<>();
           for(int i=0;i<=n;i++)
           list.add(new ArrayList<>());
           dp=new int[n+1][2];visit=new boolean[n+1];
           for(int i=0;i<n-1;i++)
           {
            int x=in.nextInt(),y=in.nextInt();
            list.get(x).add(y);list.get(y).add(x);
           }
           dfs(1,1);
           Arrays.sort(dp,(a,b)->a[0]-b[0]);int pre[]=new int[n+1];
           for(int i=1;i<n+1;i++)
           {
                if(dp[i][1]==1)pre[i]=pre[i-1]+1;
                else pre[i]=pre[i-1];
                // System.out.println(pre[i]+"o");
           }long ans=100000000;
           long s[]=new long[n+2];
           s[n+1]=0;
           s[n]=dp[n][0];
           for(int i=n-1;i>=1;i--){
           s[i]=dp[i][0]+s[i+1];}
           for(int i=1;i<=n;i++)
           {
            int left=1,right=n,index=n+1;int x=dp[i][0];
            while(left<=right)//just greater
            {
                int mid=(left+right)/2;
                if(dp[mid][0]>x)
                {index=mid;right=mid-1;}
                else left=mid+1;
            }left=1;right=n;int index2=1;
            while(left<=right)//just smaller
            {
                int mid=(left+right)/2;
                if(dp[mid][0]>=x)
                {right=mid-1;}
                else {index2=mid;left=mid+1;}
            }
            // System.out.println(index+" "+index2+" "+s[index]+" "+pre[index2]);
            ans=Math.min(ans,-(long)x*(long)(n+1-index)+s[index]+pre[index2]);
            // System.out.println(ans);
           }System.out.println(ans);
        }
    }
    public static void dfs(int parent,int c)
    {
        if(visit[parent])return;
        dp[parent][0]=c;visit[parent]=true;
        if(list.get(parent).size()==1&&parent!=1){dp[parent][1]=1;}
           for(int i:list.get(parent))
           {
            dfs(i,c+1);
           }
           return;
}}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with handling certain tree structures, particularly those where the pruning logic might not correctly account for all leaf nodes at the same distance from the root.

Let's consider a small-scale test case that can expose the bug:

1. **Tree with a single path from the root to a leaf:**
   - This can help verify if the code correctly identifies that no pruning is needed when all leaves are already at the same distance from the root.

2. **Tree with multiple levels and leaves at different distances:**
   - This can help verify if the code correctly identifies the minimum number of operations needed to make all leaves at the same distance from the root.

3. **Tree with a single leaf at a different distance:**
   - This can help verify if the code correctly identifies the need to prune a single leaf to achieve the goal.

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

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

Explanation:
1. **First Test Case:**
   - A tree with 3 nodes, where nodes 2 and 3 are leaves at distance 1 from the root.
   - Expected output: 0 (no pruning needed).

2. **Second Test Case:**
   - A tree with 7 nodes, where nodes 5, 6, and 7 are leaves at distances 2, 2, and 3 from the root.
   - Expected output: 2 (pruning nodes 5 and 7).

3. **Third Test Case:**
   - A tree with 5 nodes, where nodes 2, 3, 4, and 5 are leaves at distance 1 from the root.
   - Expected output: 0 (no pruning needed).

This test input should help expose any issues in the buggy code by covering different tree structures and leaf configurations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import sys
from collections import defaultdict
from sys import stdin, stdout

sys.setrecursionlimit(100000)

def dfs(p, v, adj, depth, max_depth_under):
    if p != v:
        depth[v] = depth[p] + 1
    else:
        depth[v] = 0

    max_depth_under[v] = depth[v]
    for c in adj[v]:
        if c == p:
            continue
        dfs(v, c, adj, depth, max_depth_under)
        max_depth_under[v] = max(max_depth_under[v], max_depth_under[c])

def main():
    t = int(input())

    for _ in range(t):
        n = int(input())

        adj = defaultdict(list)
        for _ in range(n - 1):
            a, b = map(int, input().split())
            a -= 1  # Make zero-based
            b -= 1
            adj[a].append(b)
            adj[b].append(a)

        depth = [0] * n
        max_depth_under = [0] * n

        # Perform DFS
        dfs(0, 0, adj, depth, max_depth_under)

        depth_counts = [0] * n
        max_depth_counts = [0] * n

        for i in range(n):
            depth_counts[depth[i]] += 1
            max_depth_counts[max_depth_under[i]] += 1

        depth_at_least = [0] * (n + 1)
        max_depth_less_than = [0] * (n + 1)

