**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#pragma GCC target ("avx2")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")

#define f first
#define s second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int) (x).size())
#define pb push_back
#define mp make_pair
#define int long long

using namespace std;
using namespace __gnu_pbds;

template <typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> inline bool umin(T &a, const T &b) { if(a > b) { a = b; return 1; } return 0; }
template <typename T> inline bool umax(T &a, const T &b) { if(a < b) { a = b; return 1; } return 0; }

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

const ll mod = 998244353;
const ll base = 1e6 + 9;
const ll inf = 1e18;
const int MAX = 2e5 + 42;
const int LG = 20;

random_device rd;
mt19937 gen(rd());
uniform_int_distribution<ll> dis(1, inf);

void solve() {
    int n, m;
    cin >> n >> m;
    int x, y, z;
    cin >> z >> x >> y;
    vector<vector<array<int, 3>>> g(n + 1);
    while(m--) {
        int u, v, l, r;
        cin >> u >> v >> l >> r;
        g[u].pb({v, l, r}); g[v].pb({u, l, r});
    }
    auto check = [&](int st) {
        auto dijkstra = [&](vector<pii> sources, bool flag) {
            vector<int> d(n + 1, inf);
            for(auto [v, dst] : sources) d[v] = dst;
            priority_queue<pii> q; for(int v = 1; v <= n; v++) q.push({d[v], v});
            while(sz(q)) {
                auto [dst, v] = q.top(); q.pop();
                if(d[v] != dst) continue;
                for(auto [to, a, b] : g[v]) {
                    int w = (flag? b : a);
                    if(d[to] > d[v] + w) {
                        d[to] = d[v] + w;
                        q.push({d[to], to});
                    }
                }
            }
            return d;
        };
        vector<pii> sources = {mp(1, st)};
        auto d1 = dijkstra(sources, 0);
        if(d1[n] <= x) return true;
        sources.clear();
        for(int v = 1; v <= n; v++) {
            if(d1[v] <= x) sources.pb({v, d1[v]});
        }
        if(sources.empty()) sources = {mp(1, st)};
        auto d2 = dijkstra(sources, 1);
        if(d2[n] <= y) return true;
        for(int v = 1; v <= n; v++) umax(d2[v], y);
        sources.clear();
        for(int v = 1; v <= n; v++) {
            sources.pb({v, d2[v]});
        }
        auto d = dijkstra(sources, 0);
//        cout << st << ": " << d[n] << endl;
        return d[n] <= z;
    };
    int l = 0, r = z;
    while(l <= r) {
        m = l + r >> 1;
        if(check(m)) l = m + 1;
        else r = m - 1;
    }
    cout << r << '\n';
}

signed main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int ttt = 1;
    cin >> ttt;
    while(ttt--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
100 50 70
1 2 10 30
2 3 10 30
1 3 20 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++){
            int n=sc.nextInt(),m=sc.nextInt(),t0=sc.nextInt(),t1=sc.nextInt(),t2=sc.nextInt();
            List<int[]> path[]=new List[n+5];
            for(int j=1;j<=n;j++){
                path[j]=new ArrayList<>();
            }
            for(int j=0;j<m;j++){
                int u=sc.nextInt(),v=sc.nextInt(),l1=sc.nextInt(),l2=sc.nextInt();
                path[u].add(new int[]{v,l1,l2});
                path[v].add(new int[]{u,l1,l2});
            }
            if(find(n,0,path,t1,t2)>t0){
                System.out.println("-1");
            }
            else{
                int l=0,r=(int)1e9;
                while(l<r){
                    int mid=(l+r)>>1;
                    if(find(n,mid,path,t1,t2)<=t0){
                        l=mid;
                    }
                    else{
                        r=mid-1;
                    }
                    if(l==r-1){
                        if(find(n,r,path,t1,t2)<=t0){
                            l=r;
                        }
                        break;
                    }
                }
                System.out.println(l);
            }
        }
    }
    static long find(int n,int start,List<int[]> path[],int t1,int t2){
        Queue<long[]> q=new PriorityQueue<>((a,b)->Long.compare(a[1],b[1]));
        long dis[]=new long[n+5];
        Arrays.fill(dis,(long)1e18);
        dis[1]=start;
        q.add(new long[]{1,start});
        while(!q.isEmpty()){
            long a[]=q.poll();
            if(dis[(int)a[0]]<a[1]){
                continue;
            }
            for(int b[]:path[(int)a[0]]){
                long d=a[1]+b[2];
                if(dis[b[0]]>d){
                    q.add(new long[]{b[0],d});
                    dis[b[0]]=d;
                }
                d=a[1]+b[1];
                if(!(d<=t1||a[1]>=t2)){
                    d=t2+b[1];
                }
                if(dis[b[0]]>d){
                    q.add(new long[]{b[0],d});
                    dis[b[0]]=d;
                }
            }
        }
        return dis[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:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The main issues in the code are related to the handling of the phone call time and the binary search logic. Let's consider the following scenarios:

1. **Short Phone Call Duration**: The phone call duration is very short, and it might interfere with the optimal path.
2. **Long Phone Call Duration**: The phone call duration is long, which might force the algorithm to take suboptimal paths.
3. **Multiple Paths with Different Bus and Walking Times**: The city has multiple paths with different bus and walking times, which can lead to complex decisions.
4. **Edge Cases in Binary Search**: The binary search logic might not handle edge cases correctly, especially when the start and end times are very close.

Let's create a small-scale test input that combines these scenarios:

```plaintext
5
5 5
100 20 80
1 2 10 50
2 3 10 50
3 4 10 50
4 5 10 50
1 2 30 60
5 5
100 40 60
1 2 10 50
2 3 10 50
3 4 10 50
4 5 10 50
1 2 30 60
5 5
100 80 90
1 2 10 50
2 3 10 50
3 4 10 50
4 5 10 50
1 2 30 60
5 5
100 50 60
1 2 10 50
2 3 10 50
3 4 10 50
4 5 10 50
1 2 30 60
5 5
100 90 95
1 2 10 50
2 3 10 50
3 4 10 50
4 5 10 50
1 2 30 60
```

Explanation:
- **Test Case 1**: Short phone call duration (20 to 80) with a direct path from 1 to 5 and a longer path from 1 to 2 to 5. The optimal path should be the direct path, but the phone call might interfere.
- **Test Case 2**: Medium phone call duration (40 to 60) with a direct path from 1 to 5 and a longer path from 1 to 2 to 5. The optimal path should be the direct path, but the phone call might force a suboptimal path.
- **Test Case 3**: Long phone call duration (80 to 90) with a direct path from 1 to 5 and a longer path from 1 to 2 to 5. The optimal path should be the direct path, but the phone call might force a suboptimal path.
- **Test Case 4**: Medium-long phone call duration (50 to 60) with a direct path from 1 to 5 and a longer path from 1 to 2 to 5. The optimal path should be the direct path, but the phone call might force a suboptimal path.
- **Test Case 5**: Very long phone call duration (90 to 95) with a direct path from 1 to 5 and a longer path from 1 to 2 to 5. The optimal path should be the direct path, but the phone call might force a suboptimal path.

