**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solution solver = new Solution();
        boolean multipleTC = true;
        int testCount = multipleTC ? Integer.parseInt(in.next()) : 1;
        for (int i = 1; i <= testCount; i++) solver.solve(in, out, i);
        out.close();
    }

    static class Solution {
        PrintWriter out;
        InputReader in;

        static void DFS(boolean vis[], int x, int y, Stack<Integer> stack, ArrayList<Integer>[] tree, ArrayList<Integer> path) {
            stack.add(x);
            if (x == y) {
                path.addAll(stack);
                return;
            }
            vis[x] = true;
            // if backtracking is taking place
            if (!tree[x].isEmpty()) {
                for(int j = 0; j < tree[x].size(); j++) {
                    // if the node is not visited
                    if (!vis[tree[x].get(j)]) {
                        DFS(vis, tree[x].get(j), y, stack, tree, path);
                    }
                }
            }
            stack.pop();
        }

        int mxDepth = 0;
        void mxDepth(int v, int p, int h, ArrayList<Integer>[] tree) {
            mxDepth = Math.max(mxDepth, h);
            for (int u: tree[v]) {
                if (u != p) {
                    mxDepth(u, v, h + 1, tree);
                }
            }
         }

        public void solve(InputReader in, PrintWriter out, int test) {
            this.out = out;
            this.in = in;
            int n = ni();
            Stack<Integer> st = new Stack<>();
            ArrayList<Integer>[] tree = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                tree[i] = new ArrayList<>();
            }
            int a = ni() - 1;
            int b = ni() - 1;
            for (int i = 0; i < n - 1; i++) {
                int u = ni() - 1;
                int v = ni() - 1;
                tree[u].add(v);
                tree[v].add(u);
            }
            if (n == 2) {
                pn(2);
                return;
            }
            boolean[] vis = new boolean[n];
            ArrayList<Integer> path = new ArrayList<>();
            DFS(vis, a, b, st, tree, path);
            int sz = path.size();
            int root = path.get(sz / 2);
            mxDepth = 0;
            mxDepth(root, -1, 0, tree);
            //pn(mxDepth + " " + path.size());
            pn(sz / 2 + 2L * (n - 1) - mxDepth + (1 - sz % 2));
        }

        static final Comparator<Pair> com = (o1, o2) -> {
            if (Integer.compare(o1.efficiency, o2.efficiency) != 0) {
                return Integer.compare(o1.efficiency, o2.efficiency);
            } else {
                return Integer.compare(o1.speed, o2.speed);
            }
        };
         class Pair {
            int efficiency;
            int speed;
            int id;
            Pair(int id, int efficiency, int speed) {
                this.id = id;
                this.efficiency = efficiency;
                this.speed = speed;
            }
         }

        class OncallShifts {
            int L;
            int R;
            String id;
            OncallShifts(String id, int L, int R) {
                this.id = id;
                this.L = L;
                this.R = R;
            }
        }

        String n() {
            return in.next();
        }

        int ni() {
            return in.nextInt();
        }

        long nl() {
            return in.nextLong();
        }

        void pn(long zx) {
            out.println(zx);
        }

        void pn(String sz) {
            out.println(sz);
        }

        void pn(double dx) {
            out.println(dx);
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) {
                throw new UnknownError();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new UnknownError();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

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

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

        public String next() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuffer res = new StringBuffer();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));

