**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
import java.io.*;
import java.util.*;
public class D {
	public static PrintWriter out;
	static ArrayList<ArrayList<Integer>>al;
	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 a[]=new int[n];
        	al=new ArrayList<>();
        	long sum=0;
        	for(int i=0;i<n;i++) {
        		al.add(new ArrayList<>());
        		a[i]=sc.nextInt();
        		sum+=a[i];
        	}
        	int p[]=new int[n];
        	for(int i=0;i<n;i++) {
        		p[i]=sc.nextInt()-1;
        		al.get(i).add(p[i]);
        	}
        	int min[]=new int[n];//max froms based on a_i
        	Arrays.fill(min, Integer.MAX_VALUE);
        	for(int i=0;i<n;i++) {
        		for(int end:al.get(i)) {
        			if(end>i) {
        				min[end]=Math.min(min[end], a[i]);
        			}
        		}
        	}
        	//find ending places
        	TreeSet<Integer>ts=new TreeSet<>();
        	for(int i=0;i<n;i++) {
        		ts.add(i);
        	}
        	FenwickTree ft=new FenwickTree(n);
        	ft.build(a);
        	long ans=0;
        	ArrayDeque<Integer>q=new ArrayDeque<>();
        	q.add(0);
        	long total=0;
        	while(!q.isEmpty()) {
        		int cur=q.poll();
        		//stop here
        		ans=Math.max(ans, total+ft.prefixSum(cur));//stopping here and going backwards only
        		ft.update(cur, -a[cur]);
        		ts.remove(cur);
        		Integer next=ts.floor(p[cur]);//lets try to go forwards
        		if(next==null||next<cur||(next!=null&&next>cur&&min[next]!=a[cur])) {//we cannot go forwards
        			next=ts.lower(cur);
        			if(next==null) {
        				break;
        			}else {
        				total+=a[cur];
        				q.add(next);
        			}
        		}else {
        			q.add(next);
        		}
        	}
        	out.println(ans);
        }
		out.close();
	}
	/*
1
5
50 200 100 4 10000
4 5 5 1 1

1
6
50 100 300 200 4 10000
5 6 6 6 1 1
	 */
	/*
	prefix sums:
	if i want from l to r (1 indexed) inclusive do pre[r], pre[l-1]
	*/
	/*fenwickTree
	if i want from l to r (0 indexed) inclusive do r+1, l //use this one, as it will always start from 0
	if i want from l to r (1 indexed) inclusive do r, l-1
	 */
	static class FenwickTree{
		long arr[];
		public FenwickTree(int size) {
			arr=new long[size+1];
		}
		//0 indexed
		public void update(int index, int amt) {
			index++;
			while(index<arr.length) {
				arr[index]+=amt;
				index+=(index&-index);
			}
		}
		public void build(int a[]) {
			for(int i=0;i<a.length;i++) {
				update(i, a[i]);
			}
		}
		public void fastBuild(int a[]) {
			for(int i=1;i<=a.length;i++) {
				arr[i]+=a[i-1];
				int index=i+(i&-i);//next index that it contributes to
				if(index<=a.length) {
					arr[index]+=arr[i];
				}
			}
		}
		//inclusive from 0 to index
		public long prefixSum(int index){
			long res=0;
			index+=1;
			while(index!=0) {
				res+=arr[index];
				index-=(index&-index);
			}
			return res;
		}
		//0 to n, normal prefix sums
		public long prefixRange(int index1, int index2) {
			index1--;
			index2--;
			return prefixSum(index2)-prefixSum(index1);
		}
	}
    public static long gcd(long a, long b) {
        if(a > b)
            a = (a+b)-(b=a);
        if(a == 0L)
            return b;
        return gcd(b%a, a);
    }
	static void add(HashMap<Character,Integer> A, char key){
		A.put(key,A.getOrDefault(key,0)+1);
	}
	static void remove(HashMap<Character,Integer>A, char 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; int y;
		public Pair(int x, int y) {
			this.x=x;this.y=y;
		}
		public int compareTo(Pair o) {
			if(Math.min(x, y)==Math.min(o.x, o.y)) {
				return Integer.compare(Math.max(x, y), Math.max(o.x, o.y));
			}
			return Integer.compare(Math.min(x, y), Math.min(o.x, o.y));
		}
	}
	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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight issues in the logic. Here are the steps to think through this:

1. **Understand the Bug**: The code has issues with how it handles the transitions between problems, especially when deciding whether to go forward or backward based on the `b_i` parameter.

