**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XOR Sequences
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two distinct non-negative integers $x$ and $y$. Consider two infinite sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$, where

  * $a_n = n \oplus x$;    * $b_n = n \oplus y$.

Here, $x \oplus y$ denotes the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation of integers $x$ and $y$.

For example, with $x = 6$, the first $8$ elements of sequence $a$ will look as follows: $[7, 4, 5, 2, 3, 0, 1, 14, \ldots]$. Note that the indices of elements start with $1$.

Your task is to find the length of the longest common subsegment$^\dagger$ of sequences $a$ and $b$. In other words, find the maximum integer $m$ such that $a_i = b_j, a_{i + 1} = b_{j + 1}, \ldots, a_{i + m - 1} = b_{j + m - 1}$ for some $i, j \ge 1$.

$^\dagger$A subsegment of sequence $p$ is a sequence $p_l,p_{l+1},\ldots,p_r$, where $1 \le l \le r$.

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

The only line of each test case contains two integers $x$ and $y$ ($0 \le x, y \le 10^9, x \neq y$) — the parameters of the sequences.

Output: For each test case, output a single integer — the length of the longest common subsegment.

Examples:
input:
4
0 1
12 4
57 37
316560849 14570961
output:
1
8
4
33554432

Note: In the first test case, the first $7$ elements of sequences $a$ and $b$ are as follows:

$a = [1, 2, 3, 4, 5, 6, 7,\ldots]$

$b = [0, 3, 2, 5, 4, 7, 6,\ldots]$

It can be shown that there isn't a positive integer $k$ such that the sequence $[k, k + 1]$ occurs in $b$ as a subsegment. So the answer is $1$.

In the third test case, the first $20$ elements of sequences $a$ and $b$ are as follows:

$a = [56, 59, 58, 61, 60, 63, 62, 49, 48, 51, 50, 53, 52, 55, 54, \textbf{41, 40, 43, 42}, 45, \ldots]$

$b = [36, 39, 38, 33, 32, 35, 34, 45, 44, 47, 46, \textbf{41, 40, 43, 42}, 53, 52, 55, 54, 49, \ldots]$

It can be shown that one of the longest common subsegments is the subsegment $[41, 40, 43, 42]$ with a length of $4$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	static PrintWriter out = new PrintWriter(System.out);
	static ArrayList<Integer>[] adj;
	static boolean[] visited;
	static int start = -1;
	static int end = -1;
	static int[] parent;
	public static boolean dfs(int node, int parent) {
		visited[node] = true;
		for(int child : adj[node]) {
			if(child == parent) continue;
			if(visited[child]) {
				start = child;
				end = node;
				return true;
			}
			else {
				Main.parent[child] = node;
				if(dfs(child, node))
					return true;
			}
		}
		return false;
	}
	public static void main(String[] args) {
		FastScanner in = new FastScanner();
		int t = in.nextInt();
		while(t-- > 0) {
			long x = in.nextLong();
			long y = in.nextLong();
			int count = 0;
			int max = 0;
			long last = -1;

			for(long i = 0; i <= Math.abs(x - y); i++) {
				long number = x ^ i ^ y;
				if(last != -1) {
					if(number == last + 1) {
						count++;
					}
					else {
						max = Math.max(max, count);
						count = 0;
					}
				}
				last = number;
			}
			out.println(max + 1);
		}

		out.close();
	}

	static class Pair<T, U> {
		public T first;
		public U second;

		public Pair() {

		}

		public Pair(T index, U value) {
			this.first = index;
			this.second = value;
		}

		public String toString() {
			return this.first + " " + this.second;
		}

	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

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

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

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

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

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

}

**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 create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios where it might fail.

### Analysis of the Buggy Code
The code is supposed to find the length of the longest common subsegment of two sequences defined by the XOR operation. The main issue in the code is the way it handles the comparison of consecutive numbers in the sequence. The code assumes that if the next number in the sequence is exactly one more than the previous number, it should increment the count. However, this assumption is incorrect because the sequences are not necessarily increasing by 1.

### Edge Cases to Consider
1. **Small Values of \( x \) and \( y \)**: Test cases where \( x \) and \( y \) are small integers.
2. **Large Values of \( x \) and \( y \)**: Test cases where \( x \) and \( y \) are large integers.
3. **Specific Bit Patterns**: Test cases where \( x \) and \( y \) have specific bit patterns that might cause the XOR operation to produce interesting results.
4. **Edge Case for \( x = y \)**: Although the problem states \( x \neq y \), it's good to ensure the code handles this edge case correctly.
5. **Large Difference Between \( x \) and \( y \)**: Test cases where the difference between \( x \) and \( y \) is large.

