**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <random>
#include <numeric>
#include <cmath>
#include <utility>
#include <string>
#include <iomanip>
#include <functional>
#define FOR(i, s, e, t) for((i) = (s); (i) <= (e); (i) += (t))
#define ROF(i, s, e, t) for((i) = (e); (i) >= (s); (i) -= (t))
#define REP(s, e) for(int i = (s); i <= (e); ++i)
#define PER(s, e) for(int i = (e); i >= (s); --i)
#define ALL(x) (x).begin(), (x).end()
#define LLA(x) (x).rbegin(), (x).rend()
#define SIZE(x) ((int)(x).size())
#define AMONGlr(x, l, r) ((x) >= (l) && (x) <= (r))
#define AMONG(x, l, r) ((x) > (l) && (x) < (r))
#define AMONGl(x, l, r) ((x) >= (l) && (x) < (r))
#define AMONGr(x, l, r) ((x) > (l) && (x) <= (r))
#define pb push_back
#define qb pop_back
#define eb emplace_back
#define fi first
#define se second
#define maxe max_element
#define mine min_element
#define mp make_pair
#define upb upper_bound
#define lowb lower_bound
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<string, string> pss;
typedef pair<string, ll> psl;
typedef pair<ll, string> pls;
typedef pair<string, double> psd;
typedef pair<double, string> pds;
typedef pair<ll, double> pld;
typedef pair<double, ll> pdl;
typedef pair<char, ll> pcl;
typedef pair<ll, char> plc;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pdd> vpdd;
typedef vector<pss> vpss;
typedef vector<psl> vpsl;
typedef vector<pls> vpls;
typedef vector<psd> vpsd;
typedef vector<pds> vpds;
typedef vector<pld> vpld;
typedef vector<pdl> vpdl;
typedef vector<pcl> vpcl;
typedef vector<plc> vplc;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<string> vs;
typedef vector<vs> vvs;
template<typename T> bool ckMax(T& a, const T& b) { return a < b ? a = b, true : false; }
template<typename T> bool ckMin(T& a, const T& b) { return a > b ? a = b, true : false; }
inline ll read() {
    ll x = 0;
    short flag = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')flag = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    return x * flag;
}
inline void write(ll x, char end = '\0') {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
    if (end != '\0') putchar(end);
}
ll POW(ll base, ll x, ll mod = 0) {
    ll back = 1;
    while (x > 0) {
        if (x & 1)
            if (mod) back = back * base % mod;
            else back *= base;
        if (mod) base = base * base % mod;
        else base *= base;
        x >>= 1;
    }
    if (mod) return back % mod;
    return back;
}
bool isPrime(ll x) {
    if (x < 3) return x == 2;
    REP(2, sqrt(x))
        if (x % i == 0) return false;
    return true;
}
int __init__ = []() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    return 0;
}();
const ll N = 3e5 + 10, MAX = 1e16;
vl edges[N];
ll degree[N], dp[4][N];
ll n, i, j, u, v;
void calV(ll node, ll pre = -1) {
    if (degree[node] == 1) ckMax(dp[2][node], dp[0][node] + 1);
    dp[1][node] = 1;
    for (auto& next : edges[node]) {
        if (next == pre) continue;
        calV(next, node);
        dp[0][node] += max(dp[0][next], dp[1][next]);
        dp[1][node] += dp[0][next];
    }
    for (auto& next : edges[node]) {
        if (next == pre) continue;
        ckMax(dp[2][node], dp[0][node] - max(dp[0][next], dp[1][next]) + max(dp[2][next], dp[3][next]));
        ckMax(dp[3][node], dp[1][node] - dp[0][next] + dp[2][next]);
    }
}
void solve() {
    cin >> n;
    REP(1, n) {
        edges[i].clear(), degree[i] = 0;
        dp[0][i] = dp[1][i] = dp[2][i] = dp[3][i] = 0;
    }
    REP(1, n - 1) {
        cin >> u >> v;
        edges[u].pb(v);
        edges[v].pb(u);
        degree[u] += 1;
        degree[v] += 1;
    }
    calV(1);
    ll ans = dp[0][1];
    REP(1, 3) ckMax(ans, dp[i][1]);
    cout << ans << endl;
}
int main() {
    ll t;
    cin >> t;
    while (t--) solve();
    return 0;
}

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

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

Think step by step.

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

1. **Star Graph**: A star graph is a tree where one central node is connected to all other nodes. In this case, the maximum number of leaves is the number of nodes minus one (since the central node is not a leaf).

2. **Line Graph**: A line graph is a tree where nodes are connected in a linear sequence. In this case, the maximum number of leaves is 2 (the two endpoints of the line).

3. **Balanced Binary Tree**: A balanced binary tree is a tree where each node has at most two children, and the tree is as balanced as possible. In this case, the maximum number of leaves is more complex to determine but can be a good test case.

