**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

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

#define X first
#define Y second

bool used[200005];
priority_queue<pair<int, int>> pq;

void init(vector<int>& v, int node, int start, int end)
{
    if(start == end)
        v[node] = 1;
    else
    {
        init(v, node*2, start, (start+end)/2);
        init(v, node*2+1, (start+end)/2+1, end);
        v[node] = v[node*2] + v[node*2+1];
    }
}

int query(vector<int>& v, int node, int start, int end, int left, int right)
{
    if(right < start || end < left)
        return 0;
    if(left <= start && end <= right)
        return v[node];

    int l = query(v, node*2, start, (start+end)/2, left, right);
    int r = query(v, node*2+1, (start+end)/2+1, end, left, right);
    return l + r;
}

void update(vector<int>& v, int node, int start, int end, int pos)
{
    if(pos < start || end < pos)
        return;
    if(start == end)
    {
        v[node] = 0;
        return;
    }

    update(v, node*2, start, (start+end)/2, pos);
    update(v, node*2+1, (start+end)/2+1, end, pos);
    v[node] = v[node*2] + v[node*2+1];
}

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

    int t;
    cin >> t;
    while(t--)
    {
        string s;
        cin >> s;

        int sz = (int)s.size();
        fill(used, used+(int)s.size(), 0);
        while(!pq.empty())
            pq.pop();

        for(int i = 0; i < (int)s.size(); i++)
            pq.push({s[i]-'0'-i, i});

        int h = (int)ceil(log2(sz));
        int tree_size = (1<<(h+1));
        vector<int> segtree(tree_size);

        init(segtree, 1, 0, sz-1);

        int pos = 0, cur = s[0]-'0';
        for(int i = 0; i < (int)s.size(); i++)
        {
            while(used[pq.top().Y])
            {
                update(segtree, 1, 0, sz-1, pq.top().Y);
                pq.pop();
            }
            while(used[pos])
            {
                update(segtree, 1, 0, sz-1, pos);
                pos++;
            }

            int val = pq.top().X + min(i, pq.top().Y);
            if(pq.top().Y < i)
                val -= query(segtree, 1, 0, sz-1, 0, pq.top().Y-1);
            if(s[pos]-'0' >= val)
            {
                cout << s[pos]-'0';
                used[pos] = 1;
            }
            else
            {
                cout << val;
                used[pq.top().Y] = 1;
            }
        }

        cout << '\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
1
1111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
t = int(input())
for i in range(0,t):
    st = list(map(int, input()))
    res = ""
    while len(st) > 1:
        v = st[0]
        j=0
        idx = 0
        while j < min(len(st), 8-v):
            if (st[j] - j) > v:
                v = st[j] - j
                idx = j
            j += 1
        res += str(v)
        st.pop(idx)
    res += str(st[0])
    print(res)

**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
123456789
111111111
999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
import sys
import math
from collections import deque
import random
offline=0
mod=10**9+7
rand=random.randint(1,100)
input = lambda: sys.stdin.readline().rstrip()
# Remove these 2 lines while submitting your code online
# while while outputting a list(lst) use print(*lst)
#make sure that the array is sorted before doing binary search
if(offline):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')
def fg():
    return map(int,input().split())
def fgh():
    return list(map(int,input().split()))
def binsch(n,k):
    l=0
    r=2*10**9+3
    while l+1<r:
        mid=(l+r)//2
        if con(mid,n)<=k:
            l=mid
        else:
            r=mid
    return l
def presum(l):
    pr=[0]*(len(l))
    m=0
    for i in range(len(l)):
        m+=l[i]
        pr[i]=m
    return pr
def ap(pos,arr,s1,s2):
    if pos==len(arr):
        return max(s1,s2)
    ch=ap(pos+1,arr,s1+arr[pos],s2)
    nch=ap(pos+1,arr,s1,s2+arr[pos])
    return min(ch,nch)

def con(a,b,x,y):
    return math.sqrt((a-x)**2+(b-y)**2)

def lis_end_at_i(arr, i):
    # Base case
    if i == 0:
        return 1

    # Consider all elements on the left of i,
    # recursively compute LISs ending with
    # them and consider the largest
    mx = 1
    for prev in range(i):
        if arr[prev] < arr[i]:
            mx = max(mx, lis_end_at_i(arr, prev) + 1)
    return mx

def lis(arr):
    n = len(arr)
    res = 1
    for i in range(1, n):
        res = max(res, lis_end_at_i(arr, i))
    return res

def check(l):
    sm=0
    for i in range(len(l)-1):
        if l[i]==l[i+1]:
            sm+=1
    return sm

def root(n):
    l=[1]
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            l.append(i)
    return l

def cc(n):
    if n==1:
        return 1
    else:
        return -1

def chc(l):
    return (sum(l),-max(l))

def swap(a,i,pos):
    for j in range(pos,0,1):
        a[i+j],a[i+j-1]=a[i+j-1],a[i+j]
        a[i+j-1]-=1

def main():
    _t_=int(input())
    for _ in range(_t_):
        a=list(input())
        a=list(map(int,a))
        n=len(a)
        for i in range(n):
            left=min(10,n-i-1)
            m=a[i]
            pos=-1
            for j in range(1,left+1):
                if m<a[i+j]-j:
                    m=a[i+j]-j
                    pos=j
            if pos==-1 or a[i]==0:
                continue
            for j in range(pos,0,-1):
                a[i+j],a[i+j-1]=a[i+j-1],a[i+j]
                a[i+j-1]-=1
        ans=""
        for i in range(n-1):
            if a[i]==0 and a[i+1]>1:
                a[i],a[i+1]=a[i+1],a[i]
                a[i]-=1
        for i in range(n):
            ans+=str(a[i])

        print(ans)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the string has specific patterns that could cause the algorithm to fail.
2. **Boundary Conditions**: Test cases where the string length is at the boundary (minimum or maximum).
3. **Special Characters**: Test cases with digits that are critical for the algorithm's logic, such as `1` and `9`.

