**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define ls (u << 1)
#define rs (u << 1 | 1)

#define debug(args...) { \
     string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); \
     stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); cout << '\n';}
void err(istream_iterator<string> it){}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cout << *it << " = " << a << ' ';
    err(++it, args...);
}

typedef long long LL;
typedef pair<int, int> PII;
struct FWT {
    vector<LL> tr;
    int n;
    FWT (int _n) {
        n = _n;
        tr.assign(n + 5, 0);
    }
    int lowbit(int x) {
        return x & -x;
    }
    void add(int id, int x) {
        for (int i = id; i <= n; i += lowbit(i)) {
            tr[i] += x;
        }
    }
    LL query(int id) {
        LL res = 0;
        for (int i = id; i > 0; i -= lowbit(i)) {
            res += tr[i];
        }
        return res;
    }
    LL rangeSum(int l, int r) {
        return query(r) - query(l - 1);
    }
    LL rank(int p) {
        return query(p - 1) + 1;
    }
    LL val(int rk) {
        int t = 0;
        for (int i = 19; i >= 0; i--) {
            t += 1 << i;
            if (t > n || tr[t] >= rk) t -= 1 << i;
            else rk -= tr[t];
        }
        return t + 1;
    }
    LL pre(int p) {
        return val(rank(p) - 1);
    }
    LL nxt(int p) {
        return val(rank(p + 1));
    }
};
void solve() {
    int n, m;
    cin >> n >> m;
    vector<PII> p(m + 5);
    vector id(n + 5, vector<int>());
    auto fid = id;
    int mxl = 0, mil = n, mxr = 0, mir = n;
    for (int i = 1; i <= m; i++) {
        cin >> p[i].fi >> p[i].se;
        mxl = max(mxl, p[i].fi);
        mxr = max(mxr, p[i].se);
        mil = min(mil, p[i].fi);
        mir = min(mir, p[i].se);
        id[p[i].fi].pb(p[i].se);
        fid[p[i].se].pb(p[i].fi);
    }
    int res = min({n, mxl - 1 + n - mir + 1, mxr - mil});
    vector<int> f(n + 5), fr(n + 5);
    set<int> zero;
    FWT fwt(n);
    auto fdel = [&](int l, int r) {
        auto it = zero.upper_bound(l);
        vector<int> del;
        while (it != zero.end() && (*it) <= r) {
            fwt.add(*it, 1);
            del.pb(*it);
            it++;
        }
        for (auto x : del) zero.erase(x);
    };
    for (int i = 2; i <= n; i++) zero.insert(i);
    multiset<int> nr;
    for (int i = 1; i <= m; i++) nr.insert(p[i].se);
    for (int i = n - 1; i >= 1; i--) {
        // cerr << i << "IN\n";
        for (auto r : id[i]) {
            fdel(i, r);
            nr.erase(nr.find(r));
        }
        auto it = nr.lower_bound(i + 1);
        if (it == nr.end()) fr[i] = 0;
        else fr[i] = n - (*it) - (fwt.query(n) - fwt.query(*it));
        f[i] = n - 1 - zero.size();
        // debug(i, f[i], *it, fr[i]);
    }
    multiset<int> nl; nl.insert(1);
    zero.clear();
    fwt = FWT(n);
    for (int i = 2; i <= n; i++) {
        zero.insert(i);
        for (auto r : id[i - 1]) {
            nl.insert(i - 1);
        }
        for (auto l : fid[i]) {
            fdel(l, i);
            nl.erase(nl.find(l));
        }
        int l = *nl.rbegin();

        int x = l - 1 - fwt.query(l) + fr[i] + 1;
        int y = i - 1 - (int)zero.size();
        // debug(i, l, x, y, f[i]);
        res = min(res, x + y + f[i]);
    }
    cout << res << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt = 1; cin >> tt;
    while (tt--) 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
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
// Problem: G. Penacony
// Contest: Codeforces - Codeforces Round 962 (Div. 3)
// URL: https://codeforces.com/contest/1996/problem/G
// Memory Limit: 512 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){
	return b > 0 ? gcd(b , a % b) : a;
}

LL lcm(LL a , LL b){
	return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){
	for(int i = 0 ; i <= n ; i ++){
		a[i] = 0;
	}
}
template<class Info, class Tag>
struct LazySegmentTree {
    const int n;
    std::vector<Info> info;
    std::vector<Tag> tag;
    LazySegmentTree(int n) : n(n), info(4 << std::__lg(n)), tag(4 << std::__lg(n)) {}
    LazySegmentTree(std::vector<Info> init) : LazySegmentTree(init.size()) {
        std::function<void(int, int, int)> build = [&](int p, int l, int r) {
            if (r - l == 1) {
                info[p] = init[l];
                return;
            }
            int m = (l + r) / 2;
            build(2 * p, l, m);
            build(2 * p + 1, m, r);
            pull(p);
        };
        build(1, 0, n);
    }
    void pull(int p) {
        info[p] = info[2 * p] + info[2 * p + 1];
    }
    void apply(int p, const Tag &v) {
        info[p].apply(v);
        tag[p].apply(v);
    }
    void push(int p) {
        apply(2 * p, tag[p]);
        apply(2 * p + 1, tag[p]);
        tag[p] = Tag();
    }
    void modify(int p, int l, int r, int x, const Info &v) {
        if (r - l == 1) {
            info[p] = v;
            return;
        }
        int m = (l + r) / 2;
        push(p);
        if (x < m) {
            modify(2 * p, l, m, x, v);
        } else {
            modify(2 * p + 1, m, r, x, v);
        }
        pull(p);
    }
    void modify(int p, const Info &v) {
        modify(1, 0, n, p, v);
    }
    Info rangeQuery(int p, int l, int r, int x, int y) {
        if (l >= y || r <= x) {
            return Info();
        }
        if (l >= x && r <= y) {
            return info[p];
        }
        int m = (l + r) / 2;
        push(p);
        return rangeQuery(2 * p, l, m, x, y) + rangeQuery(2 * p + 1, m, r, x, y);
    }
    Info rangeQuery(int l, int r) {
        return rangeQuery(1, 0, n, l, r);
    }
    void rangeApply(int p, int l, int r, int x, int y, const Tag &v) {
        if (l >= y || r <= x) {
            return;
        }
        if (l >= x && r <= y) {
            apply(p, v);
            return;
        }
        int m = (l + r) / 2;
        push(p);
        rangeApply(2 * p, l, m, x, y, v);
        rangeApply(2 * p + 1, m, r, x, y, v);
        pull(p);
    }
    void rangeApply(int l, int r, const Tag &v) {
        return rangeApply(1, 0, n, l, r, v);
    }
    void half(int p, int l, int r) {
        if (info[p].act == 0) {
            return;
        }
        if ((info[p].min + 1) / 2 == (info[p].max + 1) / 2) {
            apply(p, {-(info[p].min + 1) / 2});
            return;
        }
        int m = (l + r) / 2;
        push(p);
        half(2 * p, l, m);
        half(2 * p + 1, m, r);
        pull(p);
    }
    void half() {
        half(1, 0, n);
    }
};
struct Tag {
    int add = 0;
    void apply(Tag t) {
		add += t.add;
    }
};
struct Info {
    int min_cnt = 0;
    int min = 0;
    int act = 0;
    void apply(Tag t) {
        min += t.add;
    }
};
Info operator + (Info a, Info b) {
    Info c;
    if(b.act == 0){
    	return a;
    }
    if(a.act == 0){
    	return b;
    }
	if(a.min == b.min){
		c.min_cnt = a.min_cnt + b.min_cnt;
	}
	else if(a.min > b.min){
		c.min_cnt = b.min_cnt;
	}
	else{
		c.min_cnt = a.min_cnt;
	}
	c.min = min(a.min , b.min);
    c.act = a.act + b.act;
    return c;
}
void solve()
{
	int n , m;
	cin >> n >> m;
	vector<Info>v;
	for(int i = 0 ; i <= 2 * n ; i ++){
		v.pb({1 , 0 , 1});
	}
	LazySegmentTree <Info , Tag> seg(v);
	set< pair<int,int> >st;
	set< pair<int,int> >ts;
	for(int i = 0 ; i < m ; i ++){
		int l , r;
		cin >> l >> r;
		l-- , r--;
		Tag t = {1};
		seg.rangeApply(l , r , t);
		seg.rangeApply(r , l + n , t);
		ts.insert({l , r});
		st.insert({r , l});
	}
	queue<pair<int,int>>q;
	queue<pair<int,int>>p;
	for(auto it : ts){
		p.push(it);
	}
	for(auto it : st){
		q.push(it);
	}
	int mx = 0;
	for(int i = 0 ; i < 2 * n ; i ++){
		while(!q.empty()){
			pair<int,int>tmp = q.front();
			if(tmp.first <= i - n){
				q.pop();
				Tag t = {1};
				Tag tt = {-1};
				seg.rangeApply(tmp.first , tmp.second + n , tt);
				seg.rangeApply(tmp.second + n , tmp.first + n , t);