4. **Complex Tree**: A tree with a mix of different structures (e.g., a central node with multiple branches) can be a good test case to ensure the algorithm handles various scenarios correctly.

Given the constraints and the need for a small-scale test input, let's focus on a star graph and a balanced binary tree. Here is a small-scale test input that combines both:

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

Explanation:
1. **First Test Case (Star Graph)**:
   - 6 nodes.
   - Node 1 is the center, connected to nodes 2, 3, 4, 5, and 6.
   - Maximum number of leaves: 5 (nodes 2, 3, 4, 5, and 6).

2. **Second Test Case (Balanced Binary Tree)**:
   - 7 nodes.
   - Node 1 is the root, connected to nodes 2 and 3.
   - Node 2 is connected to nodes 4 and 5.
   - Node 3 is connected to nodes 6 and 7.
   - Maximum number of leaves: 4 (nodes 4, 5, 6, and 7).

This test input should be sufficient to expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
import sys
import heapq
import math
from decimal import Decimal, getcontext
sys.setrecursionlimit(200200)
MAX = 1000000000
def dfs(u):
    visited[u] = True
    ce, cne, nce,ncne = 1, 1, 0, 0
    L = []
    for v in graph[u]:
        if not visited[v]:
            cev,cnev,ncev,ncnev = dfs(v)
            L.append((cev,cnev,ncev,ncnev))
    if len(graph[u]) == 1:
        r = 0
        minval = MAX
        for a,b,c,d in L:
            ce += min(a,b,c,d)
            ncne += b
            nce += min(a, b)
            if b >= a:
                r += 1
            else:
                minval = a - b
        if r == 0:
            nce += minval
        cne = MAX
    else:
        r = 0
        rce = 0
        minvalce = MAX
        minval = MAX
        for a,b,c,d in L:
            ce += min(a,b,c,d)
            if min(b,d) >= min(a,c):
                rce += 1
            else:
                minvalce = min(a,c) - min(b,d)
            ncne += b
            nce += min(a, b)
            cne += min(b, d)
            if b >= a:
                r += 1
            else:
                minval = a - b
        if r == 0:
            nce += minval
        if rce == 0:
            ce += minvalce
    return ce, cne, nce, ncne

t = int(sys.stdin.readline())
for _ in range(t):
    n = int(sys.stdin.readline())
    graph = [[] for i in range(n)]
    visited = [False for i in range(n)]
    for _ in range(n - 1):
        u , v = map(int,sys.stdin.readline().split())
        u -= 1
        v -= 1
        graph[u].append(v)
        graph[v].append(u)
    a,b,c,d = dfs(0)
    print(n - min(a - 1,b,c - 1,d))

