**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

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

pows = [1]

for _ in range(31):
    pows.append(pows[-1]*2)

pows = set(pows)

def main():

    n = int(input())
    arr = list(map(int,input().split()))

    if n==1:
        print(1)
        return

    ans = n
    arr = [abs(arr[i+1]-arr[i]) for i in range(n-1)]
    n -= 1

    stack = []
    diff = [-1]*n

    for i in range(n):
        while stack and arr[stack[-1]]!=arr[i]:
            diff[stack.pop()] = i
        stack.append(i)

    for i in range(n):
        curr = arr[i]
        index = i
        if curr == 0:
            nextindex = diff[index]
            if nextindex < 0:
                ans += n - index
                continue
            else:
                ans += nextindex - index
                index = nextindex

        while index >= 0:
            curr = gcd(curr, arr[index])
            if curr in pows:
                ans += n - index
                break
            index = diff[index]

#        print(i,ans)

    print(ans)

T = int(input())
t = 1
while t <= T:
    main()
    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 provided buggy code, we need to focus on edge cases or scenarios that are likely to cause issues. Here are some steps to consider:

1. **Edge Case with Small Array**: Test with a small array to ensure the basic logic is correct.
2. **Edge Case with Repeated Elements**: Test with an array containing repeated elements to ensure the set operations are handled correctly.
3. **Edge Case with Large Range**: Test with an array containing elements that span a large range to ensure the algorithm can handle large numbers.
4. **Edge Case with Consecutive Elements**: Test with an array that is already consecutive to ensure the algorithm correctly identifies it.
5. **Edge Case with Non-Consecutive Elements**: Test with an array that is not consecutive but can become consecutive after operations to ensure the algorithm can handle such cases.
6. **Edge Case with Powers of Two**: Test with an array containing elements that are powers of two to ensure the algorithm correctly handles these special cases.

Given these considerations, let's create a small-scale test input that includes a mix of these edge cases:

```plaintext
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
```

This test input includes:
- A small array with repeated elements.
- An array that can become consecutive after operations.
- An array with elements that span a large range.
- An array with a single element.
- An array with repeated elements and a mix of values.
- An array with elements that are not consecutive but can become consecutive after operations.

This should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define endl "\n"
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
typedef pair<ll,ll> pii;
const ll N=400010,inf=LLONG_MAX;

string yes="Yes",no="No";
int a[N],b[N],f[30][N];
int n;

void st(){
	for(int i=1;i<=20;i++){
		int len=(1<<i);
		for(int j=1;j+len-1<n;j++){
			f[i][j]=__gcd(f[i-1][j],f[i-1][j+len/2]);
		}
	}
}

int ask(int l,int r){
	int res=0;
	for(int i=20;i>=0;i--){
		int len=(1<<i);
		if(l+len-1<=r){
			res=__gcd(res,f[i][l]);
			l+=len;
		}
	}
	return res;
}

void solve(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(int i=1;i<n;i++){
		b[i]=a[i+1]-a[i];
		f[0][i]=abs(b[i]);
	}
	st();
	int ans=0,now=0;
	for(int i=1;i<=n;i++){
		if(a[i]==a[i-1])now++;
		else now=1;
		ans+=now;
	}
//	cout<<ans<<endl;
//	cout<<ask(1,2)<<endl;
	for(int i=1;i<=n;i++){
		int l=i,r=n-1;
		while(l<=r){
			int md=l+r>>1;
			int x=ask(i,md);
//			cout<<md<<" "<<x<<endl;
			if(x!=0&&(x&-x)==x)r=md-1;
			else l=md+1;
		}
//		cout<<l<<endl;
//		cout<<l<<" "<<n-l<<endl;
		ans+=n-l;
	}
	cout<<ans<<endl;
}

int main(){
 	ios::sync_with_stdio(0);
	cin.tie(0);
	int txt=1;
	cin>>txt;
	while(txt--){
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that are likely to trigger incorrect behavior. Here are some key points to consider:

1. **Consecutive Elements**: The code should correctly handle subarrays that are already consecutive.
2. **Non-Consecutive Elements**: The code should correctly handle subarrays that can be made consecutive through the described operations.
3. **Duplicate Elements**: The code should handle duplicate elements in the array.
4. **Large GCD Values**: The code should handle cases where the GCD of differences between elements is large.

Given these points, let's construct a small-scale test input that can expose the bug:

```plaintext
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
```

This test input includes:
- A test case with all elements being the same.
- A test case with elements that can be made consecutive through the operations.
- A test case with elements that cannot be made consecutive.
- A test case with a single element.
- A test case with duplicate elements.
- A test case with a mix of elements that have large GCD values.

This should help in exposing any bugs in the code, particularly in how it handles the GCD calculations and the conditions for making subarrays consecutive.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

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

//#define int int64_t

//#define DEBUG
//#define FACTORS
#define endl '\n'

using namespace std;

const int INF = 2e9 + 2;
const int MAXN = 2e5 + 1;
const int MAXX = 2e6 + 10;
const int BS = 20;
const long double EPS = 1e-5;
const long double EPS1 = 1e-4;
const int MOD = 998244353;
const int LG = 30;

const long double DIS = 2e8;

int bp(int a, int p){
    if(p == 0){
        return 1;
    }
    if(p & 1){
        return bp(a, p - 1) * a % MOD;
    }
    int k = bp(a, p / 2);
    return k * k % MOD;
}

int bp_md(int a, int p, int mod){
    if(p == 0){
        return 1;
    }
    if(p & 1){
        return bp_md(a, p - 1, mod) * a % mod;
    }
    int k = bp_md(a, p / 2, mod);
    return k * k % mod;
}

int gcd(int a, int b){
    while(b){
        if(a >= b){
            a %= b;
        }
        swap(a, b);
    }
    return a + b;
}

int nok(int a, int b){
    return a / gcd(a, b) * b;
}

bool is_prime(int x){
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0){
            return 0;
        }
    }
    return 1;
}

#ifdef FACTORS
vector <int> fact, ofact;

int C(int k, int n){
    if(k < 0 || k > n){
        return 0;
    }
    return fact[n] * ofact[k] % MOD * ofact[n - k] % MOD;
}
#endif // FACTORS

mt19937 rnd(12);

struct node{
    int g = 0;
    int mx_l = INF;
    int mx_r = -INF;
    int cnt = 0;
    node(){

    }
};

node tr[MAXX];
vector <int> obr(MAXN);

node merg(node a, node b){
    int g = __gcd(a.g, b.g);
    if(a.mx_r != -INF && b.mx_l != INF){
        g = __gcd(b.mx_l - a.mx_r, g);
    }
    node c;
    c.g = g;
    c.mx_l = a.mx_l;
    if(a.mx_l == INF){
        c.mx_l = b.mx_l;
    }
    c.mx_r = b.mx_r;
    if(b.mx_r == -INF){
        c.mx_r = a.mx_r;
    }
    return c;