Let's think through these points:

### Edge Cases
- **String with all digits being `1`**: This could cause the algorithm to fail if it doesn't handle the case where no digit can be decreased.
- **String with a high number of `9`s**: This could test the algorithm's ability to handle the maximum digit and its decrements.
- **String with a mix of digits where the leftmost digit is `1` and the rest are high**: This could test the algorithm's ability to handle the leftmost digit constraint.

### Boundary Conditions
- **Single-digit string**: This is the smallest valid input.
- **Maximum length string**: This is the largest valid input, but since we need a small-scale test, we can use a shorter length that still tests the boundary conditions.

### Special Characters
- **Strings with `1` and `9`**: These digits are critical because they are the limits of the digit range.

### Challenging Test Input
Let's create a test input that includes these edge cases and special characters:

```plaintext
4
1
999
1999
1111111111
```

### Explanation
1. **`1`**: A single-digit string. This tests the boundary condition of the shortest possible input.
2. **`999`**: A string with all digits being `9`. This tests the algorithm's ability to handle the maximum digit and its decrements.
3. **`1999`**: A string where the leftmost digit is `1` and the rest are high. This tests the algorithm's ability to handle the leftmost digit constraint.
4. **`1111111111`**: A string with all digits being `1`. This tests the algorithm's ability to handle a string where no digit can be decreased.

This test input should expose any issues in the buggy code related to handling edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
for _ in range(int(input())):
    def fun(l):
        i = len(l) - 1
        while i >= 0:
            if i > 0 and int(l[i]) > int(l[i - 1]) + 1:
                temp = int(l[i - 1])
                l[i - 1] = str(int(l[i]) - 1)
                l[i] = str(temp)
            i -= 1
        number = int("".join(l))
        return number

    s = input()
    l = list(s)

    maxx = 0
    for _ in range(1000):
        number = fun(l)
        maxx = max(maxx, number)
    print(str(maxx))

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

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

Think step by step.

Response:
```plaintext
1
1234567899
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

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

public class D_Digital_string_maximization {
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {}
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }

        char[] nextCharArray() {
            try {
                String line = br.readLine();
                return line.toCharArray();
            } catch (IOException e) {
                e.printStackTrace();
                return new char[0];
            }
        }
    }

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner s = new FastScanner();

        int t = s.nextInt();

        while(t-- != 0) {
            char[] num = s.nextCharArray();

            for(int i = 0; i < num.length; i++) {
                if(num[i] == '0') continue;

                for(int j = 1; j <= 9; j++) {
                    if(i+j >= num.length) break;

                    int val_diff = num[i+j] - num[i];

                    if(val_diff - j > 0) {
                        for(int k = i+j; k > i; k--) {
                            int temp = num[k] - '0';
                            num[k] = num[k-1];
                            num[k-1] = ((temp - 1) + "").charAt(0);
                        }
                        i--;
                        break;
                    }
                }
            }

            for(int i = 0; i < num.length; i++) {
                out.print(num[i]);
            }
            out.println();
        }
        out.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
6
19
1709
11555
51476
9876543210
5891917899
```

This test input includes a variety of cases:
1. A simple two-digit number (`19`) to test basic operation.
2. A number with a zero in the middle (`1709`) to test handling of zero.
3. A number with multiple identical digits (`11555`) to test handling of repeated digits.
4. A number with a sequence of digits that can be rearranged (`51476`) to test more complex operations.
5. A number that is already maximized (`9876543210`) to test the edge case where no operations are needed.
6. A number with a mix of high and low digits (`5891917899`) to test the algorithm's ability to handle more complex rearrangements.

These cases are designed to cover a range of scenarios that could expose potential bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

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

public class a
{
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        int[] nextIntArray(int n)
        {
            int[] ans = new int[n];
            for(int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        Integer[] nextIntegerArray(int n)
        {
            Integer[] ans = new Integer[n];
            for(int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        long[] nextLongArray(int n)
        {
            long[] ans = new long[n];
            for(int i = 0; i < n; i++)
                ans[i] = nextLong();
            return ans;
        }

        String[] nextStringArray(int n)
        {
            String[] ans = new String[n];
            for(int i = 0; i < n; i++)
                ans[i] = next();
            return ans;
        }

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static final int MOD = (int)(1e9) + 7;
    static final long INF = (long)(1e18);
    static final int INF_INT = (int)(1e9);

    @SuppressWarnings("unused")
    private static int gcd(int a, int b)
    {
        if(a % b == 0)
            return b;
        return gcd(b, a % b);
    }

    static class Pair
    {
        int index;
        boolean wasUnioned;