            return res.toString();
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -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:
```plaintext
3
1
1 1
2
1 2
1 2
3
1 2
1 2
2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
from collections import defaultdict
import sys

sys.setrecursionlimit(200000)

def solve():
    n = int(input())

    a, b = map(int, input().split())

    mp = defaultdict(list)
    for _ in range(n - 1):
        x, y = map(int, input().split())
        mp[x].append(y)
        mp[y].append(x)
    # first meet in the middle
    ans = 0
    path = []
    mid = a

    if a != b:
        q = [a]
        vis = [False] * (n + 1)
        vis[a] = True
        level = 0
        parent = [0] * (n + 1)
        parent[a] = a

        while q:
            l = len(q)
            level += 1
            for _ in range(l):
                top = q.pop(0)
                for nxt in mp[top]:
                    if nxt == b:
                        parent[b] = top
                        ans += (level + 1) // 2
                        break
                    if vis[nxt] == False:
                        parent[nxt] = top
                        q.append(nxt)
                        vis[nxt] = True
        curr = b
        while curr != a:
            path.append(curr)
            curr = parent[curr]
        path.append(a)
        mid = path[len(path) // 2]

    # second just go through everything
    mx = 0
    vis = [False] * (n + 1)
    def dfs(node, depth):
        if vis[node]:
            return
        nonlocal mx
        vis[node] = True
        mx = max(mx, depth)
        for nxt in mp[node]:
            dfs(nxt, depth + 1)
    dfs(mid, 0)
    ans += mx
    ans += (n - 1 - mx) * 2
    print(ans)

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
4
1 4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
from collections import deque

class State:
    def __init__(self, a, b, steps, colors):
        self.a = a
        self.b = b
        self.steps = steps
        self.colors = colors

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        a, b = map(int, input().split())
        a -= 1
        b -= 1

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

        initial_colors = [0] * n
        if a == b:
            initial_colors[a] = 2
        else:
            initial_colors[a] = 1

        q = deque([State(a, b, 0, initial_colors)])
        visited = set()
        visited.add((a, b, tuple(initial_colors)))

        result = -1
        while q:
            current = q.popleft()
            pa, pb, steps, colors = current.a, current.b, current.steps, current.colors

            if all(color == 2 for color in colors):
                result = steps
                break

            for neighbor_a in adj[pa]:
                new_colors = colors[:]
                if new_colors[neighbor_a] == 0:
                    new_colors[neighbor_a] = 1
                for neighbor_b in adj[pb]:
                    temp_colors = new_colors[:]
                    if temp_colors[neighbor_b] == 1:
                        temp_colors[neighbor_b] = 2
                    new_state = (neighbor_a, neighbor_b, tuple(temp_colors))
                    if new_state not in visited:
                        visited.add(new_state)
                        q.append(State(neighbor_a, neighbor_b, steps + 1, temp_colors))

        print(result)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

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

#define ll long long
#define all(x) x.begin(), x.end()
#define endl '\n'
#define max1 first
#define max2 second

#define file(in, out)                                                        \
  freopen(in, "r", stdin);                                                     \
  freopen(out, "w", stdout)

#define FIn                                                                    \
  cin.tie(0);                                                                  \
  cout.tie(0);                                                                 \
  ios_base::sync_with_stdio(false)

int tc;
const bool MULTI_TESTS = true;
const string IN = "input.txt";
const string OUT = "output.txt";
const int N = 2e5 + 10;
const int MOD = 1e9 + 7;

void pre();
//check MAX_N

int n, a, b, u, v;
ll distRed[N], distBlue[N];
ll ans[N];
pair<ll, ll> pp[N];
vector<int> g[N];

void init(){
    for(int i = 1; i <= n; ++i) {
        g[i].clear();
        ans[i] = distRed[i] = distBlue[i] = 0;
        pp[i] = {0, 0};
    }
}

void findDist(int node, int par, bool red){
    for(auto child: g[node]){
        if(child != par) {
            if(red)
                distRed[child] = distRed[node] + 1;
            else
                distBlue[child] = distBlue[node] + 1;
            findDist(child, node, red);
        }
    }
}

ll dfsForRootAns(int node, int par){
    ll sumChild = 0;
    for(auto child: g[node]){
        if(child == par)
            continue;

        ll childAns = dfsForRootAns(child, node);
        sumChild += childAns;
        if(childAns > pp[node].max1)
            pp[node] = {childAns, pp[node].max1};
        else if(childAns > pp[node].max2)
            pp[node].max2 = childAns;

    }
    ans[node] = sumChild * 2 - pp[node].max1;
    return ans[node] + 1;
}

void dfsForUpdateAns(int node, int par){
    for(auto child : g[node]){
        if(child == par)
            continue;

        ll tmpChild;

        if(ans[child] + 1 != pp[node].max1)
            tmpChild = 1 + ans[node] - 2 * (ans[child] + 1);
        else
            tmpChild = 1 + ans[node] - pp[node].max1 - pp[node].max2;

        if(tmpChild > pp[child].max1){
            ans[child] = ans[child] + pp[child].max1 + tmpChild;
            pp[child] = {tmpChild, pp[child].max1};
        }
        else {
            ans[child] = ans[child] + 2 * tmpChild;
            if(tmpChild > pp[child].second)
                pp[child].second = tmpChild;
        }

        dfsForUpdateAns(child, node);
    }
}

void solve(){
    cin >> n;
    cin >> a >> b;
    ll rslt = LONG_LONG_MAX;
    init();

    for(int i = 1; i < n; ++i){
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    findDist(a, -1, true);
    findDist(b, -1, false);

    dfsForRootAns(1, -1);
    dfsForUpdateAns(1, -1);

    for(int i = 1; i <= n; ++i){
//        cout<<i<<": "<<ans[i]<<'\n';

        rslt = min(rslt,
                   ans[i] + max(
                            distBlue[i], distRed[i] + ((distBlue[i] & 1) ^ (distRed[i] & 1))
                        )
                );
    }

    cout<<rslt<<'\n';

}

int main() {
//    FIn;
#ifndef ONLINE_JUDGE
    file(IN.c_str(),OUT.c_str());
#endif
    tc = 1;
    if(MULTI_TESTS) cin >> tc;
    pre();
    for (int i = 1; i <= tc; i++) {
        // cout << "Case " << i <<": \n";
        solve();
    }
