**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Field Division (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.

Alice and Bob are dividing the field. The field is a rectangle of size $n \times m$ ($2 \le n, m \le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).

Bob has $k$ ($2 \le k \le 2 \cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:

  * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field.    * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob.    * Alice will own the part that includes the cell ($n, 1$).    * Bob will own the part that includes the cell ($1, m$).

Alice wants to divide the field in such a way as to get as many cells as possible.

Bob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).

Then output $k$ non-negative integers $a_1, a_2, \dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\alpha + a_i$.

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

The first line of each test case contains three integers $n$, $m$, and $k$ ($2 \le n, m \le 10^9$, $2 \le k \le 2 \cdot 10^5$) — the field sizes and the number of fountains, respectively.

Then follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \le r_i \le n$, $1 \le c_i \le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).

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

Output: For each test case, first output $\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\alpha + a_i$.

Examples:
input:
5
2 2 3
1 1
1 2
2 2
5 5 4
1 2
2 2
3 4
4 3
2 5 9
1 2
1 5
1 1
2 2
2 4
2 5
1 4
2 3
1 3
6 4 4
6 2
1 3
1 4
1 2
3 4 5
2 1
3 2
1 4
1 3
2 4
output:
1
1 0 1
11
0 1 0 4
1
0 0 1 1 0 0 0 0 0
6
15 0 0 0
1
2 3 0 0 0

Note: Below are the images for the second example:

![](CDN_BASE_URL/0e3fee1fe74ac4348d869fc256da57a3) The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.

Note that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.

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

import static java.lang.Math.min;

public class F_1980 {

  private static final int MAX = 200005;

  public static void main(String[] args) throws IOException {
    final var cin = new Reader();
    final var sb = new StringBuilder();
    int t = cin.nextInt();
    while (t-- > 0) {
      int n = cin.nextInt();
      int m = cin.nextInt();
      int k = cin.nextInt();
      ArrayList<Pair>[] tower = new ArrayList[n];
      for (int i = 0; i < n; i++) {
        tower[i] = new ArrayList<>();
      }

      for (int i = 0; i < k; i++) {
        int x = cin.nextInt();
        int y = cin.nextInt();
        tower[x-1].add(new Pair(y, i));
      }

      for (int i = 0; i < n; i++) {
        tower[i].sort(Pair::compareTo);
      }

      Integer current = m;
      int[] width = new int[n];
      int total = 0;
      int[] res = new int[k];
      Pair last = null;
      int lastAcc = 0;
      int lastWidth = 0;

      for (int i = n - 1; i >= 0; i--) {
        if (!tower[i].isEmpty()) {
          Pair rowTower = tower[i].get(0);
          int rowWidth = rowTower.x - 1;
          if (current > rowWidth) {
            if (last != null) {
              res[last.y] = lastAcc;
            }
            last = rowTower;
            if (tower[i].size() != 1) {
              lastWidth = min(current, tower[i].get(1).x - 1) - rowWidth;
            } else {
              lastWidth = current - rowWidth;
            }
            lastAcc = lastWidth;

            current = rowWidth; // required for total
          } else {
            lastWidth = min(rowWidth - current, lastWidth);
            lastAcc += lastWidth;
          }
        } else {
          lastAcc += lastWidth;
        }
        width[i] = current;
        total += current;
      }

      if (last != null) {
        res[last.y] = lastAcc;
      }

      sb.append(total);
      sb.append("\n");
      for (int i = 0; i < k; i++) {
        sb.append(res[i]).append(" ");
      }
      sb.append("\n");

    }
    System.out.println(sb);
  }

  static class Reader {
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
      din = new DataInputStream(System.in);
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
      din = new DataInputStream(
        new FileInputStream(file_name));
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException {
      byte[] buf = new byte[64]; // line length
      int cnt = 0, c;
      while ((c = read()) != -1) {
        if (c == '\n') {
          if (cnt != 0) {
            break;
          } else {
            continue;
          }
        }
        buf[cnt++] = (byte) c;
      }
      return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException {
      int ret = 0;
      byte c = read();
      while (c <= ' ') {
        c = read();
      }
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');

      if (neg)
        return -ret;
      return ret;
    }

    public long nextLong() throws IOException {
      long ret = 0;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public double nextDouble() throws IOException {
      double ret = 0, div = 1;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();

      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');

      if (c == '.') {
        while ((c = read()) >= '0' && c <= '9') {
          ret += (c - '0') / (div *= 10);
        }
      }

      if (neg)
        return -ret;
      return ret;
    }

    private void fillBuffer() throws IOException {
      bytesRead = din.read(buffer, bufferPointer = 0,
        BUFFER_SIZE);
      if (bytesRead == -1)
        buffer[0] = -1;
    }

    private byte read() throws IOException {
      if (bufferPointer == bytesRead)
        fillBuffer();
      return buffer[bufferPointer++];
    }

    public void close() throws IOException {
      if (din == null)
        return;
      din.close();
    }
  }

  static final class Pair implements Comparable<Pair> {
    int x;
    int y;

    Pair(int x, int y) {
      this.x = x;
      this.y = y;
    }

    @Override
    public String toString() {
      return "(" + x + ", " + y + ")";
    }

    @Override
    public int hashCode() {
      return x + y + (y << 16);
    }

    @Override
    public boolean equals(Object obj) {
      if (this == obj)
        return true;
      if (obj == null)
        return false;
      if (getClass() != obj.getClass())
        return false;
      Pair other = (Pair) obj;
      return x == other.x && y == other.y;
    }

    @Override
    public int compareTo(Pair o) {
      if (this.x < o.x) return -1;
      if (this.x > o.x) return 1;
      if (this.y < o.y) return -1;
      if (this.y > o.y) return 1;
      return 0;
    }
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Field Size and Fountains Placement**:
   - The field size can be large, but since we need a small-scale test input, we can use a smaller field size.
   - Place fountains in positions that can challenge the logic of the code, such as near the borders or in a way that affects the path calculation.

