**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

public class MaxGcdSumQueries {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());

        int[] a = new int[n];
        int[] b = new int[n];
        int[] c = new int[n];
        long[] d = new long[q];

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            b[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            c[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < q; i++) {
            d[i] = Long.parseLong(st.nextToken());
        }

        int gcdA = computeGCD(a);
        int gcdB = computeGCD(b);
        int initialSum = gcdA + gcdB;

        List<SwapImpact> impacts = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            int tempA = a[i];
            int tempB = b[i];
            a[i] = tempB;
            b[i] = tempA;

            int newGcdA = computeGCD(a);
            int newGcdB = computeGCD(b);
            int gain = (newGcdA + newGcdB) - initialSum;
            int cost = c[i];

            if (gain > 0) {
                impacts.add(new SwapImpact(cost, gain));
            }

            // revert back the swap
            a[i] = tempA;
            b[i] = tempB;
        }

        impacts.sort(Comparator.comparingInt(impact -> impact.cost));

        StringBuilder result = new StringBuilder();
        for (long budget : d) {
            long currentSum = initialSum;
            long currentCost = 0;
            for (SwapImpact impact : impacts) {
                if (currentCost + impact.cost <= budget) {
                    currentCost += impact.cost;
                    currentSum += impact.gain;
                } else {
                    break;
                }
            }
            result.append(currentSum).append("\n");
        }

        System.out.print(result.toString());
    }

    private static int computeGCD(int[] arr) {
        int result = arr[0];
        for (int num : arr) {
            result = gcd(result, num);
        }
        return result;
    }

    private static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    private static class SwapImpact {
        int cost;
        int gain;

        SwapImpact(int cost, int gain) {
            this.cost = cost;
            this.gain = gain;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider a few key points:

1. **Initial GCD Calculation**: The initial GCDs of arrays `a` and `b` should be relatively small to maximize the potential for improvement through swaps.
2. **Swap Impacts**: There should be some swaps that provide significant gains in GCD, but their costs should be carefully chosen to create a tricky scenario for the algorithm.
3. **Query Values**: The query values should be chosen to test the boundary conditions and the handling of multiple queries.

Let's create a test input that meets these criteria:

- **Initial Arrays**: `a` and `b` should have small GCDs.
- **Costs**: Some costs should be very high, and some should be very low to test the sorting and selection logic.
- **Query Values**: Include values that test the limits of the budget.

Here is the test input:

```plaintext
3 4
1 2 3
4 5 6
1 1000000000 1
0 1 1000000000 1000000001
```

### Explanation:
- **n = 3**: We have three elements in each array.
- **q = 4**: We have four queries.
- **Arrays**:
  - `a = [1, 2, 3]`
  - `b = [4, 5, 6]`
  - `c = [1, 1000000000, 1]`
- **Queries**:
  - `d = [0, 1, 1000000000, 1000000001]`

### Expected Output:
- For `d = 0`: No swaps are allowed, so the GCD sum is `gcd(1, 2, 3) + gcd(4, 5, 6) = 2`.
- For `d = 1`: Only the first swap can be made, so the GCD sum is `gcd(4, 2, 3) + gcd(1, 5, 6) = 3`.
- For `d = 1000000000`: Only the second swap can be made, so the GCD sum is `gcd(1, 5, 3) + gcd(4, 2, 6) = 3`.
- For `d = 1000000001`: Both swaps can be made, so the GCD sum is `gcd(4, 5, 3) + gcd(1, 2, 6) = 3`.

This test input will challenge the algorithm to correctly handle the sorting and selection of swaps based on their costs and gains, especially when the budget is very high or very low.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

public class MaxGcdSumQueries {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());

        int[] a = new int[n];
        int[] b = new int[n];
        int[] c = new int[n];
        long[] d = new long[q];

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            b[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            c[i] = Integer.parseInt(st.nextToken());
        }

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < q; i++) {
            d[i] = Long.parseLong(st.nextToken());
        }

        // Initial GCDs
        int gcdA = computeGCD(a);
        int gcdB = computeGCD(b);

        int initialSum = gcdA + gcdB;

        PriorityQueue<SwapImpact> impactHeap = new PriorityQueue<>((x, y) -> x.cost - y.cost);

        for (int i = 0; i < n; i++) {
            int[] newA = a.clone();
            int[] newB = b.clone();

            // Swap a[i] with b[i]
            int temp = newA[i];
            newA[i] = newB[i];
            newB[i] = temp;

            int newGcdA = computeGCD(newA);
            int newGcdB = computeGCD(newB);

            int gain = (newGcdA + newGcdB) - initialSum;
            int cost = c[i];

            if (gain > 0) {
                impactHeap.add(new SwapImpact(cost, gain));
            }
        }

        StringBuilder result = new StringBuilder();
        for (long budget : d) {
            int currentSum = initialSum;
            long currentCost = 0;

            PriorityQueue<SwapImpact> tempHeap = new PriorityQueue<>((x, y) -> x.cost - y.cost);

            while (!impactHeap.isEmpty() && currentCost + impactHeap.peek().cost <= budget) {
                SwapImpact impact = impactHeap.poll();
                currentCost += impact.cost;
                currentSum += impact.gain;
                tempHeap.add(impact);
            }

            result.append(currentSum).append("\n");

            while (!tempHeap.isEmpty()) {
                impactHeap.add(tempHeap.poll());
            }
        }

        System.out.print(result.toString());
    }

    private static int computeGCD(int[] arr) {
        int result = arr[0];
        for (int num : arr) {
            result = gcd(result, num);
        }
        return result;
    }

    private static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    private static class SwapImpact {
        int cost;
        int gain;

        SwapImpact(int cost, int gain) {
            this.cost = cost;
            this.gain = gain;
        }
    }
}

**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 and scenarios that can highlight the issues in the algorithm. Here are some key points to consider:

1. **Zero Swaps Allowed**: Test the case where no swaps are allowed to ensure the initial GCD sum is calculated correctly.
2. **Single Element Arrays**: Test with arrays of length 1 to ensure the GCD calculation is correct for single elements.
3. **Large Number of Swaps**: Test with a large number of possible swaps to see if the priority queue and budget handling are correct.
4. **High Cost Swaps**: Test with high-cost swaps to ensure the budget is respected and the algorithm does not exceed the budget.
5. **Equal Elements**: Test with arrays where elements are equal to ensure the GCD calculation is correct and the algorithm does not make unnecessary swaps.

Given the constraints and the need for a small-scale test input, let's create a test case that includes these scenarios:

```plaintext
5 5
1 2 3 4 5
5 4 3 2 1
100 100 100 100 100
0 100 200 300 400
```

### Explanation:
- **n = 5**: We have 5 elements in each array.
- **q = 5**: We have 5 budget values to test.
- **a = [1, 2, 3, 4, 5]**: Initial array A.
- **b = [5, 4, 3, 2, 1]**: Initial array B.
- **c = [100, 100, 100, 100, 100]**: Cost of swapping each element.
- **d = [0, 100, 200, 300, 400]**: Budget values for the queries.

### Expected Output:
- **d = 0**: No swaps allowed. The initial GCD sum is \(\gcd(1, 2, 3, 4, 5) + \gcd(5, 4, 3, 2, 1) = 1 + 1 = 2\).
- **d = 100**: One swap is allowed. The best swap is to swap the first elements, resulting in \(\gcd(5, 2, 3, 4, 1) + \gcd(1, 4, 3, 2, 5) = 1 + 1 = 2\).
- **d = 200**: Two swaps are allowed. The best swaps are to swap the first and second elements, resulting in \(\gcd(5, 4, 3, 4, 1) + \gcd(1, 2, 3, 2, 5) = 1 + 1 = 2\).
- **d = 300**: Three swaps are allowed. The best swaps are to swap the first, second, and third elements, resulting in \(\gcd(5, 4, 3, 4, 1) + \gcd(1, 2, 3, 2, 5) = 1 + 1 = 2\).
- **d = 400**: Four swaps are allowed. The best swaps are to swap the first, second, third, and fourth elements, resulting in \(\gcd(5, 4, 3, 4, 1) + \gcd(1, 2, 3, 2, 5) = 1 + 1 = 2\).

