**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

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

int main() {
    string str;
    cin >> str;

    int k = -1;
    vector<int> num;

    for (int i = 0; i < str.size(); i++) {
        if (str[i] == '(') {
            k = k + 1;
            num.push_back(k);
        } else if (str[i] == ')') {
            num.push_back(k + 1);
            k = k - 1;
        }
    }

    int maxi = *max_element(num.begin(), num.end());
    int start = 0;

    while (start <= maxi) {
        for (int i = num.size() - 1; i >= 0; i--) {
            if (num[i] == start) {
                cout << str[i];
            }
        }
        start++;
    }

    cout << endl;
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the code to fail. The main issue in the buggy code is the way it handles the prefix balance and the sorting of the positions. Specifically, the code incorrectly updates the prefix balance for closing parentheses and does not correctly handle the sorting of positions.

Here are the steps to think through the problem:

1. **Identify the Bug:**
   - The code incorrectly increments the balance for closing parentheses by `k + 1` and then decrements `k` by 1. This is incorrect because the balance should be decremented before adding the balance to the `num` vector.
   - The sorting of positions is done in a way that does not correctly handle the order of positions with the same balance.

2. **Create a Test Case:**
   - We need a test case where the balance of the sequence is such that the incorrect handling of the balance and the incorrect sorting of positions can be easily observed.
   - A good test case would be a sequence where the balance changes frequently and the positions with the same balance are in a specific order that the buggy code cannot handle correctly.

3. **Example Test Case:**
   - A sequence like `(()())` is a good candidate because it has a simple structure but still requires correct handling of the balance and positions.
   - The correct balanced shuffle of `(()())` should be `()()()`.

4. **Verify the Test Case:**
   - Run the buggy code with the test case `(()())` and observe the output.
   - The correct output should be `()()()`.

Given the above analysis, the test input that exposes the bug in the buggy code is:

```plaintext
(()())
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void solve(){
    int n,k;
    cin>>n>>k;
    if(k==1){
        if(n%2==0) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    else{
        int p=n-k+1;
        if(p%2==0 && n%2==0){
            cout<<"NO"<<endl;
        }
        else if(p%2!=0 && n%2!=0){
            cout<<"YES"<<endl;
        }
        else if(p%2!=0 && n%2==0){
            if(n==p+1){
                cout<<"NO"<<endl;
            }
            else cout<<"YES"<<endl;
        }
        else if(n%2!=0 && p%2==0){
            if(n==p+1){
                cout<<"NO"<<endl;
            }
            else cout<<"YES"<<endl;
        }
    }
}
int main(){
    int 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 given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has several conditions and branches, so we should test the following cases:

1. **Small values of \( n \) and \( k \)**: These can help us verify the basic logic.
2. **Large values of \( n \) and \( k \)**: These can test the performance and correctness of the logic for larger inputs.
3. **Boundary conditions**: Test cases where \( n \) and \( k \) are at their limits.
4. **Special cases**: Test cases where \( k = 1 \) or \( n \) is a small multiple of \( k \).

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

```plaintext
6
1 1
2 1
2 2
3 2
4 4
1000000000 1000000000
```

This test input includes:
1. The smallest possible values for \( n \) and \( k \) (1 1).
2. A case where \( n = 2 \) and \( k = 1 \) to test the even number condition.
3. A case where \( n = 2 \) and \( k = 2 \) to test the even number condition with a different \( k \).
4. A case where \( n = 3 \) and \( k = 2 \) to test the odd number condition.
5. A case where \( n = 4 \) and \( k = 4 \) to test a larger even number condition.
6. A large value for both \( n \) and \( k \) to test the boundary condition.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

int main()
{
   int t;cin>>t;
   while(t--)
   {
      int m,n,r;cin>>m>>n;
      r = m - n;
      long long int sum=0;
      for(int i=m; i>r; i--)
      {
          sum+= pow(i,i);
      }

      if(sum%2==0)cout<<"YES"<<endl;
      else cout<<"NO"<<endl;
   }
}

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

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

Think step by step.

Response:
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 can cause the code to fail. The buggy code has a few issues:

1. **Incorrect Calculation of Leaf Durability**: The code incorrectly calculates the number of leaves by iterating from `m` to `m - n` and summing `i^i` for each `i`. This is incorrect because it should consider the leaves that are still on the tree from previous years.
2. **Overflow Issues**: The code uses `long long int` for `sum`, but the value of `i^i` can be extremely large, leading to overflow.

To create a test case that exposes these issues, we can use a scenario where the number of leaves is large enough to cause overflow, or where the incorrect calculation logic leads to a wrong result.

Let's consider the following test case:

```plaintext
1
10 5
```

**Explanation:**
- **Year 10**: The tree will have leaves from years 6 to 10 (since leaves last for 5 years).
- **Leaves Calculation**:
  - Year 6: \(6^6\)
  - Year 7: \(7^7\)
  - Year 8: \(8^8\)
  - Year 9: \(9^9\)
  - Year 10: \(10^{10}\)

The sum of these values is:
\[ 6^6 + 7^7 + 8^8 + 9^9 + 10^{10} \]

This sum is extremely large and can easily cause overflow. Additionally, the incorrect logic in the buggy code will not correctly sum the leaves from the previous years.

**Expected Output:**
- The correct output should be "YES" or "NO" based on the parity of the sum of the leaves. However, the buggy code will likely produce an incorrect result due to the overflow and incorrect logic.

This test case should effectively 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: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

typedef long long ll;
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define gcd(a,b) __gcd(a,b);

const int mx = 2e5+123;
int a[mx];
vector<pair<ll, ll>>v;

bool cmp ( const pair<int,int> &p1, const pair<int, int> &p2  )
{
    if ( p1.first > p2.first ) return 1;
    if ( p1.first == p2.first ) return ( p1.second < p2.second );
    return 0;
}

int main()
{
  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
  ll t = 1;
   cin >> t;
  while(t--){
    ll n ,  k ;
    cin >> n >> k;

    ll sum =pow(n,n) ;
       for(ll i = 1 ; i < n ;i++){
            if(i-1 + k >= n){
                ll x = pow(i,i);
                sum +=x;
            }
    }
    if(sum & 1)
    {
        no;
    }else yes;

      }
    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 given code, we need to focus on edge cases and scenarios where the code might fail. The code provided has a potential issue with handling large values and the logic for determining the number of leaves in a given year.

### Key Points to Consider:
1. **Large Values**: The code uses `pow` which can overflow for large values of `n` and `i`.
2. **Edge Cases**: Test with small values of `n` and `k` to see if the logic is correct.
3. **Special Cases**: Test with `n = k` and `n > k` to ensure the logic handles these scenarios correctly.

### Test Case Analysis:
- **Small `n` and `k`**: Ensure the basic logic is correct.
- **Large `n` and `k`**: Check for overflow and correctness in handling large values.
- **`n = k`**: Ensure the logic handles the case where `n` equals `k`.
- **`n > k`**: Ensure the logic handles the case where `n` is greater than `k`.

### Challenging Test Input:
```plaintext
5
1 1
2 1
2 2
3 2
1000000000 1000000000
```

### Explanation:
1. **Test Case 1**: `1 1`
   - In year 1, there is only 1 leaf.
   - Expected output: "NO" (1 is odd).

2. **Test Case 2**: `2 1`
   - In year 2, there are `2^2 = 4` leaves.
   - Expected output: "YES" (4 is even).

3. **Test Case 3**: `2 2`
   - In year 2, there are `1 + 2^2 = 5` leaves.
   - Expected output: "NO" (5 is odd).

4. **Test Case 4**: `3 2`
   - In year 3, there are `2^2 + 3^3 = 4 + 27 = 31` leaves.
   - Expected output: "NO" (31 is odd).

5. **Test Case 5**: `1000000000 1000000000`
   - In year 1000000000, the number of leaves will be the sum of `i^i` for all `i` from 1 to 1000000000 where `i-1 + k >= n`.
   - This is a very large number and can cause overflow in the `pow` function.
   - Expected output: "NO" (The actual value is not important, but the code should handle large values correctly).

This test input will help expose potential issues with overflow and the logic for determining the number of leaves in a given year.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
t = int(input()) # набор количества данных
result = []
for i in range(t):
    n, k = list(map(int, input().split(" ")))
    cnt = []
    start = 1
    if n-k > 0:
        start = n-k
    for i in range(start, n+1):
        if i > k:
            cnt = cnt[1:]
        cnt.append(i**i)
    if sum(cnt) % 2 == 0:
        result.append("YES")
    else:
        result.append("NO")

print("\n".join(result))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the code and identify potential edge cases or scenarios that could cause issues.

### Code Analysis:
1. **Input Handling:**
   - The code reads the number of test cases `t` and then processes each test case.
   - For each test case, it reads two integers `n` and `k`.

2. **Logic for Leaf Calculation:**
   - The code initializes a list `cnt` to store the number of leaves added each year.
   - It starts from `start = n - k` and iterates up to `n`.
   - If `i > k`, it removes the first element of `cnt` to simulate the leaves falling off.
   - It appends `i^i` to `cnt` for each year `i`.
   - Finally, it checks if the sum of `cnt` is even or odd and appends the result to the `result` list.

### Potential Issues:
1. **Edge Case: Small `k` and Large `n`**:
   - If `k` is small and `n` is large, the code might not correctly handle the removal of old leaves.
   - For example, if `k = 1` and `n` is large, the code should only consider the leaves from the current year.

2. **Edge Case: `n = k`**:
   - If `n = k`, the code should consider all leaves from year 1 to year `n`.

3. **Edge Case: `n < k`**:
   - If `n < k`, the code should consider all leaves from year 1 to year `n`.

### Challenging Test Input:
To expose these potential issues, we can create a test input that includes these edge cases:

```plaintext
3
1 1
2 1
10 1
```

### Explanation:
1. **Test Case 1: `n = 1, k = 1`**:
   - The tree starts with 1 leaf in year 1.
   - Expected output: "NO" (1 is odd).

2. **Test Case 2: `n = 2, k = 1`**:
   - In year 2, the tree has 2^2 = 4 leaves.
   - Expected output: "YES" (4 is even).

3. **Test Case 3: `n = 10, k = 1`**:
   - In year 10, the tree has 10^10 leaves.
   - Expected output: "NO" (10^10 is odd).

This test input should help expose the bug in the code by testing edge cases where the number of years is small and the number of years the leaves last is also small.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
def power(x, y):

    if (y == 0):
        return 1
    elif (int(y % 2) == 0):
        return (power(x, int(y / 2)) *
                power(x, int(y / 2)))
    else:
        return (x * power(x, int(y / 2)) *
                power(x, int(y / 2)))

t=int(input())
for _ in range(t):
    n,k=map(int,input().split())
    l=0
    for i in range(n,n-k,-1):
        l+=power(i,i)
    if l%2==0:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the following aspects:

1. **Edge Cases**: Test with the smallest and largest possible values for `n` and `k`.
2. **Large Values**: Test with large values of `n` and `k` to ensure the code handles large numbers correctly.
3. **Specific Values**: Test with specific values that might cause the `power` function to behave incorrectly, such as values that result in large intermediate results.

The buggy code has a potential issue in the `power` function, which might not handle large values efficiently and could lead to overflow or incorrect results. Additionally, the loop in the main function might not correctly account for the range of years.

Here is a small-scale test input that includes edge cases and large values to maximize the failure probability:

```plaintext
5
1 1
2 2
1000000000 1000000000
3 3
1000000000 1
```

Explanation:
- **1 1**: The smallest possible values for `n` and `k`.
- **2 2**: A simple case where `k` equals `n`.
- **1000000000 1000000000**: The largest possible values for `n` and `k`.
- **3 3**: A case where `k` equals `n` but with a larger `n` than the second test case.
- **1000000000 1**: A large `n` with the smallest possible `k`.

This test input should help expose any issues with the `power` function and the main loop 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: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
from sys import stdin
def input(): return stdin.readline()[:-1]
def arin(): return map(int,input().split())

def main():
    for _ in range(int(input())):
        n, k = arin()
        if k % 4 == 1:
            print('YES') if not n % 2 else print('NO')
        elif k % 4 == 2:
            print('NO')
        elif k % 3 == 3:
            print('YES') if n % 2 else print('NO')
        else:
            print('YES')

main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has several conditions based on the values of `k % 4` and `k % 3`, which might not handle all possible cases correctly. Let's analyze the code and identify potential issues:

1. **Condition for `k % 4 == 1`**:
   - The code prints 'YES' if `n` is even and 'NO' if `n` is odd.
   - This condition might be incorrect for some values of `n` and `k`.

2. **Condition for `k % 4 == 2`**:
   - The code always prints 'NO'.
   - This might be incorrect for some values of `n` and `k`.

3. **Condition for `k % 3 == 3`**:
   - The code prints 'YES' if `n` is even and 'NO' if `n` is odd.
   - This condition might be incorrect for some values of `n` and `k`.

4. **Default case**:
   - The code always prints 'YES'.
   - This might be incorrect for some values of `n` and `k`.

To create a test input that exposes these issues, we can use the following test cases:

1. **Test Case 1**: `n = 1, k = 1`
   - Expected output: 'NO'
   - This is a basic case where the tree starts with 1 leaf.

2. **Test Case 2**: `n = 2, k = 1`
   - Expected output: 'YES'
   - The tree grows 4 leaves in the 2nd year.

3. **Test Case 3**: `n = 2, k = 2`
   - Expected output: 'NO'
   - The tree has 5 leaves in the 2nd year.

4. **Test Case 4**: `n = 3, k = 2`
   - Expected output: 'NO'
   - The tree has 31 leaves in the 3rd year.

5. **Test Case 5**: `n = 4, k = 4`
   - Expected output: 'YES'
   - The tree has 256 leaves in the 4th year.

6. **Test Case 6**: `n = 5, k = 4`
   - Expected output: 'NO'
   - The tree has 1025 leaves in the 5th year.

7. **Test Case 7**: `n = 10, k = 5`
   - Expected output: 'YES'
   - This is a more complex case to test the general behavior.

8. **Test Case 8**: `n = 10, k = 6`
   - Expected output: 'NO'
   - This is another complex case to test the general behavior.

Here is the test input:

```plaintext
8
1 1
2 1
2 2
3 2
4 4
5 4
10 5
10 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;
public class codeforceseasy {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int check = 1;
        int testcases = input.nextInt();

        while (check <= testcases) {
            int n= input.nextInt();
            int k= input.nextInt();
            //ArrayList<Integer> list = new ArrayList<>();
            int oddcount=0;
           if(k == n){
               if(n %2 ==0){
                   oddcount = n/2;
               }
               else{
                   oddcount = n/2+1;
               }
           }
           if(k<n){
               int even = (n)/2;
               int odd = n - even;
               int even1 = (n-k+1)/2;
               int odd1 = (n-k+1) - even1;
               oddcount = odd - odd1;
           }
            if(oddcount %2 == 0){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }

            check++;
        }

    }
}

**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
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner sc=new Scanner(System.in);
	    int t=sc.nextInt();
	    while(t-->0){
	        long n=sc.nextLong();
	        int k=sc.nextInt();

	        long data;

	        if(n>k+k-1){
	            System.out.println("YES");
	        }else if(n<=k){
	            data=n*(n+1)/2;
	            if(data%2==0){
	                 System.out.println("YES");
	            }else{
	                System.out.println("NO");
	            }

	        }else {
	            data=n*(n+1)/2;
	            long v=n-k;
	            data-=v*(v+1)/2;
	            if(data%2==0){
	                System.out.println("YES");
	            }else{
	                System.out.println("NO");
	            }

	        }

	    }

	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the logic might fail. The code has several potential issues:

1. **Incorrect logic for large values of `n` and `k`**: The code uses a simple if-else structure that might not correctly handle the number of leaves for large values of `n` and `k`.
2. **Incorrect handling of the sum of leaves**: The code calculates the sum of leaves using a simple arithmetic formula, which might not correctly account for the exponential growth of leaves (`i^i`).

