**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
class Graph:
    def __init__(self, sz):
        self.sz = sz
        self.adj = [[] for _ in range(sz + 1)]
        self.visited = [False] * (sz + 1)

    def add_edge(self, i, j):
        self.adj[i].append(j)
        self.adj[j].append(i)

    def bridge(self):
        self.visited = [False] * (self.sz + 1)
        self.disc = [0] * (self.sz + 1)
        self.low = [0] * (self.sz + 1)
        parent = -1
        self.time = 0
        self.bridges = []
        for i in range(1, self.sz + 1):
            if not self.visited[i]:
                self.bridge_dfs(i, parent)

    def bridge_dfs(self, u, p):
        self.visited[u] = True
        self.time += 1
        self.disc[u] = self.time
        self.low[u] = self.time
        for v in self.adj[u]:
            if v == p:
                continue
            if self.visited[v]:
                self.low[u] = min(self.low[u], self.disc[v])
            else:
                self.bridge_dfs(v, u)
                self.low[u] = min(self.low[u], self.low[v])
                if self.low[v] > self.disc[u]:
                    self.bridges.append((u, v))
class UnionFind:
    def __init__(self, n, size=None):
        self.parent_or_size = [-1] * n if size is None else [-x for x in size]

    def find(self, a):
        a = self.parent_or_size[a] if self.parent_or_size[a] >= 0 else a
        acopy = a
        while self.parent_or_size[a] >= 0:
            a = self.parent_or_size[a]
        while acopy != a:
            self.parent_or_size[acopy], acopy = a, self.parent_or_size[acopy]
        return a

    def merge(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb: return False
        if self.parent_or_size[pa] > self.parent_or_size[pb]:
            pa, pb = pb, pa
        self.parent_or_size[pa] += self.parent_or_size[pb]
        self.parent_or_size[pb] = pa
        return True

    def getSize(self, a):
        return -self.parent_or_size[self.find(a)]

from bisect import bisect_left, bisect_right
from itertools import combinations, permutations, product
from collections import defaultdict, Counter, deque
from heapq import heappush, heappop
from math import gcd, isqrt, lcm
from random import randint, shuffle
import sys

lol = lambda func=int: list(map(func, sys.stdin.readline().split()))
t = lol()[0]
for _ in range(t):
    n, m = lol()
    g = Graph(n)
    edges = []
    for _ in range(m):
        a, b = lol()
        g.add_edge(a, b)
        edges.append((a, b))
    g.bridge()
    bridge = set(g.bridges)
    if not bridge:
        print(n * (n - 1) // 2)
        continue
    uf = UnionFind(n + 1)
    for i, j in edges:
        if (i, j) in bridge or (j, i) in bridge:
            continue
        uf.merge(i, j)
    adj = defaultdict(list)
    for i, j in bridge:
        adj[uf.find(i)].append(uf.find(j))
        adj[uf.find(j)].append(uf.find(i))
    parent = -1
    terminal = uf.find(1)
    while len(adj[terminal]) != 1:
        for u in adj[terminal]:
            if u != parent:
                parent = terminal
                terminal = u
                break
    x = uf.getSize(terminal)
    y = n - x
    parent = terminal
    terminal = adj[terminal][0]
    mn = x * (x - 1) // 2 + y * (y - 1) // 2
    while len(adj[terminal]) != 1:
        for u in adj[terminal]:
            if u != parent:
                parent = terminal
                terminal = u
                x += uf.getSize(parent)
                y = n - x
                mn = min(mn, x * (x - 1) // 2 + y * (y - 1) // 2)
                break
    print(mn)

**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
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
from collections import defaultdict
# from sys import setrecursionlimit

# setrecursionlimit(100050)

def solve(n, m, edges):

    graph = defaultdict(list)
    for u, v in edges: graph[u].append(v); graph[v].append(u)

    time, llink = 0, [float('inf')]*n

    vis = {0}
    def tarjan(node, par = -1):
        nonlocal time, llink, graph
        time += 1
        llink[node] = time

        ccount = 1

        for ne in graph[node]:
            if ne == par or ne in vis: continue
            else:
                vis.add(ne)
                tarjan(ne, node)

        for ne in graph[node]:
            if ne == par: continue
            llink[node] = min(llink[node], llink[ne])

        return ccount

    tarjan(0, -1)
    bedges = {(u, v) for u, v in edges if llink[u]!=llink[v]}
    if not bedges: return n*(n-1)//2

    childs = defaultdict(int)
    vis = {0}
    def rec(root):
        cnt = 0
        for ne in graph[root]:
            if ne in vis: continue
            vis.add(ne)
            curr = rec(ne)
            if (root, ne) in bedges: childs[(root, ne)] = curr
            elif (ne, root) in bedges: childs[(ne, root)] = curr
            cnt += curr
        return cnt + 1
    rec(0)

    res = float('inf')
    for u, v in bedges:
        if (u, v) not in childs: u, v = v, u
        a = childs[(u, v)]
        b = n - a
        res = min(res, (a*(a-1)>>1)+(b*(b-1)>>1))
    return res

for _ in range(int(input())):
    n, m = map(int, input().split())
    edges = [list(map(lambda x: int(x)-1, input().split())) for _ in range(m)]
    print(solve(n, m, edges))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases where the algorithm might fail. The key points to focus on are:

1. **Graph with a single bridge**: A bridge is an edge whose removal increases the number of connected components in the graph. The algorithm should correctly identify and handle bridges.
2. **Graph with multiple bridges**: The algorithm should be able to handle graphs with multiple bridges and correctly calculate the minimum number of reachable pairs.
3. **Graph with a large number of edges**: The algorithm should handle graphs with a large number of edges efficiently without running out of time or memory.
4. **Graph with a specific structure**: A graph with a specific structure that might cause the algorithm to miscount the number of reachable pairs.

Given these considerations, let's create a test input that includes a graph with a single bridge and a graph with multiple bridges.

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

### Explanation
1. **First Test Case (4 3)**
   - This graph has 4 vertices and 3 edges.
   - The edges form a single bridge between vertices 2 and 3.
   - Removing the edge (2, 3) will split the graph into two components: {1, 2} and {3, 4}.
   - The number of reachable pairs after removing the bridge should be 2 (pairs: (1, 2) and (3, 4)).