This test case is designed to challenge the algorithm's ability to handle different budget values and ensure that the priority queue and budget constraints are correctly managed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

using namespace std;

#define ll long long
#define pb push_back
#define ff first
#define ss second
#define all(s) s.begin(),s.end()
#define rall(s) s.rbegin(),s.rend()

void calc(ll x,vector<ll>&p,vector<ll>&c,vector<ll>&d,map<ll,ll>&m){
    vector<ll>cnt;
    d.pb(1);
    for(ll i=2;i*i<=x;i++){
        if(x%i==0){
            p.pb(i);
            cnt.pb(0);
            while(x%i==0){
                cnt.back()++;
                x/=i;
            }
        }
    }
    if(x>1){
        p.pb(x);
        cnt.pb(x);
    }
    for(ll i=0;i<p.size();i++){
        c.pb(d.size());
        for(ll j=0;j<c.back()*cnt[i];j++){
            d.pb(d[j]*p[i]);
        }
    }
    for(ll i=0;i<d.size();i++){
        m[d[i]]=i;
    }
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll n,q;
    cin>>n>>q;
    ll a[n+5],b[n+5],c[n+5],d[q+5];
    for(ll i=1;i<=n;i++) cin>>a[i];
    for(ll i=1;i<=n;i++) cin>>b[i];
    for(ll i=1;i<=n;i++) cin>>c[i];
    for(ll i=1;i<=q;i++) cin>>d[i];
    ll g[n+5];
    for(ll i=1;i<=n;i++){
        g[i]=gcd(a[i],b[i]);
    }
    vector<pair<ll,ll>>v;
    for(ll it=0;it<2;it++){
        vector<ll>pa,pb,ca,cb,da,db;
        map<ll,ll>ma,mb;
        calc(a[1],pa,ca,da,ma);
        calc(b[1],pb,cb,db,mb);
        auto p=vector(da.size(),vector<ll>(db.size()));
        auto s=vector(da.size(),vector<ll>(db.size()));
        for(ll i=2;i<=n;i++){
            p[ma[gcd(a[1],a[i])]][mb[gcd(b[1],b[i])]]++;
            p[ma[gcd(a[1],b[i])]][mb[gcd(b[1],a[i])]]++;
            s[ma[gcd(a[1],b[i])]][mb[gcd(b[1],a[i])]]+=c[i];
            p[ma[gcd(a[1],g[i])]][mb[gcd(b[1],g[i])]]--;
            s[ma[gcd(a[1],g[i])]][mb[gcd(b[1],g[i])]]-=c[i];
        }
        for(ll ia=0;ia<pa.size();ia++){
            for(ll i=da.size()-1;i>=0;i--){
                if(da[i]%pa[ia]==0){
                    for(ll j=0;j<db.size();j++){
                        p[i-ca[ia]][j]+=p[i][j];
                        s[i-ca[ia]][j]+=s[i][j];
                    }
                }
            }
        }
        for(ll ib=0;ib<pb.size();ib++){
            for(ll j=db.size()-1;j>=0;j--){
                if(db[j]%pb[ib]==0){
                    for(ll i=0;i<da.size();i++){
                        p[i][j-cb[ib]]+=p[i][j];
                        s[i][j-cb[ib]]+=s[i][j];
                    }
                }
            }
        }
        for(ll i=0;i<da.size();i++){
            for(ll j=0;j<db.size();j++){
                if(p[i][j]>=n-1){
                    v.pb({s[i][j]+(it==1?c[1]:0),da[i]+db[j]});
                }
            }
        }
        swap(a[1],b[1]);
    }
    sort(all(v));
    for(ll i=1;i<v.size();i++){
        v[i].ss=max(v[i].ss,v[i-1].ss);
    }
    for(ll i=1;i<=q;i++){
        cout<<v[lower_bound(all(v),make_pair(d[i]+1,0ll))-v.begin()-1].ss<<"\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
3 3
1 2 3
4 5 6
10 10 10
0 10 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define loop(i, n) for (int i = 0; i < (n); i++)
#define all(A) A.begin(), A.end()
#define pb push_back
#define mp make_pair
#define sz(A) ((int)A.size())
using vi = std::vector<int>;
using pi = std::pair<int, int>;
using vp = std::vector<pi>;
using ll = long long;
#define popcnt(x) __builtin_popcount(x)
#define LSOne(x) ((x) & (-(x)))
#define print(A, t)                             \
  cerr << #A << ": ";                           \
  copy(all(A), ostream_iterator<t>(cerr, " ")); \
  cerr << endl
#define prArr(A, n, t)                            \
  cerr << #A << ": ";                             \
  copy(A, A + n, ostream_iterator<t>(cerr, " ")); \
  cerr << endl
#define PRESTDIO() cin.tie(0), cerr.tie(0), ios_base::sync_with_stdio(0)
#define what_is(x) cerr << #x << " is " << x << endl
#define bit_lg(x) (assert(x > 0), __builtin_ffsll(x) - 1)
template <class A, class B>
std::ostream& operator<<(std::ostream& st, const std::pair<A, B> p) {
  st << "(" << p.first << ", " << p.second << ")";
  return st;
}
void tc();
auto test_cases = []() {
  int T;
  scanf("%d", &T);
  while (T--) tc();
};
template <typename... Args>
void VLOG(int depth, bool endline, Args... args) {
#ifdef VLOG
  std::cerr << std::string(depth, '\t');
  (std::cerr << ... << args);
  if (endline) {
    std::cerr << std::endl;
  }
#endif
}
using namespace std;

const int MAX = 1 << 20;
int A[MAX], B[MAX], C[MAX];
int n, m;
map<ll, int> scores;

vi divs(int x) {
    vi ret;
    for(int i = 1; i*i <= x; i++) if(x){
        ret.push_back(i);
        if(i*i != x) ret.push_back(x/i);
    }
    return ret;
}

void all_possibilities() {
    vi order(n); iota(order.begin(), order.end(), 0);
    sort(order.begin(), order.end(), [](int a, int b) {return tie(A[a], B[a]) < tie(A[b], B[b]);});

    vi d1 = divs(A[0]), d2 = divs(B[0]);

  vector<tuple<int, int, ll>> prv, cur, aux;
  for(int a : d1) for(int b : d2) cur.emplace_back(a, b, 0);
  for(int a : d2) for(int b : d1) cur.emplace_back(a, b, 0);
  sort(cur.begin(), cur.end());
  cur.resize(unique(cur.begin(), cur.end()) - cur.begin());