}

node get(int v, int l, int r, int ll, int rr){
    if(r <= ll || l >= rr){
        return node();
    }
    if(l >= ll && r <= rr){
        return tr[v];
    }
    return merg(get(2 * v, l, l + r >> 1, ll, rr), get(2 * v + 1, l + r >> 1, r, ll, rr));
}

void add(int v, int l, int r, int ps, int x){
    if(l >= ps + 1 || r <= ps){
        return;
    }
    if(l == ps && r == ps + 1){
        tr[v].cnt += x;
        if(!tr[v].cnt){
            tr[v].mx_l = INF;
            tr[v].mx_r = -INF;
        }
        else{
            tr[v].mx_l = obr[l];
            tr[v].mx_r = obr[l];
        }
        return;
    }
    add(2 * v, l, l + r >> 1, ps, x);
    add(2 * v + 1, l + r >> 1, r, ps, x);
    tr[v] = merg(tr[2 * v], tr[2 * v + 1]);
}
int df = 0;

map <int, int> mp;

void add(int x){
    df += mp[x] == 0;
    mp[x]++;
}

void del(int x){
    mp[x]--;
    df -= mp[x] == 0;
}

map <int, bool> good;

bool check(int l, int r, int c){
    return good[get(1, 0, c, 0, c).g];
}

void solve(){
    mp.clear();
    df = 0;
    int n;
    cin >> n;
    vector <int> a;
    set <int> st;
    for(int i = 0; i < n; i++){
        int k;
        cin >> k;
        a.push_back(k);
        st.insert(k);
    }
    for(int i = 1; i <= 4 * n + 12; i++){
        tr[i] = node();
    }
    int c = 0;
    map <int, int> mp1;
    for(auto i:st){
        mp1[i] = c;
        obr[c] = i;
        c++;
    }
    for(int i = 0; i < n; i++){
        a[i] = mp1[a[i]];
    }
    int r = -1;
    int ans = 0;
    for(int l = 0; l < n; l++){
        while(r < n && (df <= 1 || !check(l, r + 1, c))){
            r++;
            if(r != n){
                add(a[r]);
                add(1, 0, c, a[r], 1);
            }
        }
        ans += n - r;
        del(a[l]);
        add(1, 0, c, a[l], -1);
    }
    for(int i = 0; i < n; ){
        int cur = a[i];
        int j = i;
        while(j < n && cur == a[j]){
            j++;
        }
        ans += (j - i) * (j - i + 1) / 2;
        i = j;
    }
    cout << ans;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    for(int i = 0; i <= LG; i++){
        good[(1 << i)] = 1;
    }
    #ifdef FACTORS
    fact.push_back(1);
    for(int i = 1; i <= MAXX; i++){
        fact.push_back(fact[i - 1] * i % MOD);
    }
    ofact.push_back(bp(fact.back(), MOD - 2));
    for(int i = MAXN; i >= 1; i--){
        ofact.push_back(ofact.back() * i % MOD);
    }
    reverse(ofact.begin(), ofact.end());
    #endif // FACTORS
    int t = 1;
    cin >> t;
    long double f = clock();
    while (t--){
        solve();
        long double s = clock();
        #ifdef time
        cout << (s - f) / CLOCKS_PER_SEC;
        #endif
        cout << endl;
    }
    return 0;
}

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

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

Think step by step.

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

1. **Duplicate Elements**: The code should handle duplicate elements correctly by only considering each unique element once in the set.
2. **Consecutive Sets**: Ensure that the subarrays can be transformed into consecutive sets.
3. **Edge Cases**: Test with small arrays, arrays with all elements the same, and arrays with a mix of small and large elements.

Given these points, a small-scale test input that can expose bugs might look like this:

```plaintext
3
5
1 2 3 4 5
4
1 1 2 2
6
1 2 4 8 16 32
```

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - This is a consecutive set, and all subarrays should be brilliant.
   - Expected output: `15` (since there are 15 subarrays in total).

2. **Test Case 2:**
   - Array: `[1, 1, 2, 2]`
   - After removing duplicates, the set is `{1, 2}`, which is consecutive.
   - Expected output: `6` (since there are 6 subarrays in total).

3. **Test Case 3:**
   - Array: `[1, 2, 4, 8, 16, 32]`
   - This array cannot be transformed into a consecutive set because the gaps between elements are too large.
   - Expected output: `0` (since no subarray can be transformed into a consecutive set).

This test input should help identify any issues with handling duplicates and checking for consecutive sets.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <queue>
#include <string>
#include <map>
#include <set>
#include<cmath>

using namespace std;
typedef  long long ll;
const int maxn =2e5 + 5;
int c[maxn];
vector<int>e[maxn];
int cnt;
int s[maxn];
int main() {
	ios::sync_with_stdio(false);
	//freopen("sample1.in", "r",stdin);
	//freopen("1.out", "w", stdout);
	cin.tie(0);
	int T;
	cin >> T;
	while (T--) {
		int n,q;
		cin >> n >> q;
		cnt = 0;
		for (int i = 1; i <= n; i++) {
			cin >> c[i];
			e[i].clear();

		}
		for (int i = 1; i <= n - 1; i++) {
			int x, y;
			cin >> x >> y;
			e[x].push_back(y);
			e[y].push_back(x);
		}
		int s1 = 0, s2 = 0, s3 = 0,s0=0;
		for (int i = 1; i <= n; i++) {
			if (c[i]) {
				for (int j = 0; j < e[i].size(); j++) {

					int v = e[i][j];
					if (!c[v])continue;
					s[i]++;
				}
				if (s[i] == 0)s0++;
				else if (s[i] == 1)s1++;
				else if (s[i] == 2)s2++;
				else s3++;
			}
		}
		for (int i = 1; i <= q; i++) {
			int u;
			cin >> u;
			if (c[u]==1) {
				c[u] = 0;
				s[u] = 0;
				if (s[u] == 0)s0--;
				else if (s[u] == 1)s1--;
				else if (s[u] == 2)s2--;
				else s3--;
				for (int j = 0; j < e[u].size(); j++) {
					int v = e[u][j];
					if (c[v]) {
						s[v]--;
						if (s[v] == 1)s0++,s1--;
						else if (s[v] == 2)s1++,s2--;
						else if (s[v] == 3)s3--,s2++;

					}

				}
			}
			else {
				c[u] = 1;

				for (int j = 0; j < e[u].size(); j++) {
					int v = e[u][j];
					if (c[v]) {
						s[u]++;
						s[v]++;
						if (s[v] == 3)s2--, s3++;
						else if (s[v] == 1)s0--, s1++;
						else if (s[v] == 2)s1--, s2++;

					}
				}
				if (s[u] == 0)s0++;
				else if (s[u] == 1)s1++;
				else if (s[u] == 2)s2++;
				else s3++;
			}
			//cout << s0 << " " << s1 << " " << s2 << " " << s3 << endl;
			if ((s0==1 && s1==0 && s2==0 && s3==0)||(s0==0 && s1==2 && s3==0))cout << "YES" << endl;
			else cout << "NO" << endl;
		}
	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key aspects to consider are:

1. **Single Black Vertex**: The code should correctly handle a single black vertex, which should be considered a chain.
2. **Multiple Black Vertices**: The code should correctly determine if the black vertices form a chain.
3. **Toggle Operations**: The code should correctly handle toggling the color of vertices and updating the chain status accordingly.