2. **Edge Cases**:
   - Consider the case where fountains are placed in a way that affects the path calculation significantly.
   - Test with fountains placed in a straight line, which can challenge the path optimization logic.
   - Test with fountains placed in the corners, which can affect the boundary conditions.

3. **Example Test Case**:
   - Let's create a test case with a 5x5 field and 4 fountains placed in a way that challenges the path calculation.

Here is the test input:

```plaintext
1
5 5 4
1 2
2 2
3 4
4 3
```

### Explanation:
- **Field Size**: 5x5
- **Number of Fountains**: 4
- **Fountains Placement**:
  - (1, 2): Top row, second column
  - (2, 2): Second row, second column
  - (3, 4): Third row, fourth column
  - (4, 3): Fourth row, third column

This placement ensures that the fountains are not in a straightforward line and are spread out in a way that can challenge the path calculation logic. The fountains are placed in positions that can affect the path Alice takes to maximize her plot size, especially when considering the constraints of the problem.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Field Division (hard version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is a hard version of the problem; it differs from the easy version only by the question. The easy version only needs you to print whether some values are non-zero or not. The hard version needs you to print the exact values.

Alice and Bob are dividing the field. The field is a rectangle of size $n \times m$ ($2 \le n, m \le 10^9$); the rows are numbered from $1$ to $n$ from top to bottom, and the columns are numbered from $1$ to $m$ from left to right. The cell at the intersection of row $r$ and column $c$ is denoted as ($r, c$).

Bob has $k$ ($2 \le k \le 2 \cdot 10^5$) fountains, all of them are located in different cells of the field. Alice is responsible for dividing the field, but she must meet several conditions:

  * To divide the field, Alice will start her path in any free (without a fountain) cell on the left or top side of the field and will move, each time moving to the adjacent cell down or right. Her path will end on the right or bottom side of the field.    * Alice's path will divide the field into two parts — one part will belong to Alice (this part includes the cells of her path), the other part — to Bob.    * Alice will own the part that includes the cell ($n, 1$).    * Bob will own the part that includes the cell ($1, m$).

Alice wants to divide the field in such a way as to get as many cells as possible.

Bob wants to keep ownership of all the fountains, but he can give one of them to Alice. First, output the integer $\alpha$ — the maximum possible size of Alice's plot, if Bob does not give her any fountain (i.e., all fountains will remain on Bob's plot).

Then output $k$ non-negative integers $a_1, a_2, \dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\alpha + a_i$.

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

The first line of each test case contains three integers $n$, $m$, and $k$ ($2 \le n, m \le 10^9$, $2 \le k \le 2 \cdot 10^5$) — the field sizes and the number of fountains, respectively.

Then follow $k$ lines, each containing two numbers $r_i$ and $c_i$ ($1 \le r_i \le n$, $1 \le c_i \le m$) — the coordinates of the cell with the $i$-th fountain. It is guaranteed that all cells are distinct and none of them is ($n, 1$).

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

Output: For each test case, first output $\alpha$ — the maximum size of the plot that can belong to Alice if Bob does not give her any of the fountains. Then output $k$ non-negative integers $a_1, a_2, \dots, a_k$, where $a_i$ is a value such that after Bob gives Alice the $i$-th fountain, the maximum size of her plot will be $\alpha + a_i$.

Examples:
input:
5
2 2 3
1 1
1 2
2 2
5 5 4
1 2
2 2
3 4
4 3
2 5 9
1 2
1 5
1 1
2 2
2 4
2 5
1 4
2 3
1 3
6 4 4
6 2
1 3
1 4
1 2
3 4 5
2 1
3 2
1 4
1 3
2 4
output:
1
1 0 1
11
0 1 0 4
1
0 0 1 1 0 0 0 0 0
6
15 0 0 0
1
2 3 0 0 0

Note: Below are the images for the second example:

![](CDN_BASE_URL/0e3fee1fe74ac4348d869fc256da57a3) The indices of the fountains are labeled in green. The cells belonging to Alice are marked in blue.

Note that if Bob gives Alice fountain $1$ or fountain $3$, then that fountain cannot be on Alice's plot.

Buggy Code:
import java.io.*;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