2. **Edge Cases**:
   - **Single Problem**: Test cases with a single problem to ensure the code handles it correctly.
   - **Multiple Problems with Specific `b_i` Values**: Test cases where the `b_i` values are set to create complex transitions.
   - **High Scores and Low Scores**: Test cases with high and low scores to ensure the code correctly maximizes the points.
   - **All Problems Skippable**: Test cases where all problems can be skipped to ensure the code handles the end of the competition correctly.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of problems (e.g., 5) to ensure it is manageable and valid.

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

```plaintext
1
5
50 200 100 4 10000
4 5 5 1 1
```

Explanation:
- **Test Case 1**:
  - **Number of Problems (n)**: 5
  - **Scores (a_i)**: [50, 200, 100, 4, 10000]
  - **Parameters (b_i)**: [4, 5, 5, 1, 1]

In this test case:
- If Prokhor skips the first problem, he will receive the fourth problem (since `b_1 = 4`).
- If Prokhor submits the fourth problem, he will receive the fifth problem (since `b_4 = 1`).
- If Prokhor submits the fifth problem, he will receive the second problem (since `b_5 = 1`).
- If Prokhor submits the second problem, the competition will end.

The optimal strategy is to skip the first problem, submit the fourth problem, submit the fifth problem, and submit the second problem, resulting in a total of 10000 + 200 = 10200 points.

This test case should help expose any issues in the logic for handling the transitions between problems and maximizing the points.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
import java.util.*;

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

        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
            }
            for(int i=0;i<n;i++){
                b[i]=sc.nextInt()-1;
            }

            long ans = helper2(a, b, n, null, 0, 0);
            System.out.println(ans);

            // System.out.println(helper(a, b, n));
        }
    }
    public static long helper(int[] a , int[] b , int n){

        if(b[0]==0){
            return a[0];
        }
        long ans = a[0];

        long curr = 0L;
        int currMax = -1;
        long currMaxValue = 0;

        TreeSet<Integer> treeSet = new TreeSet<>();
        for(int i=0;i<n;i++){
            treeSet.add(i);
        }

        int start=0;

        while(treeSet.size() >0 && start>=treeSet.first()){
            Iterator<Integer> iterator = treeSet.headSet(start, true).iterator();

            while (iterator.hasNext()) {
                Integer value = iterator.next();

                curr+=a[value];
                if(b[value] > currMax){
                    currMax=b[value];
                    currMaxValue=a[value];
                }
                iterator.remove();
            }

            ans = Math.max(ans , curr);
            start=currMax;
            curr-=currMaxValue;
        }
        return ans;
    }

    public static long helper2(int[] a , int[] b , int n , long[] prefixSum , int start , int end){

        long total = 0;

        for(int i=start;i<=end;i++){
            total+=a[i];
        }

        long ans = total;

        for(int i=start ; i<=end; i++){
            if(b[i]<=end) continue;

            long currAns = total - a[i] + helper2(a, b, n, prefixSum, end+1, b[i]);
            ans = Math.max(ans,currAns);
        }
        return ans;
    }

}

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

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

Think step by step.