}

void pre(){}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long int
#define mod 1000000007
#define ordered_set tree < int ,  null_type ,  less<int> ,  rb_tree_tag ,  tree_order_statistics_node_update >

vector<ll>adj[200001];
ll cnt=0;

vector<ll>subnode(200001);
void dfs1(ll node,ll parent){
subnode[node]=1;
for(auto i:adj[node]){
    if(i!=parent){
        dfs1(i,node);
        subnode[node]+=subnode[i];
    }
}
}

vector<ll>dist_from_a(200001);
void dfs2(ll node,ll parent,ll d){
dist_from_a[node]=d;
for(auto i:adj[node]){
    if(i!=parent){
        dfs2(i,node,d+1);
    }
}
}

vector<ll>dist_from_b(200001);
void dfs4(ll node,ll parent,ll d){
dist_from_b[node]=d;
for(auto i:adj[node]){
    if(i!=parent){
        dfs4(i,node,d+1);
    }
}
}

void dfs3(ll node,ll parent){
ll c=-1;
for(auto i:adj[node]){
    if(i!=parent){
        if(c==-1) c=i;
        else if(subnode[i]>subnode[c]){
            cnt+=2+(subnode[c]-1)*2;
            c=i;
        }
        else{
            cnt+=2+(subnode[i]-1)*2;
        }
    }
}

if(c!=-1){
    cnt++;
    dfs3(c,node);
}
}
int main(){
ios::sync_with_stdio(0);
  cin.tie(0);
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
ll t; cin>>t;
while(t--){

    ll n; cin>>n;
    for(ll i=1;i<=n;i++){
        adj[i].clear();
        subnode[i]=-1;
        dist_from_a[i]=-1;
        dist_from_b[i]=-1;
    }

    ll a,b; cin>>a>>b;

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

    dfs2(a,-1,0);
    dfs4(b,-1,0);

    ll ans=LLONG_MAX;
    for(ll i=1;i<=n;i++){
        ll add;
        if(dist_from_a[b]%2==0){
        add=max(dist_from_a[i],dist_from_b[i]);
        }
        else{
        add=max(dist_from_a[i],dist_from_b[i]-1)+1;
        }
        dfs1(i,-1);
        cnt=0;
        dfs3(i,-1);
//        cout<<i<<" "<<add<<" "<<cnt<<endl;
        ans=min(ans,add+cnt);
    }

    cout<<ans<<endl;
}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import java.util.*;

public class Main {

    static long LLONG_MAX = Long.MAX_VALUE;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            int[][] edges = new int[n - 1][2];

            for (int i = 0; i < n - 1; i++) {
                edges[i][0] = sc.nextInt();
                edges[i][1] = sc.nextInt();
            }

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

            for (int[] edge : edges) {
                adj.get(edge[0]).add(edge[1]);
                adj.get(edge[1]).add(edge[0]);
            }

            Pair<Integer, Integer> diameter = findDiameter(adj, n);
            int d1 = diameter.first;
            int d2 = diameter.second;

            long ans = LLONG_MAX;
            for (int i = 1; i <= n; i++) {
                int tmp = 2 * n - 1 - Math.max(dist(adj, i, d1), dist(adj, i, d2));
                int da = dist(adj, i, a), db = dist(adj, i, b);
                if (da <= db) {
                    ans = Math.min(ans, tmp + db - 1);
                }
            }

            System.out.println(ans);
        }

        sc.close();
    }

    static int dist(List<List<Integer>> adj, int u, int v) {
        int[] distance = new int[adj.size()];
        Arrays.fill(distance, -1);
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(u);
        distance[u] = 0;

        while (!queue.isEmpty()) {
            int curr = queue.poll();
            for (int neighbor : adj.get(curr)) {
                if (distance[neighbor] == -1) {
                    queue.offer(neighbor);
                    distance[neighbor] = distance[curr] + 1;
                }
            }
        }

        return distance[v];
    }

    static Pair<Integer, Integer> findDiameter(List<List<Integer>> adj, int n) {
        int[] dist = new int[n + 1];
        int startNode = 1;
        int farthestNode = bfs(adj, startNode, dist);
        dist = new int[n + 1];
        int secondFarthestNode = bfs(adj, farthestNode, dist);
        return new Pair<>(farthestNode, secondFarthestNode);
    }

    static int bfs(List<List<Integer>> adj, int src, int[] dist) {
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(src);
        dist[src] = 1;
        int farthestNode = src;

        while (!queue.isEmpty()) {
            int curr = queue.poll();
            for (int neighbor : adj.get(curr)) {
                if (dist[neighbor] == 0) {
                    queue.offer(neighbor);
                    dist[neighbor] = dist[curr] + 1;
                    if (dist[neighbor] > dist[farthestNode]) {
                        farthestNode = neighbor;
                    }
                }
            }
        }

        return farthestNode;
    }

    static class Pair<U, V> {
        U first;
        V second;

        public Pair(U first, V second) {
            this.first = first;
            this.second = second;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint the Tree
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a tree with $n$ vertices. Initially, all vertices are white.

There are two chess pieces called $P_A$ and $P_B$ on the tree. $P_A$ and $P_B$ are initially located on vertices $a$ and $b$ respectively. In one step, 378QAQ will do the following in order:

  1. Move $P_A$ to a neighboring vertex. If the target vertex is white, this vertex will be painted red.    2. Move $P_B$ to a neighboring vertex. If the target vertex is colored in red, this vertex will be painted blue.

Initially, the vertex $a$ is painted red. If $a=b$, the vertex $a$ is painted blue instead. Note that both the chess pieces must be moved in each step. Two pieces can be on the same vertex at any given time.

378QAQ wants to know the minimum number of steps to paint all vertices blue.

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 one integer $n$ ($1\leq n\leq 2\cdot 10^5$).

The second line of each test case contains two integers $a$ and $b$ ($1\leq a,b\leq n$).

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.

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

Output: For each test case, output the minimum number of steps to paint all vertices blue.

Examples:
input:
3
2
1 2
1 2
5
1 2
1 2
1 3
1 4
1 5
8
5 4
7 1
1 5
1 8
8 3
7 2
8 6
3 4
output:
2
8
13

Note: In the first test case, 378QAQ can paint all vertices blue in the following order:

  * Initially, $P_A$ is located on the vertex $1$, and $P_B$ is located on the vertex $2$. The vertex $1$ is painted red and the vertex $2$ is white.    * 378QAQ moves $P_A$ to the vertex $2$ and paints it red. Then 378QAQ moves $P_B$ to the vertex $1$ and paints it blue.    * 378QAQ moves $P_A$ to the vertex $1$. Then 378QAQ moves $P_B$ to the vertex $2$ and paints it blue.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) throws Exception{
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        int tests = Integer.parseInt(buffer.readLine());