public class Pro7 {

    public static void solve() throws IOException{
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int[][] points = new int[k][3];
        for (int i = 0; i < k; i++) {
            points[i][0] = i;
            points[i][1] = in.nextInt();
            points[i][2] = in.nextInt() - 1;
        }
        Arrays.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                if (o1[2] == o2[2]) return Integer.compare(o1[1], o2[1]);
                return Integer.compare(o1[2], o2[2]);
            }
        });
        long sum = 0, row = 0;
        boolean[] vis = new boolean[k];
        for (int i = 0; i < k; i++) {
            if (row >= points[i][1]) continue;
            sum += (points[i][1] - row) * points[i][2];
            row = points[i][1];
            vis[points[i][0]] = true;
        }
        sum += (n - row) * m;
        out.println(sum);
        long[] ans = new long[k];
        for (int i = 0; i < k; i++) {
            if (!vis[points[i][0]]) continue;
            int L = i - 1, R = i + 1;
            while (L >= 0 && !vis[points[L][0]]) L--;
            while (R < k && !vis[points[R][0]]) R++;
            int x1 = L < 0 ? 0 : points[L][1];
            int y1 = L < 0 ? 0 : points[L][2];
            int x2 = R >= k ? n : points[R][1];
            int y2 = R >= k ? m : points[R][2];
            long res = 0;
            res -= (points[i][1] - x1) * points[i][2];
            res -= (x2 - points[i][1]) * y2;
            for (int j = i + 1; j < R; j++) {
                if (x1 >= points[j][1]) continue;
                res += (points[j][1] - x1) * points[j][2];
                x1 = points[j][1];
            }
            res += (x2 - x1) * y2;
            ans[points[i][0]] = res;
        }
        for (long v : ans) out.print(v + " ");
        out.println();
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

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

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

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

/*

2 2 1 2 3 2 2 3
1 2 2 2 2 2 3 3

 */

**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
2
2 2 2
1 1
2 2
5 5 4
1 2
2 2
3 4
4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
#define optimize() ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

int32_t main()
{
    optimize();

    int t;
    cin>>t;

    while(t--)
    {
        int a,b,p=0;
        cin>>a>>b;

        if(a>0 && b==0)
        {
            if(a%2==0) p=1;
        }
        else if(a>0 && b>0 && a%2==0)
        {
            b+=(a/2);

            if(b%2==0) p=1;
        }

        if(p) 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
1
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

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

int l;

int main(){
	cin.tie(0)->sync_with_stdio(0);

	cin >> l;

	while (l--){
		int a, b;
		cin >> a >> b;
		if ((a == 0 && b == 1) || (a == 1 && b == 0))
			cout << "No" << "\n";
		else if (a % 2 == 0 && b % 2 == 0)
			cout << "Yes" << "\n";
		else if (a == (b / 2 * 2))
			cout << "Yes" << "\n";
		else
			cout << "No" << "\n";
	}

	return 0;
}

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

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

Think step by step.

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

**Explanation:**
1. **Case 1: (0, 0)**
   - This is an edge case where both `a` and `b` are 0. The array is empty, and the sum is trivially 0. The expected output is "Yes".
   - The buggy code might not handle this case correctly because it doesn't have a specific condition for this scenario.

2. **Case 2: (1, 2)**
   - The array is [1, 2, 2]. It is impossible to place signs to make the sum 0. The expected output is "No".
   - The buggy code might incorrectly output "Yes" because it doesn't properly handle the case where `a` is 1 and `b` is 2.

3. **Case 3: (4, 2)**
   - The array is [1, 1, 1, 1, 2, 2]. It is possible to place signs to make the sum 0 (e.g., +1 +1 -1 -1 -2 +2). The expected output is "Yes".
   - The buggy code might incorrectly output "No" because it doesn't handle the case where `a` is 4 and `b` is 2 correctly.

These test cases are designed to expose potential issues in the buggy code by focusing on edge cases and scenarios that are not handled correctly by the given logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
// Online C++ compiler to run C++ program online
#include <iostream>
using namespace std;
#include <algorithm>
#include <map>
#include <climits>
#include <vector>
#include <cmath>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <list>
#include <iomanip>
#define mod 1000000007
struct ing{
    long long  e,s,x;
};

long long gcd(long long a,long long b){
      long long temp;
    while(a > 0 && b > 0){
        temp = b;
        b = a%b;
        a = temp;

    }
    return  max(a,b);
}
    string alpha="abcdefghijklmnopqrstuvwxyz";

 bool condition(pair <int,string> a,pair<int,string>b){
     if(a.first ==b.first){
         return a.second < b.second;
     }
     return a.first>b.first;
 }

int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t = 1;
    cin >> t;
//  freopen("input.txt", "r", stdin);
  //  freopen("output.txt", "w", stdout);

    while(t--){
int a ,b;cin >> a >> b;
if(a%2!=1){
    if(b==0){cout <<"YES\n";continue;}
    b+=a/2;
    if(b%2==0)cout <<"YES\n";
    else cout <<"NO\n";
}
else cout <<"NO\n";
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the logic and identify potential edge cases:

1. **Basic Conditions:**
   - The code checks if `a` is even. If `a` is even, it proceeds to check if `b` is even after adjusting `b` by adding `a/2`.
   - If `a` is odd, it directly outputs "NO".

2. **Edge Cases:**
   - **Case 1:** `a` is even and `b` is 0.
   - **Case 2:** `a` is even and `b` is odd.
   - **Case 3:** `a` is even and `b` is even.
   - **Case 4:** `a` is odd and `b` is 0.
   - **Case 5:** `a` is odd and `b` is non-zero.

3. **Potential Bugs:**
   - The code might incorrectly handle cases where `a` is even and `b` is adjusted to be odd.
   - The code might not correctly handle cases where `a` is even and `b` is 0.

Let's create a test input that includes these edge cases:

```plaintext
5
0 0
2 0
2 1
2 2
3 1
```

Explanation:
- **Test Case 1:** `a = 0, b = 0` — The array is empty, and the sum is already 0. Expected output: "YES".
- **Test Case 2:** `a = 2, b = 0` — The array is `[1, 1]`. We can make the sum 0 by using `+1 -1`. Expected output: "YES".
- **Test Case 3:** `a = 2, b = 1` — The array is `[1, 1, 2]`. It is impossible to make the sum 0. Expected output: "NO".
- **Test Case 4:** `a = 2, b = 2` — The array is `[1, 1, 2, 2]`. We can make the sum 0 by using `+1 +1 -2 -2`. Expected output: "YES".
- **Test Case 5:** `a = 3, b = 1` — The array is `[1, 1, 1, 2]`. It is impossible to make the sum 0. Expected output: "NO".