### Chosen Test Input
Given the constraints and the need for a small-scale test input, we will choose a test case that includes a small value for \( x \) and a large value for \( y \). This will help expose any issues with the way the code handles large numbers and specific bit patterns.

```plaintext
4
0 1
12 4
57 37
316560849 14570961
```

### Explanation
1. **Test Case 1: \( x = 0 \), \( y = 1 \)**: This is a simple case where the sequences are \( a = [1, 2, 3, 4, 5, 6, 7, \ldots] \) and \( b = [0, 3, 2, 5, 4, 7, 6, \ldots] \). The longest common subsegment is of length 1.
2. **Test Case 2: \( x = 12 \), \( y = 4 \)**: This case involves small values and should produce a longer common subsegment. The sequences are \( a = [14, 12, 13, 11, 10, 9, 8, 0, 7, 6, 5, 3, 2, 1, 15, \ldots] \) and \( b = [4, 7, 6, 3, 2, 1, 15, 13, 12, 14, 10, 9, 8, 11, 5, \ldots] \). The longest common subsegment is of length 8.
3. **Test Case 3: \( x = 57 \), \( y = 37 \)**: This case involves small values with a specific bit pattern. The sequences are \( a = [57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 3


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

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

int main() {
    int t = 0;
    cin >> t;
    for(int i = 0; i < t; i++){
        int w, b;
        cin >> w >> b;
        int count = 0;
        int j = 1;
        while (true) {
            int pinsNeeded = j * (j + 1) / 2; // Total pins needed for side length j
            if (pinsNeeded > w + b) {
                break;
            }
            // Check if we can form the current side length j with the available pins
            int halfPinsNeeded = (j * (j + 1)) / 4;
            if (halfPinsNeeded <= min(w, b) || pinsNeeded <= max(w, b)) {
                count = j;
            }
            j++;
        }
        cout << count << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
2 2
0 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

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

int main()
{
    int t,w,b,r,s,l;
    cin>>t;
    while(t--)
    {
        r=0;
        cin>>w>>b;
        if(w<=b) s=w,l=b;
        else s=b,l=w;

        for(int i=1;s>0;i++)
        {
            if(s>=i)
            {
                r++;
            }
            s-=i;

        }
        for(int i=r+1;l>0;i++)
        {
            if(l>=i)
            {
                r++;
            }
            l-=i;
        }
        cout<<r<<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 buggy code, we need to consider edge cases and scenarios where the code might fail. The bug in the code is that it does not correctly handle the distribution of pins between rows, especially when the number of pins in one color is significantly larger than the other.

Let's break down the potential issues:
1. **Uneven Distribution of Pins**: The code might fail when there are a large number of pins in one color and a small number in the other.
2. **Edge Cases**: The code might fail when the number of pins in one color is just enough to form a row or just not enough.
3. **Large Input Values**: The code might fail when the input values are large, as it might not handle the large sums correctly.

Given these considerations, let's create a test input that includes these scenarios:

1. **Uneven Distribution of Pins**:
   - A large number of white pins and a small number of black pins.
   - A large number of black pins and a small number of white pins.

2. **Edge Cases**:
   - Just enough pins to form a row.
   - Just not enough pins to form a row.

3. **Large Input Values**:
   - Large values for both white and black pins.

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

```plaintext
4
100000000 1
1 100000000
10 10
100000000 100000000
```

Explanation:
- The first test case has a large number of white pins (100,000,000) and a very small number of black pins (1).
- The second test case has a large number of black pins (100,000,000) and a very small number of white pins (1).
- The third test case has a small number of both white and black pins (10 each).
- The fourth test case has a large number of both white and black pins (100,000,000 each).

This input will help expose any issues in the distribution of pins and the handling of large input values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

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

int main() {
	long long int t;
	cin>>t;
	while(t--) {
		long long int w, b;
		cin>>w>>b;
		long long int small, big;
		if (w>b) {
			small = b;
			big = w;
		} else {
			small = w;
			big = b;
		}
		long long int n = 1;
		while (small>0 && small>n) {
			small = small - n;
			n++;
		}
		while (big>0 && big>n) {

			if (n==small){
				n++;
			}

			big = big - n;
			n++;
		}
		cout<<n-1<<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 buggy code, we need to focus on edge cases where the code might fail. The bug in the code seems to be related to how it handles the distribution of pins between rows and the calculation of the maximum side length of the frame.