This test input should help expose any issues in the algorithm's handling of the phone call constraints and the binary search logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

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

#define int long long
#define ll long long

#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define deb(x...)
#endif

const int mod = 1000000007; // 998244353;
const int N = 2 * 1e5 + 5;
const int INF = 1000000000000000000;

void solve()
{
	int n, m, t0, t1, t2, u, v, l1, l2;
	cin >> n >> m >> t0 >> t1 >> t2;

	vector<vector<int>> adj[n + 1];

	for (int i = 0; i < m; i++) {

		cin >> u >> v >> l1 >> l2;
		adj[u].push_back({v, l1, l2});
		adj[v].push_back({u, l1, l2});

	}

	vector<int> dist(n + 1, -INF);
	dist[n] = t0;

	set<pair<int, int>> s;
	s.insert({t0, n});

	while (!s.empty()) {

		auto it = *s.begin();
		s.erase(it);

		int node = it.second;

		for (auto &it : adj[node]) {

			int adjNode = it[0], busTime = it[1], walkTime = it[2];

			if (dist[node] <= t1 or dist[node] - busTime >= t2) {

				// free to use bus
				if (dist[node] - busTime > dist[adjNode]) {

					if (dist[adjNode] != -INF) s.erase({dist[adjNode], adjNode});

					dist[adjNode] = dist[node] - busTime;
					s.insert({dist[adjNode], adjNode});

				}

			} else {

				int best = max(dist[node] - walkTime, t1 - busTime);

				if (best > dist[adjNode]) {

					if (dist[adjNode] != -INF) s.erase({dist[adjNode], adjNode});

					dist[adjNode] = best;
					s.insert({dist[adjNode], adjNode});

				}

			}

		}

	}

	if (dist[1] >= 0) cout << dist[1] << "\n";
	else cout << -1 << "\n";

}

signed main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t; cin >> t; while (t--) solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5+10;
int ar[N],n,m;
vector<array<int,3>> q[N];
int t0,t1,t2;
int dist[N];
inline bool check(int mid){
    for(int i=1; i<=n; i++) dist[i] = 1e15;
    dist[1] = mid;
    priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int>>> p;
    p.push({dist[1],1});
    while(!p.empty()){
        auto [zz,x] = p.top(); p.pop();
        for(auto[a,b,c]:q[x]){
            if(dist[x]>=t1&&dist[x]<t2){
                int xx = dist[x]+c;
                int yy = t2+b;
                xx = min(xx,yy);
                if(xx<dist[a]){
                    dist[a] = xx;
                    p.push({dist[a],a});
                }
            }else if(dist[x]<t1&&dist[x]+b>t1){
                int xx = dist[x]+c;
                int yy = t2+b;
                xx = min(xx,yy);
                if(xx<dist[a]){
                    dist[a] = xx;
                    p.push({dist[a],a});
                }
            }else {
                if(dist[x]+b<dist[a]){
                    dist[a] = dist[x]+b;
                    p.push({dist[a],a});
                }
            }
        }
    }
    return dist[n]<=t0;
}
inline void solve(){
    cin>>n>>m;
    cin>>t0>>t1>>t2;
    for(int i=1; i<=n; i++) q[i].clear();
    for(int i=1,a,b,c,d; i<=m; i++) {
        cin>>a>>b>>c>>d;
        q[a].push_back({b,c,d});
        q[b].push_back({a,c,d});
    }
    int l = 0,r = 1e10;
    while(l<r){
        int mid = l+r+1>>1;
        if(check(mid)) l = mid;
        else r = mid-1;
    }
    if(check(l)) cout<<l<<"\n";
    else cout<<"-1\n";
}

signed main(){
	ios::sync_with_stdio(false); cin.tie(0);
	int _ = 1;
	cin>>_;
	while(_--){
		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
1
3 3
100 50 70
1 2 20 40
2 3 20 40
1 3 40 60
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
import heapq, math
from collections import Counter, defaultdict, deque
from functools import cache

import io,os
import sys

# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

t = int(input())
ret = []
MOD = 1_000_000_007

for _ in range(t):

    # n = int(input())
    n, m = list(map(int, input().split()))
    deadline, pstart, pend = list(map(int, input().split()))
    # m = int(input())
    # s = input()
    adj = defaultdict(list)

    for _ in range(m):
        u, v, c1, c2 = list(map(int, input().split()))

        u -= 1
        v -= 1

        adj[u].append((v, c1, c2))
        adj[v].append((u, c1, c2))

    def overlap(a, b, x, y):

        return not (b < x or y < a)

    def check(start):

        h = [(start, 0)]
        vis = [math.inf] * n
        vis[0] = h[0][0]

        # print('--------------START---------------')

        while h:
            # print(h)
            time, curr = heapq.heappop(h)

            if time > deadline:
                return False

            if curr == n - 1:
                return True

            # if pstart < time < pend:
            #     time = pend

            for j, c1, c2 in adj[curr]:

                if not overlap(time, time + c1, pstart + 1, pend - 1) and vis[j] > time + c1 and time + c1 <= deadline:
                    vis[j] = time + c1
                    heapq.heappush(h, (time + c1, j))

                if time <= pend and vis[j] > pend + c1 and pend + c1 <= deadline:
                    vis[j] = pend + c1
                    heapq.heappush(h, (pend + c1, j))

                if vis[j] > time + c2 and time + c2 <= deadline:
                    vis[j] = time + c2
                    heapq.heappush(h, (time + c2, j))

    def bins(start, end):

        if start >= end:
            return -math.inf
        if start + 1 == end:
            return start if check(start) else -math.inf

        mid = (start + end) // 2

        if not check(mid):
            return bins(start, mid)
        else:
            return max(mid, bins(mid + 1, end))

    res = bins(0, deadline + 1)
    # ret.append((res if res != -math.inf else -1))
    print((res if res != -math.inf else -1))