			}
			else{
				break;
			}
		}
		while(!p.empty()){
			pair<int,int>tmp = p.front();
			if(tmp.first <= i - n){
				p.pop();
				Tag tt ={-1};
				seg.rangeApply(tmp.first , tmp.second, tt);
			}
			else{
				break;
			}
		}
		if(seg.rangeQuery(max(0 , i - n) , i).min == 0)
			mx = max(mx , seg.rangeQuery(max(0 , i - n) , i).min_cnt);
	}
	cout << n - mx << endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('YES')
def df():
    print('NO')
def cntt(L, R, k):
    ll = -1
    rr = len(d[k])
    while rr - ll > 1:
        mm = (ll + rr) // 2
        if d[k][mm] >= L:
            rr = mm
        else:
            ll = mm
    ll1 = -1
    rr1 = len(d[k])
    while rr1 - ll1 > 1:
        mm = (ll1 + rr1) // 2
        if d[k][mm] <= R:
            ll1 = mm
        else:
            rr1 = mm
    return ll1 - rr
for __ in range(fg()):
    n, m = fgh()
    a = []
    b = [0] * (n + 1)
    for _ in range(m):
        l, r = fgh()
        l -= 1
        r -= 1
        b[l] += 1
        b[r] -= 1
        a.append((l, r))
    a.sort()
    cnt = [0] * n
    for i in range(n):
        if i:
            cnt[i] = cnt[i - 1] + b[i]
        else:
            cnt[i] = b[i]
    d = [[] for i in range(m + 1)]
    for i in range(n):
        d[cnt[i]].append(i)
    ans = n - 1
    for i in range(n - 1):
        if cnt[i] == 0:
            ans -= 1
    minR = 293293293293923
    for i in range(m):
        minR = min(minR, a[i][1])
        L = a[i][0]
        R = minR + 1
        zz = L + n - R
        zz += R - L - 1 - cntt(L, R - 2, i + 1)
        ans = min(ans, zz)
    maxL = -22392932932932939
    bb = []
    for i in range(m):
        bb.append((a[i][1], a[i][0]))
    bb.sort()
    bb = bb[::-1]
    for i in range(m):
        maxL = max(maxL, bb[i][1])
        R = bb[i][0] + 1
        L = maxL
        zz = L + n - R
        zz += R - L - 1 - cntt(L, R - 2, i + 1)
        ans = min(ans, zz)
    for i in range(m):
        L = a[i][0]
        R = a[i][1] + 1
        zz = L + n - R
        zz += R - L - 1 - cntt(L, R - 2, 1)
        ans = min(ans, zz)
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
def solve():
  n, m = map(int, input().split())
  friendships = []
  for _ in range(m):
    friendships.append(list(map(int, input().split())))

  min_roads = float('inf')

  for i in range(1 << (n)):
    roads = []
    num_roads = 0
    for j in range(n):
      if (i >> j) & 1:
        roads.append((j + 1, (j + 2) % n if (j + 2) % n != 0 else n))
        num_roads += 1

    valid = True
    for a, b in friendships:
      connected = False
      visited = set()
      q = [a]
      while q:
        curr = q.pop(0)
        if curr == b:
          connected = True
          break
        visited.add(curr)
        for u, v in roads:
          if u == curr and v not in visited:
            q.append(v)
          if v == curr and u not in visited:
            q.append(u)
      if not connected:
        valid = False
        break

    if valid:
      min_roads = min(min_roads, num_roads)

  print(min_roads)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
def solve():
    n, m = map(int, input().split())
    friendships = []
    for _ in range(m):
        friendships.append(list(map(int, input().split())))

    min_roads = float('inf')

    for i in range(1 << n):
        roads = []
        num_roads = 0
        for j in range(n):
            if (i >> j) & 1:
                roads.append((j + 1, (j + 1) % n + 1))
                num_roads += 1

        valid = True
        for a, b in friendships:
            connected = False
            if a == b:
                connected = True
                continue

            visited = set()
            q = [a]
            while q:
                curr = q.pop(0)
                if curr == b:
                    connected = True
                    break
                visited.add(curr)

                for u, v in roads:
                    if u == curr and v not in visited:
                        q.append(v)
                    if v == curr and u not in visited:
                        q.append(u)

            if not connected:
                valid = False
                break

        if valid:
            min_roads = min(min_roads, num_roads)

    print(min_roads)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

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

using ll  = long long;
using ld  = long double;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using ull = unsigned long long;