        # Calculate depth_at_least
        for i in range(n - 1, -1, -1):
            depth_at_least[i] = depth_at_least[i + 1] + depth_counts[i]

        # Calculate max_depth_less_than
        for i in range(1, n + 1):
            max_depth_less_than[i] = max_depth_less_than[i -
                                                         1] + max_depth_counts[i - 1]

        # Calculate the result
        res = sys.maxsize
        for d in range(n):
            res = min(res, depth_at_least[d + 1] + max_depth_less_than[d])

        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Leaf Nodes at Different Depths**: The code needs to handle cases where leaf nodes are at different depths and the goal is to prune the tree such that all remaining leaves are at the same depth from the root.

2. **Multiple Test Cases**: The code should handle multiple test cases efficiently.

3. **Edge Cases**: Consider edge cases where the tree is already in the desired form or where the tree is a simple path.

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

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

Explanation of the test cases:
1. **First Test Case**: The tree is as described in the problem statement. By removing edges (1, 3) and (2, 5), the resulting tree has all leaves at the same distance from the root.
2. **Second Test Case**: Similar to the first test case, but with a different structure. By removing edges (1, 4) and (5, 7), the resulting tree has all leaves at the same distance from the root.
3. **Third Test Case**: A more complex tree with multiple branches. The goal is to find the minimum number of operations to make all leaves at the same distance from the root. This case should challenge the code to handle multiple levels and branches efficiently.

This test input should help expose any issues in the buggy code, especially in handling different tree structures and ensuring the correct minimum number of operations is calculated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
from collections import Counter

def dpt(node, cur_dept, tree, depth, max_depth, used):

    depth[node] = cur_dept
    used.add(node)

    if node in tree:
        depth_ = [dpt(i, cur_dept + 1, tree, depth, max_depth, used) for i in tree[node] if i not in used]

        if len(depth_) == 0:
            depth_ = [cur_dept]

        max_depth_ = max(depth_)
        max_depth[node] = max_depth_
        return max_depth_

    max_depth[node] = cur_dept

    return cur_dept

def solution_():

    n = int(input())

    tree = {}

    for i in range(n - 1):
        par, ch = map(int, input().split())
        if par in tree:
            tree[par].append(ch)
        else:
            tree[par] = [ch]

        if ch in tree:
            tree[ch].append(par)
        else:
            tree[ch] = [par]

    if n <= 2:
        return 0

    used = set()

    depth = {}
    max_depth = {}

    _ = dpt(1, 0, tree, depth, max_depth, used)

    depth_ = list(depth.values())
    depth = {}
    for i in depth_:
        if i in depth:
            depth[i] += 1
        else:
            depth[i] = 1

    max_depth_ = list(max_depth.values())
    max_depth = {}
    for i in max_depth_:
        if i in max_depth:
            max_depth[i] += 1
        else:
            max_depth[i] = 1

    depth = sorted(list(depth.items()), key = lambda s: s[0])[::-1]
    max_depth = sorted(list(max_depth.items()), key = lambda s: s[0])

    max_depth_ = []
    for i in range(len(max_depth)):
        if i == 0:
            max_depth_ += [max_depth[0]]
        else:
            max_depth_ += [(max_depth[i][0], max_depth_[i - 1][1] + max_depth[i][1])]

    depth_ = []
    for i in range(len(depth)):
        if i == 0:
            depth_ += [depth[0]]
        else:
            depth_ += [(depth[i][0], depth_[i - 1][1] + depth[i][1])]

    depth_ = depth_[::-1]

    depth_ = [(0, n + 1)] + depth_ + [(n + 1, 0)]

    l_index = 0
    r_index = 0
    max_depth_ = [(0, 0)] + max_depth_  + [(n + 1, 0)]

    index_depth_more = 0
    index_max_depth_less = 0
    results = {}

    for i in range(1, n):
        while depth_[index_depth_more][0] <= i:

            index_depth_more += 1

        while max_depth_[index_max_depth_less][0] < i:
            index_max_depth_less += 1

        index_max_depth_less -= 1

        results[i] = depth_[index_depth_more][1] + max_depth_[index_max_depth_less][1]