# print(ret)

# 1
# 4 4
# 100 40 60
# 1 2 30 100
# 2 4 30 100
# 1 3 20 50
# 3 4 20 50

# 1
# 2 1
# 12 9 10
# 2 1 6 10

# 1
# 5 5
# 100 20 80
# 1 5 30 100
# 1 2 20 50
# 2 3 20 50
# 3 4 20 50
# 4 5 20 50

**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 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
```

**Explanation:**

This test case is designed to expose the bug in the code by creating a scenario where the phone call overlaps with the optimal bus route. The key points are:

1. **Number of Test Cases**: 1
2. **Intersections and Streets**: 4 intersections and 4 streets.
3. **Event and Phone Call Times**:
   - Event starts at time 100.
   - Phone call starts at time 40 and ends at time 60.
4. **Streets**:
   - Street 1-2: 30 minutes by bus, 100 minutes by foot.
   - Street 2-4: 30 minutes by bus, 100 minutes by foot.
   - Street 1-3: 20 minutes by bus, 50 minutes by foot.
   - Street 3-4: 20 minutes by bus, 50 minutes by foot.

**Why This Test Case is Challenging:**

- **Optimal Route Without Phone Call**: The optimal route without considering the phone call would be to take the bus from 1 to 2 (30 minutes), then from 2 to 4 (30 minutes), totaling 60 minutes. This would allow you to arrive at intersection 4 at time 60, just as the phone call ends.
- **Phone Call Overlap**: The phone call from 40 to 60 overlaps with the optimal bus route. The code needs to handle this overlap correctly, either by walking or waiting until the phone call ends before taking the bus.
- **Correct Handling**: The correct solution should find a way to delay the bus trip to avoid the phone call, or use a different route that avoids the overlap. The buggy code might fail to correctly handle this overlap, leading to an incorrect result.

This test case should help expose the bug in the code by requiring careful handling of the phone call overlap.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class CallDuringTheJourney {
	public static PrintWriter out;
	public static void main(String[] args)throws IOException{
        JS sc=new JS();
        out=new PrintWriter(System.out);
        int t=sc.nextInt();
        outer: while(t-->0) {
        	int n=sc.nextInt();
        	int m=sc.nextInt();
        	int t0=sc.nextInt();
        	int t1=sc.nextInt();
        	int t2=sc.nextInt();
        	ArrayList<ArrayList<int[]>>al=new ArrayList<>();
        	for(int i=0;i<n;i++) {
        		al.add(new ArrayList<int[]>());
        	}
        	for(int i=0;i<m;i++) {
        		int a=sc.nextInt()-1;
        		int b=sc.nextInt()-1;
        		int bus=sc.nextInt();
        		int walk=sc.nextInt();
        		al.get(a).add(new int[] {b, bus, walk});
        		al.get(b).add(new int[] {a, bus, walk});
        	}
        	int dist[]=new int[n];
        	Arrays.fill(dist, (int) -1e9);
        	dist[n-1]=t0;
        	PriorityQueue<Integer>q=new PriorityQueue<>((Integer p1,Integer p2)->(dist[p1]-dist[p2]));
//        	ArrayDeque<Integer>q=new ArrayDeque<>();
        	q.add(n-1);
        	while(!q.isEmpty()) {
        		int cur=q.poll();
        		for(int[] next:al.get(cur)) {
        			int nexttime=dist[cur]-next[1];
        			if(nexttime>=t2||dist[cur]<=t1) {
        				nexttime=dist[cur]-next[1];
        			}else {
        				nexttime=dist[cur]-next[2];
        				nexttime=Math.max(nexttime, t1-next[1]);
        			}
        			if(dist[next[0]]<nexttime&&nexttime>=-20) {
        				dist[next[0]]=nexttime;
        				q.add(next[0]);
        			}
        		}
        	}
        	out.println(dist[0]>=0?dist[0]:-1);
        }
		out.close();
	}
	static void add(HashMap<Integer,Integer> A, int key){
		A.put(key,A.getOrDefault(key,0)+1);
	}
	static void remove(HashMap<Integer,Integer>A, int key){
		if(!A.containsKey(key))return;
		A.put(key,A.get(key)-1);
		if(A.get(key)==0){
			A.remove(key);
		}
	}
	static class Pair implements Comparable<Pair>{
		int x, y;
		public Pair(int x, int y) {
			this.x=x;this.y=y;
		}
		public int compareTo(Pair p) {
			if(y==p.y) {
				return Integer.compare(x, p.x);
			}
			return Integer.compare(y, p.y);
		}
	}
//	static class Triple{
//		int node, bus, walk;
//		public Triple(int node, int bus, int walk) {
//			this.node=node;this.bus=bus;this.walk=walk;
//		}
//	}
	static void sort(int a[]) {
		ArrayList<Integer>al=new ArrayList<>();
		for(int i=0;i<a.length;i++) {
			al.add(a[i]);
		}
		Collections.sort(al);
		for(int i=0;i<a.length;i++) {
			a[i]=al.get(i);
		}
	}
	static class JS {
	    public int BS = 1<<16;
	    public char NC = (char)0;
	    byte[] buf = new byte[BS];
	    int bId = 0, size = 0;
	    char c = NC;
	    double num = 1;
	    BufferedInputStream in;

	    public JS() {
	        in = new BufferedInputStream(System.in, BS);
	    }

	    public JS(String s) throws FileNotFoundException {
	        in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
	    }

	    public char nextChar(){
	        while(bId==size) {
	            try {
	                size = in.read(buf);
	            }catch(Exception e) {
	                return NC;
	            }
	            if(size==-1)return NC;
	            bId=0;
	        }
	        return (char)buf[bId++];
	    }

	    public int nextInt() {
	        return (int)nextLong();
	    }

	    public long nextLong() {
	        num=1;
	        boolean neg = false;
	        if(c==NC)c=nextChar();
	        for(;(c<'0' || c>'9'); c = nextChar()) {
	            if(c=='-')neg=true;
	        }
	        long res = 0;
	        for(; c>='0' && c <='9'; c=nextChar()) {
	            res = (res<<3)+(res<<1)+c-'0';
	            num*=10;
	        }
	        return neg?-res:res;
	    }

	    public double nextDouble() {
	        double cur = nextLong();
	        return c!='.' ? cur:cur+nextLong()/num;
	    }

	    public String next() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c>32) {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public String nextLine() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c!='\n') {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public boolean hasNext() {
	        if(c>32)return true;
	        while(true) {
	            c=nextChar();
	            if(c==NC)return false;
	            else if(c>32)return true;
	        }
	    }
	}
}

**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 2
100 40 60
1 2 10 20
2 3 10 20
```

**Explanation:**

1. **Test Case Setup:**
   - The test case has 3 intersections and 2 streets.
   - The event is scheduled at time 100, the phone call is from time 40 to 60.

2. **Street Connections:**
   - Street 1 connects intersections 1 and 2 with bus travel time 10 and walk travel time 20.
   - Street 2 connects intersections 2 and 3 with bus travel time 10 and walk travel time 20.

3. **Expected Behavior:**
   - The latest time to leave home (intersection 1) should be calculated considering the constraints of the phone call and the travel times.
   - The bug in the code might cause it to miscalculate the latest possible departure time, especially when the phone call overlaps with the travel times.