Let's construct a small-scale test input that includes these aspects:

### Test Input
```plaintext
2
3 2
1 0 0
1 2
2 3
1
2
5 4
1 0 0 0 0
1 2
1 3
1 4
1 5
1
2
3
2
```

### Explanation
1. **Test Case 1:**
   - **Tree Structure**: A simple tree with 3 vertices.
   - **Initial Colors**: Vertex 1 is black, vertices 2 and 3 are white.
   - **Queries**: Toggle vertex 1, then toggle vertex 2.
   - **Expected Output**:
     - After the first query (toggle vertex 1), the tree has no black vertices: "No".
     - After the second query (toggle vertex 2), the tree has one black vertex (vertex 2): "Yes".

2. **Test Case 2:**
   - **Tree Structure**: A more complex tree with 5 vertices.
   - **Initial Colors**: Vertex 1 is black, vertices 2, 3, 4, and 5 are white.
   - **Queries**: Toggle vertex 1, then toggle vertex 2, then toggle vertex 3, and finally toggle vertex 2 again.
   - **Expected Output**:
     - After the first query (toggle vertex 1), the tree has no black vertices: "No".
     - After the second query (toggle vertex 2), the tree has one black vertex (vertex 2): "Yes".
     - After the third query (toggle vertex 3), the tree has two black vertices (vertices 2 and 3): "Yes".
     - After the fourth query (toggle vertex 2), the tree has one black vertex (vertex 3): "Yes".

This test input covers the scenarios of having a single black vertex, multiple black vertices, and toggling operations, which should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

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

const int N = 3e5+10;

int n, q;
int c[N];
std::vector<int> h[N];
std::set<int> st[N];
int p[N], has3, has2, cnt;

void dfs(int u, int fa) {
    p[u] = fa;
    if(c[u]) st[fa].insert(u);
    if(c[u]&&!c[fa]) cnt ++;
    for(auto v:h[u]) if(v!=fa) {
        dfs(v,u);
    }

    if(c[u]&&st[u].size()>=3) has3 ++;
    if(c[u]&&st[u].size()==2) has2 ++;
}

void solve()
{
    std::cin >> n >> q;

    for(int i=1;i<=n;i++) std::cin >> c[i];

    for(int i=1;i<n;i++) {
        int u, v; std::cin >> u >> v;
        h[u].push_back(v), h[v].push_back(u);
    }

    dfs(1,0);
    //std::cout << has2 << " " << has3 << " " << cnt <<  "?\n";
    std::vector<std::array<int,2>> ops;
    for(int i=1;i<=q;i++) {
        int u; std::cin >> u;
        if(c[u]) {
            ops.push_back({u,-1});
            int fa = p[u];
            st[fa].erase(u);
            if(st[fa].size()==2) has3 --;
            if(st[fa].size()==1) has2 --;
            if(!c[fa]) cnt --;
            c[u] = 0;
        }
        else {
            ops.push_back({u,1});
            int fa = p[u];
            st[fa].insert(u);
            if(st[fa].size()==3) has3 ++;
            if(st[fa].size()==2) has2 ++;
            if(!c[fa]) cnt ++;
            c[u] = 1;
        }

        if(has3==0) {
            for(auto [u,k]:ops) {
                for(auto v:st[u]) {
                    if(k==-1) cnt ++;
                    else cnt --;
                }
            }
            ops.clear();
        }
        //std::cout << has2 << " " << has3 << " " << cnt <<  "?\n";
        if(has3==0&&has2<=1&&cnt==1) std::cout << "YES\n";
        else std::cout << "NO\n";
    }

    for(int i=0;i<=n;i++) {
        h[i].clear();
        st[i].clear();
        c[i] = p[i] = 0;
    }
    has2 = has3 = cnt = 0;
}

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

    int t = 1;
    std::cin >> t;
    for(int i=1;i<=t;i++)
        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 provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through the problem:

1. **Understanding the Bug**:
   - The code uses a set to keep track of black vertices and their connections.
   - The main issue is in handling the conditions for a chain of black vertices.

2. **Edge Cases**:
   - **Single Vertex Chain**: A single black vertex should be considered a chain.
   - **No Black Vertices**: No black vertices should not be considered a chain.
   - **Multiple Chains**: Multiple disconnected chains of black vertices should not be considered a single chain.
   - **Chain Breaking**: Toggling a vertex that breaks a chain should be handled correctly.
   - **Chain Formation**: Toggling a vertex that forms a chain should be handled correctly.

3. **Test Case Design**:
   - **Test Case 1**: Single vertex tree with one query.
   - **Test Case 2**: Tree with a chain of black vertices and queries to toggle vertices.
   - **Test Case 3**: Tree with multiple chains of black vertices and queries to toggle vertices.
   - **Test Case 4**: Tree with no black vertices and queries to toggle vertices.

4. **Generate the Test Input**:
   - **Test Case 1**: Single vertex tree with one query.
   - **Test Case 2**: Tree with a chain of black vertices and queries to toggle vertices.
   - **Test Case 3**: Tree with multiple chains of black vertices and queries to toggle vertices.
   - **Test Case 4**: Tree with no black vertices and queries to toggle vertices.

Here is the test input:

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

This test input includes:
- A single vertex tree.
- A tree with a chain of black vertices.
- A tree with multiple chains of black vertices.
- A tree with no black vertices.

Each test case is designed to challenge the code's ability to handle different scenarios correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
const int N=400005;
int n,h[N],fa[N],d[N],cnt,T,Q,op[N],ct[N],wt,num[4];
struct edge{int to,next;}w[N<<1];
void Add(int x,int y){
	cnt++;w[cnt].to=y;w[cnt].next=h[x];h[x]=cnt;
}
void DFS1(int u,int ff){
	d[u]=d[ff]+1;fa[u]=ff;
	for(int i=h[u];i;i=w[i].next){
		int e=w[i].to;
		if(e==ff)continue;
		DFS1(e,u);
	}
}
struct node{
	int u,d;
	bool operator <(node A)const{
		return (d<A.d)||(d==A.d&&u<A.u);
	}
};
set<node>S;
void Ins(int u){S.insert((node){u,d[u]});}
void Er(int u){S.erase((node){u,d[u]});}
int main(){
	int i,x,y;
	scanf("%d",&T);
	while(T--){
		cnt=0;S.clear();
		scanf("%d%d",&n,&Q);
		for(i=0;i<=n;i++)h[i]=fa[i]=d[i]=ct[i]=op[i]=0;
		for(i=0;i<=3;i++)num[i]=0;
		for(i=1;i<=n;i++)scanf("%d",&op[i]);
		for(i=1;i<n;i++)scanf("%d%d",&x,&y),Add(x,y),Add(y,x);
		DFS1(1,0);