    # print(depth_)
    # print(max_depth_)
    # print(results)

    return min(results.items(), key = lambda s: s[1])[1]

def solution():
    n = int(input())
    res = []
    for i in range(n):
        res1 = None
        try:
            res1 = solution_()
        except:
            res1 = 0
        res += [res1]

    for i in res:
        print(i)
    return

solution()

**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
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

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

#define int long long

int t, n, a, b, mx, res, cur, dpt[200010], dts[200010], dp[200010], f[200010];
vector<int> v[200010];

void dfs(int u, int fa){
	dpt[u] = dpt[fa] + 1, mx = max(mx, dpt[u]), dts[u] = dpt[u];
	for(int i : v[u])
		if(i != fa)
			dfs(i, u), dts[u] = max(dts[u], dts[i]);
	dp[dts[u]]++, f[dpt[u]]++;
}

signed main(){
	cin >> t;
	while(t--){
		cin >> n, mx = 0, res = 1e18;
		for(int i = 1; i <= n; i++)
			f[i] = dp[i] = 0, v[i].clear();
		for(int i = 2; i <= n; i++)
			cin >> a >> b, v[a].push_back(b), v[b].push_back(a);
		dfs(1, 0);
		int cnt1 = 0, cnt2 = 0;
		for(int i = 1; i < mx; i++)
			cnt1 += dp[i];
		for(int i = mx; i; i--){
			// cout << "\t" << i << ' ' << cnt1 + cnt2 << endl;
			res = min(res, cnt1 + cnt2);
			cnt1 -= dp[i - 1], cnt2 += f[i];
		}
		cout << res << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
/// PARK CHAEYOUNG IS MY LOVE
/// Author: Ngo Ba Van
/// roses_are_rosie
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
#define fi first
#define se second
#define MASK(i) (1LL<<i)
#define BIT(x,i) (x&MASK(i))
#define ONBIT(x,i) (x|MASK(i))
#define OFFBIT(x,i) (x&~MASK(i))
#define COUNT(x) (__builtin_popcount(x))
#define RVAN1102 signed main()
using ii=pair<int,int>;
const ll N=2e5+5,M=1e9+7;
ll n,m,res,depth[N],mx_depth[N],pre[N],suf[N];
vector<int>a[N];
void DFS(int i,int pa)
{
    mx_depth[i]=depth[i];
    for (auto v:a[i])
    {
        if (v==pa) continue;
        depth[v]=depth[i]+1;
        DFS(v,i);
        mx_depth[i]=max(mx_depth[v],mx_depth[i]);
    }
    pre[mx_depth[i]]++;
    suf[depth[i]]++;
}
void nhap()
{
    memset(pre,0,sizeof pre);
    memset(suf,0,sizeof suf);
    cin>>n;
    for (int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        a[x].push_back(y);
        a[y].push_back(x);
    }
    depth[1]=1;
    DFS(1,0);
    ///
    for (int i=1;i<=n;i++)
    pre[i]+=pre[i-1];
    for (int i=n;i>=1;i--)
    suf[i]+=suf[i+1];
    ///
    res=M;
    for (int i=1;i<=n;i++)
    res=min(res,pre[i-1]+suf[i+1]);
    cout<<res<<"\n";
    ///
    for (int i=1;i<=n;i++)
    a[i].clear();
}
RVAN1102 {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#define task ""
#define task "a"
if(fopen(task".inp","r"))
{
    freopen(task".inp","r",stdin);
    freopen(task".out","w",stdout);
}
int t=1;
cin>>t;
while (t--)
nhap();
}

**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 edge cases and scenarios that maximize the failure probability. The key points to focus on are:

1. **Multiple Test Cases**: Ensure the test input includes multiple test cases to check the robustness of the solution.
2. **Edge Cases**: Test with trees that have specific structures, such as linear chains, balanced trees, and unbalanced trees.
3. **Leaf Nodes**: Create scenarios where the tree has a significant number of leaf nodes at different distances from the root.
4. **Large and Small Trees**: Test with both large and small trees to ensure the solution handles all sizes correctly.