  for(int j = 0; j < n; j++) {
      int i = order[j];
    prv.swap(cur);
    cur.clear();
    for(const auto [a, b, c] : prv) {
        if(A[i]%a == 0 && B[i]%b == 0) {
            cur.emplace_back(a, b, c);
            continue;
        }
        if(B[i]%a == 0 && A[i]%b == 0) {
            cur.emplace_back(a, b, c + C[i]);
            continue;
        }
    }
  }
  vector<pair<ll, int>> scores;
  for(const auto [a, b, c] : cur) {
    scores.emplace_back(c, -(a + b));
  }
  sort(scores.begin(), scores.end());
  ll val = 0;
  for(auto [s, nv] : scores) {
    if(-nv > val) {
      val = -nv;
      ::scores[s] = val;
    }
  }
}
int main() {
#ifdef HOME
  freopen("in.in", "r", stdin);
#endif
  scanf("%d %d", &n, &m);
  for(int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
  }
  for(int i = 0; i < n; i++) {
    scanf("%d", &B[i]);
  }
  for(int i = 0; i < n; i++) {
    scanf("%d", &C[i]);
  }
  all_possibilities();
  for(int i = 0; i < m; i++) {
    ll d; scanf("%lld", &d);
    auto ptr = scores.upper_bound(d);
    ptr--;
    printf("%d ", ptr->second);
  }
  puts("");
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 3
1 2 3
4 5 6
1 2 3
0 3 5
```

**Explanation:**
- **n = 3**: The number of elements in the arrays.
- **q = 3**: The number of queries.
- **Arrays**:
  - **a**: [1, 2, 3]
  - **b**: [4, 5, 6]
  - **c**: [1, 2, 3]
- **Queries**:
  - **d**: [0, 3, 5]

**Reasoning:**
1. **Initial GCDs**:
   - \(\gcd(1, 2, 3) = 1\)
   - \(\gcd(4, 5, 6) = 1\)
   - Initial sum: \(1 + 1 = 2\)

2. **Query 1 (d = 0)**:
   - No swaps allowed.
   - The answer remains \(2\).

3. **Query 2 (d = 3)**:
   - Swapping \(a_1\) and \(b_1\) costs 1 coin:
     - New arrays: \(a = [4, 2, 3]\), \(b = [1, 5, 6]\)
     - \(\gcd(4, 2, 3) = 1\)
     - \(\gcd(1, 5, 6) = 1\)
     - Sum: \(1 + 1 = 2\)
   - Swapping \(a_2\) and \(b_2\) costs 2 coins:
     - New arrays: \(a = [4, 5, 3]\), \(b = [1, 2, 6]\)
     - \(\gcd(4, 5, 3) = 1\)
     - \(\gcd(1, 2, 6) = 1\)
     - Sum: \(1 + 1 = 2\)
   - Swapping \(a_3\) and \(b_3\) costs 3 coins:
     - New arrays: \(a = [4, 5, 6]\), \(b = [1, 2, 3]\)
     - \(\gcd(4, 5, 6) = 1\)
     - \(\gcd(1, 2, 3) = 1\)
     - Sum: \(1 + 1 = 2\)
   - The answer remains \(2\).

4. **Query 3 (d = 5)**:
   - Swapping \(a_1\) and \(b_1\) and \(a_2\) and \(b_2\) costs 3 coins:
     - New arrays: \(a = [4, 5, 3]\), \(b = [1, 2, 6]\)
     - \(\gcd(4, 5, 3) = 1\)
     - \(\gcd(1, 2, 6) = 1\)
     - Sum: \(1 + 1 = 2\)
   - Swapping \(a_1\) and \(b_1\) and \(a_3\) and \(b_3\) costs 4 coins:
     - New arrays: \(a = [4, 2, 6]\), \(b = [1, 5, 3]\)
     - \(\gcd(4, 2, 6) = 2\)
     - \(\gcd(1, 5, 3) = 1\)
     - Sum: \(2 + 1 = 3\)
   - Swapping \(a_2\) and \(b_2\) and \(a_3\) and \(b_3\) costs 5 coins:
     - New arrays: \(a = [1, 5, 6]\), \(b = [4, 2, 3]\)
     - \(\gcd(1, 5, 6) = 1\)
     - \(\gcd(4, 2, 3) = 1\)
     - Sum: \(1 + 1 = 2\)
   - The answer is \(3\).

This test input is designed to cover the edge cases and ensure that the algorithm correctly handles the constraints and edge conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

using namespace std;
#define int long long
#define app push_back
#define all(x) (x).begin(),(x).end()
#ifdef LOCAL
#define debug(...) [](auto...a){ ((cout << a << ' '), ...) << endl;}(#__VA_ARGS__, ":", __VA_ARGS__)
#define debugv(v) do {cout<< #v <<" : {"; for(int i=0;i<v.size();++i) {cout << v[i];if(i+1!=v.size()) cout << ","; }cout <<"}"<< endl;} while(0)
#else
#define debug(...)
#define debugv(v)
#endif
#ifdef LOCAL
#define __int128 long long
#endif // LOCAL
const int p=998244353;
int po(int a,int b) {if(b==0) return 1; if(b==1) return a; if(b%2==0) {int u=po(a,b/2);return (u*1LL*u)%p;} else {int u=po(a,b-1);return (a*1LL*u)%p;}}
int inv(int x) {return po(x,p-2);}
const int inf=1e17;
mt19937 rnd;
vector<int> adi(int x)
{
    vector<int> v;for(int i=1;i*i<=x;++i) {if(x%i==0) {v.app(i);v.app(x/i);}} sort(all(v));v.erase(unique(all(v)),v.end());
    return v;
}
const int maxn=1e6+5;
__int128 u[maxn];
int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n,q;cin>>n>>q;
    vector<array<int,3> > h(n);for(int i=0;i<n;++i) {cin>>h[i][0];} for(int i=0;i<n;++i) {cin>>h[i][1];} for(int i=0;i<n;++i) {cin>>h[i][2];}
    sort(all(h));
    vector<int> a(n),b(n),c(n),a1(n),b1(n),c1(n),ap(n),bp(n);for(int i=0;i<n;++i) a[i]=h[i][0]; for(int i=0;i<n;++i) b[i]=h[i][1];  for(int i=0;i<n;++i) c[i]=h[i][2];
    vector<pair<int,int> > have;
    int ad=0;
    for(int cyc=0;cyc<2;++cyc)
    {
        vector<int> v=adi(a[0]);
        vector<int> o=adi(b[0]);
        #define v2 o
        for(int i=0;i<n;++i) {a1[i]=__gcd(a[i],b[0]);ap[i]=lower_bound(all(v2),a1[i])-v2.begin();}
        for(int i=0;i<n;++i) {b1[i]=__gcd(b[i],b[0]);bp[i]=lower_bound(all(v2),b1[i])-v2.begin();}
        for(int i=0;i<n;++i) {c1[i]=lower_bound(all(v2),__gcd(a1[i],b1[i]))-v2.begin();}
        vector<__int128> answ(o.size());
        for(int d:v)
        {
            for(int i=0;i<v2.size();++i) {u[i]=0;answ[i]=0;}
            int res=0;
            int h=b[0];
            for(int j=0;j<n;++j)
            {
                if(a[j]%d!=0 && b[j]%d!=0) {res=inf;}
                else if(a[j]%d==0 && b[j]%d!=0) {u[v2.size()-1]+=inf;u[bp[j]]-=inf;}
                else if(a[j]%d!=0 && b[j]%d==0) {res+=c[j];u[v2.size()-1]+=inf;u[ap[j]]-=inf;}
                else {
                u[v2.size()-1]+=inf;u[bp[j]]-=inf;u[ap[j]]+=c[j]-inf;u[c1[j]]+=inf-c[j];}
            }
            for(int j=0;j<v2.size();++j)
            {
                for(int i=0;i<o.size();++i)
                {
                    if(o[j]%o[i]==0)
                    {
                        answ[i]+=u[j];
                    }
                }
            }
            for(int i=0;i<o.size();++i)
                {
                    if(res+answ[i]<inf){
                    have.app({d+o[i],res+answ[i]+ad});
                    }
                }
        }
        if(cyc==0) {swap(a[0],b[0]);ad+=c[0];}
    };
    vector<int> que1(q);for(int i=0;i<q;++i) {cin>>que1[i];} vector<int> que=que1;sort(all(que));
    vector<int> ans(q);
    for(auto [key,val]:have)
    {
        int pos=lower_bound(all(que),val)-que.begin();
        if(pos>=0) {ans[pos]=max(ans[pos],key);}
    }
    for(int i=0;i<q-1;++i) {ans[i+1]=max(ans[i],ans[i+1]);}
    for(int i=0;i<q;++i)
    {
        cout<<ans[lower_bound(all(que),que1[i])-que.begin()]<<' ';
    }
    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
6 15 10
10 6 15
1 2 3
0 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

int gcd(int a, int b) {
    while (b != 0) {
        int t = b;
        b = a % b;
        a = t;
    }
    return a;
}

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

    int n, q;
    cin >> n >> q;
    vector<int> a(n), b(n), c(n);
    for (int &x : a) cin >> x;
    for (int &x : b) cin >> x;
    for (int &x : c) cin >> x;

    vector<long long> budgets(q);
    for (auto &d : budgets) cin >> d;

    // Precompute initial GCDs
    int gcd_a = a[0], gcd_b = b[0];
    for (int i = 1; i < n; ++i) {
        gcd_a = gcd(gcd_a, a[i]);
        gcd_b = gcd(gcd_b, b[i]);
    }

    vector<int> results(q);
    for (int i = 0; i < q; ++i) {
        long long d = budgets[i];
        int max_gcd_sum = gcd_a + gcd_b;

        // Try all possible swaps within budget d
        for (int mask = 1; mask < (1 << n); ++mask) {
            long long cost = 0;
            int new_gcd_a = 0, new_gcd_b = 0;
            bool first_a = true, first_b = true;

            for (int j = 0; j < n; ++j) {
                if (mask & (1 << j)) {
                    cost += c[j];
                    if (first_a) {
                        new_gcd_a = b[j];
                        first_a = false;
                    } else {
                        new_gcd_a = gcd(new_gcd_a, b[j]);
                    }
                    if (first_b) {
                        new_gcd_b = a[j];
                        first_b = false;
                    } else {
                        new_gcd_b = gcd(new_gcd_b, a[j]);
                    }
                } else {
                    if (first_a) {
                        new_gcd_a = a[j];
                        first_a = false;
                    } else {
                        new_gcd_a = gcd(new_gcd_a, a[j]);
                    }
                    if (first_b) {
                        new_gcd_b = b[j];
                        first_b = false;
                    } else {
                        new_gcd_b = gcd(new_gcd_b, b[j]);
                    }
                }
            }

            if (cost <= d) {
                max_gcd_sum = max(max_gcd_sum, new_gcd_a + new_gcd_b);
            }
        }

        results[i] = max_gcd_sum;
    }

    for (int res : results) {
        cout << res << " ";
    }
    cout << 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
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
```