		for(i=1;i<=n;i++){
			if(op[i])ct[fa[i]]++;
			if(op[i]&&op[fa[i]]==0)wt++;
		}
		for(i=1;i<=n;i++){
			if(op[i])num[min(ct[i],3)]++,Ins(i);
		}
		while(Q--){
			scanf("%d",&x);
			if(op[x]==0){
				num[min(ct[fa[x]],3)]--;
				wt-=ct[x];
				if(!op[fa[x]])wt++;

				ct[fa[x]]++;
				num[min(ct[fa[x]],3)]++;
				num[min(ct[x],3)]++;op[x]=1;
				Ins(x);
			}
			else{
				num[min(ct[fa[x]],3)]--;
				if(!op[fa[x]])wt--;
				wt+=ct[x];

				ct[fa[x]]--;
				num[min(ct[fa[x]],3)]++;
				num[min(ct[x],3)]--;op[x]=0;
				Er(x);
			}
			if(wt==1){
				if(num[3])puts("NO");
				else if(num[2]>1)puts("NO");
				else if(num[2]==1){
					int u=(*S.begin()).u;
					if(ct[u]!=2)puts("NO");
					else puts("YES");
				}
				else puts("YES");
			}
			else puts("NO");
		}
	}
	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
2
5 3
1 0 0 0 0
1 2
1 3
2 4
2 5
1
2
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
//package force;

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;

public class idk  {
	public static int top;
	public static long[][] recur;
	public static long[] powers;
	public static final long mod=1000000007;
    public static void main(String[] args){
    	  top=0;
    	  //System.out.println("AAAA"+func("11000011010100000")); //18
    	  powers=new long[61];
          Scanner inp=new Scanner(System.in);
          int yuc=inp.nextInt();
          for(int pol=0;pol<yuc;pol++) {
        	 int n=inp.nextInt();
        	 int q=inp.nextInt();
        	 List<List<Integer>> nei=new ArrayList<List<Integer>>();
        	 for(int i=0;i<n;i++)
        		 nei.add(new ArrayList<Integer>());
        	 HashMap<Integer, Integer> mapc=new HashMap();
        	 for(int i=0;i<n;i++) {
        		 int col=inp.nextInt();
        		 mapc.put(i,  col);
        	 }
        	 //System.out.println(mapc.get(0)+" "+mapc.get(1));
        	  //1 =black
        	 HashMap<Integer, Integer> mapd=new HashMap();
        	 for(int i=0;i<n-1;i++) {
        		 int a=inp.nextInt(); int b=inp.nextInt();
        		 nei.get(a-1).add(b-1);
        		 nei.get(b-1).add(a-1);
        	 }
        	 boolean[] vis=new boolean[n];
        	 Arrays.fill(vis,  false);
        	 int i=0;
        	 int comps=0;
        	 List<Integer> seek=new ArrayList<Integer>();
        	 while(i<n) {
        		 while(i<n && mapc.get(i)==0) {
        			 i++;
        		 }
        		 if(i>=n)
        			 break;
        		 comps++;
        		 vis[i]=true;
        		 seek.add(i);
        		 while(seek.size()>0) {
        			 int deg=0;
        			 for(int x:nei.get(seek.get(0))) {
        				 if(mapc.get(x)==1) {
        					 deg++;
        					 if(!vis[x]) {
        						 vis[x]=true;
        						 seek.add(x);
        					 }
        				 }
        			 }
        			 mapd.put(seek.get(0), deg);
        			 seek.remove(0);
        		 }
        		 i++;
        	 }

        	 int deg1=0;
        	 int deg2=0;
        	 for(int i1=0;i1<n;i1++) {
        		 int ch=mapd.getOrDefault(i1,  0);
        		 if(ch==1)
        			 deg1++;
        		 if(ch==2)
        			 deg2++;
        	 }
        	 for(int popop=0;popop<q;popop++) {

        		 int chg=inp.nextInt()-1;
        		 if(mapc.get(chg)==0) {
        			 int deg=0;
        			 for(int x:nei.get(chg)) {
        				 if(mapc.get(x)==1) {
        					 deg++;
        					 //System.out.println(x+" "+popop+" "+mapd.get(x));
        					 int curr=mapd.get(x);
        					 if(curr==0)
        						 deg1++;
        					 if(curr==1) {
        						 deg1--;
        						 deg2++;
        					 }
        					 if(curr==2)
        						 deg2--;
        					 mapd.put(x, curr+1);
        				 }
        			 }
        			 mapd.put(chg,  deg);
        			 if(deg==1)
        				 deg1++;
        			 if(deg==2)
        				 deg2++;
        			 comps-=deg-1;
        			 mapc.put(chg,  1);
        		 }
        		 else {
        			 for(int x:nei.get(chg)) {
        				 if(mapc.get(x)==1) {
        					 int curr=mapd.get(x);
        					 if(curr==1) {
        						 deg1--;

        					 }
        					 if(curr==2) {
        						 deg2--;
        						 deg1++;
        					 }
        					 if(curr==3)
        						 deg2++;
        					 mapd.put(x, curr-1);
        				 }
        			 }
        			 comps+=mapd.get(chg)-1;
        			 mapc.put(chg,  0);
        			 mapd.remove(chg);
        		 }
        		 //System.out.println(deg1+" "+deg2+" "+comps);
        		 if(comps==1 && mapd.size()==deg2+2 && deg1==2)
        			 System.out.println("YES");
        		 else
        			 System.out.println("NO");
        	 }

          }
    }
    public static long func(String s) {
    	long pow=1;
    	long res=0;
    	for(int i=0;i<s.length();i++) {
    		if(s.charAt(s.length()-1-i)=='1')
    			res=(res+pow)%mod;
    		pow=(pow*2)%mod;
    	}
    	return res;
    }

    public static int[] bits(int n) {
    	int[] res=new int[20];
    	Arrays.fill(res,  0);
    	int ind=0;
    	while(n>0) {
    		res[ind++]=n%2;
    		n/=2;
    	}
    	return res;
    }

    public static long gcd(long a, long b) {
    	if(a==-1)
    		return b;
    	else if(b%a==0)
    		return a;
    	else
    		return gcd(b%a, a);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
def solve():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        q = int(data[index + 1])
        index += 2

        colors = list(map(int, data[index:index + n]))
        index += n

        adj = [[] for _ in range(n + 1)]
        for _ in range(n - 1):
            x = int(data[index])
            y = int(data[index + 1])
            index += 2
            adj[x].append(y)
            adj[y].append(x)

        queries = []
        for _ in range(q):
            queries.append(int(data[index]))
            index += 1

        black_vertices = set()
        for i in range(n):
            if colors[i] == 1:
                black_vertices.add(i + 1)