**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
1 3
2 4
```

This test case is designed to expose potential issues in the buggy code by creating a tree structure that is neither a simple line nor a star graph. The tree has a structure where node 1 is connected to nodes 2 and 3, and node 2 is connected to node 4. This structure can challenge the logic used in the depth-first search (DFS) function, especially in how it handles the calculation of leaves and non-leaves for different subtrees.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
from sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        for _ in range(ac.read_int()):
            n = ac.read_int()
            dct = [[] for _ in range(n)]
            degree = [0] * n
            for _ in range(n - 1):
                x, y = ac.read_list_ints_minus_one()
                dct[y].append(x)
                dct[x].append(y)
                degree[x] += 1
                degree[y] += 1
            ans = [0] * n
            dp = [(0, 0) for _ in range(n)]  # not or in
            stack = [(0, -1)]
            while stack:
                i, fa = stack.pop()
                if i >= 0:
                    stack.append((~i, fa))
                    for j in dct[i]:
                        if j != fa:
                            stack.append((j, i))
                else:
                    i = ~i
                    x = 1
                    y = 0
                    for j in dct[i]:
                        if j != fa:
                            a, b = dp[j]
                            x += a
                            y += b
                    dp[i] = (y, ac.max(x, y))
                    ans[i] = y + 1

            stack = [(0, -1, 0, 0)]
            while stack:
                i, fa, pre_a, pre_b = stack.pop()
                ans[i] += pre_b
                lst = [j for j in dct[i] if j != fa]
                aa = sum(dp[j][0] for j in lst)
                bb = sum(dp[j][1] for j in lst)
                for j in lst:
                    nex_aa = aa - dp[j][0] + pre_a + 1
                    nex_bb = bb - dp[j][1] + pre_a
                    stack.append((j, i, nex_bb, max(nex_bb, nex_aa)))
            for i in range(n):
                if degree[i] > 1:
                    ans[i] -= 1
            ac.st(max(ans))
        return

Solution().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:
```plaintext
1
4
1 2
1 3
2 4
```

### Explanation:
This test input is designed to expose potential bugs in the algorithm, especially in how it handles small trees with a specific structure. Here’s the breakdown:

1. **Single Test Case**: The input starts with `1`, indicating a single test case.
2. **Tree with 4 Nodes**: The tree has 4 nodes, which is a small and manageable size.
3. **Edges**:
   - `1 2`: Node 1 is connected to Node 2.
   - `1 3`: Node 1 is connected to Node 3.
   - `2 4`: Node 2 is connected to Node 4.

### Structure of the Tree:
- Node 1 is the root and has two children: Node 2 and Node 3.
- Node 2 has one child: Node 4.
- Node 3 and Node 4 are leaves.

### Expected Output:
The expected output for this test case is `3`. Here’s why:
- If we choose Node 1 as the root for the shuffle, the tree will have three leaves: Node 3, Node 4, and Node 2 (since Node 2 will become a leaf when Node 4 is connected to it).

### Why This Input is Challenging:
- **Small Size**: The small size of the tree ensures that the input is manageable and can be easily validated.
- **Specific Structure**: The tree has a structure that can test the algorithm's ability to handle multiple levels of children and leaves.
- **Edge Cases**: This input tests how the algorithm handles a root with multiple children and how it correctly identifies and counts leaves.

This input should help in identifying any issues in the algorithm, particularly in how it handles the dynamic programming part and the calculation of leaves.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
import io,os
from collections import deque
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def main(t):

    n = int(input())
    neigh = [[] for _ in range(n)]

    children = [[] for _ in range(n)]

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

    queue = deque()

    root = -1

    for i in range(n):
        if len(neigh[i]) > 2:
            root = i
            break

    if root < 0:
        print((n-1)//2+1)
        return

    parent = [-2]*n
    parent[root] = -1

    queue.append(root)
    seq = []

    while queue:
        index = queue.popleft()
        seq.append(index)
        for nextindex in neigh[index]:
            if parent[nextindex] > -2:  continue
            parent[nextindex] = index
            children[index].append(nextindex)
            queue.append(nextindex)

    dp = [[0,0] for _ in range(n)]

    skip = []
    for index in seq[::-1]:
        if len(children[index])!=0:  continue
        if parent[index] >= 0 and len(children[parent[index]])==1:
            dp[parent[index]] = [1,2]
            skip = [index,parent[index]]
            break

    for index in seq[::-1]:
        if index in skip:  continue

        if len(children[index])==0:
            dp[index][0] = 0
            dp[index][1] = 1
            continue

        # if we choose 1
        dp[index][1] = 1
        for nextindex in children[index]:
            dp[index][1] += dp[nextindex][0]
            dp[index][0] += dp[nextindex][1]

#    print(dp)

    ans = max(dp[root])

    print(ans)

T = int(input())
t = 1
while t<=T:
    main(t)
    t += 1

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that are likely to cause issues. The buggy code has a few potential problems:

1. **Root Selection**: The code selects a root node based on the condition `len(neigh[i]) > 2`. This might not always be the best root to maximize the number of leaves.
2. **DP Initialization and Calculation**: The dynamic programming (DP) initialization and calculation might not handle all cases correctly, especially when dealing with nodes that have only one child.

Let's consider a test case that can expose these issues:

### Test Case Description
- A tree with a structure that has a node with exactly two children, and those children have different structures to challenge the DP initialization and calculation.

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

### Explanation
- **Node 1** has two children, **Node 2** and **Node 3**.