Given these points, let's create a test input that includes a variety of tree structures and sizes:

```plaintext
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
1 2
1 3
2 4
2 5
4 6
4 7
3 8
8 9
8 10
9 11
9 12
10 13
10 14
10 15
```

### Explanation:
1. **First Test Case (7 nodes)**:
   - This tree is a simple unbalanced tree with a root node 1.
   - It has leaves at different distances: 3, 2, 1, 2, 1.
   - The expected output is 2, as removing edges (1, 3) and (2, 5) results in all leaves at the same distance from the root.

2. **Second Test Case (7 nodes)**:
   - This tree is another unbalanced tree with a root node 1.
   - It has leaves at different distances: 2, 1, 1, 2, 1.
   - The expected output is 2, as removing edges (1, 4) and (5, 7) results in all leaves at the same distance from the root.

3. **Third Test Case (15 nodes)**:
   - This tree is a more complex unbalanced tree with a root node 1.
   - It has leaves at different distances: 3, 2, 1, 2, 1, 3, 3, 3, 3.
   - The expected output is 5, as removing edges (1, 3), (2, 5), (4, 6), (4, 7), and (3, 8) results in all leaves at the same distance from the root.

This test input should help expose any bugs in the provided code by covering a range of tree structures and sizes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;

public class codeforces {

    static int dfs(int v ,int depth, List<List<Integer>> children, int[] h, int[] visited) {
        visited[v] = 1 ;

        int mx = -1 ;
        for(int c : children.get(v)) {
            if(visited[c] == 1) continue;
            mx = Math.max(mx, dfs(c, depth+1, children, h, visited) ) ;
        }

        //base case
        if(mx == -1) mx = depth ;

        h[v] = mx ;
        return mx ;
    }

    static void dfs2(int v ,int depth, List<List<Integer>> children, int[] h, int[] visited) {
        visited[v] = 1 ;

        if(children.get(v).isEmpty()) {  //leaf node
            h[v] = depth  ;
            return ;
        }

        h[v] = depth ;

        for(int c : children.get(v)) {
            if(visited[c] == 1) continue;
            dfs2(c, depth+1, children, h, visited)  ;
        }

    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        int u = 0 ;

        while (tc-- > 0) {
            u++ ;
            int n = sc.nextInt();

            List<List<Integer>> children = new ArrayList<>(); // List to store children for each vertex
            for (int i = 0; i <= n; i++) {
                children.add(new ArrayList<>());
            }

            for (int i = 0; i < n-1; i++) {
                int p = sc.nextInt() , c = sc.nextInt() ;
                children.get(p).add(c) ;
                children.get(c).add(p) ;
            }

            int[] vh = new int[n+1] , vlh = new int[n+1] ;
//            Arrays.fill(vh, -1);
//            Arrays.fill(vlh, -1);
            int[] visited = new int[n+1] ;
            dfs(1, 0, children, vh, visited) ;
            Arrays.fill(visited, 0);
            dfs2(1, 0, children, vlh, visited) ;

            int[] pre = new int[n+1] , suf = new int[n+1] ;

            int mxh = 0 ;

           // System.out.println(children);

            for (int i = 1; i < n+1; i++) {
                mxh = Math.max(mxh , vh[i]) ;
                pre[vh[i]]++ ;
                suf[vlh[i]]++ ;
            }

            for (int i = 1; i <n+1 ; i++) {
                pre[i] += pre[i-1] ;
            }

            for (int i = n-1; i >= 0 ; i--) {
                suf[i] += suf[i+1] ;
            }

//            for(int height : vh) {
//                System.out.print(height + " , ");
//            }
//            System.out.println();
//            for(int height : vlh) {
//                System.out.print(height + " , ");
//            }

            int mincut = Integer.MAX_VALUE ;

            for (int i = 1; i <= mxh ; i++) {
                mincut = Math.min(mincut, pre[i-1] + suf[i+1]) ;
            }

            System.out.println(mincut);

        }

        sc.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 generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause issues in the faulty solution. Here are the steps to think through this:

1. **Identify Potential Bugs**:
   - The code uses two DFS functions (`dfs` and `dfs2`) to calculate the heights of nodes.
   - The `dfs` function calculates the maximum depth of subtrees.
   - The `dfs2` function seems to be recalculating the depths, but it might not handle all cases correctly.
   - The final calculation of the minimum number of operations might be incorrect due to the way `pre` and `suf` arrays are used.

