**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class D {
    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());
        }

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

    static FastReader s = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    private static int[] rai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        return arr;
    }

    private static int[][] rai(int n, int m) {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextInt();
            }
        }
        return arr;
    }

    private static long[] ral(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextLong();
        }
        return arr;
    }

    private static long[][] ral(int n, int m) {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextLong();
            }
        }
        return arr;
    }

    private static int ri() {
        return s.nextInt();
    }

    private static long rl() {
        return s.nextLong();
    }

    private static String rs() {
        return s.next();
    }

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

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

    static int MOD = (int) (1e9 + 7);

    public static void primeFactors(int n) {
        // Print the number of 2s that divide n
        while (n % 2 == 0) {
            System.out.print(2 + " ");
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            // While i divides n, print i and divide n
            while (n % i == 0) {
                System.out.print(i + " ");
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            System.out.print(n);
        System.out.println();
    }

    static boolean isPrime(int val) {
        for (int i = 2; i <= Math.sqrt(val); i++) {
            if (val % i == 0) {
                return false;
            }
        }
        return true;
    }

    static long pow(long x, long y, long n) {
        if (y == 0) {
            return 1;
        }
        long val = pow(x, y / 2, n);
        if (y % 2 == 1) {
            return (((val * val) % n) * x) % n;
        } else {
            return (val * val) % n;
        }
    }

    public static void main(String[] args) {
        StringBuilder ans = new StringBuilder();
        int t = ri();
//        int t=1;
        int v = 1;
        while (t-- > 0) {
            int n=ri();
            int m =ri();
            long L = ri();

            List<Hurdle> list = new ArrayList<>();
//            if(v==106) {
//                System.out.println(n+" "+m+ " "+L);
//            }

            for(int i=0;i<n;i++) {
                long pos1= rl();
                long pos2 = rl();
//                if(v==106) {
//                    System.out.println(pos1+" "+pos2);
//                }
                list.add(new Hurdle(pos1, pos2-pos1+2));
            }

            List<PowerUp> power = new ArrayList<>();
            for(int i=0;i<m;i++) {
                long pos = rl();
                long power1 =rl();
//                if(v==106) {
//                    System.out.println(pos+" "+power1);
//                }
                power.add(new PowerUp(pos, power1));

            }
//            v++;
            int ind = 0;
            int i = 0;
            boolean can = true;

//            PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
            PriorityQueue<Long> require = new PriorityQueue<>();
            PriorityQueue<Long> current = new PriorityQueue<>(Collections.reverseOrder());

            long reqSum = 1;
            long currSum = 0;
            long pqSum = 0;

            while(ind < n) {

                Hurdle hurdle = list.get(ind);
                for(int j=i; j<m;j++) {
                    PowerUp powerUp = power.get(j);
                    if(powerUp.pos > hurdle.pos) {
                        i=j;
                        break;
                    }
                    currSum += powerUp.power;
                    current.add(powerUp.power);

                }

                while(reqSum  < hurdle.reqJump && !current.isEmpty()) {

                    long val = current.remove();
                    currSum -=val;
                    require.add(val);
                    reqSum+=val;
                }
//                if(!current.isEmpty()) {
//                    pq.addAll(current);
//                    pqSum+=currSum;
//                    current = new PriorityQueue<>();
//                    currSum = 0;
//                }
//
//                while(reqSum < hurdle.reqJump && !pq.isEmpty()) {
//
//
//                    long val = pq.remove();
//                    pqSum -=val;
//                    require.add(val);
//                    reqSum+=val;
//                }

                if(reqSum < hurdle.reqJump) {
                    can = false;
                    break;
                }

                ind++;

            }

            ans.append(can ? require.size():"-1").append("\n");

        }
        out.print(ans);
        out.flush();
    }

    static class Hurdle {
        long pos;
        long reqJump;

        public Hurdle(long pos, long reqJump) {
            this.pos = pos;
            this.reqJump = reqJump;
        }

        @Override
        public String toString() {
            return "Hurdle{" +
                    "pos=" + pos +
                    ", reqJump=" + reqJump +
                    '}';
        }
    }

    static class PowerUp{
        long pos, power;

        public PowerUp(long pos, long power) {
            this.pos = pos;
            this.power = power;
        }

        @Override
        public String toString() {
            return "PowerUp{" +
                    "pos=" + pos +
                    ", power=" + power +
                    '}';
        }
    }

    static long traverse(long[] arr, int[] b, long[] prefSum, int currInd, int prevInd, long toSub, int maxReach) {

        long currSum = prefSum[currInd] - toSub;
//        System.out.println(currInd +" "+ currSum);

        int newMaxReach = maxReach;
        int ind = -1;
        for(int i=currInd; i>prevInd; i--) {

            if(b[i]>newMaxReach) {
                newMaxReach = b[i];
                ind = i;
            } else if(b[i] == newMaxReach) {
                if(ind != -1 && arr[i] < arr[ind]) {
                    newMaxReach = b[i];
                    ind = i;
                }
            }