        Pair(int a, boolean b)
        {
            index = a;
            wasUnioned = b;
        }
    }

    static class DSU
    {
        int n;
        int[] parent;
        List<List<Integer>> ar;

        public DSU(int _n, int[] _ar)
        {
            n = _n;
            parent = new int[n];
            ar = new ArrayList<>();
            for(int i = 0; i < n; i++)
            {
                parent[i] = i;
                ar.add(new ArrayList<>());
                ar.get(i).add(_ar[i]);
            }
        }

        public int findParent(int node)
        {
            if(node == parent[node])
                return node;
            return parent[node] = findParent(parent[node]);
        }

        public boolean union(int u, int v)
        {
            int rootU = findParent(u), rootV = findParent(v);
            if(ar.get(rootU).get(ar.get(rootU).size() - 1) != ar.get(rootV).get(0) - 1 || rootU == rootV)
                return false;
            parent[rootV] = rootU;
            ar.get(rootU).add(ar.get(rootV).get(0));
            ar.get(rootV).clear();
            return true;
        }

        public boolean same(int u, int v)
        {
            return findParent(u) == findParent(v);
        }
    }

    static int[][][] dp;
    static int[][] trace;
    static Function<Integer, Integer> f1;

    public static void main(String args[]) throws IOException
    {
        FastReader fr = new FastReader();
        // Scanner fr = new Scanner(System.in);
        // Scanner fr = new Scanner(new File("C:\\Users\\ykgup\\Downloads\\test_input.txt"));
        int _t = fr.nextInt();
        // int _t = 1;
        // int _case = 1;
        StringBuilder _v = new StringBuilder();
        PrintWriter out = new PrintWriter(System.out);
        while(_t-- > 0)
        {
            char[] ar = fr.next().toCharArray();
            int n = ar.length;
            for(int i = 0; i < n; i++)
            {
                // Look into next 10 whichever gives max here
                int pos = i;
                for(int j = i+1; j < n && j <= i + 10; j++)
                {
                    if(ar[j] - (j - i) >= ar[pos])
                        pos = j;
                }
                char max = (char)(ar[pos] - (pos - i));
                for(int j = pos; j > i; j--)
                    ar[j] = ar[j-1];
                ar[i] = max;
            }
            _v.append(ar);
            _v.append("\n");
            // _v.append("Case");
            // _v.append(" ");
            // _v.append(_case++);
            // _v.append(": ");
        }
        out.print(_v);
        out.flush();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios where the code might fail. The code appears to be trying to maximize the string by moving the largest possible digit to the leftmost position, but it has some logical flaws in how it handles the operations and the constraints.

### Key Points to Consider:
1. **Handling of Leading Zeros**: The code should not allow operations that result in leading zeros.
2. **Maximum Operations**: The code should ensure that the maximum string is obtained after any number of operations.
3. **Digit Decrease and Swap**: The code should correctly handle the decrease and swap operations.

### Potential Edge Cases:
- **Single Digit Strings**: Strings with only one digit.
- **Strings with Repeated Digits**: Strings where the same digit appears multiple times.
- **Strings with Digits Close to 0**: Strings where the digits are close to 0, which can cause issues with the decrease operation.
- **Strings with Digits in Decreasing Order**: Strings where the digits are already in decreasing order, which can make it hard to find a valid operation.

### Challenging Test Input:
Let's create a small-scale test input that includes these edge cases:

```plaintext
6
1
999
1234567890
210
1111111111
9000000000
```

### Explanation:
1. **"1"**: A single digit string. The code should handle this without performing any operations.
2. **"999"**: A string with repeated digits. The code should ensure the string remains the same since no valid operations can be performed.
3. **"1234567890"**: A string with digits in increasing order. The code should maximize the string by moving the highest possible digit to the leftmost position.
4. **"210"**: A string with digits close to 0. The code should handle the decrease operation correctly without resulting in leading zeros.
5. **"1111111111"**: A string with repeated digits. The code should ensure the string remains the same since no valid operations can be performed.
6. **"9000000000"**: A string with a high leading digit followed by zeros. The code should handle the decrease and swap operations correctly without resulting in leading zeros.

This test input should help expose any bugs in the given code, especially in how it handles the constraints and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
t = int(input())
for i in range(t):
    n = int(input())
    for j in range(len(str(n))):
        maxx = 0
        ind = 0
        for k in range(min(len(str(n)) - j - 1, 10)):
            if maxx < int(str(n)[j+k+1]) - k - 1:
                ind = j+k+1
                maxx = int(str(n)[j+k+1]) - k - 1
        if maxx > int(str(n)[j]):
            n = int(str(n)[:j] + str(maxx) + str(n)[j:ind] + str(n)[ind+1:])
    print(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
11111
19999
19990
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
t = int(input())
final = []

def ans(num):
    num_list = [int(digit) for digit in str(num)]
    length = len(num_list)

    while True:
        differences = []
        for i in range(length - 1):
            if num_list[i + 1] - num_list[i] > 1:
                differences.append((num_list[i + 1] - num_list[i], i))

        if not differences:
            break

        differences.sort(reverse=True, key=lambda x: x[0])

        max_diff, index = differences[0]
        num_list[index + 1] -= 1
        num_list[index], num_list[index + 1] = num_list[index + 1], num_list[index]

    return int(''.join(map(str, num_list)))

for k in range(t):
    c = ans(input().strip())
    final.append(c)

for element in final:
    print(element)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The main issue in the buggy code is that it only considers the largest difference between adjacent digits to perform the operation. This can lead to suboptimal results, especially when there are multiple possible operations that could lead to a lexicographically larger string.