2. **Edge Cases**:
   - Trees with a single leaf at a different depth from other leaves.
   - Trees with multiple levels of nodes where the depth of leaves varies significantly.
   - Trees with a large number of leaves at the same depth but a few leaves at a different depth.

3. **Generate a Small-Scale Test Input**:
   - We will create a tree with a root node and several levels of nodes, ensuring that the leaves are at different depths to test the correctness of the depth calculations and the minimum operations calculation.

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

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

This input creates a tree with the following structure:
```
  1
 / \
2   3
/ \ / \
4 5 6 7
```

In this tree:
- Nodes 4, 5, 6, and 7 are leaves.
- The depths of the leaves are:
  - Node 4: depth 2
  - Node 5: depth 2
  - Node 6: depth 2
  - Node 7: depth 2

The correct output should be `0` because all leaves are already at the same depth. However, if the code has issues with the depth calculation or the minimum operations calculation, it might produce an incorrect result.

This test input should help identify any issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

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

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

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

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();

        while (t-- > 0) {
            int n = scan.nextInt();

            List<List<Integer>> g = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                g.add(new ArrayList<>());
            }

            for (int i = 0; i < n - 1; i++) {
                int u = scan.nextInt() - 1;
                int v = scan.nextInt() - 1;
                g.get(u).add(v);
                g.get(v).add(u);
            }

            int[] cnt = new int[n]; cnt[0] = 1;
            for (int i = 0; i < n; i++) {
                cnt[i] += g.get(i).size() - 1;
            }

            int[] p = new int[n];
            Queue<Integer> q = new LinkedList<>();
            Queue<Integer> nxt_q = new LinkedList<>();
            q.add(0);
            int m = 1, ans = n - m;
            while (!q.isEmpty()) {
                int u = q.poll();
                for (int v : g.get(u)) {
                    if (v == p[u]) continue;
                    p[v] = u;
                    nxt_q.add(v);
                    m++;
                }
                ans = Math.min(n - m, ans);
                if (q.isEmpty()) {
                    while (!nxt_q.isEmpty()) {
                        int v = nxt_q.poll();
                        boolean ok = true;
                        while (cnt[v] == 0) {
                            ok = false;
                            m--;
                            v = p[v];
                            cnt[v]--;
                        }
                        if (ok) q.add(v);
                    }
                }
            }
            System.out.println(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
3
4
1 2
1 3
1 4
5
1 2
1 3
1 4
1 5
6
1 2
1 3
1 4
1 5
1 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <string>

using namespace std;
using ll = long long;

ll maxn = 5001;
vector <vector <int>> g(maxn);
vector <ll> a(maxn);
vector <ll> dp(maxn);
ll ans = 0;

void dfs(int v, int p) {
    dp[v] = 1;
    ll sm = 0;
    ll mn = 1e9;
    ll cnt = 0;
    for (int u : g[v]) {
        if (u != p) {
            dfs(u, v);
            sm += a[u];
            mn = min(mn, dp[u]);
            cnt++;
        }
    }
    if (sm < a[v] && cnt > 0) {
        ans += mn * (a[v] - sm);
    }
    if (g[v].size() > 1) {
        dp[v] = mn + 1;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll tt = 1;
    cin >> tt;
    while (tt--) {
        ll n;
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
        }
        for (int i = 2; i <= n; i++) {
            int p;
            cin >> p;
            g[p].push_back(i);
            g[i].push_back(p);
        }
        dfs(1, 1);
        cout << ans;
        for (int i = 1; i <= n; i++) {
            g[i].clear();
            dp[i] = 0;
            a[i] = 0;
        }
        ans = 0;
        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 expose the bug in the provided code, we need to focus on scenarios where the logic for calculating the minimum number of operations might fail. Specifically, the code might not correctly handle cases where the tree has a more complex structure, such as multiple levels of children with varying values.