2. **Second Test Case (6 7)**
   - This graph has 6 vertices and 7 edges.
   - The edges form multiple bridges, including (3, 4) and (4, 6).
   - Removing the edge (3, 4) will split the graph into two components: {1, 2, 3} and {4, 5, 6}.
   - The number of reachable pairs after removing the bridge should be 6 (pairs: (1, 2), (1, 3), (2, 3), (4, 5), (4, 6), (5, 6)).

This test input is designed to challenge the algorithm's ability to correctly identify and handle bridges, and to calculate the minimum number of reachable pairs accurately.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
import os, sys
from io import BytesIO, IOBase
from array import array
from itertools import accumulate
import bisect
import math
from collections import deque

# from functools import cache
# cache cf需要自己提交 pypy3.9!
from copy import deepcopy

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

from typing import DefaultDict, List, Set, Tuple
from collections import defaultdict

# Tarjan 各个算法模板
class Tarjan:
    INF = int(1e20)

    @staticmethod
    def getSCC(
        n: int, adjMap: DefaultDict[int, Set[int]]
    ) -> Tuple[int, DefaultDict[int, Set[int]], List[int]]:

        def dfs(cur: int) -> None:
            nonlocal dfsId, SCCId
            if visited[cur]:
                return
            visited[cur] = True

            order[cur] = low[cur] = dfsId
            dfsId += 1
            stack.append(cur)
            inStack[cur] = True

            for next in adjMap[cur]:
                if not visited[next]:
                    dfs(next)
                    low[cur] = min(low[cur], low[next])
                elif inStack[next]:
                    low[cur] = min(low[cur], order[next])  # 注意这里是order

            if order[cur] == low[cur]:
                while stack:
                    top = stack.pop()
                    inStack[top] = False
                    SCCGroupById[SCCId].add(top)
                    SCCIdByNode[top] = SCCId
                    if top == cur:
                        break
                SCCId += 1

        dfsId = 0
        order, low = [Tarjan.INF] * n, [Tarjan.INF] * n

        visited = [False] * n
        stack = []
        inStack = [False] * n

        SCCId = 0
        SCCGroupById = defaultdict(set)
        SCCIdByNode = [-1] * n

        for cur in range(n):
            if not visited[cur]:
                dfs(cur)

        return SCCId, SCCGroupById, SCCIdByNode

    @staticmethod
    def getCuttingPointAndCuttingEdge(
        n: int, adjMap: DefaultDict[int, Set[int]]
    ) -> Tuple[Set[int], Set[Tuple[int, int]]]:

        def dfs(cur: int, parent: int) -> None:
            nonlocal dfsId
            if visited[cur]:
                return
            visited[cur] = True

            order[cur] = low[cur] = dfsId
            dfsId += 1

            dfsChild = 0
            for next in adjMap[cur]:
                if next == parent:
                    continue
                if not visited[next]:
                    dfsChild += 1
                    dfs(next, cur)
                    low[cur] = min(low[cur], low[next])
                    if low[next] > order[cur]:
                        cuttingEdge.add((cur, next))
                    if parent != -1 and low[next] >= order[cur]:
                        cuttingPoint.add(cur)
                    elif parent == -1 and dfsChild > 1:  # 出发点没有祖先啊，所以特判一下
                        cuttingPoint.add(cur)
                else:
                    low[cur] = min(low[cur], order[next])  # 注意这里是order

        dfsId = 0
        order, low = [Tarjan.INF] * n, [Tarjan.INF] * n
        visited = [False] * n

        cuttingPoint = set()
        cuttingEdge = set()

        for i in range(n):
            if not visited[i]:
                dfs(i, -1)

        return cuttingPoint, cuttingEdge

class UnionFind:
    def __init__(self, n: int):
        self.parent = list(range(n))
        self.size = [1] * n
        self.n = n
        # 当前连通分量数目
        self.setCount = n

    def findset(self, x: int) -> int:
        if self.parent[x] == x:
            return x
        self.parent[x] = self.findset(self.parent[x])
        return self.parent[x]

    def unite(self, x: int, y: int) -> bool:
        x, y = self.findset(x), self.findset(y)
        if x == y:
            return False
        if self.size[x] < self.size[y]:
            x, y = y, x
        self.parent[y] = x
        self.size[x] += self.size[y]
        self.setCount -= 1
        return True

    def connected(self, x: int, y: int) -> bool:
        x, y = self.findset(x), self.findset(y)
        return x == y

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

inf = float('inf')
#if "PyPy" in sys.version:
#    import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')

def solve():
    tar = Tarjan()
    _, bridges = tar.getCuttingPointAndCuttingEdge(n, g)
    bridges = list(bridges)
    q = deque([0])
    depth = [-1]*n
    depth[0] = 0
    d = 0
    while q:
        d += 1
        qn = len(q)
        for _ in range(qn):
            x = q.popleft()
            for y in g[x]:
                if depth[y] == -1:
                    depth[y] = d
                    q.append(y)

    bridges.sort(key = lambda x: -depth[x[0]])
    # print(bridges)
    dsu = UnionFind(n)

    ans = n*(n-1)//2

    vis = [0]*n

    @bootstrap
    def dfs(par, nxt):
        for y in g[nxt]:
            if y != par:

                dsu.unite(y, nxt)
                if vis[y] == 0:
                    yield dfs(nxt, y)
        yield None

    for par,nxt in bridges:
        vis[nxt] = 1
        dfs(par,nxt)
        sz = dsu.size[dsu.findset(nxt)]
        szy = n - sz
        # print(sz, szy)
        ans = min(ans, sz*(sz - 1)//2 + szy*(szy - 1)//2)

    print(ans)

T = Int()
while T:
    T -= 1
    n,m = ints()
    g = [[] for _ in range(n)]
    for _ in range(m):
        u,v = ints()
        u -= 1
        v -= 1
        g[u].append(v)
        g[v].append(u)

    solve()

# import sys
# for line in sys.stdin:
#     a,b = map(int, line.split())
#     print(a+b)

# python A.py < i.txt > o.txt

**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 for the given problem, we need to focus on edge cases that can expose bugs in the algorithm. Here are some key scenarios to consider:

1. **Single Edge Graph**: A graph with only one edge. This is a simple case but can expose issues in handling the smallest possible graph.
2. **Complete Graph**: A graph where every vertex is connected to every other vertex. This can expose issues in handling the maximum number of edges.
3. **Tree Structure**: A graph that is a tree (i.e., no cycles). This can expose issues in handling graphs with the minimum number of edges.
4. **Graph with a Bridge**: A graph where removing a specific edge can significantly change the number of reachable pairs.
5. **Graph with Multiple Components After Edge Removal**: A graph where removing an edge can split the graph into multiple components.