        # Endpoint detection and black vertex degree tracking
        def update_degrees_and_endpoints(v):
            degree = sum(1 for neighbor in adj[v] if neighbor in black_vertices)
            return degree

        degrees = {v: update_degrees_and_endpoints(v) for v in black_vertices}
        endpoints = {v for v in black_vertices if degrees[v] == 1}

        def is_chain():
            if len(black_vertices) == 0:
                return False
            if len(black_vertices) == 1:
                return True
            return len(endpoints) == 2

        for u in queries:
            if u in black_vertices:
                black_vertices.remove(u)
                neighbors = [v for v in adj[u] if v in black_vertices]
                for neighbor in neighbors:
                    if degrees[neighbor] == 1:
                        endpoints.remove(neighbor)
                    degrees[neighbor] -= 1
                    if degrees[neighbor] == 1:
                        endpoints.add(neighbor)
                if u in endpoints:
                    endpoints.remove(u)
                del degrees[u]
            else:
                black_vertices.add(u)
                degrees[u] = update_degrees_and_endpoints(u)
                neighbors = [v for v in adj[u] if v in black_vertices]
                if degrees[u] == 1:
                    endpoints.add(u)
                for neighbor in neighbors:
                    if degrees[neighbor] == 1:
                        endpoints.remove(neighbor)
                    degrees[neighbor] += 1
                    if degrees[neighbor] == 1:
                        endpoints.add(neighbor)

            if is_chain():
                results.append("YES")
            else:
                results.append("NO")

    sys.stdout.write("\n".join(results) + "\n")

solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from collections import *
import random
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

class CustomSet:
    def __init__(self, arr=[]):
        self.RANDOM = random.randrange(2**62)
        self.set = set([self.wrapper(i) for i in arr])

    def wrapper(self, num):
        return num ^ self.RANDOM

    def add(self, key):
        self.set.add(self.wrapper(key))

    def remove(self, key):
        self.set.remove(self.wrapper(key))

    def __contains__(self, key):
        return self.wrapper(key) in self.set

    def __iter__(self):
        return iter(self.wrapper(i) for i in self.set)

    def __len__(self):
        return len(self.set)

    def __repr__(self):
        return repr({self.wrapper(i) for i in self.set})

# Solver function
def solve():
    global cnt_c_2, cnt_c_3, cnt_2_c, cnt_w_p
    # root --> 0
    s = [CustomSet() for i in range(n)]
    q = deque()
    par = [-1] * n
    q.append(0)
    vis = [False] * n
    vis[0] = True
    while q:
        k = len(q)
        for i in range(k):
            node = q.popleft()
            for nbr in adj[node]:
                if(not vis[nbr]):
                    q.append(nbr)
                    vis[nbr] = True
                    par[nbr] = node
    def update(idx):
        global cnt_c_2, cnt_c_3, cnt_2_c, cnt_w_p
        if(color[idx] == 0):
            p = par[idx]
            if(p != -1):
                if(s[p].__contains__(idx)):
                    s[p].remove(idx)
                if(color[p] == 1):
                    if(s[p].__len__() == 2):
                        cnt_c_2 += 1
                        cnt_c_3 -= 1
                        cnt_2_c.add(p)
                    if(s[p].__len__() == 1):
                        cnt_c_2 -= 1
                        cnt_2_c.remove(p)
                if(s[idx].__len__() == 2):
                    cnt_c_2 -= 1
                    cnt_2_c.remove(idx)
                if(s[idx].__len__() > 2):
                    cnt_c_3 -= 1
                if(s[idx].__len__() > 0):
                    cnt_w_p.add(idx)
            else:
                if(s[idx].__len__() > 0):
                    cnt_w_p.add(idx)
                if(s[idx].__len__() == 2):
                    cnt_c_2 -= 1
                    cnt_2_c.remove(idx)
                if(s[idx].__len__() > 2):
                    cnt_c_3 -= 1
        else:
            p = par[idx]
            if(p != -1):
                s[p].add(idx)
                if(color[p] == 1):
                    if(s[p].__len__() == 2):
                        cnt_c_2 += 1
                        cnt_2_c.add(p)
                    if(s[p].__len__() == 3):
                        cnt_c_2 -= 1
                        cnt_c_3 += 1
                        cnt_2_c.remove(p)
                else:
                    cnt_w_p.add(p)
                if(s[idx].__len__() > 0):
                    if(cnt_w_p.__contains__(idx)):
                        cnt_w_p.remove(idx)
                if(s[idx].__len__() == 2):
                    cnt_c_2 += 1
                    cnt_2_c.add(idx)
                if(s[idx].__len__() >= 3):
                    cnt_c_3 += 1
            else:
                if(s[idx].__len__() > 0):
                    if(cnt_w_p.__contains__(idx)):
                        cnt_w_p.remove(idx)
                if(s[idx].__len__() == 2):
                    cnt_c_2 += 1
                    cnt_2_c.add(idx)
                if(s[idx].__len__() >= 3):
                    cnt_c_3 += 1
    def build(idx):
        global cnt_c_2, cnt_c_3, cnt_2_c, cnt_w_p
        if(color[idx] == 0):
            if(s[idx].__len__() > 0):
                cnt_w_p.add(idx)
        else:
            if(s[idx].__len__() == 2):
                cnt_c_2 += 1
                cnt_2_c.add(idx)
            if(s[idx].__len__() > 2):
                cnt_c_3 += 1
    for i in range(1, n):
        if(color[i]):
            p = par[i]
            s[p].add(i)
    for i in range(n):
        build(i)
    tot = sum(color)
    global cntt
    for i in range(qq):
        cntt+=1
        u = int(input())
        u -= 1
        color[u] ^= 1
        if(color[u] == 0):
            tot -= 1
        else:
            tot += 1
        update(u)
        if(tot == 0):
            print("NO\n")
            continue
        if(tot == 1):
            print("YES\n")
            continue
        if(color[0] == 1 and cnt_w_p.__len__() > 0):
            print("NO\n")
            continue
        if(color[0] == 1 and cnt_2_c.__len__() > 0 and cnt_2_c.__contains__(0) == False):
            print("NO\n")
            continue
        if(cnt_c_3 > 0):
            print("NO\n")
            continue
        if(cnt_c_2 > 1):
            print("NO\n")
            continue
        if(cnt_2_c.__len__() > 1):
            print("NO\n")
            continue
        if(cnt_w_p.__len__() > 1):
            if(cntt == 100002):
                assert 1 == 0
            print("NO\n")
            continue
        if(cnt_w_p.__len__() == 1):
            val = -1
            for tmp in cnt_2_c:
                val = tmp
            p = -2
            if(val != -1):
                p = par[val]
            if(p != -2 and (p == -1 or color[p] == 1)):
                print("NO\n")
                continue
        print("YES\n")
cntt=0
# Main
for _ in range(int(input())):
    n, qq = [int(x) for x in input().split()]
    color = [int(x) for x in input().split()]
    adj = [[] for i in range(n)]
    for i in range(n - 1):
        u, v = [int(x) - 1 for x in input().split()]
        adj[u].append(v)
        adj[v].append(u)
    cnt_c_2 = 0
    cnt_c_3 = 0
    cnt_w_p = CustomSet()
    cnt_2_c = CustomSet()
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
import math
import sys