Let's consider the following edge cases:

1. **Small values of `n` and `k`**: The code should handle small values correctly, but we can still test them to ensure the logic is sound.
2. **Large values of `n` and `k`**: The code might fail to handle large values correctly due to incorrect logic.
3. **Specific values where the sum of leaves is close to the edge of being even or odd**: This can help expose any issues in the modulo operation.

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

```plaintext
5
1 1
2 1
3 2
100 10
1000000000 1000000000
```

Explanation:
1. **Test Case 1 (1 1)**: Simplest case where `n = 1` and `k = 1`. The tree starts with 1 leaf, which is odd.
2. **Test Case 2 (2 1)**: `n = 2` and `k = 1`. The tree has 4 leaves in year 2, which is even.
3. **Test Case 3 (3 2)**: `n = 3` and `k = 2`. The tree has 31 leaves in year 3, which is odd.
4. **Test Case 4 (100 10)**: A larger value of `n` and `k`. This tests the code's ability to handle larger numbers and the sum of leaves.
5. **Test Case 5 (1000000000 1000000000)**: A very large value of `n` and `k`. This tests the code's ability to handle very large numbers and the sum of leaves.

This test input should help expose any issues 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: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
for _ in range(int(input())):
    n,k=map(int,input().split())
    ans=0
    s1=(n-(n-k)) #剩余的数字
    s=s1//2
    s2=s1%2
    if s1==1:
        if n%2==0:
            print("YES")
        else:
            print("NO")
    else:
        if s2!=0:
            s=s+1
        if s%2==0:
            print("YES")
        else:
            print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
n1=int(input())
for i in range(n1):
    n,k=map(int,input().split())
    if k==1:
        if n%2==0:
            print("YES")
        else:
            print("NO")

    else:
        if k==2:
            print("NO")

        else:
            if k%2==0:
             print('YES')
            else:
                if n%2==0:
                    if ((k-1)%2)%2!=0:
                        print("NO")
                    else:
                        print("YES")
                else:
                    if ((k-1)%2)%2!=0:
                        print("YES")
                    else:
                        print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;
public class codeforceseasy {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int check = 1;
        int testcases = input.nextInt();

        while (check <= testcases) {
            int n= input.nextInt();
            int k= input.nextInt();
            ArrayList<Integer> list = new ArrayList<>();
            for(int i = 1; i < n+1; i++){
                int num = (int)Math.pow(i,i);
                list.add(num);
            }
            int sum =0;
            for(int i = list.size()-k; i < list.size(); i++){
                sum += list.get(i);
            }
           // System.out.println(sum);
            if(sum%2 ==0){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }
            check++;
        }

    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios that might cause the code to fail. The main issue in the code is the inefficient and unnecessary storage of all the leaf counts from year 1 to year n. This can lead to performance issues, especially for large values of n.

Here are some edge cases to consider:
1. **Small values of n and k**: These can help verify the basic logic.
2. **Large values of n and k**: These can test the performance and correctness of the code.
3. **Specific values of n and k that might cause overflow or incorrect summation**.

Let's focus on a small-scale test input that can expose the bug:

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

This input includes:
- The smallest possible values for n and k (1 1).
- A case where n and k are both 2.
- A case where n is 3 and k is 2.
- A case where n is 4 and k is 4.

These test cases will help verify the basic logic and ensure that the code handles the transition from odd to even leaf counts correctly. If the code fails on any of these cases, it will be clear that there is a bug in the logic or performance handling.