### Edge Cases to Consider:
1. **Strings with multiple large differences**: If there are multiple large differences between adjacent digits, the code might not choose the optimal one.
2. **Strings with repeated digits**: Repeated digits can make it difficult to identify the best operation.
3. **Strings with small differences**: Small differences might be ignored, leading to suboptimal results.
4. **Strings with leading 1s**: Leading 1s can be tricky because they cannot be decreased.

### Challenging Test Input:
Let's create a test case that includes a string with multiple large differences and repeated digits, which should expose the bug in the code.

```plaintext
5
199
12345
11111
98887
59879
```

### Explanation:
1. **199**: This string has a large difference between the second and third digits. The optimal solution should be `991`, but the buggy code might not handle this correctly.
2. **12345**: This string has small differences between adjacent digits. The optimal solution should be `54321`, but the buggy code might not handle this correctly.
3. **11111**: This string has no large differences, and all digits are the same. The optimal solution should be `11111`, but the buggy code should handle this correctly.
4. **98887**: This string has multiple large differences. The optimal solution should be `98887`, but the buggy code might not handle this correctly.
5. **59879**: This string has a mix of large and small differences. The optimal solution should be `99885`, but the buggy code might not handle this correctly.

This test input should help in identifying the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
import java.util.*;

public class DigitalStringMaximization {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();
        for (int i = 0; i < t; i++) {
            String s = sc.nextLine();
            System.out.println(maximize(s));
        }
        sc.close();
    }

    private static String maximize(String s) {
        char[] arr = s.toCharArray();
        int n = arr.length;
        boolean found;
        do {
            found = false;
            for (int j = n - 1; j > 0; j--) {
                if (arr[j] > '0') {
                    char[] copy = arr.clone();
                    copy[j]--;
                    char temp = copy[j];
                    copy[j] = copy[j-1];
                    copy[j-1] = temp;
                    if (compare(copy, arr) > 0) {
                        arr = copy;
                        found = true;
                        break;
                    }
                }
            }
        } while (found);
        return new String(arr);
    }

    private static int compare(char[] a, char[] b) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] != b[i]) {
                return a[i] - b[i];
            }
        }
        return 0;
    }
}

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

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

Think step by step.

Response:
```plaintext
6
19
1709
11555
51476
9876543210
5891917899
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define ull unsigned long long
#define fr first
#define sc second
#define endl '\n'
#define rep(k, n) for (int k = 0; k < n; k++)
#define repp(k, n) for (int k = 1; k <= n; k++)
#define inn(k, n, a)           \
   for (int k = 0; k < n; k++) \
      cin >> a[k];
const ll mod = 1e9 + 7;

bool checker(char x){
   if(x == 'A' or x == 'E' or x == 'I' or x == 'O' or x == 'U') return true;
   else return false;
}

vector<pair<int,int>> v[200001];

void solve(){

   string s;
   cin >> s;
   // rep(k , s.size()){
   //    ll x = s[k] - '0';
   //    v[k].clear();
   //    for(int i = k ; i >= 0 ; i --){
   //       if(x < 0){
   //          break;
   //       }
   //       v[i].pb({k , x});
   //       x--;
   //    }
   // }
   map<int , bool> m;
   map<int , char> used;
   int rm = 0;
   string ss = s;
   deque <int> q;
   int vpos = 0;
   vector <pair<int,int>> z;
   rep(k , s.size()){
      z.pb({k , s[k] - '0' - k});
   }

   rep(k , s.size()){
      // for(auto i: z){
      //    //if(!m[i.second])
      //    cout << i.second << " " << i.first << "  // ";
      // }
      // cout << endl;
      // cout << endl;

      sort(z.begin() , z.end());

      int mx = 0;
      int pos = 0;
      int ls = 0;
      for(int i = 0 ; i < z.size() ; i++){
         used[z[i].first] = true;
         if(z[i].second > mx and !m[z[i].first]){
            mx = z[i].second;
            pos = i;
         }
      }

      cout << z[pos].second;
      m[z[pos].first] = true;
      z.erase(z.begin() + pos);

      // if(k != s.size() - 1)
      //    for(auto i:v[k + 1]){
      //       if(!used[i.first]){
      //          z.pb(i);
      //       }
      //    }

      for(int i = 0 ; i < z.size() ; i ++){

         z[i].second = min(z[i].second + 1 , (ss[z[i].first] - '0') - i);

      }

   }

   return;
}

int main(){
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   cout.tie(NULL);

   int t = 1;
   cin >> t;
   while(t--){
      solve();
      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
6
19
1709
11555
51476
9876543210
5891917899
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
import java.util.*;

public class Codeforces {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            String s = sc.next();
            StringBuilder sb = new StringBuilder(s);

            for(int j = 0;j<s.length();j++){for (int i = 1; i < sb.length(); i++) {
                if (sb.charAt(i) - '0' - 1 > sb.charAt(i-1) - '0') {
                    sb.setCharAt(i, (char) (sb.charAt(i) - 1));
                    char temp = sb.charAt(i);
                    sb.setCharAt(i, sb.charAt(i-1));
                    sb.setCharAt(i-1, temp);
                }
            }}

            System.out.println(sb.toString());
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
9999999999
1234567890
1111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

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

int main()
{
    int testcase;
    cin >> testcase;
    cin.ignore();
    vector<int> s;
    for (int i = 0 ; i < testcase ; i++)
    {
        string number;
        getline(cin, number);
        vector<int> s(number.length());
        for (int i = 0; i < number.length() ; i++)
        {
            s[i] = number[i] - '0';
        }
        for(int i = 1 ; i < s.size() ; i++)
        {
            if(s[i] - s[i-1] > 1)
            {
                s[i] = s[i] - 1;
                swap(s[i], s[i-1]);
                i = 0;
            }
        }
        for(int i = 0; i < s.size() ; i++)
        {
            cout << s[i];
        }
        cout << "\n";
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the algorithm might fail to produce the lexicographically maximum string. The main issue in the code is that it only checks if the difference between consecutive digits is greater than 1, and if so, it performs a single operation and resets the loop. This approach is too simplistic and may miss optimal operations, especially in cases where multiple operations are needed to maximize the string.

### Step-by-Step Analysis:
1. **Edge Cases:**
   - **Long sequences of identical digits:** The code might not handle sequences like "11111" correctly.
   - **Sequences with multiple possible operations:** For example, "13333" where multiple operations can be applied to maximize the string.
   - **Sequences with large gaps:** For example, "19999" where the gap between the first and subsequent digits is large.

2. **Specific Test Input:**
   - **Input: "13333"**
     - **Expected Output: "33311"**
     - **Reasoning:**
       - The correct approach would be to transform "13333" to "21333" (by decrementing the second '3' and swapping it with the '1').
       - Then, transform "21333" to "22133" (by decrementing the third '3' and swapping it with the '1').
       - Finally, transform "22133" to "32213" (by decrementing the fourth '3' and swapping it with the '1').
       - Continue to "33212" and finally "33311".

### Test Input:
```plaintext
1
13333
```

This test input should expose the bug in the provided code, as it requires multiple operations to achieve the lexicographically maximum string, which the current code does not handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

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

void work(){
	string s; cin>>s;
	for(int i = 0; i < s.size(); i++){
		int maior_pos = i;
		int maior = s[i] - '0';
		int dec = 1;
		for(int k = i+1; k <= min((int)s.size()-1, k+10); k++){
			if(s[k] - '0' - dec > maior){
				maior = (s[k] - '0') - dec;
				maior_pos = k;
			}
			dec++;
		}
		s.erase(s.begin() + maior_pos);
		s.insert(s.begin() + i, maior + '0');
	}
	cout<<s<<endl;
}

signed main(){
	int tt; cin>>tt;
	while(tt--){
		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:
To expose the bug in the given code, we need to focus on edge cases where the algorithm might fail. The key points to consider are:

1. **Handling the leftmost digit**: The leftmost digit should not be decreased.
2. **Handling the digit '0'**: The digit '0' should not be decreased.
3. **Handling long sequences of the same digit**: This can cause issues with the logic of finding the maximum digit.
4. **Handling sequences with large gaps**: The logic of finding the maximum digit and decreasing it might fail if there are large gaps between digits.

Given these points, let's create a test case that can expose the bug:

1. **Test case with a long sequence of the same digit**: This can cause issues with the logic of finding the maximum digit.
2. **Test case with a large gap between digits**: This can cause issues with the logic of finding the maximum digit and decreasing it.

Let's create a test case with a long sequence of the same digit and a large gap:

```plaintext
3
1111111111
1999999999
1000000000
```

This test case includes:
1. A string with a long sequence of the same digit ('1').
2. A string with a large gap between the first digit and the rest ('1999999999').
3. A string with a '0' in the middle ('1000000000').

This should help in exposing the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
for t in range(int(input())):
    s = input()
    l = []
    for char in s:
        l.append(char)

    ans = ""
    while len(l)!=0:
        check = 0
        max = check
        m = min(len(l), 9)
        for x in range(1,m):
            if int(l[x])-x>int(l[max])-max:
                max = x
        ans += str(int(l[max])-max)
        l = l[:max] + l[max+1:]

    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
1099999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
for t in range(int(input())):
    l = [int(char) for char in input()]

    ans = ""
    while len(l)!=0:
        max = 0
        for x in range(1,min(len(l), 9)):
            if l[x]-x>l[max]-max:
                max = x
        ans += str(l[max]-max)
        l.pop(max)

    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
1099999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.Objects;
import java.util.StringTokenizer;

public class Main {

  public static void main(String[] args)
      throws FileNotFoundException {
    FastScanner fs = new FastScanner(args);
    PrintWriter out = new PrintWriter(System.out);
    if (Solution.IS_MULTI_TESTCASE) {
      int t = fs.nextInt();
      for (int tt = 0; tt < t; tt++) {
        Solution.solve(fs, out);
      }
      out.close();
      return;
    }
    Solution.solve(fs, out);
    out.close();
  }
}

class FastScanner {
  private final BufferedReader br;
  private StringTokenizer st = new StringTokenizer("");

  public FastScanner(String[] args)
      throws FileNotFoundException {
    if (args.length == 1 && args[0].equals("LOCAL")) {
      File file = new File(Objects.requireNonNull(getClass().getClassLoader().getResource("inp.txt")).getFile());
      br = new BufferedReader(new InputStreamReader(new FileInputStream(file)));
    } else {
      br = new BufferedReader(new InputStreamReader(System.in));
    }
  }

  String next() {
    while (!st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(br.readLine());
      } catch (IOException e) {
        System.out.println(e.getMessage());
      }
    }
    return st.nextToken();
  }

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

  int[] readArray(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
      a[i] = nextInt();
    }
    return a;
  }

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

  public long[] readLongArray(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++) {
      a[i] = nextLong();
    }
    return a;
  }
}

class Solution {
  public static boolean IS_MULTI_TESTCASE = true;

  public static void solve(FastScanner scanner, PrintWriter out) {
    char[] nums = scanner.next().toCharArray();
    int N = nums.length;
    // iterate from right to left
    for (int i = 1; i < N; i++) {
      int ai = nums[i] - '0';
      int to = -1;
      // check for the best previous position to swap
      for (int j = i - 1; j >= 0; j--) {
        int aj = nums[j] - '0';
        int posDiff = i - j;
        if (ai - posDiff > aj) {
          to = j;
        }
      }
      // swap with the best previous position
      if (to != -1) {
        swap(i, to, nums);
      }
    }
    out.println(new String(nums));
  }

  static void swap(int i, int j, char[] nums) {
    int diff = i - j;
    // move the element at i to j
    char temp = nums[i];
    for (int k = i; k > j; k--) {
      nums[k] = nums[k - 1];
    }
    // decrease diff from temp
    nums[j] = (char) ((temp - '0') - diff + '0');
  }
}

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

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

Think step by step.

Response:
```plaintext
6
19
1709
11555
51476
9876543210
5891917899
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine());