### Explanation:
1. **Input Size**:
   - `n = 3` (small to ensure the test input is manageable)
   - `q = 4` (to test multiple budget scenarios)

2. **Array Values**:
   - `a = [1, 2, 3]`
   - `b = [4, 5, 6]`
   - `c = [1, 1, 1]` (costs are all 1 to simplify the budget calculation)

3. **Budgets**:
   - `d = [0, 1, 2, 3]` (to test the effect of different budgets)

### Why This Input?
- **Initial GCDs**:
  - `gcd(a) = gcd(1, 2, 3) = 1`
  - `gcd(b) = gcd(4, 5, 6) = 1`
  - Initial GCD sum: `1 + 1 = 2`

- **Budget 0**:
  - No swaps allowed, so the GCD sum remains `2`.

- **Budget 1**:
  - Swapping one element (e.g., `a[1]` and `b[1]`):
    - New `a = [1, 5, 3]`
    - New `b = [4, 2, 6]`
    - New GCDs: `gcd(1, 5, 3) = 1` and `gcd(4, 2, 6) = 2`
    - GCD sum: `1 + 2 = 3`

- **Budget 2**:
  - Swapping two elements (e.g., `a[1]` and `b[1]`, and `a[2]` and `b[2]`):
    - New `a = [1, 5, 6]`
    - New `b = [4, 2, 3]`
    - New GCDs: `gcd(1, 5, 6) = 1` and `gcd(4, 2, 3) = 1`
    - GCD sum: `1 + 1 = 2` (but this is not optimal, so we stick with the previous swap)

- **Budget 3**:
  - Swapping all elements:
    - New `a = [4, 5, 6]`
    - New `b = [1, 2, 3]`
    - New GCDs: `gcd(4, 5, 6) = 1` and `gcd(1, 2, 3) = 1`
    - GCD sum: `1 + 1 = 2` (but this is not optimal, so we stick with the previous swap)

This input is designed to test the algorithm's ability to handle different budget scenarios and ensure it correctly calculates the maximum GCD sum. The small input size ensures it is manageable and easy to verify.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
from math import gcd
def prepare(x, dx, px, cx, mx):
    cntx = []
    i = 2
    while i * i <= x:
        if x % i == 0:
            px.append(i)
            cntx.append(0)
            while x % i == 0:
                cntx[-1] += 1
                x //= i
        i += 1
    if x > 1:
        px.append(x)
        cntx.append(1)

    dx.append(1)
    for i in range(len(px)):
        cx.append(len(dx))
        for j in range(cx[-1] * cntx[i]):
            dx.append(dx[j] * px[i])

    for i in range(len(dx)):
        mx[dx[i]] = i

def solve(n, a, b, c, qu):
    k = max(max(a), max(b))
    g = [gcd(a[i], b[i]) for i in range(n)]
    v = []
    ans = []

    for it in range(2):
        da, pa, ca = [], [], []
        ma = {}
        db, pb, cb = [], [], []
        mb = {}
        prepare(a[0], da, pa, ca, ma)
        prepare(b[0], db, pb, cb, mb)

        p = [[0] * len(db) for _ in range(len(da))]
        d = [[0] * len(db) for _ in range(len(da))]
        for i in range(1, n):
            p[ma[gcd(a[0], a[i])]][mb[gcd(b[0], b[i])]] += 1
            d[ma[gcd(a[0], a[i])]][mb[gcd(b[0], b[i])]] += 0
            p[ma[gcd(a[0], b[i])]][mb[gcd(b[0], a[i])]] += 1
            d[ma[gcd(a[0], b[i])]][mb[gcd(b[0], a[i])]] += c[i]
            p[ma[gcd(a[0], g[i])]][mb[gcd(b[0], g[i])]] -= 1
            d[ma[gcd(a[0], g[i])]][mb[gcd(b[0], g[i])]] -= c[i]

        for ia in range(len(ca)):
            for i in range(len(da) - 1, -1, -1):
                if da[i] % pa[ia] == 0:
                    for j in range(len(db) - 1, -1, -1):
                        p[i - ca[ia]][j] += p[i][j]
                        d[i - ca[ia]][j] += d[i][j]

        for ib in range(len(cb)):
            for i in range(len(db) - 1, -1, -1):
                if db[i] % pb[ib] == 0:
                    for j in range(len(da) - 1, -1, -1):
                        p[j][i - cb[ib]] += p[j][i]
                        d[j][i - cb[ib]] += d[j][i]

        for i in range(len(da)):
            for j in range(len(db)):
                if p[i][j] >= n - 1:
                    v.append((d[i][j] + (c[0] if it else 0), da[i] + db[j]))

        a[0], b[0] = b[0], a[0]

    v.sort()
    # for i in range(1, len(v)):
    #     v[i] = (v[i][0], max(v[i - 1][1], v[i][1]))

    for d in qu:
        idx = next((i for i, val in enumerate(v) if val[0] > d), len(v)) - 1
        ans.append(v[idx][1])

    return ans

if __name__ == "__main__":
    n, q = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    c = list(map(int, input().split()))
    qi = list(map(int, input().split()))
    ans = solve(n, a, b, c, qi)
    print(" ".join(map(str, 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
2 3 4
4 3 2
1 2 3
0 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
import sys
import itertools
import math

def read():
  yield from sys.stdin.read().splitlines()

inp = iter(read())

def all_possiblities(A, B, C) -> int:

  mem = [{} for _ in range(len(A) + 1)]
  mem[0][(0, 0)] = 0
  for i in range(1, len(A) + 1):
    for a, b, c in (A[i-1], B[i-1], 0), (B[i-1], A[i-1], C[i-1]):
      for (ga, gb), gc in mem[i - 1].items():
        s = (math.gcd(a, ga), math.gcd(b, gb))
        if s not in mem[i]: mem[i][s] = gc + c
        else: mem[i][s] = min(mem[i][s], gc + c)
  tmp = []
  for (a, b), c in mem[-1].items():
    tmp.append((c, a + b))
  tmp.sort(key=lambda p: (p[0], -p[1]))
  ret = []
  val = 0
  for c, v in tmp:
    if v > val:
      val = v
      ret.append((c, val))
  return ret

def tc():
  n, m = map(int, next(inp).split())
  A = tuple(map(int, next(inp).split()))
  B = tuple(map(int, next(inp).split()))
  C = tuple(map(int, next(inp).split()))
  D = tuple(map(int, next(inp).split()))
  scores = all_possiblities(A, B, C)[::-1]
  queries = [(d, i) for i, d in enumerate(D)]
  queries.sort()
  ans = [0]*m
  val = 0
  for d, i in queries:
    while len(scores) and scores[-1][0] <= d:
      val = scores.pop()[1]
    ans[i] = val
  yield from ans

print(*tc())

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given code, we need to consider edge cases that can challenge the algorithm's correctness and efficiency. Here are the steps to think through the process:

1. **Understanding the Bug**: The provided code uses dynamic programming to explore all possible combinations of swaps and their costs. However, it might not handle cases where the cost of swaps is very high or very low efficiently, or it might not correctly handle the GCD calculations.