Response:
```plaintext
1
3
100 1 1000
2 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

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

#ifndef ONLINE_JUDGE
#include "template.cpp"
#else
#define debug(...)
#define debugArr(...)
#endif

vector<pair<ll, ll>> a;
vector<ll> dp;
ll dfs(ll ind, vector<bool> &ok)
{
    // debug(ind, ok);

    if (ok[ind])
    {
        if (ind == 0)
        {
            return 0;
        }
        return dfs(ind - 1, ok);
    }
    if (dp[ind] != -1)
    {
        debug(dp[ind]);
        return dp[ind];
    }
    ok[ind] = true;
    ll ans = max(a[ind].first + dfs(ind - 1, ok), dfs(a[ind].second, ok));
    ok[ind] = false;
    return dp[ind] = ans;
}

void solve()
{
    ll n;
    cin >> n;
    a = vector<pair<ll, ll>>(n);
    dp = vector<ll>(n, -1);
    vector<bool> ok(n);
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i].first;
    }
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i].second;
        a[i].second--;
    }
    ok[0] = true;
    cout << max(a[0].first, dfs(a[0].second, ok));
}
signed main()
{
    // cout << setprecision(10) << fixed;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll tt = 1;
    cin >> tt;
    for (ll i = 1; i <= tt; i++)
    {
        solve();
        cout << "\n";
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
#define endl '\n'
using namespace std;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<vector<int> > VVI;
const int N = 4e5 + 10;
const int M = 17;
const int inf = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

struct in {

	vector<int> vi (int n) {
		vector<int> v(n + 1, 0);
		for (int i = 1; i <= n; i ++) cin >> v[i];
		return v;
	}

	vector<pair<int,int> > vpii1 (int n) {
		vector<pair<int,int> > v(n + 1);
		for (int i = 1; i <= n; i ++) cin >> v[i].fi;
		for (int i = 1; i <= n; i ++) cin >> v[i].se;
		return v;
	}

	vector<pair<int,int> > vpii2 (int n) {
		vector<pair<int,int> > v(n + 1);
		for (int i = 1; i <= n; i ++) cin >> v[i].fi >> v[i].se;
		return v;
	}

};

struct print {
	void vi (vector<int> & v) {
		for (int i = 0; i < v.size(); i ++) cout << v[i] << ' ';
		cout << endl;
	}

	void vpii (vector<pair<int,int> > & v) {
		for (int i = 0; i < v.size(); i ++) cout << v[i].fi << ' ' << v[i].se << endl;
	}

	void vvi (vector<vector<int> > & v) {
		for (int i = 0; i < v.size(); i ++) {
			for (int j = 0; j < v[i].size(); j ++) {
				cout << v[i][j] << ' ';
			}cout << endl;
		}
	}
};

struct math {
	vector<int> fact, infact;
	int p;

	int qmi(int a, int b) {
		int res = 1;
		while (b) {
			if (b & 1) res = res * a % p;
			a = a * a % p;
			b >>= 1;
		}
		return res;
	}

	void init(int n, int mod) {
		fact = vector<int> (n + 1);
		infact = vector<int> (n + 1);
		p = mod;
		fact[0] = 1;
		for (int i = 1; i <= n; i ++) {
			fact[i] = fact[i - 1] * i % p;
		}
		infact[n] = qmi(fact[n], p - 2);
		for (int i = n - 1; i >= 0; i --) {
			infact[i] = infact[i + 1] * (i + 1) % p;
		}
	}

	int C(int a, int b) {
		return fact[a] * infact[b] % p * infact[a - b] % p;
	}

	void print() {
		for (int i = 0; i < fact.size(); i ++) {
			cout << fact[i] << ' ';
		}cout << endl;
		for (int i = 0; i < infact.size(); i ++) {
			cout << infact[i] << ' ';
		}cout << endl;
	}
};

//math math;
in in;
//print print;
int a[N], b[N], n, ans;
int s[N], la[N], ne[N], maxb[N];

void dfs(int x, int y, int h) {
	if (x == 0) { return; }
	la[ne[x]] = la[x];
	ne[la[x]] = ne[x];
	ans = max(ans, s[x] - y);
	if (b[x] > h) {
		dfs(b[x], y + a[x], b[x]);