def input(): return sys.stdin.readline().strip()  # fast IO
from functools import cmp_to_key

def LII():
    return list(map(int, input().split()))
def TII():
    return tuple(map(int, input().split()))
from collections import deque
def factors(x):
    i = 1
    fac = []
    while i*i <= x:
        if x % i == 0:
            fac += [i, x//i]
        i += 1
    return fac
def main():
    n,q=LII()
    graph=[[] for i in range(n)]
    color=LII()
    for i in range(n-1):
        x,y=LII()
        x,y=x-1,y-1
        graph[x].append(y)
        graph[y].append(x)
    color.append(0)
    queue=deque([0])
    visit=[0]*n
    fals=[-1]*n
    while queue:
        node=queue.popleft()
        visit[node]=1
        for son in graph[node]:
            if not visit[son]:
                fals[son]=node
                queue.append(son)
    #统计所有节点的儿子
    cnt=[0] *n
    for node in range(n):
        fa=fals[node]
        if color[node]==1:
            if fa>=0:
                cnt[fa]+=1
    set0=set()
    set1=set()
    set2=set()
    set3=set()
    setls=[set0,set1,set2,set3]
    #只统计黑色节点
    def update(node,var):
        if node==-1:
            return
        if color[node]==1:
            v=cnt[node]
            newv=cnt[node]+var
            cnt[node]=newv
            setls[v].discard(node)
            setls[newv].add(node)
        else:
            v = cnt[node]
            newv = cnt[node] + var
            cnt[node]=newv
            setls[v].discard(node)
        return
    for i in range(n):
        update(i,0)
    #print(setls)
    for j in range(q):
        u=int(input())
        u=u-1
        fa=fals[u]
        if color[u]==1:
            color[u]=0
            update(u,0)
            update(fa,-1)
        else:
            color[u]=1
            update(u,0)
            update(fa,1)
        #print(setls)
        if len(set0)==1 and len(set2)==0:
            print("YES")
        elif len(set0)==2 and len(set2)==1 :
            for node in set2:
                if color[fals[node]]==0:
                    print("YES")
                else:print("NO")
        else:
            print("NO")

for i in range(int(input())):
    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
5 5
1 0 0 0 1
1 2
1 3
1 4
1 5
1
2
3
4
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

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

public class Main {
    private static ArrayList<ArrayList<Integer>> adjacencyList;
    private static int[] parent;
    private static int[] c;
    private static int[] b;
    private static int[] cb2;
    private static int b0;
    private static int b2;
    private static int b3;
    private static int invalid;
    private static void dfs(int u, int p) {
        parent[u] = p;
        int count = 0;
        for(int v: adjacencyList.get(u)) {
            if(v != p) {
                if(c[v] == 1) {
                    count++;
                }
                dfs(v, u);
                if(b[v] == 2) {
                    cb2[u]++;
                }
            }
        }
        b[u] = count;
        if(c[u] == 1) {
            if(b[u] == 0) {
                b0++;
            } else if(b[u] == 2) {
                b2++;
                if(p != -1 && c[p] == 1) {
                    invalid++;
                }
            } else if(b[u] > 2) {
                b3++;
            }
        }
    }
    private static void update(int u) {
        if(c[u] == 0) {
            c[u] = 1;
            if(b[u] == 0) {
                b0++;
            } else if(b[u] == 2) {
                b2++;
                if(parent[u] != -1 && c[parent[u]] == 1) {
                    invalid++;
                }
            } else if(b[u] > 2) {
                b3++;
            }
            if(parent[u] != -1) {
                if(c[parent[u]] == 1) {
                    if(b[parent[u]] == 0) {
                        b0--;
                    } else if(b[parent[u]] == 1) {
                        b2++;
                        cb2[parent[u]]++;
                        if(parent[parent[u]] != -1 && c[parent[parent[u]]] == 1) {
                            invalid++;
                        }
                    } else if(b[parent[u]] == 2) {
                        b2--;
                        cb2[parent[u]]--;
                        if(parent[parent[u]] != -1 && c[parent[parent[u]]] == 1) {
                            invalid--;
                        }
                        b3++;
                    }
                }
                b[parent[u]]++;
            }
            invalid += cb2[u];
        } else {
            c[u] = 0;
            if(b[u] == 0) {
                b0--;
            } else if(b[u] == 2) {
                b2--;
                if(parent[u] != -1 && c[parent[u]] == 1) {
                    invalid--;
                }
            } else if(b[u] > 2) {
                b3--;
            }
            if(parent[u] != -1) {
                if(c[parent[u]] == 1) {
                    if(b[parent[u]] == 1) {
                        b0++;
                    } else if(b[parent[u]] == 2) {
                        b2--;
                        cb2[parent[u]]--;
                        if(parent[parent[u]] != -1 && c[parent[parent[u]]] == 1) {
                            invalid--;
                        }
                    } else if(b[parent[u]] == 3) {
                        b2++;
                        cb2[parent[u]]++;
                        if(parent[parent[u]] != -1 && c[parent[parent[u]]] == 1) {
                            invalid++;
                        }
                        b3--;
                    }
                }
                b[parent[u]]--;
            }
            invalid -= cb2[u];
        }
    }
    public static void main(String[] args) throws IOException {
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int t = Integer.parseInt(f.readLine());
        while(t-- > 0) {
            StringTokenizer st = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());
            st = new StringTokenizer(f.readLine());
            c = new int[n];
            for(int i = 0; i < n; i++) {
                c[i] = Integer.parseInt(st.nextToken());
            }
            adjacencyList = new ArrayList<>();
            for(int i = 0; i < n; i++) {
                adjacencyList.add(new ArrayList<>());
            }
            for(int i = 0; i < n-1; i++) {
                st = new StringTokenizer(f.readLine());
                int x = Integer.parseInt(st.nextToken())-1;
                int y = Integer.parseInt(st.nextToken())-1;
                adjacencyList.get(x).add(y);
                adjacencyList.get(y).add(x);
            }
            parent = new int[n];
            b = new int[n];
            cb2 = new int[n];
            b0 = 0;
            b2 = 0;
            b3 = 0;
            invalid = 0;
            dfs(0, -1);
            while(q-- > 0) {
                int u = Integer.parseInt(f.readLine())-1;
                update(u);
                if(invalid == 0 && b3 == 0 && (b2 == 0 && b0 == 1 || b2 == 1 && b0 == 2)) {
                    out.println("Yes");
                } else {
                    out.println("No");
                }
            }
        }
        f.close();
        out.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
//#include<iostream>