4. **Challenging Scenario:**
   - The phone call (40 to 60) overlaps with the travel times, which should force the algorithm to consider walking or making stops to avoid being on the bus during the call.
   - The correct answer should be 60, as you can walk from intersection 1 to intersection 3 (total time 20 + 20 = 40) and then wait until the call ends at 60, leaving you 40 minutes to reach the event at time 100.

This test case is designed to expose any issues in handling the overlap between the phone call and the travel times, as well as the correct calculation of the latest departure time.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
//package Codeforces;

import java.util.*;

public class CallDuringTheJourney {
    final static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int testcases =  sc.nextInt();
        for (int test = 0; test<testcases; ++test) {
            int n = sc.nextInt();
            int m = sc.nextInt();

            Graph g = new Graph(n, m);

            int t0 = sc.nextInt();
            int t1 = sc.nextInt();
            int t2 = sc.nextInt();

            for(int i=0; i<m; ++i){
                int u = sc.nextInt();
                int v = sc.nextInt();
                int bussTime = sc.nextInt();
                int walkTime = sc.nextInt();
                g.createEdge(u, v, bussTime, walkTime);
            }

            int result = findTimeToReach(g, t0, t1, t2);
            System.out.println(result);
        }
    }

    private static int findTimeToReach(Graph g, int t0, int t1, int t2){
        g.populateMaxTimeToReach(t0, t1, t2);
        return g.nodes.get(0).maxTimeTorReach;
    }

    private static boolean isOverlappingInterval(int busTime, int maxTimeTorReach, int t1, int t2) {
        return (t2 > busTime && t1 < maxTimeTorReach);
    }

    public static class Graph{
        List<Node> nodes;
        List<Edge> edges;
        Map<Node, List<Edge>> adjacencyList;
        static int edgeid = 0;
        int N;
        int M;

        Graph(int N, int M){
            this.N = N;
            this.M = M;
            nodes = new ArrayList<Node>();
            edges = new ArrayList<>();
            adjacencyList = new HashMap<>();
            for(int i = 0; i < N; i++){
                Node u = new Node(i);
                nodes.add(u);
                adjacencyList.put(u, new ArrayList<>());
            }
        }

        void populateMaxTimeToReach(int t0, int t1, int t2){
            nodes.get(N-1).maxTimeTorReach = t0;
            PriorityQueue<Graph.Node> pq = new PriorityQueue<>();

            pq.add(nodes.get(N-1));
            while(!pq.isEmpty()){
                Graph.Node node = pq.poll();
                for(Graph.Edge edge: adjacencyList.get(node)){
                    Graph.Node neib = edge.u;
                    if(edge.u.id == node.id){
                        neib = edge.v;
                    }
                    int walkTime = node.maxTimeTorReach - edge.walkTime;
                    int busTime = node.maxTimeTorReach - edge.busTime;

                    if(isOverlappingInterval(busTime, node.maxTimeTorReach, t1, t2)){
                        busTime = t1 - edge.busTime;
                    }
                    int mx = Math.max(walkTime, busTime);
                    if(neib.maxTimeTorReach < mx){
                        pq.add(neib);
                        neib.maxTimeTorReach = mx;
                    }
                }
            }
        }

        public static class Node implements Comparable<Node>{
            int id;
            int maxTimeTorReach;

            Node(int id){
                this.id = id;
                this.maxTimeTorReach = -1;
            }

            @Override
            public int compareTo(Node o) {
                return   this.maxTimeTorReach - o.maxTimeTorReach;
            }
        }

        public static class Edge{
            int id;
            Node u;
            Node v;
            int walkTime;
            int busTime;

            public Edge(Node u, Node v, int walkTime, int busTime) {
                this.u = u;
                this.v = v;
                this.walkTime = walkTime;
                this.busTime = busTime;
                this.id = edgeid;
                edgeid++;
            }
        }

        void createEdge(int u, int v, int busTime, int walkTime){
            u--;
            v--;
            Node uNode = nodes.get(u);
            Node vNode = nodes.get(v);

            Edge edge = new Edge(uNode, vNode, walkTime, busTime);
            adjacencyList.get(uNode).add(edge);
            adjacencyList.get(vNode).add(edge);
        }
    }

}

**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 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Call During the Journey
time_limit_ms: 4000
memory_limit_mb: 256
Description: You live in a city consisting of $n$ intersections and $m$ streets connecting some pairs of intersections. You can travel in either direction on each street. No two streets connect the same pair of intersections, and no street connects an intersection to itself. You can reach any intersection from any other, possibly passing through some other intersections.

Every minute, you can board a bus at intersection $u_i$ and travel for $l_{i1}$ minutes to intersection $v_i$. Conversely, you can travel from intersection $v_i$ to intersection $u_i$ in $l_{i1}$ minutes. You can only board and exit the bus at intersections. You can only board the bus at an intersection if you are currently there.

You can also walk along each street, which takes $l_{i2} > l_{i1}$ minutes.

You can make stops at intersections.

You live at intersection number $1$. Today you woke up at time $0$, and you have an important event scheduled at intersection number $n$, which you must reach no later than time $t_0$. You also have a phone call planned that will last from $t_1$ to $t_2$ minutes ($t_1 < t_2 < t_0$).

During the phone call, you cannot ride the bus, but you can walk along any streets, make stops, or stay at home. You can exit the bus at minute $t_1$ and board the bus again at minute $t_2$.

Since you want to get enough sleep, you became curious — how late can you leave home to have time to talk on the phone and still not be late for the event?

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 10^5, 1 \le m \le 10^5$) — the number of intersections and streets in the city.

The second line of each test case contains three integers $t_0$, $t_1$, $t_2$ ($1 \le t_1 < t_2 < t_0 \le 10^9$) — the start time of the event, the start time of the phone call, and its end time, respectively.

The next $m$ lines of each test case contain descriptions of the streets.

The $i$-th line contains four integers $u_i$, $v_i$, $l_{i1}$, $l_{i2}$ ($1 \le u_i, v_i \le n$, $u_i \neq v_i$, $1 \le l_{i1} < l_{i2} \le 10^9$) — the numbers of the intersections connected by the $i$-th street, as well as the travel time along the street by bus and on foot. It is guaranteed that no two streets connect the same pair of intersections and that it is possible to reach any intersection from any other.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $10^5$. It is also guaranteed that the sum of the values of $m$ across all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the latest time you can leave home to have time to talk on the phone and not be late for the event. If you cannot reach the event on time, output -1.