Given the constraints, a small-scale test input that is complete and valid should be generated. Here is a test input that covers these scenarios:

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

### Explanation:
1. **Single Edge Graph**:
   - **Input**: `2 1\n1 2`
   - **Output**: `0`
   - **Explanation**: Removing the only edge makes all pairs unreachable.

2. **Complete Graph**:
   - **Input**: `3 3\n1 2\n2 3\n1 3`
   - **Output**: `3`
   - **Explanation**: No matter which edge is removed, all vertices remain reachable.

3. **Tree Structure**:
   - **Input**: `5 4\n1 2\n2 3\n3 4\n4 5`
   - **Output**: `4`
   - **Explanation**: Removing any edge splits the graph into two components, reducing the number of reachable pairs.

4. **Graph with Multiple Components After Edge Removal**:
   - **Input**: `6 7\n1 2\n1 3\n2 3\n3 4\n4 5\n4 6\n5 6`
   - **Output**: `6`
   - **Explanation**: Removing the edge `(3, 4)` splits the graph into two components, reducing the number of reachable pairs.

This test input is designed to cover a range of scenarios that can help identify bugs in the given algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.StringTokenizer;

public class Proby13 {
	public static void main(String[] args) throws IOException {
		//setup
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer tokenizer = new StringTokenizer(in.readLine());
    	//reading integers
    	int t = Integer.parseInt(tokenizer.nextToken());
    	for(int i = 0; i < t; i++) {
    		//Use this every time you are reading a new line
    		tokenizer = new StringTokenizer(in.readLine());
    		int n = Integer.parseInt(tokenizer.nextToken()), m = Integer.parseInt(tokenizer.nextToken());
    		ArrayList<Integer>[] conn = new ArrayList[n + 1];
    		for(int j = 1; j <= n; j++) conn[j] = new ArrayList<Integer>();
    		for(int j = 0; j < m; j++) {
    			tokenizer = new StringTokenizer(in.readLine());
    			int x = Integer.parseInt(tokenizer.nextToken()), y = Integer.parseInt(tokenizer.nextToken());
    			conn[x].add(y);
    			conn[y].add(x);
    		}
    		ArrayList<Integer> bri = new ArrayList<Integer>();
    		findBridges(new HashSet<Integer>(), new int[n + 1], new int[n + 1], conn, 1, 0, bri);
//    		for(int j = 0; j < bri.size()/2; j++) {
//    			System.out.println(bri.get(2*j) + " " + bri.get(2*j + 1));
//    		}
//    		System.out.println();
    		int[] size = new int[n + 1];
    		dfs(1, size, conn, new boolean[n + 1]);
    		long ans = n * (n - 1)/2;
    		for(int j = 0; j < bri.size()/2; j++) {
    			int u = Math.min(size[bri.get(2*j)], size[bri.get(2*j + 1)]);
    			int g = n - u;
    			ans = Math.min(ans, u*(u-1)/2 + g*(g - 1)/2);
    		}

    		System.out.println(ans);

    	}

	}

	private static int dfs(int i, int[] size, ArrayList<Integer>[] conn, boolean[] vis) {
		if(vis[i]) return 0;
		else {
			size[i] = 1;
			vis[i] = true;
			for(int x: conn[i]) {
				size[i] += dfs(x, size, conn, vis);
			}
		}
		return size[i];
	}

	public static int findBridges(HashSet<Integer> vis, int[] disc, int[] low, ArrayList<Integer>[] conn, int c, int p, ArrayList<Integer> bri) {
		if(vis.contains(c)) {
			return low[c];
		}else {
			disc[c] = vis.size() + 1;
			low[c] = disc[c];
			vis.add(c);
			for(int x: conn[c]) {
				if(x != p) {
					int a = findBridges(vis, disc, low, conn, x, c, bri);
					if(a > low[c]) {
						bri.add(x);
						bri.add(c);
					}
					low[c] = Math.min(low[c], a);
				}
			}
			return low[c];
		}
	}

//	//finding single cycle is easy enough. We just need to run dfs until we intersect a non-parent visited node, and then mark all the parents as part of the cycle. O(n)
//	//even once we have all cycles job is still not easy, we don't know which edges are impactful
//	//We know which edges are impactful because they are the edges which connect two different cycles
//	public static void findCycles(int[] cycles, boolean[] vis, ArrayList<Integer>[] conn, int curr, int cyc, int[] parent) {
//		if(vis[curr]) {
//			cycles[curr] = cyc;
//			int y = parent[curr];
//			while(curr != y) {
//				cycles[y] = cyc;
//				y = parent[y];
//			}
//			cyc++;
//		}
//		for(int x: conn[curr]) {
//			if(parent[curr] == x) continue;
//			findCycles(cycles, vis, conn, curr, cyc, parent);
//		}
//	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code is likely related to the identification of bridges (critical edges) in the graph and the calculation of the minimum number of reachable pairs after removing one edge.

### Step-by-Step Reasoning:

1. **Identify Key Points:**
   - The code uses a depth-first search (DFS) to find bridges and then calculates the number of reachable pairs.
   - The bug might be in the bridge detection or in the calculation of the number of reachable pairs after removing a bridge.

2. **Edge Cases:**
   - **Graph with a single bridge:** A graph where removing a single edge disconnects the graph into two components.
   - **Graph with multiple bridges:** A graph where removing any one of multiple bridges can disconnect the graph.
   - **Graph with no bridges:** A graph where removing any edge does not disconnect the graph.

3. **Small-Scale Test Input:**
   - We will create a graph with a single bridge to see if the code correctly identifies the bridge and calculates the number of reachable pairs.

### Test Input:

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

### Explanation:
- **Graph Description:**
  - The graph has 5 vertices and 5 edges.
  - The edges are: (1, 2), (2, 3), (3, 4), (4, 5), (2, 4).
  - The edge (2, 4) is a bridge because removing it will disconnect the graph into two components: {1, 2} and {3, 4, 5}.

- **Expected Output:**
  - After removing the edge (2, 4), the graph will be disconnected.
  - The number of reachable pairs in the first component {1, 2} is 1 (pairs: (1, 2)).
  - The number of reachable pairs in the second component {3, 4, 5} is 3 (pairs: (3, 4), (3, 5), (4, 5)).
  - The total number of reachable pairs is 1 + 3 = 4.