#include<cstdio>
#include<algorithm>
#include<set>
#define N 200010
using namespace std;
int read()
{ char ch=getchar();int p=0;
  for(;ch<'0'||'9'<ch;ch=getchar());
  for(;'0'<=ch&&ch<='9';p=(p<<3)+(p<<1)+ch-'0',ch=getchar());
  return p;
}
int to[N<<1],nxt[N<<1],head[N],e;
void add(int x,int y)
{ to[++e]=y,nxt[e]=head[x],head[x]=e;
}
bool co[N];
int fa[N],num[N];
void dfs(int x)
{ for(int i=head[x];i;i=nxt[i])
	{ int y=to[i];
	  if(y==fa[x]) continue;
	  if(co[y]) num[x]++;
	  fa[y]=x,dfs(y);
	}
}
int g[4],c[3];
set<int> s;
signed main()
{ int T=read();
  while(T--)
	{ int n=read(),q=read();
	  for(int i=1;i<=n;i++) co[i]=read();
	  for(int i=1;i<n;i++)
		{ int x=read(),y=read(); add(x,y),add(y,x); }
	  num[0]=co[1],fa[1]=0,dfs(1);
	  for(int i=0;i<=n;i++)
		if(co[i])
		  { g[min(num[i],3)]++;
			if(num[i]==2) s.insert(i);
		  }
		else c[min(num[i],2)]++;

//	  for(int i=1;i<=n;i++)
//		cout<<num[i]<<"nmb"<<' ';
//	  cout<<endl;
//	  for(int i=0;i<4;i++)
//		cout<<g[i]<<"sb"<<' ';
//	  cout<<endl;
//	  for(int i=0;i<3;i++)
//		cout<<c[i]<<"cnm"<<' ';
//	  cout<<endl;

	  while(q--)
		{ int x=read();
		  if(co[x])
			{ if(num[x]==2) s.erase(x);
			  g[min(num[x],3)]--,c[min(num[x],2)]++;
			  if(co[fa[x]]&&num[fa[x]]==2) s.erase(fa[x]);
			  if(co[fa[x]])
				g[min(num[fa[x]],3)]--,g[min(num[fa[x]]-1,3)]++;
			  else
				c[min(num[fa[x]],2)]--,c[min(num[fa[x]]-1,2)]++;
			  num[fa[x]]--;
			}
		  else
			{ if(num[x]==2) s.insert(x);
			  g[min(num[x],3)]++,c[min(num[x],2)]--;
			  if(co[fa[x]]&&num[fa[x]]==2) s.insert(fa[x]);
			  if(co[fa[x]])
				g[min(num[fa[x]],3)]--,g[min(num[fa[x]]+1,3)]++;
			  else
				c[min(num[fa[x]],2)]--,c[min(num[fa[x]]+1,2)]++;
			  num[fa[x]]++;
			}
		  co[x]^=1;

//		  cout<<num[3]<<"sss"<<endl;

		  if(g[3]||g[2]>1||!g[0]||c[2]||c[1]>1)
			{ printf("No\n"); continue; }
		  if(g[2])
			if(co[fa[*s.begin()]]) printf("No\n");
			else printf("Yes\n");
		  else printf("Yes\n");
		}
	  for(int i=0;i<=n;i++)  head[i]=0;
	  e=0;
	  for(int i=0;i<=n;i++) fa[i]=num[i]=0;
	  g[0]=g[1]=g[2]=g[3]=0,c[0]=c[1]=c[2]=0;
	  s.clear();
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to craft a test case that challenges the logic of the algorithm, particularly focusing on edge cases and scenarios where the number of black vertices and their arrangement can lead to incorrect outputs.

### Step-by-Step Analysis:
1. **Initial Setup:**
   - We need a small tree with a specific initial coloring to test the toggling and chain formation.
   - The tree should be structured in a way that toggling certain vertices can change the chain formation status.

2. **Test Case Design:**
   - **Tree Structure:** A simple linear tree (chain) with a few additional branches to create complexity.
   - **Initial Colors:** Start with a configuration that is close to a chain but not quite, and then toggle vertices to see if the algorithm correctly identifies the chain formation.

3. **Example Test Case:**
   - **Tree with 5 vertices:**
     - Vertices: 1, 2, 3, 4, 5
     - Edges: (1, 2), (2, 3), (3, 4), (3, 5)
   - **Initial Colors:**
     - 1: Black (1)
     - 2: White (0)
     - 3: White (0)
     - 4: White (0)
     - 5: White (0)
   - **Queries:**
     - Toggle vertex 2
     - Toggle vertex 3
     - Toggle vertex 4
     - Toggle vertex 5

### Expected Output:
- After the first query (toggling vertex 2), the tree should have vertices 1 and 2 as black, forming a chain.
- After the second query (toggling vertex 3), the tree should have vertices 1, 2, and 3 as black, forming a chain.
- After the third query (toggling vertex 4), the tree should have vertices 1, 2, 3, and 4 as black, forming a chain.
- After the fourth query (toggling vertex 5), the tree should have vertices 1, 2, 3, 4, and 5 as black, not forming a chain (since 3 has two black neighbors).

### Test Input:
```plaintext
1
5 4
1 0 0 0 0
1 2
2 3
3 4
3 5
2
3
4
5
```

This test case will challenge the algorithm to correctly identify when the black vertices form a chain and when they do not. The initial configuration and the sequence of queries are designed to expose any logical flaws in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

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

class Solution {
    vector<vector<int>> adj;
    vector<int> color;
    int n;

    vector<int> findPath(int start, int end, vector<bool>& visited, vector<int>& path) {
        if (start == end) {
            path.push_back(start);
            return path;
        }

        visited[start] = true;
        path.push_back(start);

        for (int next : adj[start]) {
            if (!visited[next]) {
                vector<int> result = findPath(next, end, visited, path);
                if (!result.empty()) return result;
            }
        }

        path.pop_back();
        return vector<int>();
    }

    bool isChain() {
        vector<int> black;
        for (int i = 1; i <= n; i++) {
            if (color[i]) black.push_back(i);
        }

        if (black.empty()) return false;
        if (black.size() == 1) return true;

        for (int i = 0; i < black.size(); i++) {
            for (int j = i + 1; j < black.size(); j++) {
                vector<bool> visited(n + 1, false);
                vector<int> path;
                vector<int> result = findPath(black[i], black[j], visited, path);

                if (!result.empty()) {
                    set<int> pathSet(result.begin(), result.end());
                    set<int> blackSet(black.begin(), black.end());

                    if (pathSet == blackSet) return true;
                }
            }
        }
        return false;
    }

public:
    void solve() {
        int t;
        cin >> t;
        while (t--) {
            int q;
            cin >> n >> q;

            color.resize(n + 1);
            adj.clear();
            adj.resize(n + 1);

            for (int i = 1; i <= n; i++) {
                cin >> color[i];
            }

            for (int i = 0; i < n - 1; i++) {
                int x, y;
                cin >> x >> y;
                adj[x].push_back(y);
                adj[y].push_back(x);
            }

            while (q--) {
                int u;
                cin >> u;
                color[u] ^= 1;  // toggle color
                cout << (isChain() ? "Yes\n" : "No\n");
            }
        }
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    Solution solution;
    solution.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 provided buggy code, we need to consider edge cases and scenarios that can cause the algorithm to fail. The main issues in the buggy code are:

1. **Inefficient Path Finding**: The `findPath` function uses a depth-first search (DFS) to find a path between two black vertices, which is not efficient and can lead to incorrect results or timeouts.
2. **Incorrect Chain Check**: The `isChain` function checks all pairs of black vertices to see if they form a chain, which is computationally expensive and can fail in certain edge cases.