	}
	if (maxb[la[x]] > h) {
		dfs(la[x], y, h);
	}
	la[ne[x]] = x;
	ne[la[x]] = x;
}

void init(int n) {
	ans = 0;
	for (int i = 1; i <= n; i ++) cin >> a[i];
	for (int i = 1; i <= n; i ++) cin >> b[i];
	for (int i = 1; i <= n; i ++) {
		s[i] = s[i - 1] + a[i];
		la[i] = i - 1;
		ne[i] = i + 1;
		maxb[i] = max(maxb[i - 1], b[i]);
	}
}

void solve() {
	cin >> n;
	init(n);
	dfs(1, 0, 1);
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cout << fixed << setprecision(20);
	int T = 1;
	cin >> T;
//	math.init(N, mod);
	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
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
/**
 *  author: xing huan liu kong
 *  created: 2024-10-20 18:27:01
 *
 *  即使这个图论(world)大的无边无际
 *  我也会深度优先搜索(dfs)找到你..
**/
#include <bits/stdc++.h>
#define ll long long
#define endl '\n'
#define point(x) fixed<<setprecision(x)
using namespace std;

#define int long long
typedef pair<int, int> PII;
typedef array<int, 3> arr;

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

    vector<int> a(n), b(n);
    for (auto &i : a) {
        cin >> i;
    }
    for (auto &i : b) {
        cin >> i, i -- ;
    }

    // for (int i = 0; i < n; i ++ ) {
    //     cerr << b[i] << " ";
    // }

    vector<int> vis(n, -1);
    priority_queue<PII, vector<PII>, greater<>> q;

    q.push({a[0], 0});
    int end = -1;

    while (!q.empty()) {
        auto [s, idx] = q.top();
        q.pop();

        int sum = s - a[idx], ne = b[idx];
        if (idx > end)
        vis[idx] = sum;

        // cerr << end << endl;
        // cerr << sum << endl;
        // cerr << idx << endl;

        for (int i = end + 1; i < idx; i ++ ) {
            vis[i] = sum;
            q.push({sum + a[i], i});
        }
        end = max(end, idx);

        // for (int i = 0; i < n; i ++ ) {
        //     cerr << vis[i] << " ";
        // }
        // cerr << endl;

        if (ne > end) {
            q.push({sum + a[idx] + a[ne], ne});
        }
    }

    int res = a[0];
    for (int i = 1; i < n; i ++ ) {
        a[i] += a[i - 1];
        if (vis[i] != -1)
        res = max(res, a[i] - vis[i]);
    }

    cout << res << endl;
}

signed main(void)
{
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    int t = 1;
    cin >> t;

    while(t -- ){
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
100 50 10
1 2 3
4
100 50 100 10
1 2 3 4
5
100 50 100 10 100
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
def max_score(a, b, visited, index, cur_score):
    if  index < 0 or index >= len(a) or visited[index]: return cur_score

    submission_score = cur_score + a[index]
    visited[index] = True
    for i in range(index - 1,-1,-1):
        if visited[i] == False:
            submission_score = max_score(a, b, visited,i, submission_score)
            break
    skip_score = max_score(a, b, visited, b[index] - 1, cur_score)

    return max(submission_score, skip_score)

T = int(input())
for t in range(T):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    submission_tracker = [False for i in range(n)]
    print(max_score(a, b, submission_tracker, 0, 0))

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

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

Think step by step.

Response:
```plaintext
3
3
100 50 1
1 2 3
4
100 200 300 1000
2 3 4 1
5
10 10 10 10 1000
3 3 3 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
def fin(mp,j):
    for i in range(j,-1,-1):
        if mp[i] == 0:
            return i
    return -1