        }

        if(ind != -1) {
            long newToSub = toSub + arr[ind];
            long val = traverse(arr, b, prefSum, b[ind], ind, newToSub, newMaxReach);
            currSum = Math.max(currSum, val);
        }
//        newMaxReach = Math.max(newMaxReach, b[i]);
//        if(b[i] > maxReach){
//
//        }
//        System.out.println(currInd +" "+ currSum);
        return currSum;
    }

    static class SparseMatrix {
        private int[] arr;
        private int m;
        private int[][] minSparse;
        private int[][] minIndex;
        private int[][] maxSparse;
        private int[][] maxIndex;
        private int n;
        public SparseMatrix(int[] arr) {
            this.arr = arr;
            this.m=arr.length;
            this.n=Integer.toBinaryString(m).length();
            minSparse=new int[n][m];
            minIndex=new int[n][m];

            maxSparse=new int[n][m];
            maxIndex=new int[n][m];
//        for(int i=0;i<n;i++)
//        {
//            Arrays.fill(minSparse[i],-1);
//        }
//        for(int i=0;i<n;i++)
//        {
//            Arrays.fill(minIndex[i],-1);
//        }

            createMinSparse();
            createMaxSparse();
        }
        private void createMaxSparse()
        {
            for(int j=0;j<m;j++)
            {
                maxSparse[0][j]=arr[j];
                maxIndex[0][j]=j;
            }
            for(int i=1;i<n;i++)
            {
                for(int j=0;j+(1<<(i-1))<m;j++)
                {
                    int left=maxSparse[i-1][j];
                    int right=maxSparse[i-1][j+(1<<(i-1))];
                    maxSparse[i][j]=Math.max(left,right);

                    if(left>=right)
                    {
                        maxIndex[i][j]=maxIndex[i-1][j];
                    }
                    else
                    {
                        maxIndex[i][j]=maxIndex[i-1][j+(1<<(i-1))];
                    }
                }
            }
        }
        private void createMinSparse()
        {
            //filling the first row of sparse matrix with the values of the input array
            for(int j=0;j<m;j++)
            {
                minSparse[0][j]=arr[j];
                minIndex[0][j]=j;
            }
            //filling other rows of the sparse matrix
            for(int i=1;i<n;i++)
            {
                for(int j=0;j+(1<<(i-1))<m;j++)
                {

                    int left=minSparse[i-1][j];
                    int right=minSparse[i-1][j+(1<<(i-1))];

                    //change to min-> max to create MaxSparseMatrix
                    minSparse[i][j]=Math.min(left,right);

                    //filling index
                    if(left<=right)
                    {
                        minIndex[i][j]=minIndex[i-1][j];
                    }
                    else {
                        minIndex[i][j]=minIndex[i-1][j+(1<<(i-1))];
                    }

                }
            }
        }

        //get minimum value in range l->r inclusive
    /*
        for any range [l, r] we can find the two values and
        find their minimum. These values are defined below:

        len: length of the required range, i.e., r-l+1

        p: maximum power of 2 that can fit in len. E.g, [1,11] , len=11, thus p=3

        k: 2^p

        find the minimum between sparse[p][l] and sparse[p][r-k+1]
    */
        public int getMin(int l,int r)
        {
            int len=r-l+1;
            int p=Integer.toBinaryString(len).length()-1;
            int k=1<<p;

            int left=minSparse[p][l];
            int right=minSparse[p][r-k+1];
            return Math.min(right,left);
        }

        public int getMinIndex(int l,int r)
        {
            int len=r-l+1;
            int p=Integer.toBinaryString(len).length()-1;
            int k=1<<p;

            int left=minSparse[p][l];
            int right=minSparse[p][r-k+1];
            if (left <= right) {
                return minIndex[p][l];
            } else {
                return minIndex[p][r - k + 1];
            }
        }

        public int getMax(int l,int r)
        {
            int len=r-l+1;
            int p=Integer.toBinaryString(len).length()-1;
            int k=1<<p;

            int left=maxSparse[p][l];
            int right=maxSparse[p][r-k+1];
            return Math.max(right,left);
        }

        public int getMaxIndex(int l,int r)
        {
            int len=r-l+1;
            int p=Integer.toBinaryString(len).length()-1;
            int k=1<<p;
            int left=maxSparse[p][l];
            int right=maxSparse[p][r-k+1];
            if(left>=right)
            {
                return maxIndex[p][l];

            }
            return maxIndex[p][r-k+1];
        }
        void print()
        {
            for(int i=0;i<minSparse.length;i++)
            {
                for(int j=0;j<minSparse[i].length;j++)
                {
                    System.out.print(minSparse[i][j]+" ");
                }
                System.out.println();
            }
            System.out.println();
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    System.out.print(minIndex[i][j]+" ");
                }
                System.out.println();
            }

        }
    }

}

**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 for the given buggy code, we need to focus on edge cases that can expose potential issues in the algorithm. Here are some considerations:

1. **Hurdles and Power-ups Alignment**: Ensure that the hurdles and power-ups are positioned in a way that requires careful selection of power-ups to clear the hurdles.
2. **Jump Power Management**: Test scenarios where the initial jump power is insufficient to clear early hurdles, and the power-ups are necessary to progress.
3. **Multiple Power-ups at the Same Position**: Test scenarios with multiple power-ups at the same position to see how the code handles this.
4. **Edge Cases**: Test with the minimum and maximum values for `n`, `m`, and `L` to ensure the code handles these limits correctly.

Given these considerations, let's create a small-scale test input:

```plaintext
4
2 5 10
2 3
7 8
1 2
2 1
3 3
4 2
6 1
2 3 15
2 3
10 12
1 3
3 2
5 1
2 4 20
5 10
15 18
1 5
3 2
7 1
9 2
11 3
1 2 10
2 4
4 6
2 1
2 2
```

### Explanation:
1. **Test Case 1**:
   - Hurdles: `[2, 3]`, `[7, 8]`
   - Power-ups: `[1, 2]`, `[2, 1]`, `[3, 3]`, `[4, 2]`, `[6, 1]`
   - The initial jump power is 1, so she needs to collect power-ups to clear the hurdles.
   - The minimum number of power-ups needed is 3 (collecting from positions 2, 3, and 6).