### Final Test Input:

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
import java.util.*;

public class Solution{

    static class DSU{
        int[] parent;
        int[] size;
        int v;
        int e;
        public DSU(int v,int e){
            this.v = v;
            this.e = e;
            parent = new int[v];
            size = new int[v];
            for(int i = 0;i<v;i++){
                parent[i] = i;
                size[i] = 1;
            }
        }

        int find(int x){
            if(x==parent[x]) return x;
            parent[x] = find(parent[x]);
            return parent[x];
        }

        void union(int x, int y) {
            int parentX = find(x);
            int parentY = find(y);
            if (parentX == parentY) {
                return;
            }
            if (size[parentX] >= size[parentY]) {
                parent[parentY] = parentX;
                size[parentX] += size[parentY];
            } else {
                parent[parentX] = parentY;
                size[parentY] += size[parentX];
            }
        }

        int countConnected(ArrayList<int[]> edges,int eliminate){
            for(int i = 0;i<e;i++){
                if(i==eliminate) continue;
                union(edges.get(i)[0],edges.get(i)[1]);
            }
            int[] pcount = new int[v];
            for(int i = 0;i<v;i++){
                pcount[parent[i]]++;
            }
            int count = 0;
            for(int i = 0;i<v;i++){
                count+=(int)(((long)(pcount[i]*(pcount[i]-1)))/2);
            }

            return count;
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i = 0;i<t;i++){
            int v = sc.nextInt();
            int e = sc.nextInt();
            ArrayList<int[]> edges = new ArrayList<>();
            for(int j = 0;j<e;j++){
                edges.add(new int[]{sc.nextInt()-1,sc.nextInt()-1});
            }

            int mini = (int)(((long)(v*(v-1)))/2);
            for(int j = 0;j<e;j++){
                DSU dsu = new DSU(v,e);
                mini = Math.min(mini,dsu.countConnected(edges, j));
            }

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

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The primary issue in the code is the way it calculates the number of connected components and the pairs of reachable vertices after removing an edge. Specifically, the code might not correctly handle the scenario where the graph is a tree or has a specific structure that makes the removal of certain edges critical.

Let's consider a small-scale test input that can expose this issue:

1. **A Tree Structure**: A tree is a connected graph with no cycles. Removing any edge in a tree will disconnect the graph, which can significantly change the number of reachable pairs.
2. **A Graph with a Critical Edge**: A graph where removing a specific edge significantly reduces the number of reachable pairs.

Here is a small-scale test input that includes both scenarios:

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

**Explanation:**

1. **First Test Case (Tree Structure)**:
   - The graph is a simple tree with 4 vertices and 3 edges.
   - Removing any edge will disconnect the graph, reducing the number of reachable pairs.
   - Expected output: 0 (since removing any edge will make the graph disconnected, and no pairs will be reachable).

2. **Second Test Case (Graph with a Critical Edge)**:
   - The graph is a linear chain with 5 vertices and 4 edges.
   - Removing the edge (2, 3) will significantly reduce the number of reachable pairs.
   - Expected output: 6 (since removing the edge (2, 3) will split the graph into two components: {1, 2} and {3, 4, 5}, and the only reachable pairs will be (1, 2), (3, 4), (3, 5), (4, 5), (3, 4), (4, 5)).

This test input is designed to be small and complete, while also challenging the code to handle both a tree structure and a critical edge scenario.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define scanf scanf_s
#define umap unordered_map
#define lowbit(x) (x&(-x))
#define ls (p<<1)
#define rs (p<<1|1)
//#define int long long
//---------------------
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const ll INF = 1e18 + 10;
const int inf = 1e9 + 10;
const int mod = 998244353;

int n, m;
const int N = 1e5 + 10;
vector<int>g[N];
void add(int u, int v) {
    g[u].push_back(v);
}
int dfn[N], low[N], tim;
vector<int>ecc[N]; int tot;
map<pii, int>mp;
pii mk(int u, int v) {
    return { min(u,v),max(u,v) };
}

void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++tim;
    for (int v : g[u]) {
        if (v == fa)continue;
        if (!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if (low[v] > low[u])mp[mk(u, v)] = 1;
        }
        else low[u] = min(low[u], dfn[v]);
    }
}
int vis[N];
void dfs(int u) {
    if (vis[u])return;
    vis[u] = tot;
    ecc[tot].push_back(u);
    for (int v : g[u]) {
        if (vis[v])continue;
        if (mp[mk(u, v)])continue;
        dfs(v);
    }
}
vector<int>G[N];
void rebuild() {
    set<pii>st;
    for (int u = 1; u <= n; u++) {
        for (int v : g[u]) {
            if (vis[u] == vis[v])continue;
            st.insert(mk(vis[u], vis[v]));
        }
    }
    for (auto [u, v] : st) {
        G[u].push_back(v);
        G[v].push_back(u);
    }
}
int siz[N];
ll ans;
void dfs1(int u, int fa) {
    siz[u] += ecc[u].size();
    for (int v : G[u]) {
        if (v == fa)continue;
        dfs1(v, u);
        ll x = siz[v] * 1ll * (siz[v] - 1) / 2;
        ll y = (n - siz[v]) * 1ll * (n - siz[v] - 1) / 2;
        ans = min(ans, x + y);
        siz[u] += siz[v];
    }
}
void clean() {
    for (int i = 1; i <= n; i++) {
        g[i].clear();
        vis[i] = 0;
        dfn[i] = 0;
        low[i] = 0;
    }
    for (int i = 1; i <= tot; i++) {
        ecc[i].clear();
        G[i].clear();
        siz[i] = 0;
    }
    tim = tot = 0;
    mp.clear();
}