def dfs(mp, i, val):
    j = fin(mp, i)
    global ans, brr, arr
    if j == -1:
        ans = max(ans, val)
        return
    else:
        mp[j] = -1
        dfs(mp,j,val+arr[j])
        dfs(mp,brr[j]-1,val)

for _ in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))
    brr = list(map(int, input().split()))
    ans = 0
    dfs([0]*n,0,0)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
2
100 100
2 1
3
100 1 100
3 2 1
4
100 100 100 100
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define For(i, a, b) for (int i = (a); i <= (b); i ++)
#define foR(i, a, b) for (int i = (a); i >= (b); i --)
using namespace std;
int n, x, y;
int a[400005], b[400005], mx[1600005];
int f[400005];//f[i]：到位置 i 的最小损失
void update (int l, int r, int k) {
	if (l == r) {
		mx[k] = min (mx[k], y);
		return;
	}
	int mid = l + r >> 1;
	if (x <= mid) update (l, mid, k << 1);
	else update (mid + 1, r, k << 1 | 1);
	mx[k] = min (mx[k << 1], mx[k << 1 | 1]);
}
int query (int l, int r, int k) {
	if (x <= l && y>= r) return mx[k];
	int mid = l +r >> 1, res = 100000000000000000LL;
	if (x <= mid) res = query (l, mid, k << 1);
	if (y > mid) res = min (res, query (mid + 1, r, k << 1 | 1) );
	return res;
}
void solve () {
	cin >> n;
	For (i, 1, n) {
		cin >> a[i];
		a[i] += a[i - 1];
	}
	For (i, 1, n) cin >> b[i];
	if (b[1] == 1) {
		cout << a[1];
		return;
	}
	For (i, 1, n) f[i] = 100000000000000000LL;
	For (i, 1, 4 * n) mx[i] = 100000000000000000LL;
	f[1] = 0;
	x = b[1];
	y = a[1];
	update (1, n, 1);
	f[b[1] ] = a[1];
	For (i, 2, n) {
		if (i > b[i] - 1) continue;
		x = i, y = b[i] - 1;
		f[b[i] ] = min (f[b[i] ], query (1, n, 1) + a[i] - a[i - 1]);
		x = b[i], y = f[b[i] ];
		update (1, n, 1);
	}
	int ans = 0;
	For (i, 1, n) ans = max (ans, a[i] - f[i]);
	cout << ans << '\n';
}
signed main () {
	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
3
2
100 101
2 1
3
100 200 300
3 2 2
4
100 200 300 400
2 3 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int>PII;
const int N=1e6+7;
int e[N],ne[N],idx,h[N],c[N],a[N];//邻接表
int n,x,dist[N],f[N];
void add(int x,int y,int z)
{
	c[idx]=z;
	e[idx]=y;
	ne[idx]=h[x];
	h[x]=idx++;
}
void dijkstra()
{
	memset(dist,0x3f,sizeof dist);
	memset(f,0,sizeof f);
	priority_queue<PII,vector<PII>,greater<PII>>p;
	dist[1]=0;
	p.push({0,1});
	while(p.size())
	{
		PII t=p.top();p.pop();
		int x=t.first,y=t.second;
		if(f[y])
			continue;
		f[y]=1;
		for(int i=h[y];i!=-1;i=ne[i])
			if(f[e[i]]==0&&dist[e[i]]>x+c[i])
			{
				dist[e[i]]=x+c[i];
				p.push({x+c[i],e[i]});
			}
	}
}
void solve()
{
	memset(h,-1,sizeof h);
	idx=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		if(i>1)
			add(i,i-1,0);
	}
	for(int i=1;i<=n;i++)
	{
		cin>>x;
		add(i,x,a[i]);
	}
	dijkstra();
	int s=0,mx=0;
	for(int i=1;i<=n;i++)
	{
		s+=a[i];
		mx=max(mx,s-dist[i]);
	}
	cout<<mx<<endl;
	return ;
}
signed main()
{
    int t;
    cin>>t;
    while(t--)
    	solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases that might cause the algorithm to fail. The code uses a Dijkstra-like approach to find the maximum points, but it might have issues with the way it handles problem skipping and the selection of the next problem.

### Key Points to Consider:
1. **Problem Skipping and Next Problem Selection:**
   - The code should correctly handle the scenario where skipping a problem affects the next problem selected.
2. **Edge Cases:**
   - Small values of `n` to ensure the test case is small.
   - Cases where skipping the first problem leads to a different sequence of problems being selected.
   - Cases where the maximum points can be achieved by a specific sequence of submissions and skips.

### Test Input:
Let's create a test case where the optimal strategy is to skip the first problem and then submit the next available problem. This should highlight any issues in the way the code handles problem selection after skipping.

```plaintext
1
4
10 100 10 1000
4 3 2 1
```

### Explanation:
- **Test Case:**
  - `n = 4`
  - `a = [10, 100, 10, 1000]`
  - `b = [4, 3, 2, 1]`

- **Optimal Strategy:**
  - Skip the first problem (index 1). This means the next problem will be the one with index `b[1] = 4`.
  - Submit the fourth problem (index 4) to get 1000 points.
  - The competition ends because there are no more problems to select.

- **Expected Output:**
  - The maximum number of points Prokhor can achieve is `1000`.

This test case should expose any issues in the code's handling of problem skipping and the selection of the next problem.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

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

public class Ez {

  static IOHandler sc = new IOHandler();
  static PrintWriter out = new PrintWriter(System.out);

  public static void main(String[] args) throws IOException {
    new Thread(null, new Solution(), "solution", 1 << 26).start();
  }

  private static class Solution implements Runnable {

    public void run() {
      int testCases = sc.nextInt();
//      int testCases = 1;
      for (int i = 1; i <= testCases; ++i) {
        solve();
      }
      out.flush();
    }

    private void solve() {
      int n = sc.nextInt();
      int[] a = sc.readArray(n);
      int[] b = sc.readArray(n);
      int[] dp = new int[n];
      Arrays.fill(dp, Integer.MAX_VALUE);
      dp[0] = 0;
      int j;
      for (int i = 0; i < n; i++) {
        j = b[i];
        for (int k = i; k < j; k++) {
          dp[k] = Math.min(dp[k], dp[i] + a[i]);
        }
      }
      long ans = 0, sum = 0;
      for (int i = 0; i < n; i++) {
        sum += a[i];
        ans = Math.max(ans, sum - dp[i]);
      }
      System.out.println(ans);
    }
  }