2. **Test Case 2**:
   - Hurdles: `[2, 3]`, `[10, 12]`
   - Power-ups: `[1, 3]`, `[3, 2]`, `[5, 1]`
   - The initial jump power is 1, and the power-ups are insufficient to clear the second hurdle.
   - The output should be -1.

3. **Test Case 3**:
   - Hurdles: `[5, 10]`, `[15, 18]`
   - Power-ups: `[1, 5]`, `[3, 2]`, `[7, 1]`, `[9, 2]`, `[11, 3]`
   - The initial jump power is 1, and the power-ups are sufficient to clear the hurdles.
   - The minimum number of power-ups needed is 2 (collecting from positions 11 and 3).

4. **Test Case 4**:
   - Hurdles: `[2, 4]`
   - Power-ups: `[2, 1]`, `[2, 2]`
   - The initial jump power is 1, and the power-ups are at the same position.
   - The minimum number of power-ups needed is 1 (collecting from position 2).

This test input should help identify issues in the buggy code by covering various scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

void solve() {
    string s;
    cin >> s;
    vector<int> res;
    long long sum = 0;
    int n = s.size();

    // Calculate the sum of digits in the string
    for (int i = 0; i < n; i++) {
        int digit = s[i] - '0'; // Convert char to int
        sum += digit;
        if (digit * digit <= 10) {
            res.push_back(digit);
        }
    }

    // Sort the result vector
    sort(res.begin(), res.end());

    // Check if the sum of digits is divisible by 9
    if (sum % 9 == 0) {
        cout << "YES" << endl;
        return;
    } else {
        // Modify sum using values from res vector
        for (int i = 0; i < res.size(); i++) {
            sum += (res[i] * res[i] - res[i]);
            if (sum % 9 == 0) {
                cout << "YES" << endl;
                return;
            }
        }
    }

    cout << "NO" << endl;
}

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

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

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

Think step by step.

Response:
```plaintext
1
11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

int digitSum(string &num)
{
    int sum = 0;
    for (char c : num)
    {
        sum += c - '0';
    }
    return sum;
}
string sacar2y3(string &num)
{
    string result;
    for (char c : num)
    {
        if (c == '2' || c == '3')
        {
            result += c;
        }
    }
    return result;
}
string makeDivisibleBy9(string &num)
{
    int sum = digitSum(num);
    int remainder = abs((sum % 9) - 9);
    string numC = sacar2y3(num);
    if (remainder == 0 || remainder == 9)
    {
        return num;
    }
    int count2 = 0, count3 = 0;
    for (char c : numC)
    {
        if (c == '2')
            count2++;
        else if (c == '3')
            count3++;
    }
    if (remainder == 2 && count2 >= 1)
    {
        return "YES";
    }
    if (remainder == 3 && count3 >= 2)
    {
        return "YES";
    }

    if (remainder == 4 && count3 >= 2)
    {
        return "YES";
    }
    if (remainder == 6 && count3 >= 1 || count2 >= 3)
    {
        return "YES";
    }
    if (remainder == 8 && count2 >= 4)
    {
        return "YES";
    }
    return "NO";
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        string num;
        cin >> num;
        string result = makeDivisibleBy9(num);
        if (result == "NO")
        {
            cout << "NO\n";
        }
        else
        {
            cout << "YES\n";
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

bool divisible(const string& n) {
    int sum = 0;
    int count2 = 0, count3 = 0;

    // Calculate sum of digits and count occurrences of 2 and 3
    for (char c : n) {
        int digit = c - '0';
        sum += digit;
        if (digit == 2) count2++;
        if (digit == 3) count3++;
    }

    sum %= 9;  // Compute sum modulo 9

    // If already divisible by 9
    if (sum == 0) return true;

    // Check conditions for each remainder
    switch (sum) {
        case 7: return count2 >= 1;  // At least one '2' needed
        case 5: return count2 >= 2;  // At least two '2's needed
        case 3: return count2 >= 3 || count3 >= 1;  // Three '2's or one '3' needed
        case 6: return count3 >= 2;  // At least two '3's needed
        case 2: return count2 >= 4 && count3 >= 1;  // At least one '3' needed
        case 1: return count2 >= 3 || (count2 >= 1 && count3 >= 1);  // At least one '3' needed
        case 4: return count2 >= 7 || (count2 >= 1 && count3 >= 2) || (count3 >= 1 && count2 >= 4);  // Combination of '2' and '3'
        case 8: return count2 >= 5 || (count3 >= 1 && count2 >= 2);  // Combination of '2' and '3'
        default: return false;  // No valid transformation for this remainder
    }
}

int main() {
    int t;
    cin >> t;

    while (t--) {
        string n;
        cin >> n;
        if (divisible(n)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << 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
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
def solve(N):
    digitsum = sum(int(x) for x in str(N))
    if digitsum%9==0:
        return "YES"
    changes = [int(x)**2-int(x) for x in str(N) if int(x)<=3 and int(x)>=2]
    for i in range(1, len(changes)):
        for x in range(len(changes)-i):
            if (digitsum+int(str(N)[x:x+i]))%9==0:
                return "YES"
    else:
        return "NO"
    return changes

T = int(input())
for __ in range(T):
    N = int(input())
    print(solve(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
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
```

### Explanation:
1. **Test Case 1: 123**
   - This is a basic test case where none of the digits can be squared to make the number divisible by 9.
   - Expected output: "NO"

2. **Test Case 2: 322**
   - The digit 2 can be squared to 4, and the digit 2 can be squared to 4. The resulting number 344 is divisible by 9.
   - Expected output: "YES"