### Analysis of the Buggy Code
1. **Initialization**: The code correctly identifies the smaller and larger number of pins.
2. **First Loop**: It tries to fill the rows with the smaller number of pins, reducing `small` by `n` and incrementing `n`.
3. **Second Loop**: It tries to fill the remaining rows with the larger number of pins, reducing `big` by `n` and incrementing `n`.

The main issue is in the second loop, where the condition `if (n == small)` is used to skip a row if `n` equals `small`. This can lead to incorrect results in certain scenarios.

### Edge Cases to Consider
1. **Equal Number of Pins**: When `w` and `b` are equal, the code might not handle it correctly.
2. **Very Small Number of Pins**: When `w` and `b` are very small, the code might not handle the base cases correctly.
3. **Large Disparity in Pin Numbers**: When one type of pin is much larger than the other, the code might not correctly distribute the pins.

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

1. **Equal Number of Pins**: `w = 3, b = 3`
2. **Very Small Number of Pins**: `w = 1, b = 1`
3. **Large Disparity in Pin Numbers**: `w = 1, b = 1000000000`
4. **One Type of Pin Only**: `w = 1000000000, b = 0`
5. **Mixed Cases**: `w = 5, b = 10`

### Test Input
```plaintext
5
3 3
1 1
1 1000000000
1000000000 0
5 10
```

This test input covers a variety of scenarios to expose potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
t = int(input())

for _ in range(t):
    w, b = list(map(int, input().split()))
    if w < b:
        b, w = w, b

    i = 1
    remaining_white = w
    while remaining_white >= i:
        remaining_white -= i
        i += 1

    remaining_black = b
    j = remaining_black
    while j > 0:
        if remaining_black >= j:
            remaining_black -= j
            remaining_white += j
            if remaining_white >= i:
                remaining_white -= i
                i += 1
        j -= 1
        if j > remaining_black:
            j = remaining_black
    print(i-1)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The key points to focus on are:

1. **Equal number of white and black pins**: This can challenge the logic of the code where it might not handle the equal distribution correctly.
2. **Very large numbers of pins**: To test the limits of the algorithm and ensure it handles large inputs efficiently.
3. **Small numbers of pins**: To ensure the code works correctly for minimal inputs.
4. **One type of pin in large quantity and the other in small quantity**: To test how the code handles imbalanced distributions.

Let's create a small-scale test input that includes these scenarios:

```plaintext
5
10 10
1 1
1000000 1
1 1000000
99 1
```

Explanation:
1. **10 10**: Equal number of white and black pins.
2. **1 1**: Minimal number of pins of each type.
3. **1000000 1**: A large number of one type of pin and a small number of the other.
4. **1 1000000**: A small number of one type of pin and a large number of the other.
5. **99 1**: A moderate number of one type of pin and a small number of the other.

These test cases will help in identifying 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: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
# # from collections import defaultdict
# #
# # tcs = int(input())
# # ans = []
# # for i in range(tcs):
# #     n,k = map(int,input().split())
# #     a = list(map(int,input().split()))
# #     dfd = defaultdict()
# #     for j in a:
# #         dfd[j] = 0
# #     for j in a:
# #         dfd[j] += 1
# #     sorted_dfd = sorted(dfd.items())  # Returns a list of tuples sorted by keys
# #     maxi = 0
# #     for k in sorted_dfd:
#
# # print(sorted_dfd)
tcs = int(input())
ans = []
for i in range(tcs):
    w, b = map(int, input().split())
    if w == 1 and b == 1:
        ans.append(1)
    elif w==0 and b == 0:
        ans.append(0)
    elif w == 0:
        cb = 0
        while b >= 0:
            if b >= cb:
                b -= cb
                cb += 1
            if b < cb:
                break
        ans.append(cb - 1)
    elif b == 0:
        cw = 0
        while w >= 0:
            if w >= cw:
                w -= cw
                cw += 1
            if w < cw:
                break
        ans.append(cw - 1)
    else:
        la = [1,2,3,4,5,6]
        co = 0
        while w >= 0 and b >= 0:
            # print(co,w,b)
            if w == co:
                w -= co
                co += 1
            if b == co:
                b = 0
                co += 1
            if w >= b and w >= co:
                w -= co
                co += 1
            elif w < b and b >= co:
                b -= co
                co += 1
            if w < co and b < co:
                break
        ans.append(co - 1)
for _ in ans:
    print(_)

**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
1 1
0 0
10 10
1000000000 0
0 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    w,b=map(int, input().split())
    x=w+b
    if x==0:
        print(0)
        continue
    s=0
    for i in range(1,x):
        s+=i
        if s==x:
            # print('result',i)
            print(i)
            break
        elif s>x:
            # print('result',i-1)
            print(i-1)
            break

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

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

Think step by step.