#define X               first
#define Y               second
#define SZ(x)           int(x.size())
#define all(x)          x.begin(), x.end()
#define mins(a,b)       (a = min(a,b))
#define maxs(a,b)       (a = max(a,b))
#define pb              push_back
#define Mp              make_pair
#define lc              id<<1
#define rc              lc|1
#define mid             ((l+r)/2)
mt19937_64              rng(chrono::steady_clock::now().time_since_epoch().count());

const ll  INF = 1e9 + 23;
const ll  MOD = 1e9 + 7;
const int MXN = 2e5 + 23;
const int LOG = 23;

int n, m, a[MXN], b[MXN], pos[MXN], ans;

pii seg[MXN<<2];
int lz[MXN<<2];
pii Merge(pii x, pii y) {
    if(x.X < y.X) return x;
    if(x.X > y.X) return y;
    return Mp(x.X, x.Y+y.Y);
}
void Build(int l=0, int r=n, int id=1) {
    seg[id] = Mp(0, r-l);
    if(r-l == 1) return;
    Build(l, mid, lc);
    Build(mid, r, rc);
}
void Apply(int x, int id) {
    seg[id].X += x;
    lz[id] += x;
}
void Shift(int l, int r, int id) {
    if(r-l > 1 && lz[id]) {
        Apply(lz[id], lc);
        Apply(lz[id], rc);
    }
    lz[id] = 0;
}
void Upd(int s, int e, int x, int l=0, int r=n, int id=1) {
    Shift(l, r, id);
    if(s <= l && r <= e) {
        Apply(x, id);
        return;
    }
    if(s < mid) Upd(s,e,x, l, mid, lc);
    if(e > mid) Upd(s,e,x, mid, r, rc);
    seg[id] = Merge(seg[lc], seg[rc]);
}
pii Get(int s, int e, int l=0, int r=n, int id=1) {
    Shift(l, r, id);
    if(s <= l && r <= e) return seg[id];
    if(e <= mid) return Get(s, e, l, mid, lc);
    if(s >= mid) return Get(s, e, mid, r, rc);
    return Merge(Get(s, e, l, mid, lc), Get(s, e, mid, r, rc));
}

vector<int> op[MXN];
int del[MXN];

void solve(int s, int e) {
    vector<int> vec;
    int i=s;
    while(1) {
        vec.pb(i);
        if(i == e) break;
        i++;
        i %= n;
    }
    for(int i=0; i<n; i++) {
        pos[i] = -1;
        del[i] = 0;
        op[i].clear();
    }
    for(int i=0; i<SZ(vec); i++) pos[vec[i]] = i;
    if(SZ(vec) == 2) return;
    Build();
    multiset<int> st;
    // pii d = Get(0, SZ(vec)-1);
    // cout << SZ(vec) << '\n';
    // cout << "-2 " << d.X << ' ' << d.Y << '\n';
    for(int i=2; i<=m; i++) {
        bool A = pos[a[i]] != -1 && pos[a[i]] != 0 && pos[a[i]] != SZ(vec)-1;
        bool B = pos[b[i]] != -1 && pos[b[i]] != 0 && pos[b[i]] != SZ(vec)-1;
        // cout << i << ' ' << A << ' ' << B << ' ' << pos[a[i]] << '\n';
        if(A&B) {
            op[min(pos[a[i]], pos[b[i]])].pb(max(pos[a[i]], pos[b[i]]));
            // cout << min(pos[a[i]], pos[b[i]]) << ' ' << max(pos[a[i]], pos[b[i]]) << '\n';
            Upd(min(pos[a[i]], pos[b[i]]), max(pos[a[i]], pos[b[i]]), 1);
        }
        else if(A) st.insert(pos[a[i]]), del[pos[a[i]]]++;
        else if(B) st.insert(pos[b[i]]), del[pos[b[i]]]++;
    }
    // d = Get(0, SZ(vec)-1);
    // cout << "-1 " << d.X << ' ' << d.Y << '\n';
    for(int i=0; i<SZ(vec)-1; i++) {
        while(del[i]--) st.erase(st.find(i));
        for(int j : op[i]) {
            Upd(i, j, -1);
            st.insert(j);
            del[j]++;
        }
        if(SZ(st)) Upd(*st.begin(), SZ(vec)-1, 1);
        pii d = Get(0, SZ(vec)-1);
        // cout << i << ' ' << d.X << ' ' << d.Y << '\n';
        mins(ans, n-SZ(vec)+1 + (d.X == 0 ? SZ(vec)-1-d.Y : SZ(vec)-1));
        if(SZ(st)) Upd(*st.begin(), SZ(vec)-1, -1);
        Upd(i, i+1, 1);
    }
}

void Main() {
    cin >> n >> m;
    for(int i=1; i<=m; i++) cin >> a[i] >> b[i], a[i]--, b[i]--;
    ans = n-1;
    solve(a[1], b[1]);
    solve(b[1], a[1]);
    cout << ans << '\n';
}

int32_t main() {
    cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while(T--) Main();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to consider:

1. **Edge Cases:**
   - Test with the minimum and maximum values for `n` and `m`.
   - Test with a single friendship.
   - Test with all houses being friends with each other.
   - Test with a circular friendship chain.

2. **Complex Scenarios:**
   - Test with multiple disjoint friendship groups.
   - Test with a mix of long and short friendship chains.

3. **Small-Scale Test Input:**
   - Ensure the test input is small enough to be manageable but complex enough to expose potential bugs.

Given these considerations, let's create a small-scale test input that includes a mix of the above scenarios:

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

This test input includes:
- A small complete graph (4 houses, 4 friendships).
- A larger complete graph (5 houses, 5 friendships).
- A graph with a mix of long and short chains (8 houses, 6 friendships).

This should help in exposing any issues in the buggy code, particularly in how it handles circular friendships and multiple disjoint friendship groups.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

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

public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Reader sc=new Reader();
		int t=sc.nextInt();
		while(t-->0)
		{
		    int n=sc.nextInt();
		    int m=sc.nextInt();
		    Pair arr[]=new Pair[m];
		    ArrayList<Integer> start[]=new ArrayList[n],end[]=new ArrayList[n];
		    for(int i=0;i<n;i++){ start[i]=new ArrayList<>();end[i]=new ArrayList<>();}
		    for(int i=0;i<m;i++){
		        arr[i]=new Pair(sc.nextInt()-1,sc.nextInt()-1);
		        start[arr[i].x].add(i);
		        end[arr[i].y].add(i);
		    }
		    int arr2[]=new int[n];
		    SegTree sg=new SegTree();
		    Node tree[]=new Node[4*n];
		    sg.buildTree(arr2,0,n-1,tree,1);
		    for(int i=0;i<m;i++)
		    sg.rangeupdate(tree,0,n-1,arr[i].x,arr[i].y-1,1,1);
		     // System.out.println(sg.query(tree,0,n-1,0,n-1,1).count);
		    int min=n;
		    for(int i=0;i<n;i++){
		        for(int j:start[i]){
		            sg.rangeupdate(tree,0,n-1,arr[j].x,arr[j].y-1,-1,1);
		            sg.rangeupdate(tree,0,n-1,arr[j].y,n-1,1,1);
		            sg.rangeupdate(tree,0,n-1,0,arr[j].x-1,1,1);
		        }
		         for(int j:end[i]){
		            sg.rangeupdate(tree,0,n-1,arr[j].x,arr[j].y-1,1,1);
		            sg.rangeupdate(tree,0,n-1,arr[j].y,n-1,-1,1);
		            sg.rangeupdate(tree,0,n-1,0,arr[j].x-1,-1,1);
		        }
		           min=Math.min(min,n-sg.query(tree,0,n-1,0,n-1,1).count);
		    }
		    System.out.println(min);