2. **Edge Cases**:
   - **Zero Cost Swaps**: What happens if all swaps are free (cost = 0)? The algorithm should be able to consider all possible swaps without any cost constraints.
   - **High Cost Swaps**: What happens if the cost of swaps is very high? The algorithm should be able to handle cases where no swaps can be made within the budget.
   - **Equal Elements**: What happens if some elements in arrays A and B are equal? The GCD calculations should be straightforward in such cases.
   - **Large GCD Values**: What happens if the elements in arrays A and B are large and have large GCD values? The algorithm should handle large numbers efficiently.

3. **Test Input**:
   - **Small Scale**: We need a small-scale test input to ensure it fits within the constraints and is easy to verify.
   - **Challenging Scenario**: We will create a scenario where the cost of swaps is very high, and the GCD values are large.

Given these considerations, here is a test input that should expose potential issues in the buggy code:

```plaintext
5 3
12 15 18 21 24
10 14 16 20 22
1000000000 1000000000 1000000000 1000000000 1000000000
0 1 2000000000
```

### Explanation:
- **n = 5**: We have 5 elements in each of the arrays.
- **q = 3**: We have 3 queries for different values of \( d \).
- **A = [12, 15, 18, 21, 24]**: These are the initial values in array A.
- **B = [10, 14, 16, 20, 22]**: These are the initial values in array B.
- **C = [1000000000, 1000000000, 1000000000, 1000000000, 1000000000]**: The cost of each swap is very high.
- **D = [0, 1, 2000000000]**: The queries for the maximum GCD sum with 0, 1, and 2000000000 coins.

### Expected Output:
- For \( d = 0 \): No swaps are allowed, so the GCD sum is \(\gcd(12, 15, 18, 21, 24) + \gcd(10, 14, 16, 20, 22) = 3 + 2 = 5\).
- For \( d = 1 \): No swaps can be made with only 1 coin, so the GCD sum remains 5.
- For \( d = 2000000000 \): With enough coins, the algorithm should consider all possible swaps to maximize the GCD sum. However, since the cost is very high, no swaps will be made, and the GCD sum remains 5.

This test input should help identify issues in the algorithm's handling of high-cost swaps and large GCD values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define maxn 500005
#define mod 1000000007
using namespace std;
mt19937 rnd(time(NULL)^clock());
ll rad(int x,int y){
	return rnd()%(y-x+1)+x;
}
map<ll,ll> dp[2],tmp;
ll n,q,a[maxn],b[maxn],c[maxn];
vector<pair<ll,ll>> ans1,ans2;
void lv(int now)
{
	tmp.clear();
	for (auto t1:dp[now])
	{
		int flag=0; ll x=t1.first/mod,y=t1.first%mod,val1=t1.second;
		for (auto t2:dp[now])
		{
			ll a=t2.first/mod,b=t2.first%mod,val2=t2.second;
			if (a%x==0 && b%y==0 && (val2<val1 || (val2==val1 && t1!=t2)))
			{flag=1; break;}
		}
		if (!flag) tmp[t1.first]=t1.second;
	}
	dp[now]=tmp;
}
void init()
{
	cin>>n>>q;
	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++) cin>>c[i];
	for (int i=2;i<=n;i++)
	{
		if (min(a[i],b[i])<min(a[1],b[1]))
		{
			swap(a[i],a[1]); swap(b[i],b[1]); swap(c[i],c[1]);
		}
	}
	int now=1,nxt=0;
	dp[now][{a[1]*mod+b[1]}]=0;
	dp[now][{b[1]*mod+a[1]}]=c[1];
	for (int i=2;i<=n;i++)
	{
		for (auto tt:dp[now])
		{
			ll x=tt.first/mod,y=tt.first%mod,val=tt.second;
			ll aa=__gcd(x,a[i]),bb=__gcd(y,b[i]);
			if (!dp[nxt].count(aa*mod+bb)) dp[nxt][aa*mod+bb]=val;
			else dp[nxt][aa*mod+bb]=min(dp[nxt][aa*mod+bb],val);
			if (a[i]==b[i]) continue;
			aa=__gcd(x,b[i]),bb=__gcd(y,a[i]);
			if (!dp[nxt].count(aa*mod+bb)) dp[nxt][aa*mod+bb]=val+c[i];
			else dp[nxt][aa*mod+bb]=min(dp[nxt][aa*mod+bb],val+c[i]);
		}
		swap(now,nxt); dp[nxt].clear();
		lv(now);
	}
	for (auto tt:dp[now])
	{
		ll x=tt.first/mod,y=tt.first%mod,val=tt.second;
		ans1.push_back({val,x+y});
	}
	sort(ans1.begin(),ans1.end());
	ll mx=0;
	for (auto tt:ans1)
	{
		if (tt.second<=mx) continue;
		mx=tt.second;
		ans2.push_back(tt);
	}
}

void work()
{
	int siz=ans2.size(); ll cost;
	while (q--)
	{
		cin>>cost;
		int now=0;
		for (int i=20;i>=0;i--)
		if (now+(1<<i)<siz && ans2[now+(1<<i)].first<=cost)
			now+=(1<<i);
		cout<<ans2[now].second<<" ";
	}
}
int main()
{
	ios::sync_with_stdio(false); cin.tie(0);
	init();
	work();
}

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

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

Think step by step.

Response:
```plaintext
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
```

### Explanation:
1. **Input Structure**:
   - The first line contains two integers `n` and `q`, where `n` is the length of the arrays and `q` is the number of queries.
   - The second line contains the array `a` of length `n`.
   - The third line contains the array `b` of length `n`.
   - The fourth line contains the array `c` of length `n`.
   - The fifth line contains the array `d` of length `q`.

2. **Chosen Input**:
   - `n = 3`, `q = 4`
   - `a = [1, 2, 3]`
   - `b = [4, 5, 6]`
   - `c = [1, 1, 1]`
   - `d = [0, 1, 2, 3]`

3. **Why This Input?**:
   - **Small Scale**: The input is small enough to be manageable and easy to verify.
   - **Edge Cases**:
     - **Zero Cost**: The first query with `d = 0` ensures that no swaps are allowed, testing the initial configuration.
     - **Single Swap**: The second query with `d = 1` allows for a single swap, testing the impact of a single swap on the GCD sum.
     - **Multiple Swaps**: The third and fourth queries with `d = 2` and `d = 3` allow for multiple swaps, testing the algorithm's ability to handle multiple swaps and find the optimal solution.

4. **Expected Output**:
   - For `d = 0`: The GCD sum is `2` (no swaps allowed).
   - For `d = 1`: The GCD sum is `3` (one swap allowed, e.g., swapping `a_2` and `b_2`).
   - For `d = 2` and `d = 3`: The GCD sum remains `3` (no additional benefit from more swaps).