- **Node 2** has one child, **Node 4**.
- **Node 3** has two children, **Node 5** and **Node 6**.

This structure is designed to:
- Test the root selection logic.
- Ensure that the DP initialization and calculation handle nodes with exactly one child correctly.
- Challenge the DP calculation to ensure it correctly counts the maximum number of leaves.

### Expected Output
The expected output for this test case is `4` because:
- If we choose **Node 1** as the root, the maximum number of leaves is `4` (Nodes 4, 5, 6, and 1).

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int[][] edges = inEdges(n - 1);
    d = edges2List(edges, n);
    int root = 0;
    for (int i = 0; i < n; ++i) {
        if (d[i].length == 1) {
            root = i;
            break;
        }
    }
    f = new int[n][2];
    dive(root, -1);
    ret = 0;
    reroot(root, -1, new int[]{0, 0});
    sl(ret);
}

int ret;
int[][] f;
int[][] d;

void reroot(int cur, int fa, int[] faf) {
    int[] xfaf = new int[]{Math.max(faf[0], faf[1]), faf[0] + 1};
    int r = 0;
    for (int x : d[cur]) {
        if (x == fa) continue;
        reroot(x, cur, xfaf);
        r += Math.max(f[x][0], f[x][1]);
    }
    r += Math.max(faf[0], faf[1]);
    if (d[cur].length == 1) r++;
    ret = Math.max(ret, r);
}

void dive(int cur, int fa) {
    f[cur][1] = 1;
    for (int x : d[cur]) {
        if (x == fa) continue;
        dive(x, cur);
        f[cur][1] += f[x][0];
        f[cur][0] += Math.max(f[x][0], f[x][1]);
    }
}

int[][] inEdges(int n) {
    int[][] edges = new int[n][2];
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < 2; ++j)
            edges[i][j] = nextInt() - 1;
    return edges;
}

int[][] edges2List(int[][] edges, int n) {
    int[] g = new int[n];
    for (int[] en : edges) {
        g[en[0]]++;
        g[en[1]]++;
    }
    int[][] r = new int[n][0];
    for (int i = 0; i < n; ++i) {
        r[i] = new int[g[i]];
    }
    int[] ri = new int[n];
    for (int[] en : edges) {
        r[en[0]][ri[en[0]]++] = en[1];
        r[en[1]][ri[en[1]]++] = en[0];
    }
    return r;

}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases or specific scenarios that can challenge the algorithm. The buggy code seems to have issues with how it handles the rerooting and the calculation of the maximum number of leaves. Let's consider the following steps:

1. **Star Graph with a Central Node**: This is a simple and common edge case where one node is connected to all other nodes. The code should correctly handle this and return the number of leaves as `n-1`.

2. **Linear Graph**: A graph where nodes are connected in a line. The code should correctly handle this and return the number of leaves as 2.

3. **Balanced Tree**: A tree where the nodes are evenly distributed. This can help test the rerooting logic.

4. **Complex Tree with Multiple Branches**: A tree with multiple branches and varying degrees of nodes. This can help test the code's ability to handle more complex structures.

Let's generate a small-scale test input that includes a combination of these scenarios:

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

### Explanation:

1. **Test Case 1 (Star Graph with a Central Node)**:
   - Nodes: 5
   - Edges: (1, 2), (1, 3), (2, 4), (2, 5)
   - This forms a star graph with node 1 as the center. The maximum number of leaves should be 4.

2. **Test Case 2 (Linear Graph)**:
   - Nodes: 5
   - Edges: (1, 2), (2, 3), (3, 4), (4, 5)
   - This forms a linear graph. The maximum number of leaves should be 2.

3. **Test Case 3 (Balanced Tree)**:
   - Nodes: 6
   - Edges: (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
   - This forms a star graph with node 1 as the center. The maximum number of leaves should be 5.

4. **Test Case 4 (Complex Tree with Multiple Branches)**:
   - Nodes: 10
   - Edges: (9, 3), (8, 1), (10, 6), (8, 5), (7, 8), (4, 6), (1, 3), (10, 1), (2, 7)
   - This forms a more complex tree with multiple branches. The maximum number of leaves should be 6.

This test input should help expose any issues in the rerooting and leaf calculation logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int[][] edges = inEdges(n - 1);
    d = edges2List(edges, n);
    int root = 0;
    for (int i = 0; i < n; ++i) {
        if (d[i].length == 1) {
            root = i;
            break;
        }
    }
    f = new int[n][2];
    dive(d[root][0], root);
//    dive(root, -1);
    sl(Math.max(f[d[root][0]][0], f[d[root][0]][1]) + 1);
}

int[][] f;
int[][] d;

void dive(int cur, int fa) {
    f[cur][1] = 1;
    for (int x : d[cur]) {
        if (x == fa) continue;
        dive(x, cur);
        f[cur][1] += f[x][0];
        f[cur][0] += Math.max(f[x][0], f[x][1]);
    }
}

int[][] inEdges(int n) {
    int[][] edges = new int[n][2];
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < 2; ++j)
            edges[i][j] = nextInt() - 1;
    return edges;
}

int[][] edges2List(int[][] edges, int n) {
    int[] g = new int[n];
    for (int[] en : edges) {
        g[en[0]]++;
        g[en[1]]++;
    }
    int[][] r = new int[n][0];
    for (int i = 0; i < n; ++i) {
        r[i] = new int[g[i]];
    }
    int[] ri = new int[n];
    for (int[] en : edges) {
        r[en[0]][ri[en[0]]++] = en[1];
        r[en[1]][ri[en[1]]++] = en[0];
    }
    return r;

}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
#include <bits/stdc++.h>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define f first
#define s second
using namespace std;
using ll = long long;
using pii = pair <int, int>;
const int N = 2e5 + 5;
int n, sz[N], deg[N];
bool used[N];
vector <int> g[N];
void calc(int v, int p = 0) {
    sz[v] = 1;
    for (auto to : g[v]) {
        if (!used[to] && to != p)
            calc(to, v), sz[v] += sz[to];
    }
}
int center(int v, int nsz, int p = 0) {
    for (auto to : g[v]) {
        if (!used[to] && to != p && sz[to] > nsz / 2)
            return center(to, nsz, v);
    }
    return v;
}
void dfs(int v, int p = 0) {
    calc(v);
    v = center(v, sz[v]);
    used[v] = 1;
    if (p) {
        ++deg[v];
        ++deg[p];
    }
    for (auto to : g[v]) {
        if (!used[to])
            dfs(to, v);
    }
}
void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        g[i].clear();
        used[i] = sz[i] = deg[i]=  0;
    }
    for (int i = 1, u, v; i < n; ++i) {
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    int mda = -1;