Response:
```plaintext
3
0 0
1 0
1000000000 1000000000
```

### Explanation:
1. **Test Case 1: (0, 0)**
   - This tests the edge case where there are no pins available. The expected output is 0 because no frame can be built.

2. **Test Case 2: (1, 0)**
   - This tests the case where there is only one white pin and no black pins. The expected output is 1 because the largest frame that can be built is a single pin.

3. **Test Case 3: (1000000000, 1000000000)**
   - This tests the case with a very large number of both white and black pins. The expected output is 63245, which is the largest side length of a triangular frame that can be built with the given pins. This is a large-scale test that ensures the algorithm can handle large inputs efficiently.

These test cases are designed to cover edge cases and large input scenarios to expose any potential bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
//package CPprogramming;

import java.util.Scanner;

public class bowlingFrame {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t= sc.nextInt();
        while (t-- > 0) {
            int w = sc.nextInt();
            int b = sc.nextInt();
            int sum = w+b;
            int ans = 0;
            int left = 0;
            int right = (int) Math.floor(Math.pow(2*sum, 0.5));
            while(left<=right){
                int mid = (left+right)/2;
                if(mid*(mid+1)/2<=sum && canFormFrames(mid, w, b)){
                    left = mid+1;
                    ans = mid;
                }else{
                    right = mid-1;
                }
            }
            System.out.println(ans);
        }
    }

    private static boolean canFormFrames(int k, int w, int b){
        for (int i = 1; i < k+1; i++) {
            if(w>=i){
                w-=i;
            } else if (b>=i) {
                b-=i;
            }else{
                return false;
            }
        }
        return true;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that could cause the `canFormFrames` function to fail. Specifically, the function should ensure that each row can be formed with either all white or all black pins, and the total number of pins used should not exceed the available pins.

Here are the steps to think through the test case:

1. **Edge Case with All White Pins**: Test a scenario where there are only white pins and no black pins.
2. **Edge Case with All Black Pins**: Test a scenario where there are only black pins and no white pins.
3. **Edge Case with Mixed Pins**: Test a scenario where the number of white and black pins is balanced but the total number of pins is just enough to form a specific frame.
4. **Edge Case with Insufficient Pins**: Test a scenario where the total number of pins is just slightly less than what is needed to form a specific frame.

Let's generate a test input that covers these cases:

```plaintext
4
10 0
0 10
5 5
14 14
```

Explanation:
1. **10 0**: Only white pins, should form the largest frame with white pins.
2. **0 10**: Only black pins, should form the largest frame with black pins.
3. **5 5**: Balanced number of white and black pins, should form the largest frame possible with the available pins.
4. **14 14**: Slightly more pins, should form the largest frame possible with the available pins.

This test input should help expose any issues in the `canFormFrames` function and the overall logic of the program.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();
        while(t-->0){
            int ans=-1;
            int w=s.nextInt();
            int b=s.nextInt();
            int l=1,h=Integer.MAX_VALUE;
            while(l<=h){
                int mid=l+(h-l)/2;
                int ch=0;
                if((long)(w+b)>=(long)mid*(mid+1)/2){
                    // System.out.println(mid);
                    ans=mid;
                    l=mid+1;
                }
                else h=mid-1;
            }
            System.out.println(ans);
        }

    }
}

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

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

Think step by step.

Response:
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 maximize the failure probability. The bug in the code is in the binary search logic, where it does not correctly account for the constraint that each row must have pins of the same color.

Here's a step-by-step approach to identify a suitable test case:

1. **Identify the Bug**: The bug is in the condition where the code checks if the total number of pins is sufficient to form a frame. It does not consider the constraint that each row must have pins of the same color.

2. **Edge Cases**:
   - **Small Number of Pins**: Test with small values of `w` and `b` to see if the code handles small inputs correctly.
   - **Large Number of Pins**: Test with large values of `w` and `b` to ensure the binary search does not overflow or fail.
   - **Equal Number of Pins**: Test with `w` and `b` being equal to see if the code can handle balanced inputs.
   - **One Color Dominates**: Test with one color having a significantly larger number of pins than the other to see if the code can handle imbalanced inputs.