Examples:
input:
7
5 5
100 20 80
1 5 30 100
1 2 20 50
2 3 20 50
3 4 20 50
4 5 20 50
2 1
100 50 60
1 2 55 110
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
3 3
100 80 90
1 2 1 10
2 3 10 50
1 3 20 21
3 2
58 55 57
2 1 1 3
2 3 3 4
2 1
12 9 10
2 1 6 10
5 5
8 5 6
2 1 1 8
2 3 4 8
4 2 2 4
5 3 3 4
4 5 2 6
output:
0
-1
60
80
53
3
2

Note:

Buggy Code:
# Created by Ketan_Raut
# Copyright © 2024 iN_siDious. All rights reserved.
import sys,os
import random
from math import *
from string import ascii_lowercase,ascii_uppercase
from collections import Counter, defaultdict, deque
from itertools import accumulate, combinations, permutations
from heapq import heappushpop, heapify, heappop, heappush
from bisect import bisect_left,bisect_right
from types import GeneratorType
from random import randint
RANDOM = randint(1, 10 ** 10)
class op(int):
    def __init__(self, x):
        int.__init__(x)
    def __hash__(self):
        return super(op, self).__hash__() ^ RANDOM
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
inp = lambda: sys.stdin.buffer.readline().decode().strip()
out=sys.stdout.write
def S():
    return inp()
def I():
    return int(inp())
def F():
    return float(inp())
def MI():
    return map(int, inp().split())
def MF():
    return map(float, inp().split())
def MS():
    return inp().split()
def LS():
    return list(inp().split())
def LI():
    return list(map(int,inp().split()))
def LF():
    return list(map(float,inp().split()))
def Query(i,j):
    print3('?', i, j)
    sys.stdout.flush()
    qi=I()
    return qi
def print1(x):
    return out(str(x)+"\n")
def print2(x,y):
    return out(str(x)+" "+str(y)+"\n")
def print3(x,y,z):
    return out(str(x)+" "+str(y)+" "+str(z)+"\n")
def print_arr(arr):
    for num in arr:
        out(str(num)+" ")
    out(str("\n"))
for _ in range(I()):
    n,m=MI()
    t0,t1,t2=MI()
    graph=[[] for _ in range(n)]
    for _ in range(m):
        u,v,l1,l2=MI()
        u,v=u-1,v-1
        graph[u].append((v,l1,l2))
        graph[v].append((u,l1,l2))
    heap,dist=[],[-sys.maxsize]*n
    dist[n-1]=t0
    heappush(heap,(-t0,n-1))
    while heap:
        time,node=heappop(heap)
        time=-time
        for nei,l1,l2 in graph[node]:
            new_time=time-l1
            if not ((t1<new_time<t2) or (new_time<=t1 and time>=t2) or t1<time<t2):
                if new_time>dist[nei]:
                    dist[nei]=new_time
                    heappush(heap,(-new_time,nei))
            else:
                new_time=max(time-l2,t1-l1)
                if new_time>dist[nei]:
                    dist[nei]=new_time
                    heappush(heap,(-new_time,nei))
    print1(-1 if dist[0]<0 else dist[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
4 4
100 40 60
1 2 30 100
2 4 30 100
1 3 20 50
3 4 20 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
#include<iostream>
#include<vector>
#define int int64_t
using namespace std;
int32_t cal(int n){
    if(n<3)return 1;
    return 1+cal(n/3);
}
int32_t main(){
    int t;
    cin>>t;
    while(t--){
        int l,r;
        cin>>l>>r;
        vector<int>vc(3e5+1,0);
        vc[1]=1;
        for(int i=2;i<3e5+1;i++){
            vc[i]=vc[i-1]+cal(i);
        }
        cout<<2*cal(l)+(vc[r]-vc[l])<<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, let's analyze the code and the problem statement:

### Analysis of the Buggy Code
1. **Recursive Function `cal`**: The function `cal` is intended to calculate the number of operations needed to reduce a number to 0. However, it only considers the number of operations needed to reduce a single number, not the entire range from `l` to `r`.
2. **Precomputation Array `vc`**: The array `vc` is precomputed to store the cumulative number of operations for each number from 1 to 300,000. This precomputation is based on the `cal` function.
3. **Output Calculation**: The output for each test case is calculated using `2*cal(l) + (vc[r] - vc[l])`.

### Potential Bugs
1. **Incorrect Precomputation**: The precomputation in the array `vc` might not correctly account for the operations needed to reduce the entire range from `l` to `r`.
2. **Incorrect Output Calculation**: The formula `2*cal(l) + (vc[r] - vc[l])` might not accurately represent the minimum number of operations needed.

### Challenging Test Case
To expose the bug, we need a test case where the precomputation and the output calculation differ significantly from the correct solution. Let's consider a range where the numbers are large and the operations needed are not straightforward.

### Test Case
```plaintext
1
199999 200000
```

### Explanation
- **Input**: `l = 199999` and `r = 200000`
- **Expected Output**: The correct number of operations needed to make all numbers in the range from 199999 to 200000 equal to 0 is 36.
- **Buggy Output**: The buggy code might produce a different result due to incorrect precomputation and output calculation.

This test case is small-scale and complete, focusing on a range where the numbers are large and the operations needed are not trivial. This will help expose the bug in the precomputation and output calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
#include<bits/stdc++.h>
    #define R register
    using namespace std;
    const int MAXN=200005;
    int T,l,r,d[MAXN];
    void solve(){
    	cin>>l>>r;
    	int s=0;
    	s+=2*d[l];
    	long long tl=l+1,tr=r;
    	while(tl<=tr){
    		s+=d[tl];
    		tl++;
    		if(d[tl]!=d[tl-1])break;
    	}
    	while(tl<=tr){
    		s+=d[tr];
    		tr--;
    		if(d[tr]!=d[tr+1])break;
    	}
    	if(tl<=tr){
    		//cout<<tl<<" "<<tr<<"\n";
    		s+=(d[tl]+d[tr])*(tr-tl+1)/2;
    	}
    	cout<<s<<"\n";
    }
    int main(){
    	for(int i=1;i<MAXN;i++)d[i]=d[i/3]+1;
    	//for(int i=1;i<=10;i++)cout<<i<<" "<<d[i]<<"\n";
    	cin>>T;
    	while(T--)solve();
    	return 0;
    }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases where `l` and `r` are close to each other or where they are at the boundary of the allowed range.
2. **Large Ranges**: Test cases with a large range of numbers to ensure the algorithm handles large inputs correctly.
3. **Special Values**: Test cases with values that might cause the algorithm to break, such as values that are powers of 3 or just below powers of 3.

Given the constraints and the nature of the problem, a good test case would be one where the range is large but still within the allowed limits, and where the values of `l` and `r` are such that the operations are not straightforward.

Let's consider the following test input:

```plaintext
1
1 100
```

This test case covers a range from 1 to 100, which is large enough to potentially expose issues in the algorithm. The values are not at the boundary of the allowed range, but they are still significant to test the correctness of the operations.