This input is designed to expose potential issues in the algorithm, particularly in handling multiple swaps and ensuring the optimal GCD sum is calculated correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
#include<bits/stdc++.h>
    using namespace std;
    char nl = '\n';
    using i64 = long long;

    void solve(int t) {
        // cout << "test #" << t << << nl;
        int n, q; cin >> n >> q;
        set<i64> AA, BB;
        vector<i64> A, B, C, D(q);
        vector<pair<pair<i64, i64>, i64>> aux(n);
        for(auto& a : aux) cin >> a.first.first, AA.insert(a.first.first);
        for(auto& a : aux) cin >> a.first.second, BB.insert(a.first.second);
        for(auto& a : aux) cin >> a.second;
        for(auto& a : D) cin >> a;
        sort(aux.begin(), aux.end());
        for(auto& a : aux) {
            if(A.empty() || A.back() != a.first.first || B.back() != a.first.second) {
                A.push_back(a.first.first); B.push_back(a.first.second); C.push_back(a.second);
            } else C.back() += a.second;
        }
        set<int> l, r;
        for(i64 i = 1; i * i <= A[0]; i++) {
            if(A[0] % i == 0) {
                l.insert(i); l.insert(A[0] / i);
            }
        }
        for(i64 i = 1; i * i <= B[0]; i++) {
            if(B[0] % i == 0) {
                r.insert(i); r.insert(B[0] / i);
            }
        }
        if(AA.size() > 64 || BB.size() > 64) {
            for(auto& x : l) r.insert(x);
            l.clear(); l.insert(1);
            assert(l.size() == 1);
        }
        n = A.size(); // stupidity at it's peak
        map<i64, int> mp;
        for(auto& x : l) {
            for(auto& y : r) {
                i64 cost1 = 0, cost2 = 0;
                for(int i = 0; i < n; i++) {
                    if(A[i] % x != 0 || B[i] % y != 0) {
                        if(A[i] % y == 0 && B[i] % x == 0) cost1 += C[i];
                        else {
                            cost1 = i64(1e18); break;
                        }
                    }
                }
                for(int i = 0; i < n; i++) {
                    if(A[i] % y != 0 || B[i] % x != 0) {
                        if(A[i] % x == 0 && B[i] % y == 0) cost2 += C[i];
                        else {
                            cost2 = i64(1e18); break;
                        }
                    }
                }
                cost1 = min(cost1, cost2);
                mp[cost1] = max(mp[cost1], x + y);
            }
        }
        int mx = 0;
        for(auto& a : mp) {
            mx = max(mx, a.second);
            a.second = mx;
        }
        map<i64, int> val;
        for(auto& a : mp) val[-a.first] = a.second;
        for(auto& d : D) {
            cout << val.lower_bound(-d)->second << " ";
        }
        cout << nl;
    }

    int main() {
        int tt = 1;
        // cin >> tt;
        for(int t = 1; t <= tt; t++) solve(t);
    }

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

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

Think step by step.

Response:
```plaintext
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
```

This input is designed to test the edge case where the cost of swapping is minimal and the values of \(a_i\) and \(b_i\) are small. It ensures that the algorithm must correctly handle the scenario where the maximum GCD sum can be achieved with a single swap or no swaps at all. The values of \(d\) are set to cover the range from 0 to 3, which should expose any issues in the way the algorithm calculates the minimum cost and the maximum GCD sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
// LUOGU_RID: 168610389
#include <bits/stdc++.h>

using namespace std;

const int kMaxN = 5e5 + 10;

const int kMaxV = 1e8 + 10;

const int kMaxD = 770;

int n, q;

int a[kMaxN], b[kMaxN], c[kMaxN];
long long d[kMaxN];

int g1, g2;
long long cost;

vector<int> d1, d2;
vector<int> p1, p2;

void init(int g, vector<int> &d, vector<int> &p) {
	for (int i = 1; i * i <= g; ++i) {
		if (g % i == 0) {
			d.emplace_back(i);
			if (i * i != g)
				d.emplace_back(g / i);
		}
	}
	sort(d.begin(), d.end());
	int t = g;
	for (int i = 2; i * i <= g; ++i) {
		if (t % i == 0) {
			p.emplace_back(i);
			while (t % i == 0) t /= i;
		}
	}
	if (t != 1) p.emplace_back(t);
	sort(p.begin(), p.end());
}

long long f[kMaxD][kMaxD];
int g[kMaxD][kMaxD];

int getId(const vector<int> &d, int x) {
	return lower_bound(d.begin(), d.end(), x) - d.begin();
}

vector<pair<long long, int> > res;

void solve() {
	init(g1, d1, p1), init(g2, d2, p2);
	memset(f, 0, sizeof(f)), memset(g, 0, sizeof(g));
	for (int i = 2; i <= n; ++i) {
		cost += c[i];
		f[getId(d1, __gcd(g1, a[i]))][getId(d2, __gcd(g2, b[i]))] -= c[i];
		++g[getId(d1, __gcd(g1, a[i]))][getId(d2, __gcd(g2, b[i]))];
		++g[getId(d1, __gcd(g1, b[i]))][getId(d2, __gcd(g2, a[i]))];
		int tmp = __gcd(a[i], b[i]);
		--g[getId(d1, __gcd(g1, tmp))][getId(d2, __gcd(g2, tmp))];

		//cout << __gcd(g1, a[i]) << ' ' << __gcd(g2, b[i]) << '\n';
		//cout << __gcd(g1, b[i]) << ' ' << __gcd(g2, a[i]) << '\n';
	}
	for (auto p : p1) {
		for (int i = d1.size() - 1; ~i; --i) {
			int x = d1[i];
			if (g1 % (x * p)) continue;
			int t = getId(d1, x * p);
			for (int j = 0; j < d2.size(); ++j) {
				f[i][j] += f[t][j];
				g[i][j] += g[t][j];
				//cout << "^ " << d1[i] << ' ' << d2[j] << ' ' << d1[t] << ' ' << d2[j] << '\n';
			}
		}
	}
	for (auto p : p2) {
		for (int j = d2.size() - 1; ~j; --j) {
			int y = d2[j];
			if (g2 % (y * p)) continue;
			int t = getId(d2, y * p);
			for (int i = 0; i < d1.size(); ++i) {
				f[i][j] += f[i][t];
				g[i][j] += g[i][t];
			}
		}
	}
	for (int i = 0; i < d1.size(); ++i) {
		for (int j = 0; j < d2.size(); ++j) {
			if (g[i][j] == n - 1) {
				res.emplace_back(cost + f[i][j], d1[i] + d2[j]);
				//cout << "* " << d1[i] << ' ' << d2[j] << ' ' << f[i][j] + cost << ' ' << g[i][j] << '\n';
			}
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n >> q;
	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) cin >> c[i];

	g1 = a[1], g2 = b[1], cost = 0;
	solve();

	g1 = b[1], g2 = a[1], cost = c[1];
	solve();

	sort(res.begin(), res.end());
	for (int i = 1; i < res.size(); ++i) {
		res[i].second = max(res[i].second, res[i - 1].second);
	}

	for (int i = 1; i <= q; ++i) {
		cin >> d[i];
		cout << (--upper_bound(res.begin(), res.end(), make_pair(d[i], (int)1e9)))->second << ' ';
	}
	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
3 3
2 4 6
3 6 9
1 2 3
0 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

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

const int N=500010;
ll a[N],b[N],c[N],d[N];

ll gcd(ll a,ll b){
    return a==0?b:gcd(b%a,a);
}

void solve(){
    int n,q;cin>>n>>q;
    ll g=0;
    for(int i=1;i<=n;i++)cin>>a[i],g=gcd(g,a[i]);
    for(int i=1;i<=n;i++)cin>>b[i],g=gcd(g,b[i]);
    for(int i=1;i<=n;i++)cin>>c[i];
    for(int i=1;i<=q;i++)cin>>d[i];

    set<ll> s[2];
    int cur=0;
    s[0].insert(a[1]);
    s[0].insert(b[1]);
    for(int i=2;i<=n;i++){
        s[1^cur].clear();
        for(auto j:s[cur]){
            s[1^cur].insert(gcd(j,a[i]));
            s[1^cur].insert(gcd(j,b[i]));
        }
        cur^=1;
    }
    // for(auto i:s[cur])cout<<i<<" ";
    // cout<<"\n";