		}
	}
	static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
}
class SegTree
{
    static int lazy[]=new int[1000001];
    public static void merge(Node tree[],int ind){
        tree[ind]=new Node(0,0);
        if(tree[2*ind]==null) tree[2*ind]=new Node(Integer.MAX_VALUE,0);
        if(tree[2*ind+1]==null) tree[2*ind+1]=new Node(Integer.MAX_VALUE,0);
        if(tree[2*ind].min<tree[2*ind+1].min){
            tree[ind].min=tree[2*ind].min;
            tree[ind].count=tree[2*ind].count;
        }
        else if(tree[2*ind+1].min<tree[2*ind].min){
             tree[ind].min=tree[2*ind+1].min;
            tree[ind].count=tree[2*ind+1].count;
        }
        else{
            tree[ind].min=tree[2*ind].min;
            tree[ind].count=tree[2*ind].count+tree[2*ind+1].count;
        }

    }
    public static void buildTree(int a[],int s,int e,Node tree[],int index)
    {
        if(s==e)
        {
            tree[index]=new Node(a[s],1);
            return;
        }
        int mid=(s+e)/2;
        buildTree(a,s,mid,tree,2*index);
        buildTree(a,mid+1,e,tree,2*index+1);
        merge(tree,index);

    }
    static Node query(Node tree[],int ss,int se,int qs,int qe,int index)
    {
        if(ss>=qs && se<=qe)
        return tree[index];
        else if(qe<ss || qs>se)
        return new Node(Integer.MAX_VALUE,1);
        int mid=(ss+se)/2;
        Node low1=query(tree,ss,mid,qs,qe,2*index);
        Node low2=query(tree,mid+1,se,qs,qe,2*index+1);
        if(low1.min<low2.min) return low1;
        if(low2.min<low1.min) return low2;
        return new Node(low1.min,low1.count+low2.count);
    }
     public static void rangeupdate(Node tree[],int ss,int se,int l,int r,int inc,int index)
    {
        if(r<ss || l>se)
        return;
        if(ss==se){
        tree[index].min+=inc;return;}
        int mid=(ss+se)/2;
        rangeupdate(tree,ss,mid,l,r,inc,2*index);
        rangeupdate(tree,mid+1,se,l,r,inc,2*index+1);
        merge(tree,index);

    }
}
class Node
{
    int min,count;
    Node(int min,int count)
    {
        this.min=min;this.count=count;
    }
}
class Pair
{
    int x,y;
    Pair(int x,int y){
        this.x=x;
        this.y=y;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
#include<bits/stdc++.h>
template<typename T, typename U>
bool ckmax(T& a, U b) {
    return a < b ? a = b, true : false;
}
template<typename T, typename U>
bool ckmin(T& a, U b) {
    return b < a ? a = b, true : false;
}
#define F(i, l, r) for (int i = l; i <= r; i++)
#define DF(i, r, l) for (int i = r; i >= l; i--)
#define fir(i, l, r) for (int i = l; i < r; i++)
#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(), x.end()
#define all1(x) x.begin() + 1, x.end()
using i64 = long long;
using u64 = unsigned long long;
using lf = double;
template <typename T>
using vla = std::vector<T>;
using pii = std::pair<int, int>;
using namespace std;
int test_id;

std::mt19937 rnd(std::chrono::system_clock::now().time_since_epoch().count());
void solve() {
    int n, m; cin >> n >> m;
    vla<u64> a(n + 1);
    F (i, 1, m) {
        int l, r; cin >> l >> r;
        u64 t = rnd();
        a[l] ^= t, a[r] ^= t;
    }
    F (i, 1, n) a[i] ^= a[i - 1];
    map<u64, int> mp;
    int ans = 0;
    F (i, 1, n) {
        ckmax(ans, ++mp[a[i]]);
    }
    cout << n - ans << '\n';
}

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int T = 1;
    cin >> T;
    for (test_id = 1; test_id <= T; test_id++) {
        // cerr << "test" << test_id << ":\n";
        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 the logic used to determine the minimum number of roads that must be maintained. The code uses a random number generator to create unique identifiers for each friendship and then uses these identifiers to track the necessary roads. However, this approach can lead to incorrect results in certain scenarios.

### Key Points to Consider:
1. **Randomness and Uniqueness**: The random numbers generated may not be unique, leading to incorrect tracking of the roads.
2. **Edge Cases**: Scenarios where the friendships form a cycle or where the number of friendships is very high compared to the number of houses.

### Test Input:
We can create a test case where the number of houses and friendships is relatively small but the friendships form a cycle or a complex structure that the random number generation might fail to handle correctly.