Here's a step-by-step approach to generating a test input:

1. **Tree Structure**: Create a tree with multiple levels and multiple children for some nodes.
2. **Values on Vertices**: Assign values to the vertices such that the condition for a wonderful tree is not initially met, but the minimum number of operations required is not straightforward.
3. **Edge Case**: Ensure that the tree structure and values are such that the code's logic for calculating the minimum number of operations might fail.

Let's consider a tree with 7 vertices:
- Root (1) has two children (2 and 3).
- Child 2 has two children (4 and 5).
- Child 3 has two children (6 and 7).

Assign values to the vertices as follows:
- \( a_1 = 10 \)
- \( a_2 = 5 \)
- \( a_3 = 6 \)
- \( a_4 = 2 \)
- \( a_5 = 3 \)
- \( a_6 = 4 \)
- \( a_7 = 2 \)

This setup ensures that the root and its direct children do not satisfy the condition for a wonderful tree, and it requires multiple operations to make the tree wonderful.

Here is the test input:

```plaintext
1
7
10 5 6 2 3 4 2
1 1 2 2 3 3
```

Explanation:
- The root (1) has a value of 10, and its children (2 and 3) have values 5 and 6, respectively.
- Child 2 has children (4 and 5) with values 2 and 3.
- Child 3 has children (6 and 7) with values 4 and 2.

To make the tree wonderful:
- For vertex 1: \( 10 \le 5 + 6 \) (already satisfied).
- For vertex 2: \( 5 \le 2 + 3 \) (already satisfied).
- For vertex 3: \( 6 \le 4 + 2 \) (already satisfied).

However, the code might incorrectly calculate the minimum number of operations needed. This test input will help expose any issues in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

Buggy Code:
#include <bits/stdc++.h>
// Common file
// #include <ext/pb_ds/assoc_container.hpp>
// Including tree_order_statistics_node_update
// #include <ext/pb_ds/tree_policy.hpp>
// #include <map>
#define fors(n) for (int i = 0; i < n; i++)
#define ULL unsigned long long
#define int long long
#define F first
#define S second
#define pb push_back
#define newl cout << "\n";
#define show(c) cout << c << "\n";
#define spcshow(c) cout << c << ' ';
#define print(v) for(auto it : v){cout << it << ' ';}newl
#define vi vector<int>
#define vpi vector<pair<int, int>>
#define vvi vector<vector<int>>
using namespace std;
// using namespace __gnu_pbds;
const int m1 = 1e9 + 7;
const int m2 = 998244353;
int max(int a, int b){
    return a > b ? a : b;
}
int min(int a, int b){
    return a < b ? a : b;
}
int absol(int n){
    if (n < 0)
        return -n;
    return n;
}
int binexp(int a, int b, int m=1000000000000000000){
    a %= m;
    int res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
        // show(res)
    }
    return res;
}
void inarr(int& n,vi &a){
    int e;
    fors(n){
        cin>>e;
        a.pb(e);
    }
    return;
}
int ans;
vi dp;
vi a;
void DFS(int v,int pa,vvi &edges,map<int,int> &xm){
    int sum=0,cost=INT_MAX;
    map<int,int> m;
    for(auto it:edges[v]){
        if(it==pa) continue;
        // map<int,int> x;
        DFS(it,v,edges,m);
        cost=min(cost,dp[it]);
        sum+=a[it];
    }
    // show(m.size())
    if(cost==INT_MAX) cost=0;
    // show(v)
    // for(auto it:m){
    //     show(it.F<<' '<<it.S)
    // }
    if(a[v]>sum){
        // ans+=cost*(a[v]-sum);
        while(a[v]>sum){
            if(m.empty()){
                sum=a[v];
                continue;
            }
            // show(a[v]<<' '<<sum<<' '<<m.begin()->F<<' '<<m.begin()->S)
            if(m.begin()->S>a[v]-sum){
                ans+=m.begin()->F*(a[v]-sum);
                if(m.begin()->S!=LLONG_MAX){
                    m.begin()->S-=a[v]-sum;
                }
                sum=a[v];
            }else{
                ans+=m.begin()->F*(m.begin()->S);
                sum+=m.begin()->S;
                m.erase(m.begin());
            }
        }
    }
    else if(a[v]<sum){
        m[0]+=sum-a[v];
    }
    // dp[v]=cost+1;
    for(auto it:m){
        // show(v<<' '<<it.F+1<<' '<<it.S)
        xm[it.F+1]=it.S;
    }
    xm[cost+1]=LLONG_MAX;
    // show(v<<' '<<cost+1<<' '<<LONG_LONG_MAX)
    dp[v]=cost+1;
    // show(v<<' '<<cost+1)
    return;
}
void solve(){
    a.clear();
    int n;
    cin>>n;
    dp.clear();
    dp.resize(n);
    inarr(n,a);
    int e;
    vvi edges(n);
    fors(n-1){
        cin>>e;
        edges[i+1].pb(e-1);
        edges[e-1].pb(i+1);
    }
    ans=0;
    map<int,int> temp;
    DFS(0,-1,edges,temp);
    show(ans)
    return;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int testcase = 1;
    cin>>testcase;
    cout << std::setprecision(50);
    // int v=1;
    // f.pb(1);
    // f.pb(1);
    // fi.pb(1);
    // fi.pb(1);
    // for(int i=2;i<=1e6;i++){
    //     v=(v*i)%m2;
    //     f.pb(v);
    //     fi.pb(binexp(v,m2-2,m2));
    // }
    while (testcase--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The bug in the code seems to be related to how the algorithm handles the distribution of values and the calculation of the minimum number of operations needed to make the tree wonderful.