    for (int i = 1; i <= n; ++i) {
        if (sz(g[i]) == 1) {
            mda = i;
            used[i] = 1;
            break;
        }
    }
    calc(1);
    dfs(1);
    int ans = 0, add = 0;
    for (int i = 1; i <= n; ++i) {
        if (i == mda)
            continue;
        ans += (deg[i] == 1);
    }
    cout << ans + 1 << '\n';
}
int main() {
    ios :: sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <assert.h>
#include <bitset>
#include <time.h>
#include <math.h>
#include <deque>
#include <set>
#define in(x) freopen(#x".in","r",stdin)
#define out(x) freopen(#x".out","w",stdout)
#define make(x) freopen(#x".in","w",stdout)
#define ll long long
#define int long long
#ifdef MY_COMPUTER
#define debug(...) fprintf(stderr,##__VA_ARGS__)
#else
#define debug(...) void(0)
#endif

using namespace std;

inline int read()
{
	int s=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')
			f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		s=s*10+ch-'0';
		ch=getchar();
	}
	return s*f;
}
int t,n,col[200010],siz[200010],son[200010],g[2],ans;
vector<int>e[200010];
void dfs(int u,int f){
	siz[u]=1,son[u]=n;
	g[col[u]]++;
	for(int v:e[u]){
		if(v==f) continue;
		col[v]=col[u]^1;
		dfs(v,u);
		siz[u]+=siz[v];
		son[u]=min(son[u],siz[v]);
	}
	if(f) son[u]=min(son[u],n-siz[u]);
}
signed main()
{
	//ios::sync_with_stdio(false);
	t=read();
	while(t--){
		n=read();
		for(int i=1;i<=n;i++) e[i].clear(),siz[i]=son[i]=col[i]=0;
		ans=g[0]=g[1]=0;
		for(int i=1,u,v;i<n;i++){
			u=read(),v=read();
			e[u].push_back(v);
			e[v].push_back(u);
		}
		dfs(1,0);
		for(int i=1;i<=n;i++) ans=max(ans,g[col[i]^1]+(son[i]!=1));
		printf("%lld\n",ans);
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code uses a depth-first search (DFS) to calculate the number of leaves in the shuffled tree, but it might not correctly handle certain tree structures.

### Key Points to Consider:
1. **Tree with a single path (linear tree):** The code might not handle this correctly, especially when the tree is a straight line.
2. **Tree with a star structure:** The code might not correctly handle the case where the tree is a star (one central node with multiple leaves).
3. **Tree with a complex structure:** A tree with a mix of long paths and branching nodes can also be challenging.

### Test Input Generation:
Let's create a test case that combines these elements to ensure the code is thoroughly tested.

### Test Case:
- **Tree with a single path (linear tree):**
  - This will test the code's ability to handle a simple linear structure.
- **Tree with a star structure:**
  - This will test the code's ability to handle a central node with multiple leaves.
- **Tree with a mix of long paths and branching nodes:**
  - This will test the code's ability to handle more complex structures.

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

### Explanation:
1. **First Test Case (Linear Tree):**
   - Tree: 1-2-3-4-5
   - Expected Output: 3
   - Reason: The maximum number of leaves is achieved by shuffling the tree to have 3 leaves (nodes 1, 4, and 5).

2. **Second Test Case (Star Tree):**
   - Tree: 1 (center) - 2, 3, 4, 5, 6
   - Expected Output: 5
   - Reason: The star tree already has 5 leaves, and shuffling it will not increase the number of leaves.

3. **Third Test Case (Complex Tree):**
   - Tree: 1-2-3-4-5-6-7-8-9-10
   - Expected Output: 6
   - Reason: The maximum number of leaves is achieved by shuffling the tree to have 6 leaves (nodes 1, 3, 5, 7, 9, and 10).

This test case covers a range of tree structures and should help in identifying 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: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
from sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

def brute_force(dct, degree):
    res = 0
    n = len(dct)
    for root in range(n):
        dp = [(0, 0) for _ in range(n)]  # not or in
        stack = [(root, -1)]
        while stack:
            i, fa = stack.pop()
            if i >= 0:
                stack.append((~i, fa))
                for j in dct[i]:
                    if j != fa:
                        stack.append((j, i))
            else:
                i = ~i
                x = 1
                y = 0
                for j in dct[i]:
                    if j != fa:
                        a, b = dp[j]
                        x += a
                        y += b
                dp[i] = (y, max(x, y))
        res = max(res, max(dp[root][0] + int(degree[root] == 1), dp[root][1]))
    return res

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        for _ in range(ac.read_int()):
            n = ac.read_int()
            dct = [[] for _ in range(n)]
            degree = [0] * n
            for _ in range(n - 1):
                x, y = ac.read_list_ints_minus_one()
                dct[y].append(x)
                dct[x].append(y)
                degree[x] += 1
                degree[y] += 1
            # ac.st(brute_force(dct, degree))
            ans = [0] * n
            dp = [(0, 0) for _ in range(n)]  # not or in
            stack = [(0, -1)]
            while stack:
                i, fa = stack.pop()
                if i >= 0:
                    stack.append((~i, fa))
                    for j in dct[i]:
                        if j != fa:
                            stack.append((j, i))
                else:
                    i = ~i
                    x = 1
                    y = 0
                    for j in dct[i]:
                        if j != fa:
                            a, b = dp[j]
                            x += a
                            y += b
                    dp[i] = (y, ac.max(x, y))
                    ans[i] = y + 1 if degree[i] == 1 else y

            stack = [(0, -1, 0, 0)]
            while stack:
                i, fa, pre_a, pre_b = stack.pop()
                ans[i] += pre_b
                lst = [j for j in dct[i] if j != fa]
                aa = sum(dp[j][0] for j in lst)
                bb = sum(dp[j][1] for j in lst)
                for j in lst:
                    nex_aa = aa - dp[j][0] + pre_a + 1
                    nex_bb = bb - dp[j][1] + pre_a
                    stack.append((j, i, nex_bb, max(nex_bb, nex_aa)))
            ac.st(max(ans))
        return

Solution().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:
```plaintext
4
3
1 2
2 3
5
1 2
2 3
3 4
4 5
6
1 2
1 3
1 4
1 5
1 6
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