3. **Specific Test Case**:
   - Let's consider a case where the number of white pins is just enough to form a complete row, but the number of black pins is insufficient to form a complete row. This will test the constraint that each row must have pins of the same color.

Based on this, a suitable test case is:

```plaintext
1
5 2
```

Explanation:
- The total number of pins is 7.
- The largest frame that can be formed is of side length 3, which requires 6 pins (1 + 2 + 3).
- However, the first row must be all white (1 pin), the second row must be all white (2 pins), and the third row can be all black (3 pins).
- The input has 5 white pins and 2 black pins, which means the third row cannot be all black (only 2 black pins are available).

This test case will expose the bug because the code will incorrectly calculate the side length as 3, while the correct side length should be 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb push_back
#define loop0(i, n) for(long long i = 0; i < n; i++)
#define loop1(i, n) for(long long i = 1; i < n; i++)
#define allin(a, i, n) for(long long i=0; i<n; i++)cin>>a[i];
#define allout(a, i, n) for(long long i=0; i<n; i++)cout<<a[i]<<" ";
#define vll vector<long long>
#define pll pair<long long, long long>
#define mll unordered_map<long long, long long>
#define vpll vector<pll>
#define allr(v) v.rbegin(), v.rend()
#define all(v) v.begin(), v.end()
#define all(a) a,a+n
#define vs vector<string>
#define sum_a(a, n) accumulate(a, a+n, 0ll)
#define sum_v(v) accumulate(all(v), 0ll)

void solve(){
    ll x, y;
    cin>>x>>y;
    ll e = x+y;
    ll s = 1;
    ll m = s+(e-s)/2;
    ll ans = m;
    while(e>=s){
        if(m*(m+1ll) <= 2ll*(x+y)){
            ans = m;
            s = m+1;
        }
        else{
            e = m-1;
        }
        m = s+(e-s)/2;
    }
    cout<<ans<<endl;

    // ll a[n];
    // allin(a, i, n);
    // sort(all(a));
    // // cout<<endl;
    // ll b[n-1];
    // for(int i = 0; i < n-1; i++){
    //     b[i] = a[i]+a[i+1];
    // }
    // ll ans = lower_bound(b, b+n-1, a[n-1]) - b;
    // if(b[ans] == a[n-1]) ans++;
    // cout<<ans<<endl;

    // ll n;
    // cin>>n;
    // ll a[n];
    // allin(a, i, n);
    // sort(a, a+n);

    // ll s = *min_element(a, a+n);
    // ll e = *max_element(a, a+n);

    // ll oca[e-s+1]={0};
    // ll ocb[e-s+1]={0};
    // for(auto x:a){
    //     oca[x-s]++;
    // }
    // bool flag=true;
    // for(auto x: oca){
    //     if(x%2) flag = false;
    // }
    // if(flag){
    //     cout<<"YES"<<endl;
    //     return;
    // }
    // ll bff = oca[0]-1;
    // ll z = 0ll;
    // for(auto x: oca){
    //     if(x==0) z++;
    // }
    // z+=(e-s+1);
    // if((bff>z) && (bff-z)%2) cout<<"YES"<<endl;
    // else{
    //     cout<<"NO"<<endl;
    // ll cnt = 1ll;
    // vll oc;
    // for(int i = 1; i < n; i++){
    //     if(a[i] == a[i-1]){
    //         cnt++;
    //     }
    //     else{
    //         oc.pb(cnt);
    //         cnt = 1;
    //     }
    // }
    // oc.pb(cnt);
    // ll bff = 0ll;
    // allout(oc, i, oc.size());
    // for(auto x: oc){
    //     if(x == 2) continue;
    //     if(x > 2){
    //         bff += x-2;
    //     }
    //     else{
    //         bff -= 1;
    //     }
    // }
    // if(bff%2 || bff< 0) cout<<"NO"<<endl;
    // else cout<<"YES"<<endl;
}

int main() {
    ll t; 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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the algorithm might fail. The main issue in the buggy code is the binary search logic that calculates the maximum side length of the triangle that can be formed with the given number of white and black pins.

### Step-by-Step Analysis:

1. **Binary Search Logic**:
   - The code uses a binary search to find the maximum side length `m` such that the total number of pins used is less than or equal to `x + y`.
   - The condition `if(m*(m+1ll) <= 2ll*(x+y))` is used to check if the current side length `m` is valid.

2. **Edge Cases**:
   - **Zero Pins**: Test cases where either `x` or `y` is zero.
   - **Large Disparity**: Test cases where one type of pin is significantly more abundant than the other.
   - **Exact Fit**: Test cases where the total number of pins is exactly equal to the number of pins required for a specific side length.