3. **Test Case 3: 333333333333**
   - This number is already divisible by 9.
   - Expected output: "YES"

4. **Test Case 4: 9997**
   - The digit 9 is already 9, and 7 cannot be squared to a digit. The number cannot be made divisible by 9.
   - Expected output: "NO"

5. **Test Case 5: 5472778912773**
   - This is a large number where the digit 2 can be squared to 4. The resulting number 5474778912773 is not divisible by 9.
   - Expected output: "NO"

6. **Test Case 6: 1234567890**
   - The digit 3 can be squared to 9, making the number 12345678990, which is divisible by 9.
   - Expected output: "YES"

7. **Test Case 7: 23**
   - The digit 2 can be squared to 4, and the digit 3 can be squared to 9. The resulting number 29 is not divisible by 9.
   - Expected output: "NO"

8. **Test Case 8: 33**
   - This number is already divisible by 9.
   - Expected output: "YES"

9. **Test Case 9: 52254522632**
   - The digit 2 can be squared to 4 multiple times. The resulting number 54454544634 is divisible by 9.
   - Expected output: "YES"

These test cases cover a variety of scenarios, including numbers that are already divisible by 9, numbers that can be made divisible by 9 through the operations, and numbers that cannot be made divisible by 9. The test cases are small enough to be manageable and ensure that the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
t=int(input())
while(t>0):
    n=int(input())
    if(n%9==0 or n%11==0 or n%2==0):
        print("YES")
    else:
        print("NO")
    t=t-1

**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
123
322
333333333333
9997
5472778912773
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
for t in range(int(input())):
    num=input()
    s=0
    two,three=0,0
    for c in num:
        s+=int(c)
        if int(c)==2:
            two+=1
        elif int(c)==3:
            three+=1
    if s%9:
        while two and s%9:
            s+=2
            two-=1
        if s%9:
            while three and s%9:
                s+=6
                three-=1
        if s%9:
            print("No")
        else:
            print("Yes")
    else:
        print("Yes")

**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
1234567890
3333333333
2222222222
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

public class Main {
          public static String fun(int n, String s) {
                    int rem = 0;
                    int count2 = 0, count3 = 0;
                    for(int i = 0; i < n; i++){
                              rem += (s.charAt(i) - '0');
                              rem %= 9;

                              if(s.charAt(i) == '2') count2++;
                              else if(s.charAt(i) == '3') count3++;
                    }

                    if(rem == 0) return "YES";

                    if(rem == 7) {
                              if(count2 >= 1) return "YES";
                              return "NO";
                    }
                    if(rem == 5) {
                              if(count2 >= 2) return "YES";
                              return "NO";
                    }
                    if(rem == 3) {
                              if(count2 >= 3 || count3 >= 1) return "YES";
                              return "NO";
                    }
                    if(rem == 1) {
                              if(count2 >= 4 || (count3 >= 1 && count3 >= 1)) return "YES";
                              return "NO";
                    }

                    if(rem == 8) {
                              if(count2 >= 5 || (count3 >= 1 && count2 >= 2)) return "YES";
                              return "NO";
                    }
                    if(rem == 6) {
                              if(count2 >= 6 || (count3 >= 1 && count2 >= 3) || count3 >= 2) return "YES";
                              return "NO";
                    }
                    if(rem == 4) {
                              if(count2 >= 7 || (count3 >= 1 && count2 >= 4) || (count3 >= 2 && count2 >= 1)) return "YES";
                              return "NO";
                    }
                    if(rem == 2) {
                              if(count2 >= 8 || (count3 >= 1 && count2 >= 5) || (count3 >= 2 && count2 >= 2)) return "YES";
                              return "NO";
                    }

                    return "NO";
          }

          public static void main(String[] args) throws IOException {
                    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                    BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
                    StringTokenizer st = new StringTokenizer(br.readLine());
                    int t = Integer.parseInt(st.nextToken());
                    while(t-- > 0){
                              String s = br.readLine();

                              bw.write(fun(s.length(), s) + "\n");
                    }

                    br.close();
                    bw.flush();
                    bw.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.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
import java.util.*;

public class C_Uninteresting_Number {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t > 0) {
            long n = sc.nextLong();
            int size = 0;
            long sum = 0;
            int two = 0, three = 0;
            while(n > 0) {
                long dig = n % 10;
                if(dig == 2) {
                    two++;
                }
                else if(dig == 3) {
                    three++;
                }
                sum += dig;
                n /= 10;
            }
            sum += (two * 2) + (three * 6);
            long eight = Math.min(two, three);
            long mod = sum % 9;
            if(mod == 0) {
                System.out.println("YES");
            }
            else {
                boolean possible = false;
            for(int i=1; i<=two; i++) {
                long sum_new = sum - (2 * i);
                if(sum_new % 9 == 0) {
                    possible = true;
                    break;
                }
            }
            if(!possible) {
                for(int i=1; i<=three; i++) {
                    long sum_new = sum - (6 * i);
                    if(sum_new % 9 == 0) {
                        possible = true;
                        break;
                    }
                }
            }

            if(!possible) {
                for(int i=1; i<=eight; i++) {
                    long sum_new = sum - (8 * i);
                    if(sum_new % 9 == 0) {
                        possible = true;
                        break;
                    }
                }
            }
            if(possible) {
                System.out.println("YES");
            }
            else {
                System.out.println("NO");
            }
            }
            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
222222222222222222222222222222
333333333333333333333333333333
232323232323232323232323232323
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