### Step-by-Step Analysis:

1. **Tree Structure and Values:**
   - The tree should have a structure where the root and its children have values that are not initially in a wonderful state.
   - The values should be such that the algorithm needs to perform a significant number of operations to make the tree wonderful.

2. **Edge Cases:**
   - **Single Child Nodes:** Nodes with only one child.
   - **Multiple Child Nodes:** Nodes with multiple children.
   - **Zero Values:** Nodes with values of zero.
   - **Large Values:** Nodes with large values that require many operations to balance.

3. **Test Case Generation:**
   - We will create a small tree with a root node and a few children to ensure the test case is manageable.
   - The values will be chosen such that the root node's value is significantly smaller than the sum of its children's values, requiring multiple operations to balance.

### Chosen Test Case:
- **Tree Structure:**
  - Root node (1) with three children (2, 3, 4).
  - Each child has a value such that the sum of the children's values is much larger than the root's value.

- **Values:**
  - Root node (1): 1
  - Child nodes (2, 3, 4): 10, 10, 10

- **Edges:**
  - 2 -> 1
  - 3 -> 1
  - 4 -> 1

### Test Input:
```plaintext
1
4
1 10 10 10
1 1 1
```

This test case will force the algorithm to perform multiple operations to make the tree wonderful. The root node (1) has a value of 1, while the sum of its children's values (10 + 10 + 10 = 30) is much larger. The algorithm should need to perform 29 operations to increase the root node's value to 30, making the tree wonderful.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

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

using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define all(a) a.begin(), a.end()

const ll INF = 1e18;

vector<vector<ll>> g;
vector<ll> a, cost, updCost;

void dfs(ll v) {
    if (g[v].size() == 0) {
        updCost[v] = 1;
        return;
    }
    ll sum = 0;
    for (auto to : g[v]) {
        dfs(to);
        sum += a[to];
    }
    if (sum >= a[v]) return;
    ll mn = INF;
    for (auto to : g[v]) {
        if ((a[v] - sum) * updCost[to] < mn) {
            mn = (a[v] - sum) * updCost[to];
        }
    }
    for (auto to : g[v]) {
        updCost[v] = min(updCost[v], updCost[to] + 1);
    }
    cost[v] = mn;
}

void solve() {
    ll n;
    cin >> n;
    g.clear();
    g.resize(n);
    a.resize(n);
    cost.assign(n, 0);
    updCost.assign(n, INF);
    for (ll i = 0; i < n; ++i) {
        cin >> a[i];
    }
    for (ll i = 1; i < n; ++i) {
        ll p;
        cin >> p;
        g[p - 1].push_back(i);
    }
    dfs(0);
    ll ans = 0;
    for (auto el : cost) ans += el;
    cout << ans << "\n";
}

int main() {
    ll t = 1;
    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:
```plaintext
1
4
1 2 3 4
1 2 3
```

**Explanation:**
This test case is designed to expose a potential issue in the buggy code. Here's the reasoning:

1. **Tree Structure:**
   - The tree has 4 vertices.
   - Vertex 1 is the root.
   - Vertex 2 is a child of vertex 1.
   - Vertex 3 is a child of vertex 2.
   - Vertex 4 is a child of vertex 3.