### Challenging Test Input:
- **Test Case 1**: `x = 0, y = 10` (Only black pins)
- **Test Case 2**: `x = 10, y = 0` (Only white pins)
- **Test Case 3**: `x = 10, y = 10` (Equal number of white and black pins)
- **Test Case 4**: `x = 15, y = 5` (More white pins than black pins)
- **Test Case 5**: `x = 5, y = 15` (More black pins than white pins)
- **Test Case 6**: `x = 20, y = 20` (Large number of pins, equal distribution)

### Final Test Input:
```plaintext
6
0 10
10 0
10 10
15 5
5 15
20 20
```

This test input includes a variety of scenarios that should help expose any issues in the binary search 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: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define pub push_back
#define pob pop_back

#define pii pair<int, int>
#define vi vector<int>
#define vb vector<bool>
#define vc vector<char>
#define v2i vector<vector<int>>
#define v2c vector<vector<char>>
#define v2b vector<vector<bool>>
#define vs vector<string>
#define mpii map<int, int>
#define mpci map<char, int>
#define umpii unordered_map<int, int>
#define qi queue<int>
#define qpi queue<pii>
#define stki stack<int>
#define sti set<int>
#define stc set<char>
#define pqi priority_queue<int>
#define minpqi priority_queue<int, vi, greater<int>>

#define in(x) (cin >> x)
#define out(x) (cout << x << endl)
using namespace std;

const int MOD = 1000000007;

void Question()
{
    int w, b;
    cin >> w >> b;

    if(w > b)
        swap(w, b);

    int ans = 0, l = 0, r = b;

    while(l <= r){
        int m = l + (r - l) / 2;
        int tot = (m * (m + 1)) / 2;

        if(tot <= b)
            ans = m, l = m + 1;
        else
            r = m - 1;
    }

    b -= (ans * (ans + 1)) / 2;

    if(ans < (w + b))
        ans++;

    cout << ans << '\n';
}

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

    while (t--)
        Question();
}

**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: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    w, b = map(int, input().split())
    balls = w + b
    i = 1
    while w >= 0 or b >= 0:
        if w >= i:
            w -= i
            i += 1
        elif b >= i:
            b -= i
            i += 1
        else:
            # i -= 1
            break

    print(i - 1)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the code incorrectly calculates the side length of the largest frame. The bug in the code is that it does not correctly handle the condition where both `w` and `b` are insufficient to form the next row. The code incorrectly breaks the loop without decrementing `i` when both `w` and `b` are less than `i`.

Let's consider a few edge cases:

1. **Both `w` and `b` are small and insufficient to form even a single row.**
2. **One of `w` or `b` is very large, and the other is small.**
3. **`w` and `b` are equal and just enough to form a complete row.**
4. **`w` and `b` are such that one color can form a complete row, but the other cannot.**

We will generate a test input that maximizes the failure probability in the buggy solution. A good test case would be where both `w` and `b` are just enough to form a complete row, but not enough to form the next row.

Here is the test input:

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

Let's break down why this test input is effective:

1. **Case 1: `1 1`**
   - `w = 1`, `b = 1`
   - The first row can be formed with either color, but there are not enough pins to form the second row.
   - Expected output: `1`
   - Buggy output: `2` (incorrect)

2. **Case 2: `2 2`**
   - `w = 2`, `b = 2`
   - The first row can be formed with either color, and the second row can also be formed with the remaining pins.
   - Expected output: `2`
   - Buggy output: `3` (incorrect)

3. **Case 3: `3 3`**
   - `w = 3`, `b = 3`
   - The first row can be formed with either color, and the second row can also be formed with the remaining pins.
   - Expected output: `2`
   - Buggy output: `3` (incorrect)

This test input will clearly show the bug in the code where it incorrectly increments `i` without checking if both `w` and `b` are sufficient to form the next row.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
tc=int(input())
for i in range(tc):
    kegels=input()
    w,b= kegels.split()
    w=int(w)
    b=int(b)
    tot=w+b
    rijen=0
    for i in range(1, tot):
        if tot-i>=0:
            tot=tot-i
            rijen +=1
    print (rijen)

**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 cause the faulty logic to fail. The main issues in the buggy code are:

1. **Incorrect calculation of the total number of pins used**: The code subtracts the number of pins in each row from the total, but it does not correctly account for the requirement that all pins in a row must be of the same color.
2. **Inefficient loop**: The loop continues to subtract pins even when the remaining total is not enough to form a complete row.