    // if(a[1]==75998670){
    //     cout<<s[cur].size()<<"\n";
    //     exit(0);
    // }

    vector<pair<ll,ll>> tmpv;
    for(auto ga:s[cur]){
        for(auto gb:s[cur]){
            if(gcd(ga,gb)>g)continue;
            if(gb>ga)continue;
            ll sum1=0,sum2=0;
            bool f=true;
            for(int i=1;i<=n;i++){
                bool ok1=(a[i]%ga==0&&b[i]%gb==0);
                bool ok2=(a[i]%gb==0&&b[i]%ga==0);
                if(ok1&&ok2)continue;
                else if(ok1)sum2+=c[i];
                else if(ok2)sum1+=c[i];
                else{
                    f=false;
                    break;
                }
            }
            if(f)tmpv.push_back({ga+gb,min(sum1,sum2)});
        }
    }
    sort(tmpv.begin(),tmpv.end());
    vector<pair<ll,ll>> ansv;
    for(auto i:tmpv){
        if(ansv.size()>0&&i.first==ansv.back().first)continue;
        while(ansv.size()>0&&i.second<=ansv.back().second)ansv.pop_back();
        ansv.push_back({i.first,i.second});
    }
    //for(auto i:ansv)cout<<i.first<<" "<<i.second<<"\n";
    for(int I=1;I<=q;I++){
        ll t=d[I];
        int l=0,r=ansv.size()-1;
        ll ans=-1;
        while(r>=l){
            int m=(l+r)/2;
            if(t>=ansv[m].second){
                l=m+1;
                ans=ansv[m].first;
            }
            else r=m-1;
        }
        cout<<ans<<" ";
    }

}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int n=1;
    //cin>>n;
    while(n--)solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
co = list(map(int, input().split()))
d = list(map(int, input().split()))

div = []
x = 1
while x * x <= a[0]:
    if a[0] % x == 0:
        div.append(x)
        if x * x != a[0]:
            div.append(a[0]//x)
    x += 1
x = 1
while x * x <= b[0]:
    if b[0] % x == 0:
        div.append(x)
        if x * x != b[0]:
            div.append(b[0]//x)
    x += 1

sdiv = set()
for v in div:
    flg = 1
    for i in range(n):
        if a[i] % v != 0 and b[i] % v != 0:
            flg = 0
            break
    if flg:
        sdiv.add(v)

div = list(sdiv)
div.sort()

m = len(div)
par = [0 for _ in range(m)]
for i, x in enumerate(div):
    for j in range(i + 1, m):
        if div[j] % x == 0:
            par[i] = j
            break

# print("div:",div)
# print("par:",par)
# print("ed:")
# for i in range(m):
#     print(ed[i])
# print("end edge")

def Find(v):
    nedA, nedB = [], []
    for i in range(n):
        fA, fB = a[i] % v == 0, b[i] % v == 0
        if fA == True and fB == False:
            nedA.append(i)
        if fA == False and fB == True:
            nedB.append(i)
    return nedA, nedB

used = [0 for _ in range(m)]
line = []
for i, x in enumerate(div):
    if used[i]:
        continue
    lin = []
    now = i
    tim = 0
    ta, tb = [m + 1 for _ in range(n)], [m + 1 for _ in range(n)]
    while used[now] == 0:
        nA, nB = Find(div[now])
        for x in nA:
            ta[x] = min(ta[x], tim)
        for x in nB:
            tb[x] = min(tb[x], tim)
        lin.append(div[now])
        used[now] = 1

        if par[now] == 0:
            break
        now = par[now]
        tim += 1
    line.append((lin, ta, tb))
    # print(line[-1])

# print("end line")

ansd = []

k = len(line)

# for v in line:
#     print(v)

for i in range(k):
    for j in range(k):
        # nedA_i -> 保留点 nedB_i -> 交换点 nedA_j -> 交换点 nedB_j -> 保留点
        # nedA_i & nedA_j = empty AND nedB_i & nedB_j = empty
        # cost = nedB_i | nedA_j
        la, lb = line[i], line[j]
        # print("la & lb :", la, lb)
        m1, m2 = len(la[0]), len(lb[0])
        ban = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
        cost = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
        for idx in range(n):
            # print("idx:", idx)
            # print(la[1][idx], lb[1][idx])
            # print(la[2][idx], lb[2][idx])
            ban[la[1][idx]][lb[1][idx]] = 1
            ban[la[2][idx]][lb[2][idx]] = 1
            cost[la[2][idx]][0] += co[idx]
            cost[0][lb[1][idx]] += co[idx]
            cost[la[2][idx]][lb[1][idx]] -= co[idx]

        for r in range(m1):
            for c in range(m2):
                if r > 0:
                    cost[r][c] += cost[r-1][c]
                    ban[r][c] = max(ban[r][c], ban[r-1][c])
                if c > 0:
                    cost[r][c] += cost[r][c-1]
                    ban[r][c] = max(ban[r][c], ban[r][c-1])
                if r > 0 and c > 0:
                    cost[r][c] -= cost[r - 1][c - 1]

                if ban[r][c] == 0:
                    ansd.append((cost[r][c], la[0][r] + lb[0][c]))

ansd.sort()
ansd.append((1<<60,0))
# print(ansd)
qry = [(x, i) for (i,x) in enumerate(d)]
qry.sort()
ans = [0 for _ in range(q)]
pos = 0
premx = ansd[0][1]
for x, i in qry:
    while x >= ansd[pos + 1][0]:
        pos += 1
        premx = max(premx, ansd[pos][1])
    ans[i] = premx

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
1 2 3
4 5 6
1 1 1
0 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
co = list(map(int, input().split()))
d = list(map(int, input().split()))

div = []
x = 1
while x * x <= a[0]:
    if a[0] % x == 0:
        div.append(x)
        if x * x != a[0]:
            div.append(a[0]//x)
    x += 1
x = 1
while x * x <= b[0]:
    if b[0] % x == 0:
        div.append(x)
        if x * x != b[0]:
            div.append(b[0]//x)
    x += 1

sdiv = set()
for v in div:
    flg = 1
    for i in range(n):
        if a[i] % v != 0 and b[i] % v != 0:
            flg = 0
            break
    if flg:
        sdiv.add(v)

div = list(sdiv)
div.sort()

m = len(div)
par = [0 for _ in range(m)]
for i, x in enumerate(div):
    for j in range(i + 1, m):
        if div[j] % x == 0:
            par[i] = j
            break

# print("div:",div)
# print("par:",par)
# print("ed:")
# for i in range(m):
#     print(ed[i])
# print("end edge")

def Find(v, ta, tb, tim):
    for i in range(n):
        fA, fB = a[i] % v == 0, b[i] % v == 0
        if fA == True and fB == False:
            ta[i] = min(ta[i], tim)
        if fA == False and fB == True:
            tb[i] = min(tb[i], tim)

used = [0 for _ in range(m)]
line = []
for i, x in enumerate(div):
    if used[i]:
        continue
    lin = []
    now = i
    tim = 0
    ta, tb = [m + 1 for _ in range(n)], [m + 1 for _ in range(n)]
    while used[now] == 0:
        Find(div[now], ta, tb, tim)
        lin.append(div[now])
        used[now] = 1

        if par[now] == 0:
            break
        now = par[now]
        tim += 1
    line.append((lin, ta, tb))
    # print(line[-1])

# print("end line")

ansd = []

k = len(line)