        StringBuilder sb = new StringBuilder();

        for (int t = 0; t < T; t++) {

            char[] chars = br.readLine().toCharArray();
            int N = chars.length;

            boolean changed = true;

            while (changed) {
                changed = false;

                for (int i = N - 1; i > 0; i--) {

                    int crt = chars[i] - '0';
                    int left = chars[i - 1] - '0';

                    if (crt - 1 > left) {
                        chars[i] = (char) ((crt - 1) + '0');

                        char temp = chars[i - 1];
                        chars[i - 1] = chars[i];
                        chars[i] = temp;

                        changed = true;
                        break;
                    }
                }
            }

            sb.append(chars).append("\n");
        }

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

**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
11111
99999
19999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital string maximization
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$, consisting of digits from $$$0$$$ to $$$9$$$. In one operation, you can pick any digit in this string, except for $$$0$$$ or the leftmost digit, decrease it by $$$1$$$, and then swap it with the digit left to the picked.

For example, in one operation from the string $$$1023$$$, you can get $$$1103$$$ or $$$1022$$$.

Find the lexicographically maximum string you can obtain after any number of operations.

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

Each test case consists of a single line consisting of a digital string $$$s$$$ ($$$1 \le |s| \le 2\cdot 10^5$$$), where $$$|s|$$$ denotes the length of $$$s$$$. The string does not contain leading zeroes.

It is guaranteed that the sum of $$$|s|$$$ of all test cases doesn't exceed $$$2\cdot 10^5$$$.

Output: For each test case, print the answer on a separate line.

Examples:
input:
6
19
1709
11555
51476
9876543210
5891917899
output:
81
6710
33311
55431
9876543210
7875567711

Note: In the first example, the following sequence of operations is suitable: $$$19 \rightarrow 81$$$.

In the second example, the following sequence of operations is suitable: $$$1709 \rightarrow 1780 \rightarrow 6180 \rightarrow 6710$$$.

In the fourth example, the following sequence of operations is suitable: $$$51476 \rightarrow 53176 \rightarrow 53616 \rightarrow 53651 \rightarrow 55351 \rightarrow 55431$$$.

Buggy Code:
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e5+9;
char s[N];
int a[N];
int read(){
	int x=0,f=0;
	char c=getchar();
	while(c>'9'||c<'0'){
		f=c=='-';
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=x*10+c-'0';
		c=getchar();
	}
	return f?-x:x;
}
int main(){
	int t=read();
	while(t--){
		scanf("%s",s);
		int n=strlen(s);
		for(int i=0;i<n;i++) a[i+1]=s[i]-'0';
		for(int i=1;i<=n;i++){
			int k=i;
			for(int j=i+1;j<=i+8&&j<=n;j++){
				if(a[j]-(j-i)>a[k]-(k-i)){
					k=j;
				}
			}
			if(k!=i){
				int tmp=a[k];
				for(int j=k-1;j>=i;j--) a[j+1]=a[j];
				a[i]=tmp-(k-i);
			}
			printf("%d",a[i]);
		}
		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
1
199
```


**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> U;
        vector<i64> A, B, C, D(q);
        vector<pair<pair<i64, i64>, i64>> aux(n);
        for(auto& a : aux) cin >> a.first.first, U.insert(a.first.first);
        for(auto& a : aux) cin >> a.first.second, U.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);
            }
        }
        for(int i = 0; i < n; i++) {
            vector<int> aux;
            for(auto& a : l) {
                if(A[i] % a != 0 && B[i] % a != 0) aux.push_back(a);
            }
            for(auto& a : aux) l.erase(a);
            aux.clear();
            for(auto& a : r) {
                if(A[i] % a != 0 && B[i] % a != 0) aux.push_back(a);
            }
            for(auto& a : aux) r.erase(a);
        }
        i64 d = i64(U.size()) / 2;
        i64 ub = i64(1e8);
        n = A.size();
        map<i64, int> mp;
        for(auto& x : l) {
            for(auto& y : r) {
                if(ub / x + 100 < d && ub / y + 100 < d) continue;
                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;
                        }
                    }
                }
                mp[cost1] = max(mp[cost1], x + y);
                mp[cost2] = max(mp[cost2], 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
```

### Explanation:

1. **Input Size and Constraints:**
   - The input size is small (n = 3, q = 4) to ensure it fits within the output limits.
   - The values of `a`, `b`, and `c` are chosen to be simple and within the constraints.