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v; cin >> u >> v;
        add(u, v);
        add(v, u);
    }
    for (int i = 1; i <= n; i++) {
        if (!dfn[i])tarjan(i, 0);
    }
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            ++tot;
            dfs(i);
        }
    }
    rebuild();
    ans = n * (n - 1) / 2;
    dfs1(1, 0);
    cout << ans << endl;
    clean();
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
    //init();
    int t = 1; cin >> t;
    for (int i = 1; i <= t; i++) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
#include<bits/stdc++.h>
#define M_M ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
#define all(a) a.begin(),a.end()
#define rall(v) (v).begin(),(v).end(),greater<>()
#define en "\n"
#define __ ll _; cin>>_; while(_--)
using namespace std;
unordered_map<ll,vector<ll>> g;
vector<pair<ll, ll>> vv;
ll bfs(ll s, pair<ll,ll> e) {
    queue<ll> q;
    unordered_set<ll> f;
    q.push(s);
    f.insert(s);
    ll size = 0;
    while (!q.empty()) {
        ll n = q.front();
        q.pop();
        size++;
        for (ll i:g[n]) {
            if ((n==e.first&&i==e.second)||(n==e.second&&i==e.first))continue;
            if (f.find(i)==f.end()) {
                f.insert(i);
                q.push(i);
            }
        }
    }

    return size;
}
ll minno(ll n) {
    ll cntr=n*(n-1)/2;
    ll ans=cntr;
    for (const auto& i : vv) {
        ll x=bfs(i.first, i);
        ll y=bfs(i.second, i);
        ll k=(x*(x-1))/2+(y*(y-1))/2;
        ans = min(ans,k);
    }

    return ans;
}

void solve()
{
ll n, m;
cin>>n>>m;
g.clear();
vv.clear();
for (int i=0;i<m;i++) {
ll u,v;
cin>>u>>v;
vv.emplace_back(u, v);
g[u].push_back(v),g[v].push_back(u);
}
cout << minno(n) << endl;
}
int main() {
M_M
__
{
solve();
}
return 0;
}

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

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

Think step by step.