        for (int test = 0; test < tests; test ++){
            StringTokenizer token = new StringTokenizer(buffer.readLine());
            int n = Integer.parseInt(token.nextToken());
            token = new StringTokenizer(buffer.readLine());
            int a = Integer.parseInt(token.nextToken());
            int b = Integer.parseInt(token.nextToken());
            HashMap<Integer, List<Integer>> map = new HashMap<>();
            // # of moves originally is (n-1) * 2
            for (int x = 0; x < n - 1; x ++){
                token = new StringTokenizer(buffer.readLine());
                int j = Integer.parseInt(token.nextToken());
                int k = Integer.parseInt(token.nextToken());
                map.putIfAbsent(j, new ArrayList<>());
                map.putIfAbsent(k, new ArrayList<>());
                map.get(j).add(k); map.get(k).add(j);
            }
            if (n == 1) {System.out.println("0"); continue;}
            // System.out.println(map);
            Queue<List<Integer>> queue = new LinkedList<>(); queue.add(new ArrayList<>(Arrays.asList(a)));
            int meeting = 0;
            int time = 0;
            while (queue.size() > 0){
                List<Integer> next = queue.poll();
                int last = next.get(next.size() - 1);
                if (last == b) {time = next.size() / 2; meeting = next.get((next.size() - 1) / 2); break;}
                for (Integer e: map.get(last)){
                    if (next.size() == 1 || next.get(next.size() - 2) != e) {
                        next.add(e); queue.add(new ArrayList<>(next)); next.remove(e);
                    }
                }
            }
            // System.out.println(meeting + " " + time);
            // find minimal time starting at meeting
            int[] depth = new int[n + 1];
            depth[meeting] = 1;
            Queue<Integer> queue2 = new LinkedList<>(); queue2.add(meeting);
            int maxdepth = 1;
            while(queue2.size() > 0){
                int next = queue2.poll();
                maxdepth = Math.max(maxdepth, depth[next]);
                for (Integer e: map.get(next)){
                    if (depth[e] == 0) {depth[e] = depth[next] + 1; queue2.add(e);}
                }
            }
            // System.out.println(maxdepth);
            System.out.println(time + 2*n - 1 - maxdepth);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
#include <iostream>

using namespace std;

int main () {
    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        int x; cin >> x;
        n--;
        int ans = 1;
        int rem = 0;
        while (n--) {
            int x; cin >> x;
            int f = min (x, 8 - rem);
            rem += f;
            x -= f;

            if (rem % 8 == 0 && x % 8 == 0)
                ans++;

            rem %= 8;
            rem += x;
            rem %= 8;
        }

        cout << ans << '\n';
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

int main() {
    int a;
    cin >> a;
    for (int i = 0; i < a; i++) {
        int b;
        cin >> b;
        int cn = 0;
        int d = 0;

        for (int j = 0; j < b; j++) {
            int c;
            cin >> c;
            cn += c;

            if (cn == 1) {
                d = 1;
            }
            else if ((cn - 1) % 8 == 0) {
                d++;
            }
        }
        cout << d << 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 provided code, we need to focus on scenarios where the logic in the code might fail. The main issue in the code is the condition `if (cn == 1) { d = 1; }` and the condition `else if ((cn - 1) % 8 == 0) { d++; }`. These conditions are not correctly handling the layers and the number of pieces in each layer.

### Key Points to Consider:
1. **Initial Piece**: The first piece is always placed on the first day.
2. **Layer Completion**: Each layer is completed by adding a specific number of pieces. The first layer has 1 piece, the second layer has 8 pieces, the third layer has 16 pieces, and so on.
3. **Happy Days**: Alyona is happy if at the end of the day, no layer is started but unfinished.

### Challenging Test Input:
We need a test case where the code incorrectly counts the number of happy days. A good test case would be one where the total number of pieces added on a day is just enough to complete a layer, but the code might miscount due to the incorrect conditions.

### Test Case:
```plaintext
1
4
1 7 1 8
```

### Explanation:
- **Day 1**: Alyona places 1 piece (the central piece). She is happy.
- **Day 2**: Alyona places 7 pieces. This completes the second layer (1 + 7 = 8 pieces). She is happy.
- **Day 3**: Alyona places 1 piece. This starts the third layer but does not complete it. She is not happy.
- **Day 4**: Alyona places 8 pieces. This completes the third layer (8 + 8 = 16 pieces). She is happy.

### Expected Output:
```plaintext
3
```

### Why This Test Case is Challenging:
- The code might incorrectly count the number of happy days because it uses the condition `(cn - 1) % 8 == 0` to determine if a layer is completed. This condition is not robust enough to handle the transition between layers correctly.
- The test case includes a scenario where a layer is started but not completed, and then completed on the next day, which the code might not handle correctly.

This test case should expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
// Alyona and a Square Jigsaw Puzzle

#include<stdio.h>
#include<iostream>
#include<vector>

using namespace std;

int main()
{
    int t;
    cin>>t;

    while(t--)
    {
        int n,tot=0,day=1,a;
        cin>>n;
        vector<int> v(n);
        cin>>v[0];
        for(int i=1;i<n;i++)
        {
            cin>>a;
            v[i]=a;
            tot=tot+a;
            if(tot%8==0)
            day++;
        }

        cout<<day<<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 provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issue with the buggy code is that it incorrectly checks if the total number of pieces assembled so far is a multiple of 8, which is not the correct condition for determining if a layer is complete.

Here are the key points to consider:
1. **Layer Completion**: A layer is complete when the total number of pieces added up to that day is a perfect square.
2. **Edge Cases**: Scenarios where the total pieces added do not align with the 8-piece layer completion condition but should still be considered as completing a layer.

Let's construct a test case that highlights these issues:

1. **Test Case**: A scenario where the total pieces added up to a day do not align with the 8-piece condition but should be considered as completing a layer.
2. **Expected Output**: The correct number of days Alyona is happy.

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