2. **Edge Cases:**
   - **No Swaps Allowed (d = 0):** The initial GCD sum is calculated without any swaps.
   - **Single Swap (d = 1):** The cost of each swap is 1, so one swap can be made.
   - **Multiple Swaps (d = 2, 3):** Multiple swaps can be made, but the optimal solution should be checked to ensure it is correct.

3. **Expected Behavior:**
   - For `d = 0`, the GCD sum is `gcd(1, 2, 3) + gcd(4, 5, 6) = 1 + 1 = 2`.
   - For `d = 1`, swapping `a_2` and `b_2` gives `gcd(1, 5, 3) + gcd(4, 2, 6) = 1 + 2 = 3`.
   - For `d = 2`, the GCD sum remains the same as `d = 1` because no additional swaps improve the result.
   - For `d = 3`, the GCD sum remains the same as `d = 1` and `d = 2`.

4. **Challenging the Buggy Code:**
   - The input is designed to test the logic of the buggy code, especially the part where it calculates the GCD sum and the cost of swaps.
   - The small input size ensures that any logical errors in the code are more likely to be exposed.

This test input is crafted to be simple yet effective in exposing potential bugs in the given code.


**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
import heapq

def read_ints():
    return list(map(int, input().split()))

def compute_initial_gcd(arr):
    result = arr[0]
    for num in arr[1:]:
        result = gcd(result, num)
    return result

def main():
    n, q = read_ints()
    a = read_ints()
    b = read_ints()
    c = read_ints()
    budgets = read_ints()

    # Initial GCD calculations
    initial_gcd_a = compute_initial_gcd(a)
    initial_gcd_b = compute_initial_gcd(b)
    initial_sum_gcd = initial_gcd_a + initial_gcd_b

    # Prepare swaps
    swaps = []
    for i in range(n):
        new_gcd_a = compute_initial_gcd([a[j] if j != i else b[i] for j in range(n)])
        new_gcd_b = compute_initial_gcd([b[j] if j != i else a[i] for j in range(n)])
        gain = (new_gcd_a + new_gcd_b) - initial_sum_gcd
        if gain > 0:
            heapq.heappush(swaps, (-gain, c[i]))  # Using negative gain for max heap

    answers = []
    for d in budgets:
        current_sum_gcd = initial_sum_gcd
        available_swaps = swaps[:]

        while available_swaps and d >= available_swaps[0][1]:
            gain, cost = heapq.heappop(available_swaps)
            if d >= cost:
                current_sum_gcd -= gain
                d -= cost

        answers.append(current_sum_gcd)

    print(' '.join(map(str, answers)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

### Explanation:
1. **Initial Setup:**
   - The arrays `a`, `b`, and `c` are set up as follows:
     - `a = [1, 2, 3]`
     - `b = [4, 5, 6]`
     - `c = [1, 2, 3]`
   - The budgets are: `d = [0, 1, 2, 3]`

2. **Initial GCD Calculation:**
   - `gcd(a) = gcd(1, 2, 3) = 1`
   - `gcd(b) = gcd(4, 5, 6) = 1`
   - `initial_sum_gcd = 1 + 1 = 2`

3. **Swaps and Gains:**
   - For `i = 0`:
     - Swap `a[0]` and `b[0]`: `a = [4, 2, 3]`, `b = [1, 5, 6]`
     - `new_gcd_a = gcd(4, 2, 3) = 1`
     - `new_gcd_b = gcd(1, 5, 6) = 1`
     - `gain = (1 + 1) - 2 = 0`
     - No gain, so no swap is added to the heap.
   - For `i = 1`:
     - Swap `a[1]` and `b[1]`: `a = [1, 5, 3]`, `b = [4, 2, 6]`
     - `new_gcd_a = gcd(1, 5, 3) = 1`
     - `new_gcd_b = gcd(4, 2, 6) = 2`
     - `gain = (1 + 2) - 2 = 1`
     - Add `(-1, 2)` to the heap.
   - For `i = 2`:
     - Swap `a[2]` and `b[2]`: `a = [1, 2, 6]`, `b = [4, 5, 3]`
     - `new_gcd_a = gcd(1, 2, 6) = 1`
     - `new_gcd_b = gcd(4, 5, 3) = 1`
     - `gain = (1 + 1) - 2 = 0`
     - No gain, so no swap is added to the heap.

4. **Processing Budgets:**
   - For `d = 0`:
     - No swaps can be made.
     - `current_sum_gcd = 2`
   - For `d = 1`:
     - No swaps can be made (cost of the only available swap is 2).
     - `current_sum_gcd = 2`
   - For `d = 2`:
     - One swap can be made.
     - `current_sum_gcd = 2 - (-1) = 3`
   - For `d = 3`:
     - One swap can be made.
     - `current_sum_gcd = 2 - (-1) = 3`

### Expected Output:
```plaintext
2 2 3 3
```

This test input is designed to expose the bug in the buggy code by focusing on the edge case where the gain from a single swap is positive, but the cost of the swap is higher than some of the budget values. This ensures that the algorithm must correctly handle the scenario where it can only perform certain swaps based on the available budget.