  private static class IOHandler {
    BufferedReader br;
    StringTokenizer st;

    public IOHandler() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

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

    int[] readArray(int n) {
      int[] res = new int[n];

      for (int i = 0; i < n; ++i)
        res[i] = nextInt();

      return res;
    }

    int[] readArray2(int n) {
      int[] res = new int[n + 1];

      for (int i = 1; i <= n; ++i)
        res[i] = nextInt();

      return res;
    }

    int[][] readGraph(int n, int m, int c) {
      int[][] graph = new int[n + 1][];

      int[] count = new int[n + 1];
      int[][] arr = new int[m][1 + c];

      int a, b;

      for (int i = 0; i < m; ++i) {
        a = sc.nextInt();
        b = sc.nextInt();

        arr[i][0] = a;
        arr[i][1] = b;

        ++count[a];
        ++count[b];

        for (int j = 2; j <= c; ++j) {
          arr[i][j] = sc.nextInt();
        }
      }

      for (int i = 0; i <= n; ++i) {
        graph[i] = new int[count[i] * c];
        count[i] = 0;
      }
      int swap;

      for (int[] tArr : arr) {
        for (int j = 1; j < tArr.length; ++j) {
          graph[tArr[0]][count[tArr[0]]++] = tArr[j];
        }

        swap = tArr[0];
        tArr[0] = tArr[1];
        tArr[1] = swap;

        for (int j = 1; j < tArr.length; ++j) {
          graph[tArr[0]][count[tArr[0]]++] = tArr[j];
        }
      }

      return graph;
    }

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

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

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }

  }

}

class Pair {
  int a;
  int b;