### Test Case:
```plaintext
1
5 5
1 2
2 3
3 4
4 5
5 1
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Houses (n)**: 5
- **Number of Friendships (m)**: 5
- **Friendships**:
  - (1, 2)
  - (2, 3)
  - (3, 4)
  - (4, 5)
  - (5, 1)

### Why This Test Case?
- This test case forms a complete cycle where every house is connected to the next and the last house is connected back to the first house.
- The code should output 5 because all roads must be maintained to ensure all friendships are connected.
- However, the random number generation might lead to incorrect tracking, potentially causing the code to output a smaller number of roads, exposing the bug.

### Expected Output:
```plaintext
5
```

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define F first
#define S second
#define FOR(i,a,b) for (auto i = (a); i <= (b); ++i)
#define NFOR(i,a,b) for(auto i = (a); i >= (b); --i)
#define all(x) (x).begin(), (x).end()
#define sz(x) lli(x.size())
#define mp(i,a) make_pair(i,a)
#define pb(a) push_back(a)
#define bit(x,b) (x&(1LL<<b))

typedef long long int lli;
typedef pair <int,int> ii;
typedef pair <ii,lli> iii;
typedef pair <ii,ii> iiii;
typedef vector <lli> vi;
typedef tree<lli,null_type,less<lli>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
ii segt[1600002];
int lazy[1600002];
void up(int node,int l,int r,int s,int e,int v)
{
  if(lazy[node]!=0)
  {
    segt[node].F+=lazy[node];
    lazy[2*node+1]+=lazy[node];
    lazy[2*node+2]+=lazy[node];
    lazy[node]=0;
  }
  if(l>e||r<s)
    return;
  if(l>=s&&r<=e)
  {
    segt[node].F+=v;
    lazy[2*node+1]+=v;
    lazy[2*node+2]+=v;
    return;
  }
  int m=(l+r)/2;
  up(2*node+1,l,m,s,e,v);
  up(2*node+2,m+1,r,s,e,v);
  segt[node]=(segt[2*node+1].F==segt[2*node+2].F)?mp(segt[2*node+1].F,segt[2*node+1].S+segt[2*node+2].S):min(segt[2*node+1],segt[2*node+2]);
  if(segt[node].S>r-l+1)
    exit(100);
}
ii query(int node,int l,int r,int s,int e)
{
  if(lazy[node]!=0)
  {
    segt[node].F+=lazy[node];
    lazy[2*node+1]+=lazy[node];
    lazy[2*node+2]+=lazy[node];
    lazy[node]=0;
  }
  if(l>e||r<s)
    return {10000000,0};
  if(l>=s&&r<=e)
    return segt[node];
  int m=(l+r)/2;
  ii q1=query(2*node+1,l,m,s,e);
  ii q2=query(2*node+2,m+1,r,s,e);
  ii re=(q1.F==q2.F)?mp(q1.F,q1.S+q2.S):min(q1,q2);
    if(re.S>e-s+1)
    exit(101);
  return re;
}
void init(int node,int l,int r)
{
  lazy[node]=0;
  segt[node]=mp(0,r-l+1);
  if(l==r)
    return;
  int m=(l+r)/2;
  init(2*node+1,l,m);
  init(2*node+2,m+1,r);
}
int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  lli test=1;
  cin>>test;
  for(;test>0;--test)
  {
    int n,m;
    cin>>n>>m;
    init(0,0,2*n);
    int l,r;
    vi al[2*n+2];
    for(int i=0;i<m;++i)
    {
      cin>>l>>r;
      up(0,0,2*n,l-1,r-2,1);
      al[l-1].pb(r-1);
    }
    int ans=0;
    l=0;
    r=n-1;
    for(int i=0;i<n-1;++i)
    {
      ii cu=query(0,0,2*n,l,r);
      if(cu.F==0)
        ans=max(ans,cu.S);
      if(cu.S>n)
        exit(i+1);
      for(int j=0;j<al[i].size();++j)
      {
        up(0,0,2*n,i,al[i][j]-1,-1);
        al[al[i][j]].pb(r+1);
      }
        l++;
        r++;
        for(int j=0;j<al[i].size();++j)
        up(0,0,2*n,al[i][j],r-1,1);
    }
      ii cu=query(0,0,2*n,l,r);
      if(cu.F==0)
        ans=max(ans,cu.S);
      if(cu.S>n)
        exit(-1);
    cout<<n-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 buggy code, we need to focus on edge cases and scenarios that can stress the algorithm. Here are some key points to consider:

1. **Cycle Structure**: The problem involves a cycle of houses, and the roads form a circular structure. This means that the input should include cases where the friendships span the entire cycle or parts of it.

2. **Minimum Roads**: The goal is to find the minimum number of roads that must be maintained to keep all friendships connected. The input should test cases where the minimum number of roads is not obvious.

3. **Edge Cases**: Test cases with the minimum and maximum values of `n` and `m` should be included to ensure the algorithm handles them correctly.

4. **Friendship Patterns**: The input should include cases where friendships are concentrated in certain parts of the cycle and cases where they are spread out.

Given these considerations, here is a small-scale test input that is complete and valid:

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

### Explanation of the Test Input:
1. **Test Case 1**:
   - `n = 3`, `m = 3`
   - Friendships: (1, 2), (2, 3), (3, 1)
   - This is a simple cycle where all houses are connected. The minimum number of roads to maintain is 3.

2. **Test Case 2**:
   - `n = 4`, `m = 3`
   - Friendships: (1, 2), (2, 3), (3, 4)
   - This is a linear path where the minimum number of roads to maintain is 3.

3. **Test Case 3**:
   - `n = 5`, `m = 2`
   - Friendships: (1, 3), (3, 5)
   - This tests a case where the friendships are not adjacent. The minimum number of roads to maintain is 3.

4. **Test Case 4**:
   - `n = 10`, `m = 4`
   - Friendships: (1, 10), (2, 9), (3, 8), (4, 7)
   - This tests a larger cycle with friendships that span the entire cycle. The minimum number of roads to maintain is 7.

This test input should help expose any issues in the buggy code, especially in handling cycles and maintaining the minimum number of roads.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
from sys import *
input = stdin.readline
inputlist = lambda: map(int, stdin.readline().split())

class SegmentTree:
    # Initialize with 1-indexed array
    def __init__(self, array):
        self.n = len(array)-1
        self.treemin = [0] * (4 * self.n)
        self.treefreq = [0] * (4 * self.n)
        self.lazy = [0] * (4 * self.n)
        # stack contains (node, l, r, 0/1)
        stack = [(1, 1, self.n, 0)]
        while stack:
            node, l, r, operation = stack.pop()
            if operation == 1:
                self.treemin[node] = min(self.treemin[node*2], self.treemin[node*2+1])
                self.treefreq[node] = 0
                if self.treemin[node*2] == self.treemin[node]: self.treefreq[node] += self.treefreq[node*2]
                if self.treemin[node*2+1] == self.treemin[node]: self.treefreq[node] += self.treefreq[node*2+1]
                continue
            if l == r:
                self.treemin[node] = array[l]
                self.treefreq[node] = 1
            else:
                mid = (l+r)//2
                stack.append((node, l, r, 1))
                stack.append((node*2, l, mid, 0))
                stack.append((node*2+1, mid+1, r, 0))

    def propogate(self, node, l, r):
        self.treemin[node] += self.lazy[node]
        if l != r:
            self.lazy[2*node] += self.lazy[node]
            self.lazy[2*node+1] += self.lazy[node]
        self.lazy[node] = 0

    # adds val to range [x, y]
    def update(self, x, y, val):
        stack = [(1, 1, self.n, 0)]
        while stack:
            node, l, r, operation = stack.pop()
            if operation == 1:
                self.treemin[node] = min(self.treemin[node*2], self.treemin[node*2+1])
                self.treefreq[node] = 0
                if self.treemin[node*2] == self.treemin[node]: self.treefreq[node] += self.treefreq[node*2]
                if self.treemin[node*2+1] == self.treemin[node]: self.treefreq[node] += self.treefreq[node*2+1]
                continue
            if self.lazy[node] != 0: self.propogate(node, l, r)
            if y < x or x > r or y < l: continue
            if l >= x and r <= y:
                self.treemin[node] += val
                if l != r:
                    self.lazy[node*2] += val
                    self.lazy[node*2+1] += val
                continue
            mid = (l+r)//2
            stack.append((node, l, r, 1))
            stack.append((2*node, l, mid, 0))
            stack.append((2*node+1, mid+1, r, 0))

    # query on range [x, y]
    def query(self, x, y):
        minval = float("inf")
        minfreq = 0
        stack = [(1, 1, self.n)]
        while stack:
            node, l, r = stack.pop()
            if self.lazy[node]: self.propogate(node, l, r)
            if y < x or x > r or y < l: continue
            if l >= x and r <= y:
                if self.treemin[node] < minval:
                    minval = self.treemin[node]
                    minfreq = self.treefreq[node]
                elif self.treemin[node] == minval:
                    minfreq += self.treefreq[node]
                continue
            mid = (l+r)//2
            stack.append((2*node, l, mid))
            stack.append((2*node+1, mid+1, r))
        return minval, minfreq

def solve():
    n, m = inputlist()
    graph = {}
    tree = SegmentTree([None] + [0] * n)
    for _ in range(m):
        a, b = inputlist()
        if a > 1: tree.update(a, b-1, 1)
        else: tree.update(b, n, 1)
        if a in graph: graph[a].append(b)
        else: graph[a] = [b]
        if b in graph: graph[b].append(a)
        else: graph[b] = [a]