This test input will help expose any issues in the logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
for _ in range(int(input())):
    a,b = map(int, input().split())
    if (a==0 and b%2 ==0) or (a%2==0 and b==0):
        print("YES")
    elif a%2==0 and abs(((a/2)-b))%2==0:
        print("YES")
    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
5
0 2
1 1
2 2
3 3
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
n = int(input())
for _ in range(n):
    a,b = map(int,input().split())
    if a%2==1 :         # a is odd
        print("NO")
    else:               # a is even
        if (a//2+b)%2==0  or  b==0:
            print("YES")
        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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the logic in the code might fail. The buggy code has the following logic:

1. If `a` is odd, it immediately prints "NO".
2. If `a` is even, it checks if `(a//2 + b) % 2 == 0` or `b == 0`.

The key points to consider are:
- What happens when `a` is even and `b` is odd?
- What happens when `a` is even and `b` is even?
- What happens when `a` is zero and `b` is non-zero?

Let's construct a test input that covers these scenarios:

1. `a = 2`, `b = 1` (a is even, b is odd)
2. `a = 2`, `b = 2` (a is even, b is even)
3. `a = 0`, `b = 2` (a is zero, b is even)
4. `a = 0`, `b = 1` (a is zero, b is odd)

These test cases will help us verify the behavior of the code in different scenarios.

Here is the test input:

```plaintext
4
2 1
2 2
0 2
0 1
```

Explanation:
- For `a = 2`, `b = 1`: The array is `[1, 1, 2]`. It is impossible to make the sum zero.
- For `a = 2`, `b = 2`: The array is `[1, 1, 2, 2]`. It is possible to make the sum zero.
- For `a = 0`, `b = 2`: The array is `[2, 2]`. It is possible to make the sum zero.
- For `a = 0`, `b = 1`: The array is `[2]`. It is impossible to make the sum zero.

This test input should expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
a = int(input())
for _ in range(a):
    b, c = map(int,input().split())
    if b % 2 != 0:
        print('NO')
    elif b == 0 and c == 1:
        print('NO')
    elif b % 2 == 0 and b // 2 <= c and (b // 2 + c) % 2 == 0:
        print('YES')
    elif b % 2 == 0 and b // 2 > c and (b - c * 2) % 2 == 0:
        print('YES')
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can highlight the issues in the logic. The main issues in the buggy code are:

1. Incorrect handling of the case where `b` is 0 and `c` is an odd number.
2. Incorrect handling of the case where `b` is even and the combination of `b` and `c` does not satisfy the conditions for the sum to be zero.

Let's consider the following edge cases:

1. **Case 1: `b = 0` and `c` is an odd number.**
   - The code incorrectly prints 'NO' for `b = 0` and `c = 1`, but it should handle other odd values of `c` correctly.

2. **Case 2: `b` is even and the combination of `b` and `c` does not satisfy the conditions for the sum to be zero.**
   - The code has conditions that might not cover all possible valid combinations.

Let's construct a test input that includes these edge cases:

```plaintext
5
0 1
0 3
2 0
2 3
4 2
```

Explanation:
1. **Test Case 1: `0 1`**
   - The array is `[2]`. It is impossible to add signs to get a sum of 0. Expected output: `NO`.
2. **Test Case 2: `0 3`**
   - The array is `[2, 2, 2]`. It is impossible to add signs to get a sum of 0. Expected output: `NO`.
3. **Test Case 3: `2 0`**
   - The array is `[1, 1]`. It is possible to add signs to get a sum of 0 (`+1-1`). Expected output: `YES`.
4. **Test Case 4: `2 3`**
   - The array is `[1, 1, 2, 2, 2]`. It is possible to add signs to get a sum of 0 (`+1+1-2-2+2`). Expected output: `YES`.
5. **Test Case 5: `4 2`**
   - The array is `[1, 1, 1, 1, 2, 2]`. It is possible to add signs to get a sum of 0 (`+1+1+1+1-2-2`). Expected output: `YES`.

This test input should expose the bug in the buggy code by covering the edge cases where the conditions in the code might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.*;
public class Main {
public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    for(int i=0;i<t;i++){
        int a=sc.nextInt();
        int b=sc.nextInt();
        int target = a + 2 * b;
            if (target % 2 == 0) {
                int x = target / 2;
                if (x >= 0 && x < a + b) {
                    System.out.println("Yes");
                } else {
                    System.out.println("No");
                }
            } else {
                System.out.println("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:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The code checks if the sum of the array can be zero by verifying if the total sum of the elements (after adding signs) can be split equally. However, it has a logical flaw in how it checks the feasibility of the solution.

### Analysis of the Buggy Code:
1. **Total Sum Calculation:**
   - The code calculates the total sum of the array as `a + 2 * b`.
   - It then checks if this sum is even (`if (target % 2 == 0)`).

2. **Feasibility Check:**
   - If the sum is even, it calculates `x = target / 2`.
   - It then checks if `x` is within the range `[0, a + b)`.
   - If `x` is within this range, it prints "Yes"; otherwise, it prints "No".