Buggy Code:
import sys
rl = sys.stdin.readline
N = int(rl())

def dfs(v, p):
    for u in g[v]:
        if u == p:
            continue
        dfs(u, v)

def cnt_leaf(v, p):
    if len(g[v]) == 1:
        return 1
    ret = 0
    for u in g[v]:
        if u == p:
            continue
        ret += cnt_leaf(u, v)
    return ret

for _ in range(N):
    n = int(rl())
    g = [[] for _ in range(n+1)]
    cnt = [0] * (n+1)
    for _ in range(n-1):
        x, y = map(int, rl().split())
        g[x].append(y)
        g[y].append(x)

    ans = cnt_leaf(1, 0)

    for x in range(1, n+1):
        for nx in g[x]:
            if len(g[nx]) == 2:
                for nnx in g[nx]:
                    if nnx != x:
                        if len(g[nnx]) == 1:
                            continue
                cnt[x] += 1

    ans += max(cnt)
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int[][] edges = inEdges(n - 1);
    d = edges2List(edges, n);
    int root = 0;
    for (int i = 0; i < n; ++i) {
        if (d[i].length == 1) {
            root = i;
            break;
        }
    }
    f = new int[n][2];
    dive(root, -1);
    sl(Math.max(f[root][0] + 1, f[root][1]));
}

int[][] f;
int[][] d;

void dive(int cur, int fa) {
    f[cur][1] = 1;
    for (int x : d[cur]) {
        if (x == fa) continue;
        dive(x, cur);
        f[cur][1] += f[x][0];
        f[cur][0] += Math.max(f[x][0], f[x][1]);
    }
}

int[][] inEdges(int n) {
    int[][] edges = new int[n][2];
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < 2; ++j)
            edges[i][j] = nextInt() - 1;
    return edges;
}

int[][] edges2List(int[][] edges, int n) {
    int[] g = new int[n];
    for (int[] en : edges) {
        g[en[0]]++;
        g[en[1]]++;
    }
    int[][] r = new int[n][0];
    for (int i = 0; i < n; ++i) {
        r[i] = new int[g[i]];
    }
    int[] ri = new int[n];
    for (int[] en : edges) {
        r[en[0]][ri[en[0]]++] = en[1];
        r[en[1]][ri[en[1]]++] = en[0];
    }
    return r;

}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

**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 code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The provided code seems to have a logical flaw in how it handles the tree structure and the shuffling process. Specifically, it might not correctly handle cases where the tree has multiple levels of branching and where the choice of the root node significantly affects the outcome.

### Step-by-Step Reasoning:

1. **Single Leaf Node:**
   - A tree with only one leaf node (and a root node) is trivial and should not expose any bugs. However, it's a good starting point to ensure the basic logic works.

2. **Star Graph:**
   - A star graph (a central node connected to all other nodes) should have a straightforward solution. The maximum number of leaves is `n-1` if the root is the central node.

3. **Line Graph:**
   - A line graph (a chain of nodes) is another simple case. The maximum number of leaves is `2` if the root is any internal node.

4. **Balanced Binary Tree:**
   - A balanced binary tree with multiple levels of branching can expose issues in the shuffling logic. The root choice can significantly affect the number of leaves.

5. **Unbalanced Tree:**
   - An unbalanced tree with a deep structure can also be challenging. The shuffling process might not handle deep branches correctly.

### Challenging Test Input:
Let's create a test input for a balanced binary tree with 7 nodes. This tree has a root node with two children, each of which has two children, and each of those children has one child. This structure should challenge the algorithm to correctly handle multiple levels of branching.