  public Pair(int a, int b) {
    this.a = a;
    this.b = b;
  }

  long sum() {
    return a + b;
  }
}

**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
100 200 300 1000
2 3 4 1
```

This test case is designed to expose the issue in the buggy code. Here's the reasoning:

1. **Problem Structure**:
   - The first problem (index 1) has a score of 100 and a parameter of 2.
   - The second problem (index 2) has a score of 200 and a parameter of 3.
   - The third problem (index 3) has a score of 300 and a parameter of 4.
   - The fourth problem (index 4) has a score of 1000 and a parameter of 1.

2. **Optimal Strategy**:
   - If Prokhor skips the first problem, he will receive the second problem (index 2).
   - He can submit the second problem to get 200 points.
   - After submitting the second problem, he will receive the third problem (index 3).
   - He can submit the third problem to get 300 points.
   - After submitting the third problem, he will receive the fourth problem (index 4).
   - He can submit the fourth problem to get 1000 points.

3. **Expected Output**:
   - The total points Prokhor can achieve is 200 + 300 + 1000 = 1500.

4. **Buggy Code Behavior**:
   - The buggy code uses a dynamic programming approach that incorrectly updates the `dp` array.
   - The loop `for (int k = i; k < j; k++)` is not correctly handling the problem parameters and indices.
   - This can lead to incorrect updates in the `dp` array, resulting in a suboptimal solution.

By using this test case, we can observe if the buggy code correctly handles the problem parameters and indices, and whether it correctly calculates the maximum points Prokhor can achieve.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

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

public class Ez {

  static IOHandler sc = new IOHandler();
  static PrintWriter out = new PrintWriter(System.out);

  public static void main(String[] args) throws IOException {
    new Thread(null, new Solution(), "solution", 1 << 26).start();
  }

  private static class Solution implements Runnable {

    public void run() {
      int testCases = sc.nextInt();
//      int testCases = 1;
      for (int i = 1; i <= testCases; ++i) {
        solve();
      }
      out.flush();
    }

    private void solve() {
      int n = sc.nextInt();
      int[] a = sc.readArray(n);
      int[] b = sc.readArray(n);
      long[] dp = new long[n];
      Arrays.fill(dp, Long.MAX_VALUE);
      dp[0] = 0;
      int j;
      for (int i = 0; i < n; i++) {
        j = b[i];
        for (int k = i; k < j; k++) {
          dp[k] = Math.min(dp[k], dp[i] + a[i]);
        }
      }
      long ans = 0, sum = 0;
      for (int i = 0; i < n; i++) {
        if(dp[i]==Long.MAX_VALUE){
          break;
        }
        sum += a[i];
        ans = Math.max(ans, sum - dp[i]);
      }
      System.out.println(ans);
    }
  }

  private static class IOHandler {
    BufferedReader br;
    StringTokenizer st;

    public IOHandler() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

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

    int[] readArray(int n) {
      int[] res = new int[n];

      for (int i = 0; i < n; ++i)
        res[i] = nextInt();

      return res;
    }

    int[] readArray2(int n) {
      int[] res = new int[n + 1];

      for (int i = 1; i <= n; ++i)
        res[i] = nextInt();

      return res;
    }

    int[][] readGraph(int n, int m, int c) {
      int[][] graph = new int[n + 1][];

      int[] count = new int[n + 1];
      int[][] arr = new int[m][1 + c];

      int a, b;

      for (int i = 0; i < m; ++i) {
        a = sc.nextInt();
        b = sc.nextInt();

        arr[i][0] = a;
        arr[i][1] = b;

        ++count[a];
        ++count[b];

        for (int j = 2; j <= c; ++j) {
          arr[i][j] = sc.nextInt();
        }
      }

      for (int i = 0; i <= n; ++i) {
        graph[i] = new int[count[i] * c];
        count[i] = 0;
      }
      int swap;

      for (int[] tArr : arr) {
        for (int j = 1; j < tArr.length; ++j) {
          graph[tArr[0]][count[tArr[0]]++] = tArr[j];
        }

        swap = tArr[0];
        tArr[0] = tArr[1];
        tArr[1] = swap;

        for (int j = 1; j < tArr.length; ++j) {
          graph[tArr[0]][count[tArr[0]]++] = tArr[j];
        }
      }

      return graph;
    }

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

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

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }

  }

}

class Pair {
  int a;
  int b;

  public Pair(int a, int b) {
    this.a = a;
    this.b = b;
  }

  long sum() {
    return a + b;
  }
}

**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
4
2
1 2
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
long long bfs[800005][2],sum[400005],a[400005],b[400005],book[800005],mx[400005];
int main()
{
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)book[i]=1e18;
        for(int i=1;i<=n;i++){
            cin>>a[i];
            sum[i]=sum[i-1]+a[i];
        }
        for(int i=1;i<=n;i++){
            cin>>b[i];
            if(b[mx[i-1]]<b[i]||a[mx[i-1]]>a[i])
                mx[i]=i;
            else
                mx[i]=mx[i-1];
        }
        int head=0,tail=0;
        book[1]=0;
        bfs[tail++][0]=1;
        while(tail>head){
            long long int x=bfs[head][0],w=bfs[head][1];
            if(book[b[x]]>w+a[x]&&b[x]>x){
                book[b[x]]=w+a[x];
                if(mx[x]==x){
                    bfs[tail][1]=w+a[x];
                    bfs[tail++][0]=b[x];
                }
            }
            if(x-1>0&&book[x-1]>w){
                book[x-1]=w;
                bfs[tail][1]=w;
                bfs[tail++][0]=x-1;
            }
            head++;
        }
        long long int ans=a[1];
        for(int i=1;i<=n;i++)
            ans=max(sum[i]-book[i],ans);

        cout<<ans<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
10 20 30 40 50
2 3 4 5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
import sys
def input()   : return sys.stdin.readline().strip()
def getints() : return map(int,sys.stdin.readline().strip().split())

from heapq import heappush, heappop
from collections import deque