// to find n in n*(n+1)/2<=x;
// n = (sqrt(1+ 8*x) -1)/2;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<pii> vpii;
typedef vector<vi> vvi;
typedef unordered_map<int, int> mapii;
typedef unordered_set<int> setii;
typedef unordered_map<ll, ll> mapll;
typedef unordered_set<ll, ll> setll;

#define F first
#define S second
#define pb push_back
#define MP make_pair
#define all(x) x.begin(), x.end()
#define fr(i, n) for (int i = 0; i < n; i++)
#define fr1(i, n) for (int i = 1; i < n; i++)
#define YES cout << "Yes" << endl
#define NO cout << "No" << endl

const int INF = 1e9;
const ll MOD = 1e9 + 7;
const ll LLINF = 1e18;

#define N 1e7 + 5
vector<bool> prime(N, true);
vector<int> prime_nos;

void sieve()
{
    prime[0] = false;
    prime[1] = false;
    for (int i = 2; i < N; i++)
    {
        if (prime[i])
        {
            prime_nos.push_back(i);
            for (int j = 2 * i; j < N; j += i)
            {
                prime[j] = false;
            }
        }
    }
}

ll mod_exp(ll base, ll exp, ll mod = MOD)
{
    ll result = 1;
    base = base % mod;
    while (exp > 0)
    {
        if (exp % 2 == 1)
            result = (result * base) % mod;

        exp = exp >> 1;
        base = (base * base) % mod;
    }
    return result;
}

ll divide_mod(ll a, ll b, ll M)
{
    ll b_inv = mod_exp(b, M - 2, M);
    return (a * b_inv) % M;
}

bool is_prime(int n)
{
    if (n <= 2)
        return false;
    if (n % 2 == 0)
        return false;
    for (int i = 3; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}

int most_sig_bit(int num)
{
    int totalBits = sizeof(num) * 8;
    int msbIndex = totalBits - __builtin_clz(num) - 1;
    return msbIndex;
}

string itob(int x)
{
    string ans;
    while (x)
    {
        char c = (x % 2 + '0');
        ans = c + ans;
        x >>= 1;
    }
    return ans;
}

struct pair_hash
{
    template <class T1, class T2>
    std::size_t operator()(const std::pair<T1, T2> &pair) const
    {
        return std::hash<T1>()(pair.first) ^ std::hash<T2>()(pair.second);
    }
};

void solve()
{
   string s;
   cin>>s;
   ll sum = 0;
    int n = s.size();
    int a = 0, b= 0;

    fr(i,n){
        sum+= s[i]-'0';
        if(s[i]=='2') a++;
        if(s[i]=='3') b++;
    }

    int req = (9 - sum%9)%9;
    // cout<<req<<endl;

    if(req==0){
        YES;
        return;
    }

    // int x = req/6;
    // if(b<=x){
    //     req-=b*6;
    // }
    // else{
    //     req%=6;
    // }

    // if(2*a<=req){
    //     req -= 2*a;
    // }
    // else{
    //     req%=2;
    // }

    // if(req){
    //     NO;
    // }
    // else{
    //     YES;
    // }

    if(req == 1){
        if(b && a>=2){
            YES;
        }
        else{
            NO;
        }
    }
    else if(req == 2){
        if(a){
            YES;
        }
        else{
            NO;
        }
    }
    else if(req == 3){
        if((a>=6 || b>=2) || (a>=3 && b)){
            YES;
        }
        else{
            NO;
        }
    }
    else if(req==4){
        if(a>=4){
            YES;
        }
        else NO;
    }
    else if(req==5){
        if(a>=7){
            YES;
        }
        else NO;
    }
    else if(req==6){
        if(b || a>=3){
            YES;
        }
        else NO;
    }
    else if(req==7){
        if(a>=8 || (b>=2 && a>=2) || (b && a>=5)){
            YES;
        }
        else NO;
    }
    else if(req==8){
        if((b && a) || a>=4){
            YES;
        }
        else NO;
    }

}
int main()
{
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
for _ in range(int(input())):
    n = input()
    dsum = sum(map(int, n))
    if dsum % 9 == 0:
        print('YES')
        continue

    flag = False
    for i in range(len(n)):
        if n[i] not in '23':
            continue

        t = list(n).copy()
        t[i] = '49'[t[i]=='3']
        for j in range(len(n)):
            if sum(map(int, t)) % 9 == 0:
                flag = True
                break

            if t[j] in '23':
                t[j] = '49'[t[j]=='3']

        if flag == True:
            break

    print('YNEOS'[not flag::2])

#for _ in range(int(input())):
#    s = input()
#    n = int(s)
#    if n % 9 == 0:
#        print('YES')
#        continue
#
#    s = list(s)
#    for i in range(len(s)):
#        if s[i] not in '23':
#            continue
#
#        t = s.copy()
#        t[i] = '4' if t[i] == '2' else '9'
#        flag = False
#        for j in range(len(s)):
#            if int(''.join(t)) % 9 == 0:
#                flag = True
#                break
#
#            if t[j] in '23' and i != j:
#                t[j] = '4' if t[j] == '2' else '9'
#
#        if flag == True:
#            print('YES')
#            break
#    else:
#        print('NO')

**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
222
333
23232323232323232323
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
two_remainders = [0, 5, 1, 6, 2, 7, 3, 8, 4]
n = int(input())
while n > 0:
    s = input()
    sum = 0
    two_cnt = 0
    three_cnt = 0
    for c in s:
        sum += int(c)
        if (c == '2'):
            two_cnt += 1
        if (c == '3'):
            three_cnt += 1
    if sum % 9 == 0:
        print("YES")
        n -= 1
        continue
    value_to_add = 9 - sum % 9
    done = False
    for i in range(0, value_to_add + 1):
        if two_cnt < two_remainders[i]:
            continue
        if value_to_add - i == 0:
            print("YES")
            done = True
            break
        if value_to_add - i == 3 and three_cnt >= 2:
            print("YES")
            done = True
            break
        if value_to_add - i == 6 and three_cnt >= 1:
            print("YES")
            done = True
            break
    if not done:
        print("NO")
    n -= 1