```plaintext
1
5
1 3 2 1 2
```

Explanation:
- **Day 1**: Alyona places 1 piece (first layer is complete).
- **Day 2**: Alyona places 3 pieces (total 4 pieces, second layer is complete).
- **Day 3**: Alyona places 2 pieces (total 6 pieces, second layer is still complete).
- **Day 4**: Alyona places 1 piece (total 7 pieces, second layer is still complete).
- **Day 5**: Alyona places 2 pieces (total 9 pieces, third layer is complete).

Expected Output:
- **Day 1**: Happy (first layer complete)
- **Day 2**: Happy (second layer complete)
- **Day 3**: Not happy (second layer still complete)
- **Day 4**: Not happy (second layer still complete)
- **Day 5**: Happy (third layer complete)

So, the correct output should be:
```plaintext
3
```

This test case should expose the bug in the provided code because it checks for a condition that is not correct for layer completion.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
n=int(input())
final=[]
for k in range(n):
    a=int(input())
    b=input()
    arr1=[]
    arr1=b.split()
    arr2=[]
    arr2 =[int(x) for x in arr1]
    arr2.pop(0)
    s,i,j=0,1,1
    for m in arr2:
        s=s+m
        if s%8==0 and s!=0:
            s=0
            j=j+1

    final.append(j)
for finals in final:
    print(finals)

**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
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
# Vaibhav Varshney
import math

def li():
    return list(map(int,input().split()))
def mi():
    return map(int,input().split())

def solve():
    n = int(input())
    a = li()
    adder = 0
    cnt = 0
    adder += a[0]
    if a[0] == 1:
        cnt+=1

    for i in range(1,n):
        adder += a[i]
        if ((adder-1) % 8 == 0):
            cnt+=1
    print(cnt)

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

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
t=int(input())
for j in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    x=0
    y=0
    for i in a:
        x+=i
        for k in range(1,n+10,2):
            if x == k**2:
                y+=1
    print(y)

**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
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc  = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int a[] = new int[n];
            int c = 1;
            int s = 0;
            for(int i = 0;i<n;i++){
               a[i] = sc.nextInt();
               if(i!=0){
                   s+=a[i];
                   if(s%8==0){
                       c++;
                   }
               }

            }
            System.out.println(c);
        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

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

public class A_Alyona_and_a_Square_Jigsaw_Puzzle {

    // Fast Input and Output
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer st;

    public static void main(String[] args) throws IOException {
        int t = nextInt();
        while (t-- > 0) {
            int n = nextInt();
            int res[] = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextInt();
            }
            int count = 0;
            int assemble = 0;
            if (n == 1) {
                if (res[0] == 1) {
                    System.out.println(1);
                } else {
                    System.out.println(0);
                }
                continue;
            }
            if (res[0] >= 1) {
                if (res[0] >= 1) {
                    assemble += res[0] - 1;
                } else {
                    assemble += res[0];
                }
                count++;
            }
            int sum = 1;
            for (int i = 1; i < n; i++) {
                int cal = 8 * i;
                assemble += res[i];
                sum += cal;
                // System.out.println(sum + " " + assemble);
                if ((assemble) % 8 == 0) {
                    count++;
                }
            }
            System.out.println(count);
        }
        out.close();
    }

    // Fast Input Utility
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

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

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

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

    // Utility Methods
    // 1. Bit Manipulation Utilities
    static int countSetBits(int n) {
        return Integer.bitCount(n);
    }

    static int highestOneBit(int n) {
        return Integer.highestOneBit(n);
    }

    static int lowestOneBit(int n) {
        return Integer.lowestOneBit(n);
    }

    static int leadingZeros(int n) {
        return Integer.numberOfLeadingZeros(n);
    }

    static int trailingZeros(int n) {
        return Integer.numberOfTrailingZeros(n);
    }

    static int reverseBits(int n) {
        return Integer.reverse(n);
    }

    static int rotateLeft(int n, int d) {
        return Integer.rotateLeft(n, d);
    }

    static int rotateRight(int n, int d) {
        return Integer.rotateRight(n, d);
    }

    // 2. Math Utilities
    static int gcd(int a, int b) {
        return BigInteger.valueOf(a).gcd(BigInteger.valueOf(b)).intValue();
    }

    static int absExact(int x) {
        return Math.absExact(x);
    }

    static int multiplyExact(int x, int y) {
        return Math.multiplyExact(x, y);
    }

    static int floorDiv(int x, int y) {
        return Math.floorDiv(x, y);
    }

    static BigInteger bigGCD(BigInteger a, BigInteger b) {
        return a.gcd(b);
    }

    static BigInteger modPow(BigInteger base, BigInteger exp, BigInteger mod) {
        return base.modPow(exp, mod);
    }

    // 3. Array Utilities
    static void fillArray(int[] arr, int val) {
        Arrays.fill(arr, val);
    }

    static int[] copyArray(int[] arr, int start, int end) {
        return Arrays.copyOfRange(arr, start, end);
    }

    static int arraySum(int[] arr) {
        return Arrays.stream(arr).sum();
    }

    // 4. Collection Utilities
    static <T extends Comparable<T>> T min(Collection<T> col) {
        return Collections.min(col);
    }

    static <T extends Comparable<T>> T max(Collection<T> col) {
        return Collections.max(col);
    }

    static <T> int frequency(Collection<T> col, T obj) {
        return Collections.frequency(col, obj);
    }

    static <T> boolean disjoint(Collection<T> col1, Collection<T> col2) {
        return Collections.disjoint(col1, col2);
    }

    static <T> void rotate(List<T> list, int distance) {
        Collections.rotate(list, distance);
    }

    static <T> void swap(List<T> list, int i, int j) {
        Collections.swap(list, i, j);
    }

    // 5. String Utilities
    static int indexOf(String s, int ch, int fromIndex) {
        return s.indexOf(ch, fromIndex);
    }

    static String substring(String s, int beginIndex, int endIndex) {
        return s.substring(beginIndex, endIndex);
    }

    static String replace(String s, char oldChar, char newChar) {
        return s.replace(oldChar, newChar);
    }

    static boolean matches(String s, String regex) {
        return s.matches(regex);
    }

    // 6. Priority Queue Utilities
    static PriorityQueue<Integer> minHeap() {
        return new PriorityQueue<>();
    }

    static PriorityQueue<Integer> maxHeap() {
        return new PriorityQueue<>(Collections.reverseOrder());
    }

    // 7. Deque Utilities (Useful for Sliding Window problems)
    static Deque<Integer> createDeque() {
        return new ArrayDeque<>();
    }

    // 8. TreeMap and TreeSet Utilities
    static TreeMap<Integer, Integer> createTreeMap() {
        return new TreeMap<>();