Response:
```plaintext
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
from  collections import  defaultdict,Counter
from math import inf
from itertools import  accumulate
import sys
sys.setrecursionlimit(10**5+1)

class Union:
    def __init__(self,n):
        self.root=[i for i in range(n)]
    def find(self,x):
        if x==self.root[x]:
            return x
        self.root[x]=self.find(self.root[x])
        return self.root[x]
    def union(self,x,y):
        rx,ry=self.find(x),self.find(y)
        self.root[rx]=ry

class TARJAN:
    def __init__(self, n: int, g):
        # 记录结点的当前时间戳和最早时间戳,从根开始的一条路径上dfn 严格递增，low 严格非降
        self.dfn = [-1] * n
        self.low = [-1] * n
        self.bridge = []  # 桥
        self.flag = [False] * n  # 是否是割点
        self.g = g

    def tarjan(self, o: int, f: int, t: int) -> None:
        self.dfn[o] = t
        self.low[o] = t
        c = 0
        for child in self.g[o]:
            if child != f:
                if self.dfn[child] == -1:
                    c += 1
                    self.tarjan(child, o, t + 1)
                    self.low[o] = min(self.low[o], self.low[child])
                    # 找到割点, 非root,有儿子
                    if self.dfn[o] <= self.low[child] and f != -1 and not self.flag[o]:
                        self.flag[o] = True
                    # 找到桥
                    if self.dfn[o] < self.low[child]:
                        self.bridge.append([o, child])
                else:
                    self.low[o] = min(self.low[o], self.dfn[child])
        # root点 儿子数大于等于2
        if f == -1 and c >= 2 and not self.flag[o]:
            self.flag[o] = True

def criticalConnections(n: int, connections):
    g=[[] for _ in range(n)]
    for x,y in connections:
        g[x].append(y)
        g[y].append(x)
    res=[]
    dfn=[-1]*n
    low=[-1]*n
    def tarjan(x,pa,depth):
        dfn[x]=depth
        low[x]=depth
        for y in g[x]:
            if y==pa:
                continue
            if dfn[y]==-1:
                tarjan(y,x,depth+1)
                # 如果当前节点仍然比子节点小，说明找到桥
                if dfn[x]<low[y]:
                    res.append([x,y])
            # 更新当前节点的最小时间戳
            low[x]=min(low[x],low[y])
    tarjan(0,-1,0)
    return res
def solve():
    n, m = map(int,input().split())

    edges=[]
    for _ in range(m):
        x,y=map(int,input().split())
        x-=1
        y-=1
        edges.append([x,y])

    bridge=criticalConnections(n,edges)

    if not bridge:
        print(n*(n-1)//2)
        return
    d=set()
    for x,y in bridge:
        d.add((x,y))

    uf=Union(n)
    for x, y in edges:
        if (x, y) in d or (y, x) in d:
            continue
        uf.union(x,y)
    a=[uf.find(i) for i in range(n)]
    size=Counter(a)

    g=[[] for _ in range(n)]
    res=n*(n-1)//2
    for x,y in d:
        g[uf.find(x)].append(uf.find(y))
        g[uf.find(y)].append(uf.find(x))

    def dfs(x,pa):
        nonlocal res
        c=size[x]
        for y in g[x]:
            if y!=pa:
                c+=dfs(y,x)
        res=min(res,c*(c-1)//2+(n-c)*(n-c-1)//2)
        return c
    dfs(uf.find(0),-1)
    print(res)

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
3
3 3
1 2
2 3
1 3
4 4
1 2
2 3
3 4
4 1
5 5
1 2
2 3
3 4
4 5
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
import sys
import collections
input = sys.stdin.read

def find_bridges(n, edges):
    # Prepare graph adjacency list
    graph = [[] for _ in range(n + 1)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # Helper variables for Tarjan's Bridge Finding
    ids = [-1] * (n + 1)
    low = [0] * (n + 1)
    visited = [False] * (n + 1)
    id_counter = 0
    bridges = []

    def dfs(at, parent):
        nonlocal id_counter
        visited[at] = True
        ids[at] = low[at] = id_counter
        id_counter += 1

        for to in graph[at]:
            if to == parent:
                continue
            if not visited[to]:
                dfs(to, at)
                low[at] = min(low[at], low[to])
                if ids[at] < low[to]:
                    bridges.append((at, to))
            else:
                low[at] = min(low[at], ids[to])

    # Find all bridges
    for i in range(1, n + 1):
        if not visited[i]:
            dfs(i, -1)

    return bridges

def calculate_component_size(graph, start, excluded_edge, n):
    visited = [False] * (n + 1)
    stack = [start]
    visited[start] = True
    size = 0

    while stack:
        node = stack.pop()
        size += 1
        for neighbor in graph[node]:
            if not visited[neighbor] and (node, neighbor) != excluded_edge and (neighbor, node) != excluded_edge:
                visited[neighbor] = True
                stack.append(neighbor)

    return size

def process_test_case(n, edges):
    # Find bridges in the graph
    bridges = find_bridges(n, edges)
    if not bridges:
        # No bridges, meaning we cannot split the graph into two components
        return n * (n - 1) // 2

    # Prepare graph adjacency list for component size calculation
    graph = [[] for _ in range(n + 1)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # Try removing each bridge and calculate the resulting component sizes
    min_pairs = float('inf')
    for u, v in bridges:
        size_u = calculate_component_size(graph, u, (u, v), n)
        size_v = n - size_u
        pairs = (size_u * (size_u - 1)) // 2 + (size_v * (size_v - 1)) // 2
        min_pairs = min(min_pairs, pairs)

    return min_pairs

def main():
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    results = []

    for _ in range(t):
        n = int(data[idx])
        m = int(data[idx + 1])
        idx += 2
        edges = []
        for __ in range(m):
            u = int(data[idx])
            v = int(data[idx + 1])
            edges.append((u, v))
            idx += 2
        results.append(str(process_test_case(n, edges)))

    print("\n".join(results))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the algorithm to fail. Specifically, we need to consider scenarios where the graph structure can lead to incorrect identification of bridges or incorrect calculation of component sizes.

### Key Considerations:
1. **Graph with a Bridge that is Not a Simple Edge:**
   - The code uses Tarjan's algorithm to find bridges. A common mistake is not correctly identifying bridges in more complex graph structures, such as those with multiple paths between vertices.

2. **Graph with a High Density of Edges:**
   - The code should handle graphs with a high number of edges efficiently. A dense graph can cause the algorithm to fail if it does not correctly manage the component sizes.

3. **Graph with Multiple Components After Removing a Bridge:**
   - The code should correctly handle the calculation of the number of reachable pairs after removing a bridge, especially when the graph splits into multiple components.

### Test Input:
Let's create a test input with a graph that has a bridge and multiple paths between vertices to test the bridge identification and component size calculation.

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

### Explanation:
- **Graph Structure:**
  - The graph is connected and has 5 vertices and 6 edges.
  - The edges are: (1, 2), (2, 3), (3, 4), (4, 5), (1, 3), (3, 5).
  - The edge (3, 4) is a bridge because removing it will split the graph into two components: {1, 2, 3} and {4, 5}.

- **Expected Output:**
  - After removing the edge (3, 4), the graph will be split into two components:
    - Component 1: {1, 2, 3} with 3 vertices.
    - Component 2: {4, 5} with 2 vertices.
  - The number of reachable pairs in each component:
    - Component 1: 3 * (3 - 1) / 2 = 3 pairs.
    - Component 2: 2 * (2 - 1) / 2 = 1 pair.
  - Total pairs: 3 + 1 = 4.

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

This test input should expose the bug in the given code if it fails to correctly identify the bridge (3, 4) or if it incorrectly calculates the component sizes after removing the bridge.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

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

public class MinimizeReachablePairs {

    static class Graph {
        int n, m;
        List<Integer>[] adj;
        List<int[]> edges;
        int[] discovery, low, size, parent;
        int time;
        List<int[]> bridges;

        public Graph(int n, int m) {
            this.n = n;
            this.m = m;
            adj = new ArrayList[n + 1];
            edges = new ArrayList<>();
            for (int i = 0; i <= n; i++) {
                adj[i] = new ArrayList<>();
            }
            discovery = new int[n + 1];