**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
22222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
import java.util.Scanner;

public class QB {

    static String solve(String s){
        long sum = 0;
            int c2 = 0, c3  = 0;
            for(int i = 0 ; i < s.length() ; i ++){
                if(s.charAt(i) == '2') c2++;
                else if(s.charAt(i) == '3') c3++;
                sum += s.charAt(i)  - '0';
            }
            if(sum % 9 == 0) return "YES";
            long rem = sum % 9;
            if(sum > 9 && rem % 2 == 1) return "NO";

            for (int i = 0; i <= c2; i++) {
                for (int j = 0; j <= c3; j++) {
                    long newRemainder = (rem + i * 2 + j * 6) % 9;
                    if (newRemainder == 0) {
                        return "YES";
                    }
                }
            }

            return "NO";
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int tc = sc.nextInt();
        while(tc-- != 0){
            String s = sc.next();

           System.out.println(solve(s));
        }

        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
23
32
22
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

public class A {
    static FastRead sc;
    static {
        try {
            sc = new FastRead(System.in);
        } catch (Exception ignored) {}
    }

    static PrintWriter out;
    static {
        try {
            out = new PrintWriter(System.out);
        } catch (Exception ignored) {
        }
    }
    private static class FastRead {
        private final InputStream stream;private final byte[] buf = new byte[1<<18];private int curChar, numChars;public FastRead (InputStream stream) {this.stream = stream;}
        public FastRead (String file) throws IOException {this.stream = new FileInputStream (file);}
        void run() {try {PrintStream fs = new PrintStream("error.txt");System.setErr(fs);} catch (FileNotFoundException e) {e.printStackTrace();}} //error.txt handler
        public int cscan () throws IOException {
            if (curChar >= numChars) {curChar = 0;numChars = stream.read (buf);}if (numChars == -1) return numChars;return buf[curChar++];}
        public int nextInt () throws IOException {
            int c = cscan (), sgn = 1;while (space (c)) c = cscan ();if (c == '-') {sgn = -1;c = cscan ();}
            int res = 0;do {res = (res << 1) + (res << 3);res += c - '0';c = cscan ();} while (!space (c));return res * sgn;}
        public String nextString () throws IOException {
            int c = cscan ();while (space (c)) c = cscan ();StringBuilder res = new StringBuilder ();do {res.appendCodePoint (c);c = cscan ();} while (!space (c));
            return res.toString ();}
        public double nextDouble () throws IOException {
            int c = cscan (), sgn = 1;while (space (c)) c = cscan ();if (c == '-') {sgn = -1;c = cscan ();}double res = 0;
            while (!space (c) && c != '.') {if (c == 'e' || c == 'E') return res * exp(10, nextInt ()); res *= 10;res += c - '0';c = cscan ();}
            if (c == '.') {c = cscan ();double m = 1;while (!space (c)) {if (c == 'e' || c == 'E') return res * exp(10, nextInt ());m /= 10;res += (c - '0') * m;c = cscan ();}}
            return res * sgn;}
        public long nextLong () throws IOException {
            int c = cscan (), sgn = 1;while (space (c)) c = cscan ();if (c == '-') {sgn = -1;c = cscan ();}long res = 0;do {res = (res << 1) + (res << 3);res += c - '0';c = cscan ();}
            while (!space (c));return res * sgn;}
        public boolean space (int c) {return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;}
    }
    static final int mod = 1_000_000_007;
    static long mul(long a, long b) {return (a * b) % mod;}
    static long exp(long base, long exp) {if (exp == 0) return 1;long half = exp(base, exp / 2); if (exp % 2 == 0) return mul(half, half);return mul(half, mul(half, base));}
    static void debug(Object... o){if(o.length != 0) System.err.println(Arrays.deepToString(o));else System.err.println();}
    static long expMod(long base, long exp, long modulo){ // T.C: O(log exp)
        long res = 1; base = base % modulo;if (base == 0) return 0;while (exp > 0) {if ((exp & 1) != 0) res = (res * base) % modulo;exp = exp >> 1;base = (base * base) % modulo;}return res;}
    static long modInv(long a, long modulo){return expMod(a , modulo-2 , modulo);}
    static long mod_add(long a, long b, long modulo) {a = a % modulo; b = b % modulo; return (((a + b) % modulo) + modulo) % modulo;}
    static long mod_mult(long a, long b, long modulo) {a = a % modulo; b = b % modulo; return (((a * b) % modulo) + modulo) % modulo;}
    static long mod_sub(long a, long b, long modulo) {a = a % modulo; b = b % modulo; return (((a - b) % modulo) + modulo) % modulo;}
    static long mod_div(long a, long b, long modulo) {a = a % modulo; b = b % modulo; return (mod_mult(a, modInv(b, modulo), modulo) + modulo) % modulo;}
    static long[] factorials = new long[2_000_001];
    static long[] invFactorials = new long[2_000_001];
    static void preCompFacts() {
        powerUp = true;
        factorials[0] = invFactorials[0] = 1;
        for (int i = 1; i < factorials.length; i++)
            factorials[i] = mod_mult(factorials[i - 1], i,mod);
        invFactorials[factorials.length - 1] = modInv(factorials[factorials.length - 1],mod);
        for (int i = invFactorials.length - 2; i >= 0; i--)
            invFactorials[i] = mod_mult(invFactorials[i + 1], i + 1,mod);
    }
    static boolean powerUp = false;
    static long nCk(int n, int k) {
        if(!powerUp) preCompFacts();
        return mod_mult(factorials[n], mod_mult(invFactorials[k], invFactorials[n - k],mod),mod);
    }