To expose these issues, we can create a test case where the tree structure and the sequence of queries make it difficult for the algorithm to correctly determine if the black vertices form a chain. Here's a step-by-step approach to generate such a test case:

1. **Tree Structure**: Create a tree where the black vertices are not directly connected, but the path between them involves multiple white vertices.
2. **Initial Colors**: Start with a configuration where some vertices are black and some are white.
3. **Queries**: Design queries that toggle the colors in a way that the algorithm might incorrectly determine the chain.

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

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

### Explanation:
- **Tree Structure**:
  - Vertex 1 is connected to vertex 2.
  - Vertex 2 is connected to vertex 3.
  - Vertex 3 is connected to vertex 4.
  - Vertex 4 is connected to vertex 5.
- **Initial Colors**:
  - Vertices 1 and 5 are black.
  - Vertices 2, 3, and 4 are white.
- **Queries**:
  - Toggle vertex 1 (now white).
  - Toggle vertex 2 (now black).
  - Toggle vertex 3 (now black).
  - Toggle vertex 4 (now black).
  - Toggle vertex 5 (now white).

### Expected Output:
- After the first query, the black vertices are 5, so the output should be "Yes".
- After the second query, the black vertices are 2 and 5, and they form a chain, so the output should be "Yes".
- After the third query, the black vertices are 2, 3, and 5, and they form a chain, so the output should be "Yes".
- After the fourth query, the black vertices are 2, 3, and 4, and they form a chain, so the output should be "Yes".
- After the fifth query, the black vertices are 2, 3, and 4, and they form a chain, so the output should be "Yes".

This test case should expose the inefficiency and potential incorrectness in the `isChain` function, as it needs to check all pairs of black vertices and the path between them. The algorithm might fail to correctly determine the chain in such a scenario.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chain Queries
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a tree of $n$ vertices numbered from $1$ to $n$. Initially, all vertices are colored white or black.

You are asked to perform $q$ queries:

  * "u" — toggle the color of vertex $u$ (if it was white, change it to black and vice versa).

After each query, you should answer whether all the black vertices form a chain. That is, there exist two black vertices such that the simple path between them passes through all the black vertices and only the black vertices. Specifically, if there is only one black vertex, they form a chain. If there are no black vertices, they do not form a chain.

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

The first line of each test case contains two integers $n$ and $q$ ($1\leq n,q\leq 2\cdot 10^5$).

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($c_i \in \\{ \mathtt{0}, \mathtt{1} \\}$) — the initial color of the vertices. $c_i$ denotes the color of vertex $i$ where $\mathtt{0}$ denotes the color white, and $\mathtt{1}$ denotes the color black.

Then $n - 1$ lines follow, each line contains two integers $x_i$ and $y_i$ ($1 \le x_i,y_i \le n$), indicating an edge between vertices $x_i$ and $y_i$. It is guaranteed that these edges form a tree.

The following $q$ lines each contain an integer $u_i$ ($1 \le u_i \le n$), indicating the color of vertex $u_i$ needs to be toggled.

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

Output: For each query, output "Yes" if the black vertices form a chain, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
2
2 1
1 0
1 2
1
5 4
1 0 0 0 0
1 2
1 3
1 5
3 4
4
3
2
5
output:
No
No
Yes
Yes
No

input:
4
5 3
1 1 1 1 1
3 5
2 5
3 4
1 5
1
1
1
4 4
0 0 0 0
1 2
2 3
1 4
1
2
3
2
1 1
1
1
1 1
0
1
output:
Yes
No
Yes
Yes
Yes
Yes
No
No
Yes

Note: In the second test case, the color of the vertices are as follows:

The initial tree:

![](CDN_BASE_URL/b4c312b1023ec449df862310fa3c507b)

The first query toggles the color of vertex $4$:

![](CDN_BASE_URL/f7f4726f86d3ba27bbee2f70be9c2f51)

The second query toggles the color of vertex $3$:

![](CDN_BASE_URL/61a2241ad4b2defae2bf8781f718c6d1)

The third query toggles the color of vertex $2$:

![](CDN_BASE_URL/b888921fda4dbca3670c3f99ac8553ac)

The fourth query toggles the color of vertex $5$:

![](CDN_BASE_URL/bab22014ef15ff4590580c6b5c7239cb)

Buggy Code:
def solve():
  n, q = map(int, input().split())
  c = list(map(int, input().split()))
  black_vs = set()
  local_black_es = []
  is_chain = True

  for i in range(0, n):
    if c[i] == 1:
      black_vs.add(i)

    local_black_es.append(0)

  g = dict()

  for i in range(0, n):
    g[i] = []

  for i in range(0, n - 1):
    u, v = map(int, input().split())

    g[u - 1].append(v - 1)
    g[v - 1].append(u - 1)

  black_es = 0
  for v in black_vs:
    for t in g[v]:
      if c[t] == 1:
        local_black_es[v] += 1

    black_es += local_black_es[v]

    if local_black_es[v] > 2: is_chain = False

  if len(black_vs) == 1: is_chain = True
  elif black_es / 2 != len(black_vs) - 1: is_chain = False
  else: is_chain = True

  # print(f'c: {c}, black_vs: {black_vs}, local_black_es: {local_black_es}, black_es: {black_es}, is_chain: {is_chain}')

  for _ in range(q):
    u = int(input()) - 1

    c[u] = 1 - c[u]
    mod = 1 if c[u] == 1 else -1

    if c[u] == 1:
      black_vs.add(u)
    else:
      black_vs.remove(u)

    for t in g[u]:
      if c[u] == 1 and c[t] == 1:
        local_black_es[u] += 1
        local_black_es[t] += 1
        black_es += 2
      elif c[u] == 0 and c[t] == 1:
        local_black_es[u] -= 1
        local_black_es[t] -= 1
        black_es -= 2

      if local_black_es[u] > 2: is_chain = False
      elif local_black_es[t] > 2: is_chain = False
      elif len(black_vs) == 1: is_chain = True
      elif black_es / 2 != len(black_vs) - 1: is_chain = False
      else: is_chain = True

    if len(black_vs) == 1: is_chain = True
    elif n == 1: is_chain = False

    # print(f'c: {c}, black_vs: {black_vs}, local_black_es: {local_black_es}, black_es: {black_es}, is_chain: {is_chain}')
    if is_chain: print("YES")
    else: print("NO")

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

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

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

Think step by step.

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