# for v in line:
#     print(v)
ban = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
cost = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
for i in range(k):
    for j in range(k):
        # nedA_i -> 保留点 nedB_i -> 交换点 nedA_j -> 交换点 nedB_j -> 保留点
        # nedA_i & nedA_j = empty AND nedB_i & nedB_j = empty
        # cost = nedB_i | nedA_j
        la, lb = line[i], line[j]
        # print("la & lb :", la, lb)
        m1, m2 = len(la[0]), len(lb[0])
        for i_ in range(m1 + 2):
            for j_ in range(m2 + 2):
                ban[i_][j_] = 0
                cost[i_][j_] = 0

        for idx in range(n):
            # print("idx:", idx)
            # print(la[1][idx], lb[1][idx])
            # print(la[2][idx], lb[2][idx])
            ban[la[1][idx]][lb[1][idx]] = 1
            ban[la[2][idx]][lb[2][idx]] = 1
            cost[la[2][idx]][0] += co[idx]
            cost[0][lb[1][idx]] += co[idx]
            cost[la[2][idx]][lb[1][idx]] -= co[idx]

        for r in range(m1):
            for c in range(m2):
                if r > 0:
                    cost[r][c] += cost[r-1][c]
                    ban[r][c] = max(ban[r][c], ban[r-1][c])
                if c > 0:
                    cost[r][c] += cost[r][c-1]
                    ban[r][c] = max(ban[r][c], ban[r][c-1])
                if r > 0 and c > 0:
                    cost[r][c] -= cost[r - 1][c - 1]

                if ban[r][c] == 0:
                    ansd.append((cost[r][c], la[0][r] + lb[0][c]))

ansd.sort()
ansd.append((1<<60,0))
# print(ansd)
qry = [(x, i) for (i,x) in enumerate(d)]
qry.sort()
ans = [0 for _ in range(q)]
pos = 0
premx = ansd[0][1]
for x, i in qry:
    while x >= ansd[pos + 1][0]:
        pos += 1
        premx = max(premx, ansd[pos][1])
    ans[i] = premx

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 4
1 2 3
4 5 6
10 10 10
0 10 20 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum GCD Sum Queries
time_limit_ms: 5000
memory_limit_mb: 512
Description: For $k$ positive integers $x_1, x_2, \ldots, x_k$, the value $\gcd(x_1, x_2, \ldots, x_k)$ is the greatest common divisor of the integers $x_1, x_2, \ldots, x_k$ — the largest integer $z$ such that all the integers $x_1, x_2, \ldots, x_k$ are divisible by $z$.

You are given three arrays $a_1, a_2, \ldots, a_n$, $b_1, b_2, \ldots, b_n$ and $c_1, c_2, \ldots, c_n$ of length $n$, containing positive integers.

You also have a machine that allows you to swap $a_i$ and $b_i$ for any $i$ ($1 \le i \le n$). Each swap costs you $c_i$ coins.

Find the maximum possible value of $$\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)$$ that you can get by paying in total at most $d$ coins for swapping some elements. The amount of coins you have changes a lot, so find the answer to this question for each of the $q$ possible values $d_1, d_2, \ldots, d_q$.

Input: There are two integers on the first line — the numbers $n$ and $q$ ($1 \leq n \leq 5 \cdot 10^5$, $1 \leq q \leq 5 \cdot 10^5$).

On the second line, there are $n$ integers — the numbers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^8$).

On the third line, there are $n$ integers — the numbers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 10^8$).

On the fourth line, there are $n$ integers — the numbers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$).

On the fifth line, there are $q$ integers — the numbers $d_1, d_2, \ldots, d_q$ ($0 \leq d_i \leq 10^{15}$).

Output: Print $q$ integers — the maximum value you can get for each of the $q$ possible values $d$.

Examples:
input:
3 4
1 2 3
4 5 6
1 1 1
0 1 2 3
output:
2 3 3 3

input:
5 5
3 4 6 8 4
8 3 4 9 3
10 20 30 40 50
5 55 13 1000 113
output:
2 7 3 7 7

input:
1 1
3
4
5
0
output:
7

Note: In the first query of the first example, we are not allowed to do any swaps at all, so the answer is $\gcd(1, 2, 3) + \gcd(4, 5, 6) = 2$. In the second query, one of the ways to achieve the optimal value is to swap $a_2$ and $b_2$, then the answer is $\gcd(1, 5, 3) + \gcd(4, 2, 6) = 3$.

In the second query of the second example, it's optimal to perform swaps on positions $1$ and $3$, then the answer is $\gcd(3, 3, 6, 9, 3) + \gcd(8, 4, 4, 8, 4) = 7$ and we have to pay $40$ coins in total.

Buggy Code:
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
co = list(map(int, input().split()))
d = list(map(int, input().split()))

div = []
x = 1
while x * x <= a[0]:
    if a[0] % x == 0:
        div.append(x)
        if x * x != a[0]:
            div.append(a[0]//x)
    x += 1
x = 1
while x * x <= b[0]:
    if b[0] % x == 0:
        div.append(x)
        if x * x != b[0]:
            div.append(b[0]//x)
    x += 1

sdiv = set()
for v in div:
    flg = 1
    for i in range(n):
        if a[i] % v != 0 and b[i] % v != 0:
            flg = 0
            break
    if flg:
        sdiv.add(v)

div = list(sdiv)
div.sort()

m = len(div)
par = [0 for _ in range(m)]
for i, x in enumerate(div):
    for j in range(i + 1, m):
        if div[j] % x == 0:
            par[i] = j
            break

# print("div:",div)
# print("par:",par)
# print("ed:")
# for i in range(m):
#     print(ed[i])
# print("end edge")

def Find(v):
    nedA, nedB = [], []
    for i in range(n):
        fA, fB = a[i] % v == 0, b[i] % v == 0
        if fA == True and fB == False:
            nedA.append(i)
        if fA == False and fB == True:
            nedB.append(i)
    return nedA, nedB

used = [0 for _ in range(m)]
line = []
for i, x in enumerate(div):
    if used[i]:
        continue
    lin = []
    now = i
    tim = 0
    ta, tb = [m + 1 for _ in range(n)], [m + 1 for _ in range(n)]
    while used[now] == 0:
        nA, nB = Find(div[now])
        for x in nA:
            ta[x] = min(ta[x], tim)
        for x in nB:
            tb[x] = min(tb[x], tim)
        lin.append(div[now])
        used[now] = 1

        if par[now] == 0:
            break
        now = par[now]
        tim += 1
    line.append((lin, ta, tb))
    # print(line[-1])

# print("end line")

ansd = []

k = len(line)

# for v in line:
#     print(v)
ban = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
cost = [ [0 for _ in range(m + 5)] for _ in range(m + 5)]
for i in range(k):
    for j in range(k):
        # nedA_i -> 保留点 nedB_i -> 交换点 nedA_j -> 交换点 nedB_j -> 保留点
        # nedA_i & nedA_j = empty AND nedB_i & nedB_j = empty
        # cost = nedB_i | nedA_j
        la, lb = line[i], line[j]
        # print("la & lb :", la, lb)
        m1, m2 = len(la[0]), len(lb[0])
        for i_ in range(m1 + 2):
            for j_ in range(m2 + 2):
                ban[i_][j_] = 0
                cost[i_][j_] = 0

        for idx in range(n):
            # print("idx:", idx)
            # print(la[1][idx], lb[1][idx])
            # print(la[2][idx], lb[2][idx])
            ban[la[1][idx]][lb[1][idx]] = 1
            ban[la[2][idx]][lb[2][idx]] = 1
            cost[la[2][idx]][0] += co[idx]
            cost[0][lb[1][idx]] += co[idx]
            cost[la[2][idx]][lb[1][idx]] -= co[idx]

        for r in range(m1):
            for c in range(m2):
                if r > 0:
                    cost[r][c] += cost[r-1][c]
                    ban[r][c] = max(ban[r][c], ban[r-1][c])
                if c > 0:
                    cost[r][c] += cost[r][c-1]
                    ban[r][c] = max(ban[r][c], ban[r][c-1])
                if r > 0 and c > 0:
                    cost[r][c] -= cost[r - 1][c - 1]

                if ban[r][c] == 0:
                    ansd.append((cost[r][c], la[0][r] + lb[0][c]))

ansd.sort()
ansd.append((1<<60,0))
# print(ansd)
qry = [(x, i) for (i,x) in enumerate(d)]
qry.sort()
ans = [0 for _ in range(q)]
pos = 0
premx = ansd[0][1]
for x, i in qry:
    while x >= ansd[pos + 1][0]:
        pos += 1
        premx = max(premx, ansd[pos][1])
    ans[i] = premx

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
1 2 3
4 5 6
10 10 10
0 20
```