for _ in range(int(input())):
    n = int(input())
    a = list(getints())
    b = list(getints())

    h = []
    cost = [float('inf')]*n
    cost[0] = 0

    for j in range(1,b[0]):
        cost[j] = a[0]
        if b[j] > b[0]: heappush(h,(a[j],b[j],cost[j]))
    j = b[0]

    while h:
        x = heappop(h)
        while j < x[1]:
            cost[j] = x[2]+x[0]
            if b[j] > x[1]: heappush(h,(a[j],b[j],cost[j]))
            j += 1

    s = list(a)
    for i in range(1,n): s[i] += s[i-1]

    ans = 0
    for i in range(n): ans = max(ans,s[i]-cost[i])
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
3
100 50 1
3 2 2
4
100 200 300 1000
2 3 4 1
5
10 20 30 40 50
3 4 5 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

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

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

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

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

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
import java.util.PriorityQueue;
import java.util.Scanner;

public class D {
    static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
    }
    public static void solve() {

        int n = in.nextInt();
        int[] a = readArray(n);
        int[] b = readArray(n);

        long[] sum = new long[n+1];
        int currmax = 0;

        for(int i=1; i<=n; i++) {
            currmax = Math.max(currmax , b[i]);
            sum[i] = a[i]+sum[i-1];
            if (currmax == i) break;
        }

        long ans = a[1];
        boolean[] visited = new boolean[currmax+1];
        visited[0] = true;
        visited[1] = true;
        long[] dist = new long[currmax+1];

        PriorityQueue<int[]> pq = new PriorityQueue<>((x,y) -> x[2]-y[2]);

        pq.add(new int[]{1 , b[1] , a[1]});
        while (!pq.isEmpty()) {
            int[] arr = pq.poll();
            int t = arr[1];
            if (visited[t]) continue;
            visited[t] = true;
            dist[t] = dist[arr[0]] + arr[2];
            ans = Math.max(ans , sum[t]-dist[t]);
            if (t == currmax) {
                break;
            }
            if(!visited[b[t]]) pq.add(new int[] {t , b[t] , a[t]} );
            for (int i = t-1; i > 0; i--) {
                if (visited[i]) break;
                visited[i] = true;
                dist[i] = dist[t];
                if (!visited[b[i]]) pq.add(new int[] {i , b[i] , a[i]});
            }
        }
        System.out.println(ans);
    }

    public static void printArray(int[] arr) {
        StringBuilder sb = new StringBuilder();
        for (int x : arr) {
            sb.append(x).append(" ");
        }
        System.out.println(sb);
    }

    public static int[] readArray(int n) {
        int[] arr = new int[n+1];
        for (int i = 1; i <= n; i++) {
            arr[i] = in.nextInt();
        }
        return arr;
    }

    public static int[] freq(String s) {
        int[] arr = new int[26];
        for (char c : s.toCharArray()) {
            arr[c - 'a']++;
        }
        return arr;
    }

    public static long gcd(long a, long b) {
        if (a > b)
            a = (a + b) - (b = a);
        if (a == 0L)
            return b;
        return gcd(b % a, a);
    }

    public static boolean isPrime(long n) {
        if (n < 2) return false;
        if (n == 2 || n == 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        long sqrtN = (long) Math.sqrt(n) + 1;
        for (long i = 6L; i <= sqrtN; i += 6) {
            if (n % (i - 1) == 0 || n % (i + 1) == 0) return false;
        }
        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:
To expose the bug in the given code, we need to create a test case that challenges the logic used in the `solve` method. The code seems to have issues with how it handles the `currmax` and the `visited` array, particularly when dealing with the problem indices and the skipping logic.

### Step-by-Step Analysis:
1. **Initialization of `currmax` and `sum`**:
   - The `currmax` is updated to the maximum value of `b[i]` up to the current index `i`.
   - The `sum` array is supposed to accumulate the scores, but the loop breaks early if `currmax == i`.

2. **Priority Queue and Visited Array**:
   - The priority queue is used to manage the problems, but the logic for adding and checking the `visited` array might be flawed.
   - The code might not correctly handle the scenario where the participant skips a problem and then the next problem to be given is not the one with the maximum index.

3. **Edge Case Consideration**:
   - Consider a scenario where the first problem has a high score and a high `b` value, but subsequent problems have lower scores and lower `b` values.
   - Another edge case is where the first problem has a low score and a high `b` value, and subsequent problems have higher scores and lower `b` values.

### Test Case:
Let's create a test case where the first problem has a low score and a high `b` value, and subsequent problems have higher scores and lower `b` values. This should challenge the logic of the code to correctly handle the skipping and selecting of problems.

```plaintext
1
5
10 100 100 100 1000
5 1 1 1 1
```

### Explanation:
- **Test Case 1**:
  - **n = 5**: There are 5 problems.
  - **a = [10, 100, 100, 100, 1000]**: The scores of the problems.
  - **b = [5, 1, 1, 1, 1]**: The parameters of the problems.

### Expected Output:
- The participant should skip the first problem with a score of 10 and a `b` value of 5. This will lead to the participant receiving the problem with the maximum index that is not yet given, which is the second problem with a score of 100 and a `b` value of 1.
- The participant should submit the second problem and receive 100 points.
- The participant will then receive the third problem, submit it, and receive another 100 points.
- The participant will receive the fourth problem, submit it, and receive another 100 points.
- Finally, the participant will receive the fifth problem, submit it, and receive 1000 points.

### Total Points:
- The total points should be 1300.

This test case should expose the bug in the given code, as it requires the correct handling of skipping and selecting problems based on the `b` values.