To expose these issues, we can create test cases where:
- The number of white and black pins is such that forming a row with all pins of the same color is not possible.
- The total number of pins is just enough to form a specific number of rows but not enough for the next row.

Here are some test cases that can expose these issues:

1. **Test Case 1**: Small number of pins, where the total is just enough for a few rows but not enough for the next row.
2. **Test Case 2**: Large number of pins, where the total is just enough for a large number of rows but not enough for the next row.
3. **Test Case 3**: Equal number of white and black pins, where forming rows with the same color is challenging.
4. **Test Case 4**: One color has a very small number of pins compared to the other, making it impossible to form many rows.

Let's generate a small-scale test input that covers these scenarios:

```plaintext
4
1 2
3 2
3 3
12 0
```

This test input includes:
- **1 2**: Just enough to form 2 rows (1 white + 1 black).
- **3 2**: Just enough to form 2 rows (1 white + 2 white + 1 black).
- **3 3**: Just enough to form 3 rows (1 white + 2 white + 3 black).
- **12 0**: Enough to form 4 rows (1 white + 2 white + 3 white + 4 white).

These test cases will help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Set;
import java.util.StringTokenizer;

public class B2041 {

    private static final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static final PrintWriter writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public static void main(String[] args) throws IOException {
        int testCases = Integer.parseInt(reader.readLine());
        while (testCases-- > 0) {
            solveTestCase();
        }
        writer.close();
    }

    private static void solveTestCase() throws IOException {
        StringTokenizer st = new StringTokenizer(reader.readLine());
        int w = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());

        int w_1 = w;
        int w_2 = w;

        int b_1 = b;
        int b_2 = b;

        // testing white first
        int temp_1 = 1;
        while (w_1 >= temp_1) {
            w_1 -= temp_1;
            temp_1++;
        }

        temp_1--;
        while (b_1 >= temp_1) {

            temp_1++;
            w_1 -= temp_1;
        }
        temp_1--;

        // testing black first
        int temp_2 = 1;
        while (b_2 >= temp_2) {
            b_2 -= temp_2;
            temp_2++;
        }
        temp_2--;
        while (w_2 >= temp_2) {
            w_2 -= temp_2;
            temp_2++;
        }
        temp_2--;

        System.out.println(
                temp_1 >= temp_2 ? temp_1 : temp_2);

    }

    private static int[] readArray(int size) throws IOException {
        StringTokenizer st = new StringTokenizer(reader.readLine());
        int[] arr = new int[size];
        for (int i = 0; i < size; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }
        return arr;
    }

}

**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
10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import java.util.Scanner;

public class Bawling {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t;
        t = sc.nextInt();
        while (t > 0)
        {
            int w, b;
            w = sc.nextInt();
            b = sc.nextInt();
            int n = w + b;
            int s = n / 2  + 1;

            char[][] m = new char[s][s];
            if(w == 0){
                for(int i = 0; i < s; i++){
                    for(int j = 0; j <= i; j++){
                        if(b > 0){
                            m[i][j] = 'b';
                            b--;
                        }
                    }
                }

            }else{

                while (b > 0)
                {
                    int bIndex = b / 2;
                    for(int i = 0; i <= bIndex; i++)
                        m[bIndex][i] = 'b';
                    b -= bIndex + 1;
                }
                int i = 0;
                while(i < s && w > 0){
                    if(m[i][0] == 'b'){
                        i++;
                    }else{
                        for(int j = 0; j < i + 1; j++){
                            if(w > 0)
                                m[i][j] = 'w';
                            w--;
                        }
                        i++;
                    }
                }
            }

        /* for(int i = 0; i < s; i++){
            for(int j = 0; j < s; j++){
                cout << m[i][j] << " ";
            }
            cout << '\n';
        } */
            int count = 0;
            for(int i = 0; i < s; i++){
                if(m[i][i] == 'b' || m[i][i] == 'w')
                    count++;
            }
           System.out.println(count);
            t--;
        }

    }
}

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

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

Think step by step.

Response:
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 cause the code to fail. Here are some considerations:

1. **Edge Case: Zero Pins of One Color**:
   - When one of the colors has zero pins, the code should handle this case correctly. For example, if there are no black pins, the code should only use white pins.

2. **Edge Case: Equal Number of Pins**:
   - When the number of white and black pins is equal, the code should be able to build a frame with alternating rows of each color.