### Edge Cases to Consider:
1. **Zero Elements:**
   - Cases where either `a` or `b` is zero.
2. **Odd and Even Combinations:**
   - Cases where the sum of `a + 2 * b` is even but the distribution of `1`s and `2`s makes it impossible to achieve a zero sum.
3. **Maximum Values:**
   - Cases where `a` and `b` are at their maximum values (9).

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

```plaintext
5
0 0
1 0
0 1
9 9
1 8
```

### Explanation:
1. **(0, 0):**
   - The array is empty. The sum is already zero.
   - Expected output: "Yes".

2. **(1, 0):**
   - The array is `[1]`. It is impossible to make the sum zero.
   - Expected output: "No".

3. **(0, 1):**
   - The array is `[2]`. It is impossible to make the sum zero.
   - Expected output: "No".

4. **(9, 9):**
   - The array is `[1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2]`.
   - The total sum is 27, which is odd. The code should output "No".
   - Expected output: "No".

5. **(1, 8):**
   - The array is `[1, 2, 2, 2, 2, 2, 2, 2, 2]`.
   - The total sum is 17, which is odd. The code should output "No".
   - Expected output: "No".

This test input covers a variety of edge cases and should help expose the logical flaws in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.*;
public class main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int one = sc.nextInt();
            int two = sc.nextInt();
            int sum1=1*one;
            int sum2=2*two;
            int possible=0;
            if(one>0 && one%2!=0) possible=0;
            else if(one%2==0){
                if(two==0) possible=1;
                else{
                    int diff=two-one/2;
                    if(Math.abs(diff)%2==0) possible=1;
                }
            }
            if(possible==1) System.out.println("YES");
            else System.out.println("NO");
        }
        sc.close();
    }
}

/*
LOGIC ---
*/

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

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

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

    int t;
    cin>>t;
    while(t--){
        int a,b;
        cin>>a>>b;

        if(a%2!=0) cout<<"NO\n";

        else if(a%2==0 && b%2==0) cout<<"YES\n";

        else if(a==0 && b%2==0) cout<<"YES\n";

        else if (a==0 && b%2!=0) cout<<"NO\n";

        else if(b%2!=0){
            if((a/2) %2!=0) cout<<"YES\n";

        }

        else cout<<"NO\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
10
0 0
1 0
0 1
2 0
0 2
2 1
1 2
3 0
0 3
3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

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

const int P = 239017;
const int MOD_X = 1e9 + 7;
const int MOD_Y = 1e9 + 9;

struct H {
    int x, y;
    H() = default;
    H(int _x): x(_x), y(_x) {}
    H(int _x, int _y): x(_x), y(_y) {}

    inline H operator+(const H& h) const {
        return H(add(x, h.x, MOD_X), add(y, h.y, MOD_Y));
    }
    inline H operator-(const H& h) const {
        return H(sub(x, h.x, MOD_X), sub(y, h.y, MOD_Y));
    }
    inline H operator*(const H& h) const {
        return H(mul(x, h.x, MOD_X), mul(y, h.y, MOD_Y));
    }
    inline bool operator==(const H& h) const {
        return x == h.x && y == h.y;
    }

    static inline int add(int a, int b, int mod) {
        int res = a + b;
        return res >= mod ? res - mod : res;
    }

    static inline int sub(int a, int b, int mod) {
        int res = a - b;
        return res < 0 ? res + mod : res;
    }

    static inline int mul(int a, int b, int mod) {
        return (1LL * a * b) % mod;
    }
};

const int N = 1e5 + 5;
H p[N], h[N];

inline H get(int l, int r) {
    return h[r] - h[l] * p[r - l];
}

void init(const string& s) {
    int n = s.size();
    p[0] = H(1);
    for (int i = 0; i < n; i++) {
        h[i + 1] = h[i] * P + s[i];
        p[i + 1] = p[i] * P;
    }
}

//first check whether input is taken correctly or not
//then check the function which you are calling is called with a correct name
//int to long long

void csolve() {
//     int n;
//     cin >> n;
//     vector<int>v(n);
//     for (int i = 0 ; i < n ; i++) {

//     }
    int a, b;
    cin >> a >> b;
    if (a == 0) {
        if (b % 2 == 0) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
        return ;
    }
    if (b == 0) {
        if (a % 2 == 0) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
        return ;
    }
    if (a & 1) {
        cout << "NO\n";
        return ;
    }
    a /= 2;
    b += a;
    if (b % 2 == 0) {
        cout << "YES\n";
    } else {
        cout << "NO\n";
    }
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        csolve();
    }
    // string s = "examplesfdexample";
    // init(s);

    // // Example: Get the hash of the substring "exam"
    // int l = 0, r = 6; // "exam" corresponds to the substring [0, 4)
    // H hash_value = get(l, r);
    // cout << "Hash value of substring 'exam': (" << hash_value.x << ", " << hash_value.y << ")" << endl;
    // H hash_value2 = get(10, 16);
    // cout << "Hash value of substring 'exam': (" << hash_value2.x << ", " << hash_value2.y << ")" << 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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might not be handled correctly. The code seems to have a logical error in how it handles the distribution of '1's and '2's to achieve a sum of 0. Let's analyze the code and identify potential issues:

1. **Edge Case: Zero Elements**:
   - The code handles the case where either `a` or `b` is zero, but it might not handle the case where both `a` and `b` are zero.

2. **Odd and Even Distributions**:
   - The code checks if `a` is odd and immediately returns "NO" if it is. However, it should also consider the combined effect of `a` and `b` to achieve a sum of 0.