Here is the test input:

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

public class Main{
	public static void main (String[] args) throws java.lang.Exception{
		Scanner scnr = new Scanner(System.in);
		int cases = scnr.nextInt();
		while(cases-- > 0){
		    int n = scnr.nextInt();
		    int k = scnr.nextInt();
		    boolean leavesOdd = false;
		    if (k%2 == 0 && k>2) {
		        System.out.println("YES");
		        continue;
		    }

		    int first = n-k+1;
		    if (k==1 && first %2 ==0){
		        		        System.out.println("YES");
		        continue;

		    }
		    if (k==1 && first %2 ==1){
		        		        System.out.println("NO");
		        continue;

		    }
		    if (k==2 && first % 2 == 1){
		        System.out.println("NO");
		        continue;
		    }

		    if (first % 2 ==0){
		        System.out.println("NO");
		        continue;
		    }		}
    }

}

**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, let's analyze the code and identify the potential issues:

### Analysis of the Buggy Code:
1. **Edge Cases and Special Conditions:**
   - The code has specific checks for `k == 1`, `k == 2`, and `k % 2 == 0 && k > 2`.
   - The code does not handle the general case correctly, especially for larger values of `k`.

2. **General Case Handling:**
   - The code only checks if `first % 2 == 0` and prints "NO" in that case. This is incorrect because the parity of the number of leaves depends on the sum of `i^i` for `i` from `n-k+1` to `n`.

### Challenging Test Input:
To expose the bug, we need a test case that:
- Forces the code to evaluate the general case where `k` is not 1 or 2.
- Ensures that the sum of `i^i` from `n-k+1` to `n` results in an even or odd number, which the code might not handle correctly.

### Example Test Case:
Let's choose `n = 5` and `k = 3`:
- In year 5, the leaves are from `5-3+1 = 3` to `5`.
- The leaves are `3^3 + 4^4 + 5^5`.
- Calculate the sum:
  - `3^3 = 27`
  - `4^4 = 256`
  - `5^5 = 3125`
  - Sum = `27 + 256 + 3125 = 3408`
- The sum `3408` is even, so the correct output should be "YES".

### Test Input:
```plaintext
1
5 3
```

This test case should expose the bug in the general case handling of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

int main()
{
    int t;
    cin >> t;
    while(t--){
      int n,k;
      cin >> n >> k;
      int c=0;
      for(int i=0;i<k;i++){
        c+=n;
        n-=1;
      }
      if(c%2==0) cout << "yes" << endl;
      else cout << "no" << endl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

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

    while (t--)
    {
        int n, k;
        cin >> n >> k;

        int tl = 0;

        for (int i = 1; i <= n; ++i)
        {
            tl += i;

            if (i > k)
            {
                tl -= (i - k);
            }
        }

        if (tl % 2 == 0)
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's analyze the problem and the code in detail.

### Problem Analysis
- The tree grows \(i^i\) new leaves in the \(i\)-th year.
- Leaves last for \(k\) years.
- We need to determine if the number of leaves in year \(n\) is even.

### Buggy Code Analysis
- The code incorrectly accumulates the total number of leaves by adding \(i\) instead of \(i^i\).
- The code also incorrectly subtracts \((i - k)\) instead of the leaves that actually fell off.

### Challenging Test Input
To expose the bug, we need to create a test case where the difference between the correct and incorrect leaf count is significant. A good candidate is a year \(n\) where \(i^i\) grows rapidly, and the value of \(k\) is relatively small.

Let's choose:
- \(n = 5\) (a year where \(i^i\) is significant)
- \(k = 2\) (a small \(k\) to ensure leaves fall off quickly)

### Test Input
```plaintext
1
5 2
```

### Explanation
- **Year 1**: \(1^1 = 1\) leaf (1 leaf)
- **Year 2**: \(2^2 = 4\) leaves (1 + 4 = 5 leaves)
- **Year 3**: \(3^3 = 27\) leaves (5 + 27 = 32 leaves, 1 leaf falls off, 31 leaves)
- **Year 4**: \(4^4 = 256\) leaves (31 + 256 = 287 leaves, 4 leaves fall off, 283 leaves)
- **Year 5**: \(5^5 = 3125\) leaves (283 + 3125 = 3408 leaves, 27 leaves fall off, 3381 leaves)

The correct number of leaves in year 5 is 3381, which is odd. The buggy code will incorrectly accumulate and subtract leaves, leading to a different result.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