    _, maxfreq = tree.query(1, n)
    for edge in range(2, n+1):
        if edge not in graph: continue
        for friendship in graph[edge]:
            if friendship > edge:
                tree.update(edge, friendship-1, -1)
                tree.update(1, edge-1, 1)
                tree.update(friendship, n, 1)
            else:
                tree.update(friendship, edge-1, 1)
                tree.update(edge, n, -1)
                tree.update(1, friendship-1, -1)
        _, newfreq = tree.query(1, n)
        maxfreq = max(maxfreq, newfreq)

    return n - maxfreq

for _ in range(int(input())):
    print(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
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
class lazy_segtree():
    def update(self, k):
        self.d[k] = self.op(self.d[2 * k], self.d[2 * k + 1])

    def all_apply(self, k, f):
        self.d[k] = self.mapping(f, self.d[k])
        if (k < self.size): self.lz[k] = self.composition(f, self.lz[k])

    def push(self, k):
        self.all_apply(2 * k, self.lz[k])
        self.all_apply(2 * k + 1, self.lz[k])
        self.lz[k] = self.identity

    def __init__(self, V, OP, E, MAPPING, COMPOSITION, ID):
        self.n = len(V)
        self.log = (self.n - 1).bit_length()
        self.size = 1 << self.log
        self.d = [E for i in range(2 * self.size)]
        self.lz = [ID for i in range(self.size)]
        self.e = E
        self.op = OP
        self.mapping = MAPPING
        self.composition = COMPOSITION
        self.identity = ID
        for i in range(self.n): self.d[self.size + i] = V[i]
        for i in range(self.size - 1, 0, -1): self.update(i)

    def set(self, p, x):
        assert 0 <= p and p < self.n
        p += self.size
        for i in range(self.log, 0, -1): self.push(p >> i)
        self.d[p] = x
        for i in range(1, self.log + 1): self.update(p >> i)

    def get(self, p):
        assert 0 <= p and p < self.n
        p += self.size
        for i in range(self.log, 0, -1): self.push(p >> i)
        return self.d[p]

    def prod(self, l, r):
        assert 0 <= l and l <= r and r <= self.n
        if l == r: return self.e
        l += self.size
        r += self.size
        for i in range(self.log, 0, -1):
            if (((l >> i) << i) != l): self.push(l >> i)
            if (((r >> i) << i) != r): self.push(r >> i)
        sml, smr = self.e, self.e
        while (l < r):
            if l & 1:
                sml = self.op(sml, self.d[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.d[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def all_prod(self):
        return self.d[1]

    def apply_point(self, p, f):
        assert 0 <= p and p < self.n
        p += self.size
        for i in range(self.log, 0, -1): self.push(p >> i)
        self.d[p] = self.mapping(f, self.d[p])
        for i in range(1, self.log + 1): self.update(p >> i)

    def apply(self, l, r, f):
        assert 0 <= l and l <= r and r <= self.n
        if l == r: return
        l += self.size
        r += self.size
        for i in range(self.log, 0, -1):
            if (((l >> i) << i) != l): self.push(l >> i)
            if (((r >> i) << i) != r): self.push((r - 1) >> i)
        l2, r2 = l, r
        while (l < r):
            if (l & 1):
                self.all_apply(l, f)
                l += 1
            if (r & 1):
                r -= 1
                self.all_apply(r, f)
            l >>= 1
            r >>= 1
        l, r = l2, r2
        for i in range(1, self.log + 1):
            if (((l >> i) << i) != l): self.update(l >> i)
            if (((r >> i) << i) != r): self.update((r - 1) >> i)

    def max_right(self, l, g):
        assert 0 <= l and l <= self.n
        assert g(self.e)
        if l == self.n: return self.n
        l += self.size
        for i in range(self.log, 0, -1): self.push(l >> i)
        sm = self.e
        while (1):
            while (l % 2 == 0): l >>= 1
            if not (g(self.op(sm, self.d[l]))):
                while (l < self.size):
                    self.push(l)
                    l = (2 * l)
                    if (g(self.op(sm, self.d[l]))):
                        sm = self.op(sm, self.d[l])
                        l += 1
                return l - self.size
            sm = self.op(sm, self.d[l])
            l += 1
            if (l & -l) == l: break
        return self.n

    def min_left(self, r, g):
        assert (0 <= r and r <= self.n)
        assert g(self.e)
        if r == 0: return 0
        r += self.size
        for i in range(self.log, 0, -1): self.push((r - 1) >> i)
        sm = self.e
        while (1):
            r -= 1
            while (r > 1 and (r % 2)): r >>= 1
            if not (g(self.op(self.d[r], sm))):
                while (r < self.size):
                    self.push(r)
                    r = (2 * r + 1)
                    if g(self.op(self.d[r], sm)):
                        sm = self.op(self.d[r], sm)
                        r -= 1
                return r + 1 - self.size
            sm = self.op(self.d[r], sm)
            if (r & -r) == r: break
        return 0

################################
INF = 1 << 31
# opの単位元 op(data1, e) = data1
e = [INF,1]

# 区間演算
def op(x, y):
    if x[0] == y[0]: return [x[0],x[1]+y[1]]
    if x[0] < y[0]: return x
    return y
# 上のlazy→下のdata
def mapping(lazy_upper, data_lower):

    return [lazy_upper+data_lower[0], data_lower[1]]
# 上のlazy→下のlazy
def composition(lazy_upper, lazy_lower):
    return lazy_upper + lazy_lower
# mapping(_id, data_lower) = data_lower
_id = 0
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n,m = map(int, input().split())
    A = [[0,1] for _ in range(n)]
    F = [[] for _ in range(n)]
    R = [[] for _ in range(n)]
    seg=lazy_segtree(A,op,e,mapping,composition,_id)
    for _ in range(m):
        a,b = map(int,input().split())
        a-=1;b-=1
        F[a].append(b)
        R[b].append(a)
        seg.apply(a,b,1)
    s = 0
    for i in range(n):
        s = max(s,seg.all_prod()[1])
        for j in F[i]:
            seg.apply(i,j,-1)
            seg.apply(0,i,1)
            seg.apply(j,n,1)
        for j in R[i]:
            seg.apply(j,i,1)
            seg.apply(0,j,-1)
            seg.apply(i,n,-1)
    print(n-s)

**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 buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some key considerations:

1. **Small-scale but challenging**: The test input should be small enough to fit within the constraints but complex enough to challenge the algorithm.
2. **Edge cases**: Consider scenarios with the minimum and maximum values for `n` and `m`.
3. **Special configurations**: Test configurations that might cause the lazy segment tree to fail, such as when the friends are distributed in a way that requires specific road maintenance.

Given the problem constraints, a good test input would be one where the friends are distributed in a way that requires a specific number of roads to be maintained, and the segment tree operations might fail to handle it correctly.

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

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

### Explanation:
- **Test Case 1**:
  - **n = 5**: There are 5 houses.
  - **m = 4**: There are 4 pairs of friendships.
  - **Friendships**:
    - House 1 is friends with House 2.
    - House 2 is friends with House 3.
    - House 3 is friends with House 4.
    - House 4 is friends with House 5.

### Expected Output:
- **Output**: 5
- **Explanation**: To maintain all the friendships, all 5 roads must be maintained.

This test case is designed to challenge the segment tree operations and ensure that the algorithm correctly identifies the minimum number of roads required. The friendships are linear, which might expose issues in the segment tree's lazy propagation or update mechanisms.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

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

public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Reader sc=new Reader();
		int t=sc.nextInt();
		while(t-->0)
		{
		    int n=sc.nextInt();
		    int m=sc.nextInt();
		    Pair arr[]=new Pair[m];
		    ArrayList<Integer> start[]=new ArrayList[n],end[]=new ArrayList[n];
		    for(int i=0;i<n;i++){ start[i]=new ArrayList<>();end[i]=new ArrayList<>();}
		    for(int i=0;i<m;i++){
		        arr[i]=new Pair(sc.nextInt()-1,sc.nextInt()-1);
		        start[arr[i].x].add(i);
		        end[arr[i].y].add(i);
		    }
		    int arr2[]=new int[n];
		    SegTree sg=new SegTree();
		    Node tree[]=new Node[4*n];
		    sg.buildTree(arr2,0,n-1,tree,1);
		    for(int i=0;i<m;i++)
		    sg.rangeupdate(tree,0,n-1,arr[i].x,arr[i].y-1,1,1);
		     // System.out.println(sg.query(tree,0,n-1,0,n-1,1).count);
		    int min=n;
		    for(int i=0;i<n;i++){
		        for(int j:start[i]){
		            sg.rangeupdate(tree,0,n-1,arr[j].x,arr[j].y-1,-1,1);
		            sg.rangeupdate(tree,0,n-1,arr[j].y,n-1,1,1);