Here's the reasoning:
- The range from 1 to 100 includes a variety of numbers, including powers of 3 and numbers just below powers of 3.
- The algorithm needs to handle a large number of operations efficiently.
- The output should be the minimum number of operations needed to make all numbers on the board equal to 0.

This test case should help in identifying any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <bit>
using namespace std;
#define hash #
#define remr %
#define cmp ==
#define ll long long int
#define ld long double
#define get1(n) \
    ll n;       \
    cin >> n;
#define get2(a, b) \
    ll a, b;       \
    cin >> a >> b;
#define get3(a, b, c) \
    ll a, b, c;       \
    cin >> a >> b >> c;
#define rep(i, k, n) for (ll i = k; i < n; i++)
#define rn(i, n) for (ll i = 0; i < n; i++)
#define repr(i, k, n, r) for (ll i = k; i < n; i += r)
#define drep(i, k, n) for (ll i = k; i >= n; i--)
#define gets(s) \
    string s;   \
    cin >> s;
#define get(a, n)              \
    for (ll i = 0; i < n; i++) \
    {                          \
        cin >> a[i];           \
    }
#define getv(v, n) \
    vil v(n);      \
    get(v, n);
#define getp(v, n)    \
    rep(i, 0, n)      \
    {                 \
        cinn v[i].ff; \
        cinn v[i].ss; \
    }
#define getpi(v, n)   \
    rep(i, 0, n)      \
    {                 \
        cinn v[i].ff; \
        v[i].ss = i;  \
    }
#define showv(v, n)       \
    for (auto c : v)      \
    {                     \
        cout << c << " "; \
    }                     \
    coutt nl;
#define showp(v, n)                          \
    for (auto c : v)                         \
    {                                        \
        cout << c.ff << " " << c.ss << endl; \
    }                                        \
    coutt nl;
#define YES          \
    cout << "YES\n"; \
    return;
#define NO          \
    cout << "NO\n"; \
    return;
#define cinn cin >>
#define coutt cout <<
#define nl '\n'
#define ln << '\n'
#define tin >>
#define lc <<          // line continue
#define spb << " "     // space bar
#define spbc << " " << // space bar
#define prt(x) coutt x ln;
#define show(ans) \
    coutt ans ln; \
    return;
#define pb push_back
#define ppb pop_back
#define ub upper_bound
#define lb lower_bound
#define sg string
#define qi queue<int>
#define ql queue<ll>
#define qc queue<char>
#define ski stack<int>
#define skl stack<ll>
#define skc stack<char>
#define usi unordered_set<int>
#define usl unordered_set<ll>
#define upsi unordered_map<sg, int>
#define upi unordered_map<int, int>
#define upl unordered_map<ll, ll>
#define upsl map<ll, ll>
#define mpsi map<sg, int>
#define mpi map<int, int>
#define mpl map<ll, ll>
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vp vector<pair<int, int>>
#define vpl vector<pair<ll, ll>>
#define vi vector<int>
#define vil vector<ll>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define vc vector<char>
#define vs vector<string>
#define vb vector<bool>
#define mp(a, b) make_pair(a, b)
#define seti set<int>
#define setl set<ll>
#define bs(k, n) bitset<k>(n)
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define sortit(v) sort(all(v));
#define reverseit(v) reverse(all(v));
#define tos(x) to_string(x)
#define toint(x) stoi(x)
#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
//............................................ SOLUTION STARTS HERE ...................................................//

const int mod = 1e9 + 7;

// An optimised power function.
int power(int x, long long n)
{ // O(log n)
    int ans = 1 % mod;
    while (n > 0)
    {
        if (n & 1)
        {
            ans = 1LL * ans * x % mod;
        }
        x = 1LL * x * x % mod;
        n >>= 1;
    }
    return ans;
}

bool is_int(ld num)
{
    return std::floor(num) == num; // Check if the number is equal to its floor value
}

vil powers_of_3 = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441};

void solve()
{
    get2(l, r);
    ll ans(0), pos, pos1;
    pos = lb(all(powers_of_3), l) - powers_of_3.begin();
    pos1 = lb(all(powers_of_3), r) - powers_of_3.begin();
    if (pos == pos1)
    {
        show(pos * (r - l + 2));
    }
    if (l == 1)
        ans += 3;
    else if (l != powers_of_3[pos])
        ans += (pos) * (powers_of_3[pos] - l + 1);

    // prt(ans);
    rep(i, pos, pos1 - 1)
    {
        ans += (i + 1) * (powers_of_3[i + 1] - powers_of_3[i]);
        // prt(ans);
    }
    if (r != powers_of_3[pos1])
    {
        ans += pos1 * (r - powers_of_3[pos1 - 1] + 1);
    }
    else
    {
        ans += pos1 + 1;
    }

    show(ans);
}
int main()
{
    fast;
    ll t(1);
    cinn 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
1
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
import math
def solve(l, r):
    ans = 0
    done = False
    num = -1
    if r - l + 1 == 2:
        ans += (math.floor(math.log(l, 3)) + 1) * 2
        ans += math.floor(math.log(r, 3)) + 1
        print(ans)
        return
    for i in range(l, r + 1):
        if i % 3 == 2 and not done:
            num = i
            ans += (math.floor(math.log(i, 3)) + 1) * 2
            break
    for i in range(l, r + 1):
        if i == num:
            continue
        ans += math.floor(math.log(i, 3)) + 1
    print(ans)

t = int(input())
for _ in range(t):
    inp = [int(x) for x in input().split()]
    l, r = inp
    solve(l, r)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))

def operations(x):
    res = 0
    while x != 0:
        x //= 3
        res += 1

    return res

def solve():
    l, r = invr()

    numOps = []
    numOpsSum = 0
    minNum = float("inf")

    for x in range(l, r + 1):
        nx = operations(x)
        numOps.append(nx)
        numOpsSum += nx
        minNum = min(minNum, nx)

    print(numOpsSum + minNum)

for _ in range(inp()):
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and the problem statement:

### Problem Analysis:
- The goal is to determine the minimum number of operations needed to make all numbers on the board equal to 0.
- The operation involves picking two numbers \( x \) and \( y \), erasing them, and writing \( 3x \) and \( \lfloor \frac{y}{3} \rfloor \).

### Buggy Code Analysis:
- The function `operations(x)` calculates the number of operations needed to reduce a single number \( x \) to 0.
- The `solve` function iterates over the range from \( l \) to \( r \), calculates the number of operations for each number, and sums them up.
- The final result is the sum of the operations plus the minimum number of operations among the range.

### Potential Bug:
- The code incorrectly adds the minimum number of operations to the total sum, which is not the correct approach. The correct approach should involve a more complex strategy to minimize the total number of operations.