3. **Specific Combinations**:
   - The code might not correctly handle specific combinations of `a` and `b` that should theoretically be solvable.

Let's generate a test input that includes these edge cases:

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

Explanation:
- **Test Case 1: (0, 0)**: Both `a` and `b` are zero. The array is empty, and the sum is already 0. This should return "YES".
- **Test Case 2: (1, 1)**: One '1' and one '2'. This should return "NO" because it's impossible to balance the sum to 0.
- **Test Case 3: (2, 2)**: Two '1's and two '2's. This should return "YES" because we can use signs to balance the sum to 0 (e.g., +1+1-2-2).
- **Test Case 4: (3, 3)**: Three '1's and three '2's. This should return "NO" because the sum of '1's and '2's cannot be balanced to 0.
- **Test Case 5: (4, 4)**: Four '1's and four '2's. This should return "YES" because we can use signs to balance the sum to 0 (e.g., +1+1+1+1-2-2-2-2).
- **Test Case 6: (5, 5)**: Five '1's and five '2's. This should return "NO" because the sum of '1's and '2's cannot be balanced to 0.

This test input should help identify and expose any bugs in the code's logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import sys

lines = sys.stdin.readlines()
lines.remove(lines[0])

for line in lines:
    t = [int(p) for p in line.split(" ")]

    if (((t[0]/2 + t[1]) % 2 == 0) or t[1] % 2 == 0) and (t[0]%2 == 0) and (t[1] + t[0] != 1):
        print("YES")
    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:
To expose the bug in the given buggy code, we need to focus on edge cases that can make the code's logic fail. The code checks several conditions to determine if the sum of the array can be zero. Let's break down the conditions and identify potential edge cases:

1. **Condition 1**: `(((t[0]/2 + t[1]) % 2 == 0) or t[1] % 2 == 0) and (t[0]%2 == 0) and (t[1] + t[0] != 1)`
   - This condition checks if the sum of the array can be zero by ensuring that certain parity conditions are met.

2. **Edge Cases to Consider**:
   - **Case 1**: Both `a` and `b` are zero. This should trivially be "Yes" because the sum of an empty array is zero.
   - **Case 2**: `a` is zero and `b` is odd. This should be "No" because it's impossible to make the sum zero with an odd number of 2s.
   - **Case 3**: `a` is zero and `b` is even. This should be "Yes" because it's possible to make the sum zero with an even number of 2s.
   - **Case 4**: `a` is even and `b` is zero. This should be "Yes" because it's possible to make the sum zero with an even number of 1s.
   - **Case 5**: `a` is even and `b` is odd. This should be "No" because it's impossible to make the sum zero with an odd number of 2s.
   - **Case 6**: `a` is odd and `b` is even. This should be "No" because it's impossible to make the sum zero with an odd number of 1s.
   - **Case 7**: `a` is odd and `b` is odd. This should be "No" because it's impossible to make the sum zero with an odd number of 1s and 2s.

Let's create a test input that includes these edge cases:

```plaintext
7
0 0
0 1
0 2
2 0
2 1
1 2
1 1
```

This test input covers the edge cases mentioned above and should help expose any bugs in the faulty solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
for test in range(int(input())):
    a,b=map(int,input().split())
    if a!=0 and a%2==0 :
        print("YES")
    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
5
0 0
1 1
2 2
3 3
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.Scanner;

public class GS {
    public static String func(int a,int b){
        if(a*2==b )return "YES";
        if((a==0 && b%2==0) || (b==0 && a%2==0) )return "YES";
        if( b%2==1 && a==2)return "YES";
        return "NO";
    }

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        for(int i=0;i<n;i++){
            int a= sc.nextInt();
            int b=sc.nextInt();
            System.out.println(func(a,b));
        }

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.*;

 public class mine {
    public static void main(String[] args) {
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();
        for(int i=1;i<=t;i++)
        {
            int a=s.nextInt();
            int b=s.nextInt();
            if((a>0 && a%2==0) && (b%2==0 || (a+(2*b))%2==0))
            {
                System.out.println("Yes");
            }
            else
            {
                System.out.println("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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can potentially cause the code to fail. The buggy code has a condition that checks if `a` is greater than 0 and even, and if `b` is even or if the sum of `a` and `2*b` is even. This condition is not always correct for determining if the sum can be zero.

Let's analyze the conditions and find a test case that can expose the bug:

1. **Case 1: `a` is 0 and `b` is odd**
   - The sum of the array will always be positive and cannot be zero.
   - Example: `a = 0`, `b = 1` (already given in the example, output should be "No").

2. **Case 2: `a` is even and `b` is even**
   - The sum can be zero if the number of 1s and 2s can be balanced.
   - Example: `a = 2`, `b = 2` (output should be "Yes").

3. **Case 3: `a` is even and `b` is odd**
   - The sum can be zero if the number of 1s and 2s can be balanced.
   - Example: `a = 2`, `b = 1` (output should be "No").

4. **Case 4: `a` is odd and `b` is even**
   - The sum cannot be zero because the total number of elements is odd.
   - Example: `a = 1`, `b = 2` (output should be "No").

5. **Case 5: `a` is odd and `b` is odd**
   - The sum cannot be zero because the total number of elements is even, but the sum of 1s and 2s cannot be balanced.
   - Example: `a = 1`, `b = 1` (output should be "No").