            low = new int[n + 1];
            size = new int[n + 1];
            parent = new int[n + 1];
            Arrays.fill(discovery, -1);
            Arrays.fill(low, -1);
            Arrays.fill(parent, -1);
            time = 0;
            bridges = new ArrayList<>();
        }

        public void addEdge(int u, int v) {
            adj[u].add(v);
            adj[v].add(u);
            edges.add(new int[]{u, v});
        }

        public void findBridges(int u) {
            discovery[u] = low[u] = time++;
            for (int v : adj[u]) {
                if (discovery[v] == -1) {
                    parent[v] = u;
                    findBridges(v);
                    low[u] = Math.min(low[u], low[v]);
                    if (low[v] > discovery[u]) {
                        bridges.add(new int[]{u, v});
                    }
                } else if (v != parent[u]) {
                    low[u] = Math.min(low[u], discovery[v]);
                }
            }
        }

        public int calcSize(int u) {
            size[u] = 1;
            for (int v : adj[u]) {
                if (size[v] == 0 && v != parent[u]) {
                    size[u] += calcSize(v);
                }
            }
            return size[u];
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int t = Integer.parseInt(br.readLine().trim());
        StringTokenizer st;
        List<Integer> results = new ArrayList<>();

        for (int i = 0; i < t; i++) {
            st = new StringTokenizer(br.readLine().trim());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            Graph graph = new Graph(n, m);

            for (int j = 0; j < m; j++) {
                st = new StringTokenizer(br.readLine().trim());
                int u = Integer.parseInt(st.nextToken());
                int v = Integer.parseInt(st.nextToken());
                graph.addEdge(u, v);
            }

            graph.findBridges(1);
            graph.calcSize(1);

            if (graph.bridges.isEmpty()) {
                results.add(n * (n - 1) / 2);
            } else {
                int minPairs = Integer.MAX_VALUE;
                for (int[] bridge : graph.bridges) {
                    int u = bridge[0];
                    int v = bridge[1];
                    int subtreeSize = graph.parent[v] == u ? graph.size[v] : graph.size[u];
                    int x = Math.min(subtreeSize, n - subtreeSize);
                    int y = n - x;
                    int pairs = x * (x - 1) / 2 + y * (y - 1) / 2;
                    minPairs = Math.min(minPairs, pairs);
                }
                results.add(minPairs);
            }
        }

        for (int result : results) {
            out.println(result);
        }

        out.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases that might cause the algorithm to fail. One such case is a graph that has a cycle and a bridge, where the bridge is crucial for maintaining the connectivity of the graph. If the algorithm incorrectly identifies the bridge or miscalculates the sizes of the connected components, it might produce incorrect results.

Let's construct a small-scale test input that includes a cycle and a bridge:

1. **Graph with a Cycle and a Bridge:**
   - Vertices: 5
   - Edges: 5
   - The graph will form a cycle with an additional edge (bridge) connecting two vertices in the cycle.

2. **Input Data:**
   - Number of test cases: 1
   - Number of vertices (n): 5
   - Number of edges (m): 5
   - Edges: (1, 2), (2, 3), (3, 4), (4, 1), (1, 5)

This input should force the algorithm to correctly identify the bridge and calculate the minimum number of reachable pairs after removing the bridge.

Here is the test input:

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

Explanation:
- The graph forms a cycle with vertices 1, 2, 3, and 4.
- The edge (1, 5) is a bridge.
- Removing the edge (1, 5) should disconnect vertex 5 from the cycle, reducing the number of reachable pairs.

Expected output:
- The minimum number of reachable pairs after removing the edge (1, 5) is 4 (pairs: (1, 2), (1, 4), (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: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

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

public class Main {

	static int N, M;
	static ArrayList<Integer>[] graph;
	static long best;
	static boolean[] visited;
    static int[] tin, low;
    static int timer;
	static int[] size;

	public static void main(String[] args) throws IOException {
		FastIO io = new FastIO();
		int T = io.nextInt();

		for (int t = 0; t < T; t++) {
			N = io.nextInt();
			M = io.nextInt();

			graph = new ArrayList[N + 1];
			for (int i = 0; i <= N; i++) {
				graph[i] = new ArrayList<>();
			}

			for (int i = 0; i < M; i++) {
				int u = io.nextInt();
				int v = io.nextInt();
				graph[u].add(v);
				graph[v].add(u);
			}

			visited = new boolean[N + 1];
			size = new int[N + 1];
			sizeCalc(1);

			visited = new boolean[N + 1];
			tin = new int[N + 1];
			low = new int[N + 1];

			Arrays.fill(tin, -1);
			Arrays.fill(low, -1);

			best = 0;
			dfs(1, -1);

			long ans = N * (N - 1) / 2;
			ans -= best * (N - best);
			io.println(ans);
		}

		io.close();
	}

	static void sizeCalc(int node) {
		visited[node] = true;
		for (int child : graph[node]) {
			if (visited[child]) continue;
			sizeCalc(child);
			size[node] += size[child];
		}
		size[node]++;
	}

    static void dfs(int v, int p) {
        visited[v] = true;
        timer++;
        tin[v] = timer;
		low[v] = timer;
        boolean parentSkipped = false;
        for (int to : graph[v]) {
            if (to == p && !parentSkipped) {
                parentSkipped = true;
                continue;
            }
            if (visited[to]) {
                low[v] = Math.min(low[v], tin[to]);
            } else {
                dfs(to, v);
                low[v] = Math.min(low[v], low[to]);
                if (low[to] > tin[v]) {
					int minSub = Math.min(size[to], N - size[to]);
					best = Math.max(best, minSub);
                }
            }
        }
    }
}

class FastIO extends PrintWriter {

	private InputStream stream;

	private byte[] buf = new byte[1 << 16];

	private int curChar;

	private int numChars;

	// standard input

	public FastIO() { this(System.in, System.out); }

	public FastIO(InputStream i, OutputStream o) {

		super(o);

		stream = i;

	}

	// file input

	public FastIO(String i, String o) throws IOException {

		super(new FileWriter(o));

		stream = new FileInputStream(i);

	}

	// throws InputMismatchException() if previously detected end of file

	private int nextByte() {

		if (numChars == -1) { throw new InputMismatchException(); }

		if (curChar >= numChars) {

			curChar = 0;

			try {

				numChars = stream.read(buf);

			} catch (IOException e) { throw new InputMismatchException(); }

			if (numChars == -1) {

				return -1;  // end of file

			}

		}

		return buf[curChar++];

	}

	// to read in entire lines, replace c <= ' '

	// with a function that checks whether c is a line break

	public String next() {

		int c;

		do { c = nextByte(); } while (c <= ' ');

		StringBuilder res = new StringBuilder();

		do {

			res.appendCodePoint(c);

			c = nextByte();

		} while (c > ' ');

		return res.toString();

	}

	public int nextInt() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		int res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

	public long nextLong() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		long res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

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

}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or specific scenarios where the algorithm might fail. The code is designed to find the minimum number of pairs of vertices that remain reachable after removing exactly one edge. The main issue often arises when the graph has a specific structure that the algorithm might not handle correctly, such as a graph with a bridge or a graph with multiple components that become disconnected after removing an edge.

### Step-by-Step Analysis

1. **Bridge in the Graph:**
   - A bridge is an edge whose removal increases the number of connected components in the graph.
   - If the graph has a bridge, removing it should significantly reduce the number of reachable pairs.

2. **Graph with a Single Bridge:**
   - Consider a graph where there is a single bridge connecting two large components.
   - Removing the bridge should disconnect the two components, and the number of reachable pairs should be minimized.

3. **Graph with Multiple Bridges:**
   - Consider a graph with multiple bridges.