		            sg.rangeupdate(tree,0,n-1,0,arr[j].x-1,1,1);
		        }
		         for(int j:end[i]){
		            sg.rangeupdate(tree,0,n-1,arr[j].x,arr[j].y-1,1,1);
		            sg.rangeupdate(tree,0,n-1,arr[j].y,n-1,-1,1);
		            sg.rangeupdate(tree,0,n-1,0,arr[j].x-1,-1,1);
		        }
		           min=Math.min(min,n-sg.query(tree,0,n-1,0,n-1,1).count);
		    }
		    System.out.println(min);

		}
	}
	static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
}
class SegTree
{
    //static int lazy[]=new int[1000001];
    public static void merge(Node tree[],int ind){
        tree[ind]=new Node(0,0);
        if(tree[2*ind]==null) tree[2*ind]=new Node(Integer.MAX_VALUE,0);
        if(tree[2*ind+1]==null) tree[2*ind+1]=new Node(Integer.MAX_VALUE,0);
        if(tree[2*ind].min<tree[2*ind+1].min){
            tree[ind].min=tree[2*ind].min;
            tree[ind].count=tree[2*ind].count;
        }
        else if(tree[2*ind+1].min<tree[2*ind].min){
             tree[ind].min=tree[2*ind+1].min;
            tree[ind].count=tree[2*ind+1].count;
        }
        else{
            tree[ind].min=tree[2*ind].min;
            tree[ind].count=tree[2*ind].count+tree[2*ind+1].count;
        }

    }
    public static void buildTree(int a[],int s,int e,Node tree[],int index)
    {
        if(s==e)
        {
            tree[index]=new Node(a[s],1);
            return;
        }
        int mid=(s+e)/2;
        buildTree(a,s,mid,tree,2*index);
        buildTree(a,mid+1,e,tree,2*index+1);
        merge(tree,index);

    }
    static Node query(Node tree[],int ss,int se,int qs,int qe,int index)
    {
        if(ss>=qs && se<=qe)
        return tree[index];
        else if(qe<ss || qs>se)
        return new Node(Integer.MAX_VALUE,1);
        int mid=(ss+se)/2;
        Node low1=query(tree,ss,mid,qs,qe,2*index);
        Node low2=query(tree,mid+1,se,qs,qe,2*index+1);
        if(low1.min<low2.min) return low1;
        if(low2.min<low1.min) return low2;
        return new Node(low1.min,low1.count+low2.count);
    }
     public static void rangeupdate(Node tree[],int ss,int se,int l,int r,int inc,int index)
    {
        if(r<ss || l>se)
        return;
        if(ss==se){
        tree[index].min+=inc;return;}
        int mid=(ss+se)/2;
        rangeupdate(tree,ss,mid,l,r,inc,2*index);
        rangeupdate(tree,mid+1,se,l,r,inc,2*index+1);
        merge(tree,index);

    }
}
class Node
{
    int min,count;
    Node(int min,int count)
    {
        this.min=min;this.count=count;
    }
}
class Pair
{
    int x,y;
    Pair(int x,int y){
        this.x=x;
        this.y=y;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause the algorithm to fail. The algorithm is supposed to find the minimum number of roads that must be maintained to ensure all friendships have a path. Here are the steps to think through this:

1. **Small-scale Test Case**: Ensure the test case is small enough to be manageable but large enough to cover edge cases.
2. **Edge Cases**:
   - **Linear Friendships**: All friendships are between consecutive houses.
   - **Cyclic Friendships**: Friendships that form a cycle.
   - **Isolated Friendships**: Friendships that do not form a continuous path.
3. **Special Cases**:
   - **Single Friendship**: Only one friendship exists.
   - **Full Connectivity**: All houses are friends with each other.
   - **Minimal Maintenance**: Only a few roads need to be maintained.

Given these considerations, let's create a test case that includes a mix of these scenarios:

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

### Explanation
- **Test Case 1**: Linear friendships (1-2, 2-3, 3-4) in a small cycle of 8 houses. This tests the algorithm's ability to handle linear relationships.
- **Test Case 2**: Cyclic friendships (1-2, 2-3, 3-4, 4-5, 5-6) in a larger cycle of 10 houses. This tests the algorithm's ability to handle cyclic relationships.
- **Test Case 3**: Full connectivity in a small cycle of 3 houses. This tests the algorithm's ability to handle full connectivity.
- **Test Case 4**: Minimal maintenance (1-2, 2-3, 3-4, 4-1) in a small cycle of 4 houses. This tests the algorithm's ability to handle minimal maintenance.
- **Test Case 5**: Complex relationships (1-3, 2-4) in a small cycle of 4 houses. This tests the algorithm's ability to handle more complex relationships.

This test case should help expose any bugs in the provided code by covering a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

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

public class Main
{
    public static int f(int x, int y, int[] A, int[] B, int N, int M) {
        int[] array = new int[N];
        for (int i = 0; i < M; i++) {
            int a = Math.min(A[i], B[i]), b = Math.max(A[i], B[i]);
            if (x == 0 && y == N - 1) {
                array[a] += 1;
                array[b] -= 1;
                continue;
            }
            if (a <= x && b >= y) {
                array[b] += 1;
                array[a] -= 1;
            } else {
                array[a] += 1;
                array[b] -= 1;
            }
        }
//        System.out.println(x + " " + y);
//        System.out.println(Arrays.toString(array));
        int j = y;
        while (j == x || j == y) {
            j = (j + 1) % N;
        }
        for (int i = 0; i < N; i++, j = (j + 1) % N) {
//            System.out.println(j + " " + (((j - 1) + N) % N));
            array[j] += array[((j - 1) + N) % N];
        }
        int count = 0;
        for (int i = 0; i < N; i++) {
            if (array[i] != 0) count++;
        }
//        System.out.println(Arrays.toString(array));
//        System.out.println(count);
//        System.out.println("------");
        return count;
    }

    public static void main (String[] args) throws java.lang.Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int TESTS = Integer.parseInt(br.readLine());
        StringTokenizer tokenizer;
        for (long test = 0; test < TESTS; test++) {
            tokenizer = new StringTokenizer(br.readLine());
            int N = Integer.parseInt(tokenizer.nextToken()), M = Integer.parseInt(tokenizer.nextToken());
            int[] A = new int[M], B = new int[M];
            for (int i = 0; i < M; i++) {
                tokenizer = new StringTokenizer(br.readLine());
                int a = Integer.parseInt(tokenizer.nextToken()) - 1, b = Integer.parseInt(tokenizer.nextToken()) - 1;
                A[i] = a;
                B[i] = b;
            }
            int result = N;
            for (int i = 0; i < N - 1; i++) {
                int x = i, y = i + 1;
                result = Math.min(result, f(x, y, A, B, N, M));
            }
            result = Math.min(result, f(0, N - 1, A, B, N, M));
            System.out.println(result);
        }
        bw.flush();
    }
}

**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
3 1
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: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
#include <iostream>
using namespace std;
int main() {
    int t,n,a,b;
    cin>>t;
    while(t--){
         cin>>n>>a>>b;
         if(n!=1 && n==a==b){
             cout<<b+(n-2)*a<<endl;}
         else if(n==1){
             cout<<a<<endl;
         }
         else if(n*a<(b+((n-2)*a))){
         cout<<n*a<<endl;
         }
         else{cout<<b+(n-2)*a<<endl;}
        }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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

using namespace std;

int n;
int m,x,y;
int main()
{
    cin>>n;

    while(n--){

        cin>>m>>x>>y;
        int t1=m*x;
        int t2=(m+1)/2*y;
        int t3=m/2*y+x;
        if(t1<t2&&t1<t3){
            cout<<t1<<endl;
        }else if(t3<t1&&t3<t2){
            cout<<t3<<endl;
        }else{
            cout<<t2<<endl;
        }
    }
}

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

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

Think step by step.

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

1. **Odd and Even Values of \( n \)**: The code handles the number of yogurts differently based on whether \( n \) is odd or even.
2. **Promotion Price vs. Regular Price**: The code calculates different costs based on the promotion price and the regular price.
3. **Boundary Conditions**: Test with boundary values for \( n \), \( a \), and \( b \).

Let's consider the following test case:

- \( n = 3 \), \( a = 5 \), \( b = 10 \)

This test case is particularly challenging because:
- The number of yogurts \( n \) is odd.
- The promotion price \( b \) is exactly double the regular price \( a \).

For this test case:
- If Maxim buys 3 yogurts at the regular price, the cost is \( 3 \times 5 = 15 \) burles.
- If Maxim buys 2 yogurts at the promotion price and 1 at the regular price, the cost is \( 10 + 5 = 15 \) burles.
- The code should correctly output 15 burles.

However, the buggy code might not handle this case correctly due to the way it calculates the costs.

Here is the test input:

```plaintext
1
3 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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

int main()
{
    long long  t, a, n,b,sum=0,sum2=0;
    cin >> t ;

    for (int i=0; i<t; i++)
    {
        cin >> n >> a >> b;
        sum = n*a;
        if (n==1)
        {
            if (a > b)
            {
                cout << a << endl ;
            }
            else{
                cout << b << endl ;
            }
        }
        else if(n%2==0)
        {
            if (a*2 > b)
            {
                n/=2;
                cout << n*b << endl ;
            }
            else
            {
                cout << n *a << endl;
            }
        }
        else
        {
            if (a*2 > b)
            {
                n/=2;
                if(a < b)
                {
                    sum = n*b;
                    sum +=a;
                    cout << sum << endl;
                }
                else
                {
                    cout << n*b << endl ;
                }
            }
            else
            {
                cout << n *a  << endl;
            }
        }
        sum =0;
        sum2 = 0;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1 10 20
2 10 20
3 10 20
4 10 20
5 10 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
t = int(input())
for i in range(t):
    n, a, b = list(map(int, input().split()))
    mul = n * a
    if n > 2:
        if mul < b + ((n - 2) * a):
            print(mul)
        else:
            print(b + ((n - 2) * a))
    else:
        if b < a * n:
            print(b)
        else:
            print(a * n)

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

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

Think step by step.

Response:
```plaintext
5
1 5 9
2 5 9
3 5 9
4 5 9
5 5 9
```