3. **Edge Case: Large Number of Pins**:
   - The code should handle large numbers of pins correctly, ensuring that the side length is calculated accurately.

4. **Edge Case: Small Number of Pins**:
   - When the total number of pins is small, the code should still be able to build the largest possible frame.

5. **Edge Case: Odd and Even Number of Pins**:
   - The code should handle both odd and even numbers of pins correctly.

Based on these considerations, here is a small-scale test input that covers these edge cases:

```plaintext
5
0 10
10 0
5 5
1 1
1000000000 1000000000
```

This test input includes:
- A case with zero black pins.
- A case with zero white pins.
- A case with an equal number of white and black pins.
- A case with a small number of pins.
- A case with a large number of pins.

These cases should help expose any bugs 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: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

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

int find(int a, int b, int r, vector<vector<int>>&dp) {
	if (a < r && b < r) {
		return (r-1);
	}

	if (dp[a][b] != -1 ) return dp[a][b];

	int x = INT_MIN, y = INT_MIN;
	if (a >= r) x = find(a - r, b, r+1, dp);
	if (b >= r) y = find(a, b - r, r+1, dp);

	return dp[a][b] = max(x, y);
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin >> t;
	while (t--) {
		int a, b;
		cin >> a >> b;
		vector<vector<int>>dp(a+1, vector<int>(b+1, -1));
		int res = find(a, b, 1, dp);
		cout << res << 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 buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key aspects to consider are:

1. **Small and Medium Values of `w` and `b`**: These can help in testing the basic functionality and edge cases.
2. **Equal Values of `w` and `b`**: This can test the scenario where both colors are available in equal quantities.
3. **One Color Dominates**: This can test scenarios where one color significantly outnumbers the other.
4. **Zero Values**: This can test scenarios where one or both colors are not available.

Given these considerations, a small-scale test input that covers these scenarios is:

```plaintext
5
1 0
0 1
2 2
10 10
1000000000 1
```

Explanation of the test cases:
1. **1 0**: Only one white pin, no black pins. The largest frame side length is 1.
2. **0 1**: Only one black pin, no white pins. The largest frame side length is 1.
3. **2 2**: Equal number of white and black pins. The largest frame side length is 2.
4. **10 10**: Equal number of white and black pins. The largest frame side length is 5.
5. **1000000000 1**: A large number of white pins and only one black pin. The largest frame side length is 1000000000.

This test input should help in exposing any issues with the dynamic programming approach used 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: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

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

bool output(int w,int b,int & s){
    if(w<0 || b<0) return false;
    if(s>w && s>b) return false;
    // cout<<"surya"<<endl;
    if(((s*(s+1))/2)>w+b) return false;
    if(s==1 && (w>0 || b>0)) return true;
    int total=w+b;
    int side=(-1+sqrt(1+8*total))/2;
    while(side>0){
        int x=side-1;
        s=side;
        if(output(w-side,b,x) || output(w,b-side,x)) return true;
        side--;
    }
    return false;

}

int main(){
    int t;
    cin>>t;
    while(t--){
        int w,b;
        cin>>w>>b;
        int s=0;
        output(w,b,s);
        cout<<s<<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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that are likely to cause the code to fail. Here are the steps to identify such cases:

1. **Edge Cases with Small Numbers**: The code might fail with small values of `w` and `b` where the side length `s` is 1 or 2.
2. **Equal Numbers of Pins**: When `w` and `b` are equal, the code should handle this correctly.
3. **Large Disparity in Pin Counts**: When one type of pin is significantly more than the other, the code should still find the correct side length.
4. **Specific Cases Where the Code Might Fail**: The code uses a recursive approach to check possible side lengths, which might fail in certain configurations.

Let's consider the following test cases:

1. **Small Numbers with Side Length 1 or 2**:
   - `w = 1, b = 2` (Expected output: 2)
   - `w = 3, b = 2` (Expected output: 2)
   - `w = 3, b = 3` (Expected output: 3)

2. **Equal Numbers of Pins**:
   - `w = 10, b = 10` (Expected output: 5)

3. **Large Disparity in Pin Counts**:
   - `w = 12, b = 0` (Expected output: 4)
   - `w = 0, b = 15` (Expected output: 5)

4. **Specific Cases**:
   - `w = 5, b = 7` (Expected output: 4)
   - `w = 6, b = 8` (Expected output: 4)

Combining these cases, we get a small-scale test input that is complete and valid:

```plaintext
6
1 2
3 2
3 3
10 10
12 0
0 15
```

This test input should help expose any issues in the buggy code.