2. **Initial Values:**
   - \(a_1 = 1\)
   - \(a_2 = 2\)
   - \(a_3 = 3\)
   - \(a_4 = 4\)

3. **Check for Wonderful Tree:**
   - For vertex 1: \(a_1 = 1\) and the sum of its children's values is \(a_2 = 2\). Since \(1 \le 2\), vertex 1 is already wonderful.
   - For vertex 2: \(a_2 = 2\) and the sum of its children's values is \(a_3 = 3\). Since \(2 \le 3\), vertex 2 is already wonderful.
   - For vertex 3: \(a_3 = 3\) and the sum of its children's values is \(a_4 = 4\). Since \(3 \le 4\), vertex 3 is already wonderful.
   - For vertex 4: It has no children, so it is trivially wonderful.

4. **Expected Output:**
   - Since the tree is already wonderful, the expected number of operations is 0.

5. **Potential Issue:**
   - The buggy code initializes `updCost` with `INF` and sets `updCost[v] = 1` for leaf nodes. This might lead to incorrect calculations when propagating the costs up the tree.
   - The test case ensures that the tree is already wonderful, which should result in no operations being needed. This helps in verifying if the code correctly handles the already wonderful tree scenario.

By using this test case, you can check if the buggy code correctly handles a tree that is already wonderful and if it avoids unnecessary operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

Buggy Code:
import math
M=1000000007

a=[]
p=[]
dis=[]
ans=0

def dfs(x):
    global ans
    if(len(p[x])==0):dis[x]=1
    else:
        dis[x]=5010
        sum=0
        for y in p[x]:
            dfs(y)
            sum+=a[y]
            dis[x]=min(dis[x],dis[y]+1)
        # print(x,a[x]-sum)
        ans+=max((a[x]-sum)*(dis[x]-1),0)
        a[x]=max(a[x],sum)

def solve():
    global a,dis,ans
    n=int(input())

    a=list(map(int,input().split(" ")))
    b=list(map(int,input().split(" ")))
    a=[0]+a
    b=[0,0]+b

    p.clear()
    for i in range(n+1):p.append([])
    for i in range(2,n+1):
        p[b[i]].append(i)

    dis=[0]*(n+1)
    ans=0

    dfs(1)
    print(ans)

n=1
n=int(input())
while(n>0):
    n-=1
    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 expose the bug in the provided code, we need to focus on scenarios that might cause the algorithm to fail. The key issues in the buggy code are:

1. **Incorrect handling of the `dis` array**: The `dis` array is used to store the minimum distance from a node to its farthest leaf, but it is initialized and used incorrectly.
2. **Incorrect calculation of the number of operations**: The formula `ans+=max((a[x]-sum)*(dis[x]-1),0)` is likely incorrect and might overestimate the number of operations needed.

To create a challenging test case, we should consider the following scenarios:
- A tree with a single root and multiple children where the root value is significantly higher than the sum of its children.
- A tree where the values are such that the tree is already wonderful, but the code might still perform unnecessary operations.
- A tree with a deep structure to test the recursive depth and correctness of the `dis` array.

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

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

Explanation:
1. **Test Case 1**: A tree with a root node (1) having a value of 10 and three children (2, 3, 4) each with a value of 1. The root value is significantly higher than the sum of its children, which is 3. The correct answer should be 7 (10 - 3 = 7).
2. **Test Case 2**: A tree with a root node (1) having a value of 10 and two children (2, 3) each with a value of 1. The root value is significantly higher than the sum of its children, which is 2. The correct answer should be 8 (10 - 2 = 8).
3. **Test Case 3**: A tree with a root node (1) having a value of 10, two children (2, 3) each with a value of 1, and two more children (4, 5) of node 2 each with a value of 1. The root value is significantly higher than the sum of its children, which is 4. The correct answer should be 6 (10 - 4 = 6).

These test cases should help expose the bug in the provided code by highlighting issues with the calculation of the number of operations and the handling of the `dis` array.