### Challenging Test Input:
To expose the bug, we need a test case where the sum of operations plus the minimum number of operations does not yield the correct result. A good candidate is a range where the numbers have different lengths in terms of the number of operations required to reduce them to 0.

### Example Test Input:
Let's choose a small range where the numbers have different lengths in terms of operations required:
```plaintext
1
1 4
```

### Explanation:
- For \( l = 1 \) and \( r = 4 \):
  - `operations(1) = 1`
  - `operations(2) = 2`
  - `operations(3) = 2`
  - `operations(4) = 3`
- The sum of operations is \( 1 + 2 + 2 + 3 = 8 \).
- The minimum number of operations is \( 1 \).
- The buggy code will output \( 8 + 1 = 9 \), which is incorrect.

The correct minimum number of operations for this range is 6, as shown in the example output.

### Final Test Input:
```plaintext
1
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
import math
t = int(input())
for _ in range(t):
    l, r = list(map(int, input().split()))

    y,x =l, l+1

    dp = [-1]*(r-l+1)
    count = 0
    while y != 0:
        y = math.floor(y/3)
        count += 1
    dp[0] = count
    count = 0
    while x != 0:
        x = math.floor(x/3)
        count += 1
    dp[1] = count + dp[0]

    for x in range(l+2, r+1):
        i = x
        if i-l-1 == 1 and math.floor(x/3**(dp[1] - dp[0])) == 0:
            dp[x-l] = dp[1] -dp[0]
        elif i-l-1 >= 0 and math.floor(x/3**dp[i-l-1]) == 0:
            dp[x-l] = dp[i-l-1]
        else:
            dp[x-l] = 1+dp[i-l-1]
    count = 0
    for i in dp:
        count += i
    print(count)

**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
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
import java.util.*;

public class Codechef {
 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int l = sc.nextInt(), r = sc.nextInt();
   double tans = (Math.log(l) / Math.log(3));
   long ans = 0;
   // long min = Long.MAX_VALUE;
   long min = (long) (Math.log(l) / (double) Math.log(3) + 1);
   for (long i = l; i <= r; i++) {
    tans = (Math.log(i) / (double) Math.log(3));
    // min = (long) Math.min(min, tans + 1);
    //
    long imin = (long) Math.pow(3, (long) (tans + 1)) - 1;
    imin = (long) Math.min(imin, r);
    imin = (long) Math.max(imin, i);
    ans += (long) (tans + 1) * (long) (imin - i + 1);
    i = imin;
    //
    // ans += (long) (tans + 1);
    //
    // System.out.println(tans + " " + ans);
   }
   System.out.println(ans + min);
  }
 }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic of the code might fail. The code uses logarithmic calculations and iterative operations to determine the minimum number of operations. Let's consider the following scenarios:

1. **Small Range with Specific Values**: Test cases where the range is small but the values of `l` and `r` are such that the logarithmic calculations might not work as expected.
2. **Large Range with Specific Values**: Test cases where the range is large, and the values of `l` and `r` are such that the code's logic might break down.
3. **Boundary Conditions**: Test cases where `l` and `r` are at the boundaries of the allowed range.

Let's construct a small-scale test input that can expose the bug:

```plaintext
4
1 2
2 3
1 3
199999 200000
```

### Explanation:
1. **Test Case 1: (1, 2)**
   - This is a small range with `l = 1` and `r = 2`.
   - The expected number of operations is 3:
     - 1, 2 → 3, 0 → 1, 0 → 0, 0 (3 operations).

2. **Test Case 2: (2, 3)**
   - This is a small range with `l = 2` and `r = 3`.
   - The expected number of operations is 4:
     - 2, 3 → 0, 1 → 0, 0 → 0, 0 (4 operations).

3. **Test Case 3: (1, 3)**
   - This is a small range with `l = 1` and `r = 3`.
   - The expected number of operations is 5 (as given in the example).

4. **Test Case 4: (199999, 200000)**
   - This is a large range with `l = 199999` and `r = 200000`.
   - The expected number of operations is 36 (as given in the example).

These test cases cover a variety of scenarios, including small and large ranges, and should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.security.KeyStore.Entry;
import java.util.*;
import java.util.regex.Pattern;
import java.io.*;

public class test {

    static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

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

    static boolean issorted(int[] a) {
        for (int i = 0; i < a.length - 1; i++) {
            if (a[i] > a[i + 1])
                return false;
        }
        return true;
    }

    public static int lowerBound(int[] arr, int n, int x) {
        int low = 0, high = n - 1;
        int ans = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] <= x) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return ans;
    }

    public static int upperBound(int[] arr, int n, int x) {
        int low = 0, high = n - 1;
        int ans = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] >= x) {
                ans = arr[mid];
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    static ArrayList<Integer> list = new ArrayList<>();

    static void init() {
        for (int i = 2; i <= 32; i++) {
            String s = Integer.toBinaryString(i);
            list.add(Integer.parseInt(s));
        }
        Collections.sort(list);
    }

    public static long modularExponentiation(long base, long exp, long mod) {
        long result = 1;
        base = base % mod;
        while (exp > 0) {
            if ((exp & 1) == 1) {
                result = (result * base) % mod;
            }
            exp = exp >> 1;
            base = (base * base) % mod;
        }
        return result;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);

        int[] arr = new int[200001];
        for (int i = 1; i < 200001; i++) {
            int x = (int)Math.floor(Math.log(i) / Math.log(3));
            arr[i] = x+1;
            arr[i] = arr[i-1]+arr[i];
        }

        int tc = sc.nextInt();
        while (tc-- > 0) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            long ans = 0;
            ans+=((arr[l]-arr[l-1])*2L);
            ans+=(arr[r]-arr[l]);
            System.out.println(ans);
        }

        /*
         * Collections.sort(list, (p1, p2) -> {
         * if (p1.v1 != p2.v1) {
         * return Long.compare(p2.v1, p1.v1); // Sort by v1 in descending order
         * } else {
         * return Long.compare(p1.v2, p2.v2); // Sort by v2 in ascending order if v1 are
         * equal
         * }
         * });
         */
    }

    static class Pair implements Comparable<Pair> {
        long v1; // Power of the monster
        long v2; // Health of the monster

        public Pair(long v1, long v2) {
            this.v1 = v1;
            this.v2 = v2;
        }

        @Override
        public int compareTo(Pair other) {
            if (this.v1 == other.v1) {
                return Long.compare(this.v2, other.v2);
            }
            return Long.compare(this.v1, other.v1);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause incorrect results. The main issue in the code seems to be in the precomputation of the `arr` array and the calculation of the number of operations. Let's analyze the code and identify potential edge cases:

1. **Initialization of `arr` Array**:
   - The code initializes the `arr` array using the formula `arr[i] = (int)Math.floor(Math.log(i) / Math.log(3)) + 1;` and then updates it with `arr[i] = arr[i-1] + arr[i];`.
   - This precomputation might be incorrect, especially for small values of `i`.

2. **Calculation of Operations**:
   - The code calculates the number of operations using `ans += ((arr[l] - arr[l-1]) * 2L);` and `ans += (arr[r] - arr[l]);`.
   - This might not correctly handle the operations needed to reduce all numbers to zero.

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

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

### Explanation
1. **Test Case 1: \( l = 1, r = 2 \)**
   - The numbers on the board are 1 and 2.
   - The correct minimum number of operations to make all numbers zero is 3.
   - The sequence of operations can be:
     - \( 1, 2 \xrightarrow[x=1,\,y=2]{} 3, 0 \)
     - \( 3, 0 \xrightarrow[x=0,\,y=3]{} 1, 0 \)
     - \( 1, 0 \xrightarrow[x=0,\,y=1]{} 0, 0 \)

2. **Test Case 2: \( l = 1, r = 3 \)**
   - The numbers on the board are 1, 2, and 3.
   - The correct minimum number of operations to make all numbers zero is 5.
   - The sequence of operations can be:
     - \( 1, 2, 3 \xrightarrow[x=1,\,y=2]{} 3, 0, 3 \)
     - \( 3, 0, 3 \xrightarrow[x=0,\,y=3]{} 1, 0, 3 \)
     - \( 1, 0, 3 \xrightarrow[x=0,\,y=3]{} 1, 0, 1 \)
     - \( 1, 0, 1 \xrightarrow[x=0,\,y=1]{} 0, 0, 1 \)
     - \( 0, 0, 1 \xrightarrow[x=0,\,y=1]{} 0, 0, 0 \)

3. **Test Case 3: \( l = 2, r = 4 \)**
   - The numbers on the board are 2, 3, and 4.
   - The correct minimum number of operations to make all numbers zero is 6.
   - The sequence of operations can be:
     - \( 2, 3, 4 \xrightarrow[x=2,\,y=3]{} 6, 1, 4 \)
     - \( 6, 1, 4 \xrightarrow[x=1,\,y=4]{} 6, 1, 1 \)
     - \( 6, 1, 1 \xrightarrow[x=1,\,y=1]{} 6, 0, 1 \)
     - \( 6, 0, 1 \xrightarrow[x=0,\,y=1]{} 6, 0, 0 \)
     - \( 6, 0, 0 \xrightarrow[x=0,\,y=6]{} 2, 0, 0 \)
     - \( 2, 0, 0 \xrightarrow[x=0,\,y=2]{} 0, 0, 0 \)

This test input should expose the bug in the precomputation and the calculation of the number of operations in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ln '\n'
#define ld long double
using namespace std;
const int M = 2e5 + 10;
const ll mod = 1e9 + 7;
int main() {
	int t;
	cin >>t;
	while(t--){
		ll l,r,sum=0;
		cin >>l>>r;
		ll x=l;
		while(x){
			sum++;
			x/=3;

		}
		for (int i=l+1;i<=r;i++){
			x=i;
			if (i==l+1){
				x=(ll)i*pow(3,sum);
			}
			while(x){
				sum++;
				x/=3;
				//cout <<x<<"**"<<ln;
			}
		}
		cout <<sum<<ln;
	}
	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
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triple Operations
time_limit_ms: 1000
memory_limit_mb: 256
Description: On the board Ivy wrote down all integers from $l$ to $r$, inclusive.

In an operation, she does the following:

  * pick two numbers $x$ and $y$ on the board, erase them, and in their place write the numbers $3x$ and $\lfloor \frac{y}{3} \rfloor$. (Here $\lfloor \bullet \rfloor$ denotes rounding down to the nearest integer).

What is the minimum number of operations Ivy needs to make all numbers on the board equal $0$? We have a proof that this is always possible.

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

The only line of each test case contains two integers $l$ and $r$ ($1 \leq l < r \leq 2 \cdot 10^5$).

Output: For each test case, output a single integer — the minimum number of operations needed to make all numbers on the board equal $0$.

Examples:
input:
4
1 3
2 4
199999 200000
19 84
output:
5
6
36
263

Note: In the first test case, we can perform $5$ operations as follows: $$ 1,2,3 \xrightarrow[x=1,\,y=2]{} 3,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,3 \xrightarrow[x=0,\,y=3]{} 1,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,1 \xrightarrow[x=0,\,y=1]{} 0,0,0 .$$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N =2e5 + 5;
//int mp[N];
void solve(){
	int l,r;
	cin >> l >> r;
	int ans = 0;
	int k = 1;
	int cnt = 1;
	while(k < l){
		k *= 3;
		cnt ++;
	}
//	cout << cnt << '\n';
	int nxt;
	if(k == l) nxt = k * 3;
	else cnt --,nxt = k;
//	cout << ans << '\n';

 	ans = cnt;
	ans *= 2;
	for(int i = l + 1;i <= r;i ++){
		if(i < nxt) ans += cnt;
		else{
			cnt ++;
			ans += cnt;
			nxt *= 3;
		}
	}
	cout << ans << '\n';
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);

//	int now = 1;
//	int cnt = 1;
//	while(now < N){
//		mp[now] = cnt;
//		cnt ++;
//		now *= 3;
//	}
//	for(int i = 2;i < N;i ++){
//		if(!mp[i]) mp[i] = mp[i - 1];
//	}

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios where the code's logic might fail. The code's primary logic is to determine the number of operations needed to reduce all numbers to zero by repeatedly applying the operations \(3x\) and \(\lfloor \frac{y}{3} \rfloor\).

### Key Points to Consider:
1. **Edge Cases:**
   - When \(l\) and \(r\) are very close to each other.
   - When \(l\) and \(r\) are both powers of 3.
   - When \(l\) is just below a power of 3 and \(r\) is just above a power of 3.
   - When \(l\) and \(r\) are large but close to each other.

2. **Scenarios:**
   - When the range \([l, r]\) includes numbers that are very close to a power of 3.
   - When the range \([l, r]\) includes numbers that are very far from a power of 3.

### Challenging Test Input:
Let's choose a test case where \(l\) is just below a power of 3 and \(r\) is just above a power of 3. This will test the code's ability to handle the transition between different powers of 3 correctly.

```plaintext
1
8 10
```

### Explanation:
- \(l = 8\) and \(r = 10\).
- The number 9 is a power of 3 (\(3^2 = 9\)).
- This range includes numbers just below and just above 9, which should test the code's handling of the transition between different powers of 3.

This input is small and complete, and it should expose any issues in the code's logic for handling transitions between powers of 3.