   - Removing any of the bridges should also disconnect the graph, but the impact on the number of reachable pairs might vary depending on the size of the components.

4. **Graph with a Cycle:**
   - Consider a graph with a cycle.
   - Removing any edge from the cycle should not significantly affect the number of reachable pairs, as the cycle ensures that all vertices remain connected.

### Challenging Test Input

Let's create a small-scale test input that includes a graph with a single bridge. This should expose the bug if the algorithm does not handle bridges correctly.

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

### Explanation

- **Graph Structure:**
  - Vertices: 1, 2, 3, 4, 5
  - Edges: (1, 2), (2, 3), (3, 4), (4, 5), (1, 3), (3, 5)
  - The edge (3, 4) is a bridge. Removing this edge will disconnect vertex 4 and vertex 5 from the rest of the graph.

- **Expected Output:**
  - After removing the edge (3, 4), the graph will be split into two components: {1, 2, 3} and {4, 5}.
  - The number of reachable pairs in the first component is 3 (1, 2, 3).
  - The number of reachable pairs in the second component is 1 (4, 5).
  - The total number of reachable pairs is 3 + 1 = 4.

- **Correct Output:**
  - The correct output should be 4, as removing the edge (3, 4) minimizes the number of reachable pairs.

This test input should expose the bug if the algorithm does not correctly identify and handle the bridge in the graph.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
// this code works, i have done it myself, but exceeds the time limit

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

int timer =1;
void dfs_bridge(int node, int parent, vector<int> & vis, vector<int> adj[], int tin[], int low[], int & ans,vector<int> & sub, int &T, int &n){
    vis[node]=1;
    tin[node] = low[node] = timer;
    timer++;
    for(auto it: adj[node]){
        if (it==parent) continue;
        if(vis[it]==0){
            dfs_bridge(it, node, vis, adj, tin, low, ans,sub,T,n);
            sub[node] +=sub[it];
            low[node] = min(low[node],low[it]);
            // parent can borrow from child its lowest step, not vice versa
            if(low[it] > tin[node]) {
                ans = min(ans , T- (sub[it] * (n-sub[it]) ) );
            }
        }
        else{
            low[node] = min(low[node],low[it]);
        }
    }
}

int n,m;
int t;
int u,v;
int main(){
    cin >> t;
    while(t--){
        cin >> n >> m;
        vector<int> adj[n];
        for (int i=0;i<m;i++) {
            cin >> u >> v;
            u--;v --;
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<int> vis(n, 0);
        int tin[n];
        int low[n];
        int ans = INT_MAX;
        int T = n*(n-1)/2;
        vector<int> sub(n,1);
        dfs_bridge(0, -1, vis, adj, tin, low,ans,sub,T,n);
        if(ans ==INT_MAX){
            cout << T << endl;
            continue;
        }
        cout << ans << 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 edge cases and scenarios that maximize the failure probability. The code is designed to find the smallest number of pairs of reachable vertices after removing exactly one edge. The main issue is that it might exceed the time limit for large inputs, so we need to create a small-scale test input that can still expose this inefficiency.

### Step-by-Step Analysis:
1. **Graph Structure**: The graph should be connected and have a structure that maximizes the number of edges and vertices.
2. **Edge Removal Impact**: The test case should be such that removing a specific edge significantly changes the number of reachable pairs.
3. **Small Scale**: The test input should be small enough to be manageable but large enough to expose inefficiencies.

### Chosen Test Case:
- **Number of Test Cases**: 1
- **Number of Vertices (n)**: 10
- **Number of Edges (m)**: 14
- **Graph Structure**: A graph that forms a cycle with additional edges to ensure connectivity and to test the impact of edge removal.

### Test Input:
```plaintext
1
10 14
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
1 10
2 4
4 6
6 8
8 10
```

### Explanation:
- **Vertices and Edges**: The graph has 10 vertices and 14 edges.
- **Cycle Structure**: The graph forms a cycle from 1 to 10, with additional edges (2-4, 4-6, 6-8, 8-10) to ensure connectivity and to test the impact of removing edges.
- **Impact of Edge Removal**: Removing any of the additional edges (2-4, 4-6, 6-8, 8-10) should significantly reduce the number of reachable pairs, as it breaks the cycle and isolates parts of the graph.

This test input is small enough to be manageable but large enough to expose potential inefficiencies in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

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

using namespace std;

const int MAXN = 200005;
vector<int> adj[MAXN];
int disc[MAXN], low[MAXN], parent[MAXN];
bool visited[MAXN];
vector<pair<int, int>> bridges;
int time_counter;

void dfs(int u) {
    visited[u] = true;
    disc[u] = low[u] = ++time_counter;

    for (int v : adj[u]) {
        if (!visited[v]) {
            parent[v] = u;
            dfs(v);

            low[u] = min(low[u], low[v]);

            if (low[v] > disc[u]) {
                bridges.push_back({u, v});
            }
        } else if (v != parent[u]) {
            low[u] = min(low[u], disc[v]);
        }
    }
}

int countNodes(int u, vector<bool>& visited) {
    visited[u] = true;
    int count = 1;

    for (int v : adj[u]) {
        if (!visited[v]) {
            count += countNodes(v, visited);
        }
    }

    return count;
}

void solve() {
    int n, m;
    cin >> n >> m;

    for (int i = 1; i <= n; ++i) {
        adj[i].clear();
    }

    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    fill(disc, disc + n + 1, -1);
    fill(low, low + n + 1, -1);
    fill(parent, parent + n + 1, -1);
    fill(visited, visited + n + 1, false);
    time_counter = 0;
    bridges.clear();

    for (int i = 1; i <= n; ++i) {
        if (!visited[i]) {
            dfs(i);
        }
    }

    int max_disconnected_pairs = 0;
    pair<int, int> best_bridge = {-1, -1};

    for (auto bridge : bridges) {
        int u = bridge.first;
        int v = bridge.second;

        vector<bool> local_visited(n + 1, false);
        adj[u].erase(remove(adj[u].begin(), adj[u].end(), v), adj[u].end());
        adj[v].erase(remove(adj[v].begin(), adj[v].end(), u), adj[v].end());

        int size1 = countNodes(u, local_visited);
        int size2 = n - size1;

        int disconnected_pairs = size1 * size2;
        if (disconnected_pairs > max_disconnected_pairs) {
            max_disconnected_pairs = disconnected_pairs;
            best_bridge = bridge;
        }

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    int initial_pairs = n * (n - 1) / 2;
    int result = initial_pairs - max_disconnected_pairs;

    // cout << "Best bridge to remove: " << best_bridge.first << " " << best_bridge.second << endl;
    // cout << "Initial number of pairs: " << initial_pairs << endl;
    // cout << "Max disconnected pairs: " << max_disconnected_pairs << endl;
    cout <<  result << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might fail. The code uses a depth-first search (DFS) to find bridges in the graph and then calculates the number of disconnected pairs when each bridge is removed. The bug might be exposed in the following scenarios:

1. **Graphs with multiple bridges**: The code might not handle the case where there are multiple bridges correctly.
2. **Graphs with a single bridge**: The code might not handle the removal of a single bridge correctly.
3. **Complete graphs**: The code might not handle the case where the graph is a complete graph (every vertex is connected to every other vertex).
4. **Linear graphs**: The code might not handle the case where the graph is a linear chain (a simple path).

Let's generate a small-scale test input that focuses on a graph with multiple bridges, as this is a complex scenario that can expose the bug.

### Test Input
We will create a graph with 5 vertices and 6 edges, forming a structure with multiple bridges. The graph will look like this:

```
1 - 2 - 3
|     |
4 - 5 - 6
```

This graph has the following edges:
- (1, 2)
- (2, 3)
- (2, 5)
- (1, 4)
- (4, 5)
- (5, 6)

The edges (1, 2), (2, 3), and (5, 6) are bridges in this graph.

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

This test input will help expose the bug in the code by requiring it to correctly identify and handle multiple bridges. The expected output for this test input is the number of pairs of reachable vertices after removing one of the bridges, which should be minimized.