6. **Case 6: `a` is even and `b` is such that `a + 2*b` is even**
   - The sum can be zero if the number of 1s and 2s can be balanced.
   - Example: `a = 2`, `b = 3` (already given in the example, output should be "Yes").

7. **Case 7: `a` is even and `b` is such that `a + 2*b` is odd**
   - The sum cannot be zero.
   - Example: `a = 2`, `b = 4` (output should be "No").

Let's choose a test case that is not covered in the examples and can expose the bug:

```plaintext
5
0 1
2 2
2 1
1 2
2 4
```

This test case includes:
- `0 1` (should be "No")
- `2 2` (should be "Yes")
- `2 1` (should be "No")
- `1 2` (should be "No")
- `2 4` (should be "No")

This will help to expose the bug in the condition checking of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main() {
    int N;
    cin >> N;
    for(int i=0; i<N; i++) {
        int a, b;
        cin >> a >> b;
        if (a > 2 * b) {
            a = a - 2 *b;
            b = 0;
        }else{
            b = b - a / 2;
            a = a % 2;
        }
        a = a % 2;
        b = b % 2;
        if(a==0 && b == 0) 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
0 0
1 0
0 2
3 2
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

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

#define forn(iter,size) for(int iter = 0; iter < int(size); ++iter)

typedef long long ll;
typedef double du;
const int
    INF = (int)1e9 + 7,
    MOD = (int)1e9 + 7;

void solve(){
    int a,b;cin >> a >> b;
    if(b * 2 >= a){
        b -= a / 2;
        if(b % 2 != 0){cout << "NO\n";return;}
    }else if(a > b * 2){
        a -= b * 2;
        if(a % 2 != 0){cout << "NO\n";return;}
    }
    cout << "YES\n";
}

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

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

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

Think step by step.

Response:
```plaintext
1
5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
t = int(input())
for i in range(t):
    a, b = map(int, input().split())
    array = [1 for i in range(a)] + [2 for j in range(b)]
    if a==0 and b%2!=0 or a%2!=0 and b==0:
        print('NO')
    elif a==0 and b%2==0 or a%2==0 and b==0:
        print('YES')
    elif a%2==0 and (b-a//2)%2==0:
        print('YES')
    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
5
0 2
1 1
2 2
3 3
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
x=int(input())
mylist=[]

for _ in range(x):
    y = input()
    (a,b) = tuple(y.split())
    a = int(a)
    b = int(b)

    if a%2 == 1:
        print('NO')
    else:
        if b%2 == 0:
            print('YES')
        else:
            if a%4 == 2:
                print('YES')
            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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the code might fail. The buggy code has a logical error in its conditions for determining whether the sum can be zero. Specifically, it incorrectly handles cases where the number of ones and twos are not evenly divisible in a way that allows the sum to be zero.

Let's analyze the conditions:
1. If `a` is odd, the code always prints "NO".
2. If `a` is even and `b` is even, the code always prints "YES".
3. If `a` is even and `b` is odd, the code prints "YES" if `a % 4 == 2`, otherwise "NO".

We need to find a case where these conditions lead to an incorrect result. Consider the following scenarios:

1. **Case where `a` is even and `b` is odd, but the sum can be zero**:
   - Example: `a = 4`, `b = 1`
   - Array: `[1, 1, 1, 1, 2]`
   - Possible sum: `+1 + 1 - 1 - 1 - 2 = 0`
   - Expected output: "YES"
   - Buggy code output: "NO" (because `a % 4 != 2`)

2. **Case where `a` is even and `b` is even, but the sum cannot be zero**:
   - Example: `a = 2`, `b = 2`
   - Array: `[1, 1, 2, 2]`
   - Possible sum: No combination of signs can make the sum zero.
   - Expected output: "NO"
   - Buggy code output: "YES" (because `a % 4 != 2` and `b` is even)

3. **Case where `a` is odd, but the sum can be zero**:
   - Example: `a = 3`, `b = 2`
   - Array: `[1, 1, 1, 2, 2]`
   - Possible sum: `+1 + 1 + 1 - 2 - 2 = 0`
   - Expected output: "YES"
   - Buggy code output: "NO" (because `a` is odd)

Given these scenarios, the most challenging test input that exposes the bug in the code is:

```plaintext
3
4 1
2 2
3 2
```

This input includes:
1. A case where `a` is even and `b` is odd, but the sum can be zero.
2. A case where `a` is even and `b` is even, but the sum cannot be zero.
3. A case where `a` is odd, but the sum can be zero.

These cases will help identify the logical errors in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.*;
public class A_Sakurako_s_Exam {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int tc=sc.nextInt();
        for(int io=0;io<tc;io++){
            int n=sc.nextInt(),m=sc.nextInt();
            if(n%2!=0){
                System.out.println("NO");
            }else if((m%2)*2 == n){
                System.out.println("YES");
            }else if(m*2==n){
                System.out.println("YES");
            }else if(m%2==0 && n%2==0)  {
                System.out.println("YES");
            }else{
                System.out.println("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
1
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Exam
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, Sakurako has a math exam. The teacher gave the array, consisting of $a$ ones and $b$ twos.

In an array, Sakurako must place either a '+' or a '-' in front of each element so that the sum of all elements in the array equals $0$.

Sakurako is not sure if it is possible to solve this problem, so determine whether there is a way to assign signs such that the sum of all elements in the array equals $0$.

Input: The first line contains a single integer $t$ ($1\le t\le 100$)  — the number of test cases.

The only line of each test case contains two integers $a$ and $b$ ($0\le a,b<10$)  — the number of '1's and the number of '2's in the array.

Output: For each test case, output "Yes" if you can make the sum of the entire array equal to $0$, 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:
5
0 1
0 3
2 0
2 3
3 1
output:
NO
NO
YES
YES
NO

Note: 1. $a=0$, $b=1$: This means the array is $[2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   2. $a=0$, $b=3$: This means the array is $[2, 2, 2]$ — it is impossible to add the signs '+' or '-' to get $0$ as a result;   3. $a=2$, $b=0$: This means the array is $[1, 1]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1-1=0$);   4. $a=2$, $b=3$: This means the array is $[1, 1, 2, 2, 2]$ — it is possible to add the signs '+' or '-' to get $0$ as a result ($+1+1-2-2+2=0$);

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    while (t-- > 0) {
        int ones = in.nextInt();
        int twos = in.nextInt();
        if (twos == 0 && ones % 2 == 0) {
            System.out.println("Yes");
        }
        else if (twos % 2 == 0 && ones== 0){
            System.out.println("YES");
        }
        else{
            int sum = ones + (2 * twos);
            if (sum % 2 != 0) {
                System.out.println("NO");
            } else {
                ones /= 2;
                int total = ones + twos;
                if (total % 2 == 0) {
                    System.out.println("YES");
                } else {
                    System.out.println("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
1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

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

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define max(A,B) (A>B? A:B)
#define min(A,B) (A<B? A:B)
#define yes cout<<"Yes\n"
#define no cout<<"No\n"
#define pir pair<ll,ll>
#define mpr(A,B) make_pair(A,B)
using namespace std;

const int N=2e5+10;
const int mod=998244353;

int mul(int a,int b){
    return (ll)a*b%mod;
}
int pls(int a,int b){
    if(a+b>=mod) return a+b-mod;
    if(a+b<0) return a+b+mod;
    return a+b;
}
int qpow(int a,int b){
    int ret=1;
    while(b){
        if(b&1) ret=mul(ret,a);
        a=mul(a,a);
        b>>=1;
    }
    return ret;
}

int n,a[N];
ll s[N];

void slv(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    s[0]=0;
    for(int i=1;i<=n;i++) s[i]=s[i-1]+a[i];
    ll c=0;
    for(int i=0;i<=n;i++) c=max(c,abs(s[i])+s[n]-s[i]);
    // cout<<"c="<<c<<'\n';
    int ans=0,cnt=0;
    for(int i=0;i<=n;i++){
        if(s[i]>=0&&i>0) cnt++;
        if(c==s[n]-2*s[i]){
            ans=pls(ans,qpow(2,n-i+cnt));
            // cout<<" pls: i="<<i<<" p="<<n-i+cnt<<'\n';
        }
    }
    cout<<ans<<'\n';
    return;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin>>T;
    while(T--){
        slv();
    }
    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
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find the number of unique procedures that result in $c = k$. Two procedures are different if at any index $i$, one procedure chose option $1$ and another chose option $2$, even if the value of $c$ is equal for both procedures after that turn.

Since the answer may be large, output it modulo $998\,244\,353$.

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

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 2 \cdot 10^5$).

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

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the number of unique procedures that result in $c = k$, modulo $998\,244\,353$.

Examples:
input:
5
4
2 -5 3 -3
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
12
256
1
8
16

Note: In the first test case, it can be shown that our maximal final value of $c$ is $3$. There are $12$ ways to achieve this because in order to get $3$, we have to take absolute value at indices $2$ or $4$, or both, resulting in $3$ ways. For the other two indices, it doesn't change the value whether we take absolute value or not, so we have $2 \cdot 2 = 4$ ways for them. In total, we have $3 \cdot 4 = 12$ ways.

In the second test case, taking the absolute value will never change anything, so we can either take absolute value or not, for every index. This gives us $2^8 = 256$ possible ways.

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

#include <iostream>
#include <vector>
using namespace std;

void solve(){
    int n;
    cin>>n;
    vector<long long> v;
    int a;
    cin>>a;
    v.push_back(a);
    for(int i=1;i<n;i++){
        long long b;
        cin>>b;
        v.push_back(b);
    }
    long long ans=0;
    long long nAns=0;
    long long ansWays=1;
    long long nWays=1;
    for(int i=0;i<v.size();i++){
        if(v[i]==0){
            ansWays*=2;
            nWays*=2;
            ansWays%=998244353;
            nWays%=998244353;
        }else if(v[i]<0){
            if(ansWays+v[i]>=0){
                ansWays*=2;
                ansWays%=998244353;
            }
            if(nAns+v[i]>=0){
                nWays*=2;
                nWays%=998244353;
            }
            if(abs(nAns+v[i])==ans+v[i]){
                ansWays+=nWays;
                ansWays%=998244353;
            }else if(abs(nAns+v[i])>abs(ans+v[i])){
                ansWays=nWays;
            }
            nAns+=v[i];
            ans=max(ans+v[i],abs(nAns));
        }else{
            nAns+=v[i];
            if(nAns>=0){
                nWays*=2;
                nWays%=998244353;
            }
            ans=ans+v[i];
            if(ans>=0){
                ansWays*=2;
                ansWays%=998244353;
            }
        }
    }
    if(abs(nAns)>ans){
        cout<<nWays<<"\n";//"---"<<endl;
    }else{
        cout<<ansWays<<"\n";//"---"<<endl;
    }
    //cout<<max(abs(nAns),ans)<<"\n";
}

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
4
-1 1 -1 1
```