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

### Explanation:
- The tree structure is as follows:
  - Node 1 is the root.
  - Node 1 is connected to nodes 2 and 3.
  - Node 2 is connected to nodes 4 and 5.
  - Node 3 is connected to nodes 6 and 7.
- The maximum number of leaves is 4, which can be achieved by selecting node 1 as the root and shuffling the subtrees.

This test input should expose any issues in the algorithm's handling of multiple levels of branching and the correct calculation of the maximum number of leaves.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shuffle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Two hungry red pandas, Oscar and Lura, have a tree $T$ with $n$ nodes. They are willing to perform the following shuffle procedure on the whole tree $T$ exactly once. With this shuffle procedure, they will create a new tree out of the nodes of the old tree.

  1. Choose any node $V$ from the original tree $T$. Create a new tree $T_2$, with $V$ as the root.    2. Remove $V$ from $T$, such that the original tree is split into one or more subtrees (or zero subtrees, if $V$ is the only node in $T$).    3. Shuffle each subtree with the same procedure (again choosing any node as the root), then connect all shuffled subtrees' roots back to $V$ to finish constructing $T_2$.

After this, Oscar and Lura are left with a new tree $T_2$. They can only eat leaves and are very hungry, so please find the maximum number of leaves over all trees that can be created in exactly one shuffle.

Note that leaves are all nodes with degree $1$. Thus, the root may be considered as a leaf if it has only one child.

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

The first line of every test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the number of nodes within the original tree $T$.

The next $n - 1$ lines each contain two integers $u$ and $v$ ($1 \leq u, v \leq n$) — an edge within the original tree $T$. The given edges form a tree.

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of leaves achievable with exactly one shuffle procedure on the whole tree.

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

Note: In the first test case, it can be shown that the maximum number of leaves is $4$. To accomplish this, we can start our shuffle with selecting node $3$ as the new root.

![](CDN_BASE_URL/038cb83999c75e319cd6897cdfe03b7b) Next, we are left only with one subtree, in which we can select node $2$ to be the new root of that subtree.  ![](CDN_BASE_URL/85eb09dd4b3b98aa44c0450fc1904edf) This will force all $3$ remaining nodes to be leaves, and once we connect them back to our new root, the shuffled subtree looks like this: ![](CDN_BASE_URL/71ffcc64012b118124db081f8a9493b2) We connect the shuffled subtree back to our new root of our new tree. Our final tree has four leaves (including the root), and looks like this: ![](CDN_BASE_URL/882e22ac95139c8917e71dee39d6a32c)

In our second test case, we have a line of five nodes. It can be shown that the maximum number of leaves after one shuffle is $3$. We can start off with node $2$, which forces node $1$ to become a leaf. Then, if we select node $4$ on the right side, we will also have nodes $3$ and $5$ as leaves.

The third test case is a star graph with six nodes. The number of leaves cannot increase, thus our answer will be $5$ (if we start the shuffling with the original root node).

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

public class Main implements Runnable {

  BufferedReader in;
  PrintWriter out;
  StringTokenizer tok = new StringTokenizer("");

  int[] connectedness = new int[200005];

  public static void main(String[] args) {
    new Thread(null, new Main(), "", 256 * (1L << 20)).start();
  }

  public void run() {
    try {
      long t1 = System.currentTimeMillis();
      if (System.getProperty("ONLINE_JUDGE") != null) {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
      } else {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
      }
      Locale.setDefault(Locale.US);
      solve();
      in.close();
      out.close();
      long t2 = System.currentTimeMillis();
      System.err.println("Time = " + (t2 - t1));
    } catch (Throwable t) {
      t.printStackTrace(System.err);
      System.exit(-1);
    }
  }

  //Util Functions
  String readString() throws IOException {
    while (!tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

  int readInt() throws IOException {
    return Integer.parseInt(readString());
  }

  long readLong() throws IOException {
    return Long.parseLong(readString());
  }

  double readDouble() throws IOException {
    return Double.parseDouble(readString());
  }

  ArrayList<Integer> readIntArrayList(int length) throws IOException {
    ArrayList array = new ArrayList();
    for(int input = 0; input < length; input++){
      array.add(readInt());
    }
    return array;
  }

  int[] readIntList(int length) throws IOException {
    int[] array = new int[length];
    for(int input = 0; input < length; input++){
      array[input] = readInt();
    }
    return array;
  }

  void readIntList(int[] array,int length) throws IOException {
    for(int input = 0; input < length; input++){
      array[input] = readInt();
    }
  }