    public static void main(String[] args)throws IOException {
        int t =sc.nextInt();
        mm: while(t--  >0){
            String s = sc.nextString();
            int n=s.length();
            int sum = 0,two = 0,three = 0;
            for(int i=0;i<n;i++){
                if(s.charAt(i) == '2') ++two;
                else if(s.charAt(i) == '3') ++three;
                sum += s.charAt(i)-48;
            }
            if(sum%9 == 0) out.println("YES");
            else {
//                debug(two , three);
                int rem = 9 - sum%9;
                if(rem%2==1) {
                    if(rem == 3){
                        if(three >= 2 || (two>=3&&three>=1)) out.println("YES");
                        else out.println("NO");
                    }else{
                        if((9+rem)/2 >= two || (three>0 && (9+rem-6)>two)) {
                            out.println("YES");
                        }else out.println("NO");
                    }
                }
                else {
                    if(two >= rem/2) out.println("YES");
                    else {
                        for(int i=0;i<=two;i++){
                            for(int j=0;j<=three;j++){
                                debug(i,j,(2*i + 6*j + sum));
                                if((2*i + 6*j + sum)%9 == 0) {
                                    out.println("YES");
                                    out.flush();
                                    continue mm;
                                }
                            }
                        }
                        out.println("NO");
                    }
                }
            }
            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:
```plaintext
5
23333333333333333333
22222222222222222222
33333333333333333333
12345678901234567890
23456789012345678901
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
#include <iostream>
using namespace std;
typedef long long ll;

int main() {
    ll t;
    cin >> t;
    while (t--) {
        ll n, twos=0, threes=0, sum=0;
        cin >> n;

        while (n > 0) {
            if (n % 10 == 2) {
                twos++;
            } else if (n % 10 == 3) {
                threes++;
            }
            sum += n % 10;
            sum %= 9;
            n /= 10;
        }

        bool out = false;
        for (ll i=0; i<=twos; i++) {
            for (ll j=0; j<=threes; j++) {
                if ((sum + (i*2) + (j*6)) % 9 == 0) {
                    out = true;
                    break;
                }
            }
            if (out) {
                break;
            }
        }

        if (out) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

#define int long long

bool canSolve(int a, int n) {
    for (int y = 0; y <= n; y++) {
        int x = (a - 6*y) / 2;
        if ((a - 6*y) % 2 == 0 && x >= 0 && x <= n) {
            return true;
        }
    }
    return false;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t;
    cin>>t;

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

      int n = s.size();

      int sum=0;
      int two=0,three=0;
      for(int i=0;i<n;i++)
      {
        if(s[i]=='2')
        {
          two++;
        }
        else if(s[i]=='3')
        {
          three++;
        }

        sum += (s[i]-'0');
      }

      if(sum%9==0)
      {
        cout << "YES" << endl;
        continue;
      }

      bool reb = false;
      for(int i=0;i<=three;i++)
      {
        int val = sum + 6*i;
        int x = val%9;
        if(x==0)
        {
          reb = true;
          break;
        }

        val += (two*2);
        int val1 = val%9;

        if(val1%2==0)
        {
          if(two>=(val1/2))
          {
            reb = true;
            break;
          }
        }
      }

      if(reb)
      {
        cout << "YES" << endl;
      }
      else
      {
        cout << "NO" << endl;
      }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1234567890
2345678901
1111111111111111111111111111111111111111111111111111111111111111111111111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

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

/*------------------------- MACROS ------------------------------*/
using ll = long long;
using str = string;
const ll M = 1e9+7;
#define int long long
#define forp(i, a, b) for (ll i = a; i <= b; i++)
#define forn(i, a, b) for (int i = a; i >= b; i--)
#define even(x) (x % 2 == 0)
#define pb push_back
#define mk make_pair
/* ------------------------ CONTAINERS --------------------------*/
#define vi vector<int>
#define vl vector<long>
#define vll vector<long long>
#define mi map<int, int>
/*-------------------------- FUNCTIONS -------------------------*/
bool isPowerOfTwo(int n) {
    if (n <= 0)
        return false;
    return (n & (n - 1)) == 0;
}

bool isSquare(int x) {
    int y = sqrt(x);
    return y*y == x;
}

ll ceil(ll x, ll y){
    return ((x + y - 1) / y);
}

//vll prefix(n+1, 0);
//forp(i, 1, n+1){
//    prefix[i] = prefix[i - 1] + v[i - 1];
//}

/*------------------------ Solution --------------------------*/

void solve() {
    int n;
    cin>>n;
    int sum=0,c2=0,c3=0;
    while(n>0){
        int a = n%10;
        sum += a;
        if(a==2) c2++;
        else if(a==3) c3++;
        n /= 10;
    }
    forp(i,0,c2){
        forp(j,0,c3){
        if((sum+2*i+6*j)%9 == 0){
            cout<<"YES"<<endl;
            return;
        }
    }
    }
    cout<<"NO"<<endl;
}

/*--------------------------------------------------------------*/

int32_t main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
return 0;
}

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

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

Think step by step.

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

1. **Understanding the Bug**: The code checks if the sum of the digits, plus the effect of squaring 2s and 3s, can be made divisible by 9. However, it doesn't account for the fact that the digits 2 and 3 can only be squared if the result is a single digit (0-9). The code assumes that squaring 2 and 3 will always be valid, which is incorrect.

2. **Edge Cases**:
   - Numbers with digits 2 and 3 that cannot be squared (e.g., 22, 33).
   - Numbers with digits that are already close to being divisible by 9 but need specific operations to achieve it.
   - Numbers with digits that cannot be changed to make the sum divisible by 9.

3. **Small-Scale Test Input**:
   - A number with digits that cannot be squared to achieve the required sum.
   - A number that is already divisible by 9.
   - A number that requires specific operations to become divisible by 9.

Given these considerations, here is a small-scale test input that should expose the bug:

```plaintext
3
22
33
1234567890
```

### Explanation:
1. **Test Case 1: 22**
   - The digits 2 and 2 cannot be squared to form a single digit (2^2 = 4, which is valid, but 4^2 = 16, which is not a single digit).
   - The sum of the digits is 4, and there is no way to make it divisible by 9 by squaring the digits.

2. **Test Case 2: 33**
   - The digits 3 and 3 cannot be squared to form a single digit (3^2 = 9, which is valid, but 9^2 = 81, which is not a single digit).
   - The sum of the digits is 6, and there is no way to make it divisible by 9 by squaring the digits.

3. **Test Case 3: 1234567890**
   - This number is already divisible by 9 (sum of digits = 45, which is divisible by 9).
   - The code should output "YES" for this case.

This test input should help expose the bug in the provided code by testing edge cases where the digits cannot be squared to achieve the required sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
from functools import lru_cache

t = int(input())
for _ in range(t):
    n = input()
    dic = {2: 0, 3: 0}
    t = 0
    for i in n:
        if 2 <= int(i) <= 3:
            dic[int(i)] += 1
        t += int(i)
    if t % 9 == 0:
        print("yes")
    else:
        tw = dic[2]
        th = dic[3]
        t %= 9
        if t == 1:
            if tw >= 4 or (tw >= 1 and th >= 1):
                print("Yes")
                continue
        elif t == 2:
            if (tw >= 5 and th >= 1) or (tw >= 8):
                print("Yes")
                continue
        elif t == 3:
            if (tw >= 3 or th >= 1):
                print("Yes")
                continue
        elif t == 4:
            if (tw >= 1 and th >= 2) or (tw >= 7):
                print("Yes")
                continue
        elif t == 5:
            if (tw >= 2):
                print("Yes")
                continue
        elif t == 6:
            if (th >= 2 or tw >= 6):
                print("yes")
                continue
        elif t == 7:
            if (tw >= 1):
                print("Yes")
                continue
        elif t == 8:
            if (tw >= 2 and th >= 1) or (tw >= 5):
                print("Yes")
                continue
        print("No")

**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
1234567890
3333333333
2222222222
322
9999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Uninteresting Number
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a number $$$n$$$ with a length of no more than $$$10^5$$$.

You can perform the following operation any number of times: choose one of its digits, square it, and replace the original digit with the result. The result must be a digit (that is, if you choose the digit $$$x$$$, then the value of $$$x^2$$$ must be less than $$$10$$$).

Is it possible to obtain a number that is divisible by $$$9$$$ through these operations?

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

The only line of each test case contains the number $$$n$$$, without leading zeros. The length of the number does not exceed $$$10^5$$$.

It is guaranteed that the sum of the lengths of the numbers across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible to obtain a number divisible by $$$9$$$ using the described operations, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
9
123
322
333333333333
9997
5472778912773
1234567890
23
33
52254522632
output:
NO
YES
YES
NO
NO
YES
NO
YES
YES

Note: In the first example, from the integer $$$123$$$, it is possible to obtain only $$$123$$$, $$$143$$$, $$$129$$$, and $$$149$$$, none of which are divisible by $$$9$$$.

In the second example, you need to replace the second digit with its square; then $$$n$$$ will equal $$$342 = 38 \cdot 9$$$.

In the third example, the integer is already divisible by $$$9$$$.

Buggy Code:
t = int(input())

for i in range(0,t):
    n = int(input())
    df = {}
    df[2] = str(n).count('2')
    df[3] = str(n).count('3')
    r = n % 9
    found = False
    for i in range(0, df[2] + 1):
        for j in range(0, df[3] + 1):
            change = 2*i + 6*j
            if (change + r) % 9 == 0:
                print('yes')
                found = True
                break
            elif j == df[3]:
                break
            else:
                pass
        if found == False and i == df[2]:
            print('no')
            break

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the code to fail. The buggy code only considers the digits '2' and '3', and it checks if squaring these digits can make the number divisible by 9. However, it ignores other digits that can also be squared to form a digit (i.e., 0, 1, 4, 5, 6, 7, 8, 9).

Let's consider the following scenarios:
1. Numbers that contain digits other than '2' and '3' which can be squared to form a digit (e.g., '4', '6', '9').
2. Numbers that are already divisible by 9.
3. Numbers that contain digits '2' and '3' but cannot be made divisible by 9 through the operations.

Here is a small-scale test input that includes these scenarios:

```plaintext
5
123
322
469
999
237
```

Explanation:
1. `123`: This number contains the digit '3', and the code might fail to recognize that it cannot be made divisible by 9.
2. `322`: This number contains the digits '2' and '3', and the code should recognize that it can be made divisible by 9.
3. `469`: This number contains the digits '4', '6', and '9', which can be squared to form digits, and the code might fail to recognize that it can be made divisible by 9.
4. `999`: This number is already divisible by 9.
5. `237`: This number contains the digits '2' and '3', and the code should recognize that it can be made divisible by 9.