    }

    static TreeSet<Integer> createTreeSet() {
        return new TreeSet<>();
    }

    // 9. Random Utilities
    static int randomInt(int origin, int bound) {
        return ThreadLocalRandom.current().nextInt(origin, bound);
    }

    // 10. System Utilities
    static void copyArray(Object src, int srcPos, Object dest, int destPos, int length) {
        System.arraycopy(src, srcPos, dest, destPos, length);
    }

    static long nanoTime() {
        return System.nanoTime();
    }

    // Example Utility Functions for CP Problems
    static void swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    static int gcdArray(int[] arr) {
        int result = arr[0];
        for (int i = 1; i < arr.length; i++) {
            result = gcd(result, arr[i]);
            if (result == 1)
                return 1; // No need to continue if gcd becomes 1
        }
        return result;
    }
}

**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 specific scenarios that the code might handle incorrectly. The buggy code seems to have issues with correctly calculating the number of complete layers and the days Alyona is happy.

### Key Points to Consider:
1. **Initialization of the First Day**: The first day is always handled correctly since Alyona always places the central piece on the first day.
2. **Layer Completion**: The code should correctly identify when a layer is completed and when a new layer starts.
3. **Edge Cases**: Scenarios where the number of pieces added per day is not a multiple of 8 (since each layer after the first requires 8 pieces to complete).

### Test Case Analysis:
- **Test Case 1**: Single day with 1 piece.
- **Test Case 2**: Two days with 1 piece on the first day and 8 pieces on the second day.
- **Test Case 3**: Multiple days with varying pieces, ensuring that some days complete layers and some do not.
- **Test Case 4**: Edge case where the pieces added do not complete a layer on the last day.

### Challenging Test Input:
```plaintext
4
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 1
```

### Explanation:
1. **Test Case 1**:
   - **Input**: 1
   - **Output**: 1
   - **Explanation**: Alyona is happy on the first day because she completes the first layer.

2. **Test Case 2**:
   - **Input**: 2
   - **Output**: 2
   - **Explanation**: Alyona is happy on both days because the first day completes the first layer, and the second day completes the second layer.

3. **Test Case 3**:
   - **Input**: 5
   - **Output**: 2
   - **Explanation**: Alyona is happy on the first and last days. The first day completes the first layer, and the last day completes the second layer.

4. **Test Case 4**:
   - **Input**: 7
   - **Output**: 2
   - **Explanation**: Alyona is happy on the first and fourth days. The first day completes the first layer, and the fourth day completes the second layer. The last day does not complete a layer.

This test input should help expose the bug in the provided code by testing various scenarios where layer completion is critical.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
/* --- 2047A ---*/
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define endl "\n"
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define mod (ll)(pow(10,9)+7)
#define minus cout << -1 << endl;
#define nxt cout << endl
int main()
{
  	ios::sync_with_stdio(false);
  	cin.tie(0);
	int tc=1;
	cin >> tc;
	for(int i=-1;++i<tc;nxt)
	{
		ll n;
		cin >> n;
		vector<ll>a(n);
		for(int j=-1;++j<n;cin >> a[j]);
		ll c = 1, sum =0, pos = 0;
		for(int j=0;++j<n;)
		{
			sum+=a[j];
			if(sum%8==0 && sum/8>pos)
			{
				c++;
				pos = sum/8;
			}
		}
		cout << c << endl;
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the logic and identify potential edge cases or scenarios where the code might fail.

### Analysis of the Buggy Code
The buggy code calculates the number of days Alyona is happy by checking if the sum of pieces placed so far is a multiple of 8 and if the number of layers completed is greater than the previous count. However, this logic has a flaw: it does not correctly handle the case where Alyona completes a layer and immediately starts the next one on the same day.

### Edge Cases to Consider
1. **Single Layer Completion:** Alyona completes a single layer on the first day.
2. **Multiple Layers in One Day:** Alyona completes multiple layers on the same day.
3. **Partial Layers:** Alyona starts a layer but does not complete it on the same day.
4. **Sequential Layer Completion:** Alyona completes layers sequentially over multiple days.
5. **Unusual Piece Counts:** Alyona places an unusual number of pieces on a day, such as a large number of pieces on the first day.

### Challenging Test Input
Let's create a test input that includes the above edge cases to expose the bug in the code.

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

### Explanation of the Test Input
1. **First Test Case:**
   - `1 1`
   - Alyona completes the first layer on the first day. She is happy on this day.

2. **Second Test Case:**
   - `3 1 8 7`
   - On the first day, Alyona completes the first layer (1 piece).
   - On the second day, she completes the second layer (8 pieces).
   - On the third day, she completes the third layer (7 pieces).
   - Alyona is happy on the first and second days but not on the third day because she starts a new layer on the same day.

3. **Third Test Case:**
   - `4 1 3 3 1`
   - On the first day, Alyona completes the first layer (1 piece).
   - On the second day, she places 3 pieces, which does not complete a layer.
   - On the third day, she places 3 more pieces, completing the second layer (6 pieces in total).
   - On the fourth day, she places 1 piece, which does not complete a layer.
   - Alyona is happy on the first and third days.

4. **Fourth Test Case:**
   - `6 1 1 1 1 1 4`
   - On the first day, Alyona completes the first layer (1 piece).
   - On the second, third, fourth, and fifth days, she places 1 piece each, which does not complete a layer.
   - On the sixth day, she places 4 pieces, completing the second layer (4 pieces in total).
   - Alyona is happy on the first and sixth days.

5. **Fifth Test Case:**
   - `7 1 2 3 4 5 6 7`
   - On the first day, Alyona completes the first layer (1 piece).
   - On the second day, she places 2 pieces, which does not complete a layer.
   - On the third day, she places 3 pieces, which does not complete a layer.
   - On the fourth day, she places 4 pieces, completing the second layer (10 pieces in total).
   - On the fifth day, she places 5 pieces, which does not complete a layer.
   - On the sixth day, she places 6 pieces, completing the third layer (15 pieces in total).
   - On the seventh day, she places 7 pieces, completing the fourth layer (22 pieces in total).
   - Alyona is happy on the first, fourth, sixth, and seventh days.

This test input should expose the bug in the provided code by challenging its ability to handle multiple layers completed in one day and sequential layer completion.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

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

using namespace std;

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

    long long t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vector<int> a(n);
        for(int i=0;i<n;i++){
            cin >> a[i];
        }
        int r=1;
        int v=0;
        for(int i=1;i<n;i++){
            v+=a[i];
            if(v%8==0){
                r+=1;
            }
        }
        cout<<r<<"\n";

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
1 8 1
4
1 8 1 7
6
1 8 1 7 1 7
```