  // Helper functions

  int[] bitsRep(int n){
    int[] bits = new int[22];
    int i = 0;
    while(n > 0){
      bits[i++] = n % 2;
      n = n / 2;
    }
    return bits;
  }

  boolean[] bitsBool(int[] intArray){
    boolean[] bits = new boolean[intArray.length];
    for(int i = 0; i < intArray.length; i++)
      bits[i] = intArray[i] > 0;

    return bits;
  }

  int[] arrayTopla(int[] a, int[] b){
    int[] sum = new int[Math.max(a.length,b.length)];
    for(int i = 0; i < sum.length; i++){
      if(i < a.length){
        sum[i] += a[i];
      }
      if(i < b.length){
        sum[i] += b[i];
      }
    }
    return sum;
  }

  int[] arrayCikar(int[] a, int[] b){
    int[] fark = new int[Math.max(a.length,b.length)];
    for(int i = 0; i < fark.length; i++){
      if(i < a.length){
        fark[i] += a[i];
      }
      if(i < b.length){
        fark[i] -= b[i];
      }
    }
    return fark;
  }
  // Helper Class

  int binarySearch(List<Integer> list, int eleman) {
    int l = 0;
    int r = list.size();
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (list.get(m) > eleman) {
        r = m;
      } else {
        l = m;
      }
    }
    if(list.size() <= l){
      return -1;
    }
    if(list.get(l) <= eleman){
      if(list.size() == l + 1)
        return -1;
      else{
        return l + 1;
      }
    }
    return l;
  }

  void solve() throws IOException {
    int t = readInt();
    for (int problem = 0; problem < t; problem++) {
      ArrayList<Set<Integer>> edges = new ArrayList<Set<Integer>>();
      int n = readInt();
      for(int i = 0; i < n; i++){
        edges.add(new HashSet<>());
        connectedness[i] = 0;
      }
      for(int i = 0; i < n - 1; i++){
        int source = readInt() - 1;
        int destination = readInt() - 1;
        if(edges.get(source) == null){
          edges.set(source, new HashSet<Integer>());
        }
        if(edges.get(destination) == null){
          edges.set(destination, new HashSet<Integer>());
        }
        edges.get(source).add(destination);
        edges.get(destination).add(source);
        connectedness[source]++;
        connectedness[destination]++;
      }
      boolean flag = true;
      boolean leafKestik = false;
      boolean firstCutMade = false;

      for(int i = 0; i < n; i++){
        if(connectedness[i] == 1){
          int neigbour = (int) edges.get(i).toArray()[0];
          if(connectedness[neigbour] == 2){
            int sum = (int) edges.get(neigbour).toArray()[0] + (int) edges.get(neigbour).toArray()[1] ;
            int parent = sum - i;
            edges.get(i).remove(neigbour);
            edges.get(i).add(parent);
            edges.get(neigbour).remove(i);
            connectedness[neigbour]--;
            edges.get(parent).add(i);
            connectedness[parent]++;
            break;
          }
        }
      }
      while(flag) {
        flag = false;
        for (int i = 0; i < n; i++) {
          if (connectedness[i] == 2) {
            for (int j = 0; j < 2; j++) {
              int neighbour = (int) edges.get(i).toArray()[j];
              if (connectedness[neighbour] == 2) {
                int sum =  (int) edges.get(neighbour).toArray()[0] + (int) edges.get(neighbour).toArray()[1];
                int parent = sum - i;
                edges.get(neighbour).remove(parent);
                connectedness[neighbour]--;
                edges.get(parent).remove(neighbour);
                edges.get(parent).add(i);
                edges.get(i).add(parent);
                connectedness[i]++;
                flag = true;
              }
            }
          }
        }
        firstCutMade = true;
      }
      int leafCount = 0;
      for(int i = 0; i < n; i++){
        if(connectedness[i] == 1){
          leafCount++;
        }
      }

      out.println(leafCount);
    }
  }
}

**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 might cause the algorithm to fail. The bug in the code seems to be related to how it handles the shuffling and reconnection of subtrees, especially in cases where the tree structure is complex or has specific patterns.

Here are the steps to create a challenging test input:

1. **Identify the Bug**: The bug in the code is that it incorrectly handles the reconnection of subtrees, especially when the tree has nodes with degree 2 or more. It might not correctly identify the optimal way to shuffle the subtrees to maximize the number of leaves.

2. **Create a Test Case**: We need a small-scale test case that highlights the issue. A good candidate is a tree where the optimal shuffle involves a specific node with a degree of 2 or more, and where the reconnection of subtrees can significantly change the number of leaves.

3. **Example Test Case**: Consider a tree with 7 nodes where the optimal shuffle can significantly increase the number of leaves. For example, a tree with a structure that forces the code to make suboptimal decisions.

Here is the test input:

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

### Explanation:
- **Tree Structure**: The tree is a simple path from node 1 to node 7.
- **Optimal Shuffle**: The optimal shuffle involves selecting node 4 as the root. This will create two subtrees: one with nodes 1, 2, 3 and the other with nodes 5, 6, 7. Both subtrees can be shuffled to have 3 leaves each, making a total of 4 leaves (including node 4).
- **Expected Output**: The expected output is 4, but the buggy code might not correctly handle the reconnection of subtrees, leading to a suboptimal number of leaves.

This test case is small and complete, and it should expose the bug in the provided code.