**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
from heapq import heappush, heappop

# Function to compute gcd of an array
def compute_gcd(arr):
    g = arr[0]
    for num in arr[1:]:
        g = gcd(g, num)
    return g

def solve(n, q, a, b, c, d):
    # Initial gcd computations
    G_a = compute_gcd(a)
    G_b = compute_gcd(b)

    # Store initial value of gcd sum
    initial_sum = G_a + G_b

    # Swaps to consider
    diffs = []

    # For each swap option, consider the new possible GCDs
    for i in range(n):
        new_a = a[:]
        new_b = b[:]
        new_a[i], new_b[i] = new_b[i], new_a[i]

        new_G_a = compute_gcd(new_a)
        new_G_b = compute_gcd(new_b)

        new_sum = new_G_a + new_G_b

        if new_sum > initial_sum:
            heappush(diffs, (c[i], new_sum))

    # Process each budget query
    results = []
    for budget in d:
        max_sum = initial_sum
        total_cost = 0
        considered_sums = initial_sum

        # Use a min-heap to consider the cheapest swaps first
        temp_heap = []

        while diffs and total_cost + diffs[0][0] <= budget:
            cost, new_sum = heappop(diffs)
            total_cost += cost
            considered_sums = max(considered_sums, new_sum)
            heappush(temp_heap, (cost, new_sum))

        results.append(considered_sums)

        # Push back the considered swaps
        while temp_heap:
            heappush(diffs, heappop(temp_heap))

    return results

# Read input values
n, q = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
d = list(map(int, input().split()))

# Solve the problem
results = solve(n, q, a, b, c, d)

# Print results
for result in results:
    print(result)

**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 20 30
0 25 50
```

### Explanation:
1. **Initial Setup:**
   - Arrays: `a = [1, 2, 3]`, `b = [4, 5, 6]`, `c = [10, 20, 30]`
   - Queries: `d = [0, 25, 50]`

2. **Initial GCDs:**
   - `G_a = gcd(1, 2, 3) = 1`
   - `G_b = gcd(4, 5, 6) = 1`
   - Initial sum: `G_a + G_b = 1 + 1 = 2`

3. **Swaps to Consider:**
   - For `i = 0`: Swap `a[0]` and `b[0]` -> `a = [4, 2, 3]`, `b = [1, 5, 6]`
     - New GCDs: `G_a = gcd(4, 2, 3) = 1`, `G_b = gcd(1, 5, 6) = 1`
     - New sum: `1 + 1 = 2` (No improvement)
   - For `i = 1`: Swap `a[1]` and `b[1]` -> `a = [1, 5, 3]`, `b = [4, 2, 6]`
     - New GCDs: `G_a = gcd(1, 5, 3) = 1`, `G_b = gcd(4, 2, 6) = 2`
     - New sum: `1 + 2 = 3` (Improvement)
   - For `i = 2`: Swap `a[2]` and `b[2]` -> `a = [1, 2, 6]`, `b = [4, 5, 3]`
     - New GCDs: `G_a = gcd(1, 2, 6) = 1`, `G_b = gcd(4, 5, 3) = 1`
     - New sum: `1 + 1 = 2` (No improvement)

4. **Queries:**
   - For `d = 0`: No swaps allowed, so the result is `2`.
   - For `d = 25`: Only the swap at `i = 1` is within budget, so the result is `3`.
   - For `d = 50`: The swap at `i = 1` is within budget, so the result is `3`.

This test case is designed to expose the bug in the code where it might not correctly handle the initial and new GCD calculations, especially when the swap does not improve the 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:
import sys
from math import gcd
from functools import reduce

def calculate_initial_gcd(arr):
    return reduce(gcd, arr)

def main():
    input = sys.stdin.read
    data = input().split()

    idx = 0

    # Read n and q
    n = int(data[idx])
    q = int(data[idx+1])
    idx += 2

    # Read array a
    a = list(map(int, data[idx:idx+n]))
    idx += n

    # Read array b
    b = list(map(int, data[idx:idx+n]))
    idx += n

    # Read array c
    c = list(map(int, data[idx:idx+n]))
    idx += n

    # Read array d
    d = list(map(int, data[idx:idx+q]))

    # Initial gcd calculations
    initial_gcd_a = calculate_initial_gcd(a)
    initial_gcd_b = calculate_initial_gcd(b)

    # Calculate delta gcd for each swap
    swaps = []
    for i in range(n):
        new_a = a[:]
        new_b = b[:]

        new_a[i], new_b[i] = new_b[i], new_a[i]

        new_gcd_a = calculate_initial_gcd(new_a)
        new_gcd_b = calculate_initial_gcd(new_b)

        delta_gcd = (new_gcd_a + new_gcd_b) - (initial_gcd_a + initial_gcd_b)

        if delta_gcd > 0:
            swaps.append((delta_gcd / c[i], delta_gcd, c[i]))

    # Sort swaps by efficiency (delta_gcd/cost)
    swaps.sort(reverse=True, key=lambda x: x[0])

    results = []

    for di in d:
        current_budget = di
        max_value = initial_gcd_a + initial_gcd_b

        for _, delta_gcd, cost in swaps:
            if current_budget >= cost:
                max_value += delta_gcd
                current_budget -= cost
            else:
                break

        results.append(max_value)

    print(" ".join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
5 5
1 2 3 4 5
6 7 8 9 10
1 1 1 1 1
0 1 2 3 4
```