**Explanation:**
1. **Test Case 1:**
   - Input: `3 1 8 1`
   - Expected Output: `2`
   - Explanation: On the first day, Alyona finishes the first layer. On the second day, she finishes the second layer. On the third day, she places 1 piece, which does not complete any layer. She is happy on the first and second days.

2. **Test Case 2:**
   - Input: `4 1 8 1 7`
   - Expected Output: `2`
   - Explanation: On the first day, Alyona finishes the first layer. On the second day, she finishes the second layer. On the third day, she places 1 piece, which does not complete any layer. On the fourth day, she places 7 pieces, completing the third layer. She is happy on the first and fourth days.

3. **Test Case 3:**
   - Input: `6 1 8 1 7 1 7`
   - Expected Output: `3`
   - Explanation: On the first day, Alyona finishes the first layer. On the second day, she finishes the second layer. On the third day, she places 1 piece, which does not complete any layer. On the fourth day, she places 7 pieces, completing the third layer. On the fifth day, she places 1 piece, which does not complete any layer. On the sixth day, she places 7 pieces, completing the fourth layer. She is happy on the first, fourth, and sixth days.

These test cases are designed to expose the bug in the buggy code by focusing on scenarios where the number of pieces placed on a day is just enough to complete a layer or not, and where multiple layers are completed over multiple days.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    arr=list(map(int,input().split(" ")))
    if arr[0]==1:
        layer=1
        c=1
        k=0
        for i in range(1,n):
            k+=arr[i]
            if k%(8*layer)==0:
                c+=1
                k=0
                layer+=1
    else:
        layer=1
        c=0
        k=-1
        for i in range(n):
            k+=arr[i]
            if k%(8*layer)==0:
                c+=1
                k=0
                layer+=1
    print(c)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alyona and a Square Jigsaw Puzzle
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alyona assembles an unusual square Jigsaw Puzzle. She does so in $$$n$$$ days in the following manner:

- On the first day, she starts by placing the central piece in the center of the table.
- On each day after the first one, she places a certain number of pieces around the central piece in clockwise order, always finishing each square layer completely before starting a new one.

For example, she places the first $$$14$$$ pieces in the following order:

The colors denote the layers. The third layer is still unfinished.

Alyona is happy if at the end of the day the assembled part of the puzzle does not have any started but unfinished layers. Given the number of pieces she assembles on each day, find the number of days Alyona is happy on.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of days.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 100$$$, $$$a_1 = 1$$$), where $$$a_i$$$ is the number of pieces Alyona assembles on the $$$i$$$-th day.

It is guaranteed in each test case that at the end of the $$$n$$$ days, there are no unfinished layers.

Output: For each test case, print a single integer: the number of days when Alyona is happy.

Examples:
input:
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
output:
1
2
2
2
3

Note: In the first test case, in the only day Alyona finishes the only layer.

In the second test case, on the first day, Alyona finishes the first layer, and on the second day, she finishes the second layer.

In the third test case, she finishes the second layer in a few days.

In the fourth test case, she finishes the second layer and immediately starts the next one on the same day, therefore, she is not happy on that day. She is only happy on the first and last days.

In the fifth test case, Alyona is happy on the first, fourth, and last days.

Buggy Code:
def precompute_complete_layers():
    complete_layers = []
    k = 1
    while True:
        pieces = 8 * k - 7
        if pieces > 10000:
            break
        complete_layers.append(pieces)
        k += 1
    return complete_layers

def count_happy_days(n, pieces_per_day):
    complete_layers = precompute_complete_layers()
    cumulative_sum = 0
    happy_days = 0

    for pieces in pieces_per_day:
        cumulative_sum += pieces
        if cumulative_sum in complete_layers:
            happy_days += 1

    return happy_days

def main():
    import sys
    input = sys.stdin.read
    data = input().splitlines()

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

    for _ in range(t):
        n = int(data[index])
        index += 1
        pieces_per_day = list(map(int, data[index].split()))
        index += 1

        happy_days = count_happy_days(n, pieces_per_day)
        results.append(happy_days)

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
5
1
1
2
1 8
5
1 3 2 1 2
7
1 2 1 10 2 7 2
14
1 10 10 100 1 1 10 1 10 2 10 2 10 1
```