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

int calculate(int x,int y);

int main() {
	int testNum = 0;
	cin >> testNum;
	int x = 0, y = 0;
	for (int i = 0; i < testNum; ++i) {
		cin >> x >> y;
		if (calculate(x,y) % 2 == 0) {
			cout << "YES" << endl;
		}
		else
			cout << "NO" << endl;
	}
}

int calculate(int x,int y) {
	int sum = 0;
	int remainLeaf = x - y + 1;
	for (int i = remainLeaf; i <= x; ++i) {
		int temp = 1;
		for (int j = 1; j < i; ++j) {
			temp *= i;
		}
		sum += temp;
	}
//	cout << sum << endl;
	return sum;
}

**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 given buggy code, let's analyze the code and identify potential issues:

1. **Edge Cases:**
   - **Small Values:** Test cases with small values of `n` and `k` to ensure the basic logic works.
   - **Large Values:** Test cases with large values of `n` and `k` to test the performance and correctness of the algorithm.
   - **Boundary Conditions:** Test cases where `n` is just larger than `k` or `n` is equal to `k`.

2. **Specific Scenarios:**
   - **`k = 1`**: This should be a straightforward case where the number of leaves is just the sum of powers of `i` from `1` to `n`.
   - **`k = n`**: This should be another straightforward case where all leaves from year `1` to year `n` are still on the tree.
   - **`k > 1`**: This should test the logic for multiple years of leaf persistence.

3. **Performance Issues:**
   - The code uses nested loops to calculate `i^i` for each year from `x - y + 1` to `x`. This can be very slow for large values of `x` and `y`.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
5
1 1
2 1
2 2
3 2
1000000000 1
```

### Explanation:
- **Test Case 1: `1 1`**
  - `n = 1`, `k = 1`: Only 1 leaf in the first year.
  - Expected output: `NO` (1 is odd).

- **Test Case 2: `2 1`**
  - `n = 2`, `k = 1`: In the second year, there are `2^2 = 4` leaves.
  - Expected output: `YES` (4 is even).

- **Test Case 3: `2 2`**
  - `n = 2`, `k = 2`: In the second year, there are `1 + 2^2 = 1 + 4 = 5` leaves.
  - Expected output: `NO` (5 is odd).

- **Test Case 4: `3 2`**
  - `n = 3`, `k = 2`: In the third year, there are `2^2 + 3^3 = 4 + 27 = 31` leaves.
  - Expected output: `NO` (31 is odd).

- **Test Case 5: `1000000000 1`**
  - `n = 1000000000`, `k = 1`: In the billionth year, there are `1000000000^1000000000` leaves.
  - Expected output: `YES` (1000000000 is even, and any even number raised to an even power is even).

This test input includes a mix of small and large values, edge cases, and specific scenarios to thoroughly test the code and expose any bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
#include <iostream>
#include <vector>
#include <unistd.h>
#include <unistd.h>
#include <algorithm>
#include<map>
#include<cmath>
#include<cstring>
#include<sstream>
#include <set>
#include <unordered_set>
#include <cstring>
#include <string.h>
#include <queue>

using namespace std;
#define mid(s,e)            (s+(e-s)/2)
#define ALL(x)              (x).begin(), (x).end()
#define endl                '\n'
#define MAXE(vec) *max_element(ALL(vec))
#define MINE(vec) *min_element(ALL(vec))
#define valid(a,b,c) (a >= b && a < c ? 1 : 0)
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
#define ll long long

ll cc(ll m) {
    return (m * (m + 1) * (2 * m + 1)) / 6;
}

void solve() {
    ll n, k;
    cin >> n >> k;
    ll start_year = max(1LL, n - k + 1);
    ll total_leaves = cc(n) - cc(start_year - 1);
    if (total_leaves % 2 == 0) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Large Values of \( n \) and \( k \)**: The code uses the formula for the sum of squares, which can quickly grow very large. We need to ensure that the values of \( n \) and \( k \) are large enough to test the limits of the code.

2. **Special Cases for \( k \)**: Values of \( k \) that are close to \( n \) or exactly \( n \) can be particularly tricky, as they affect the range of years considered.

3. **Odd and Even Patterns**: The code determines if the total number of leaves is even or odd. We need to create test cases where the total number of leaves can be both even and odd to ensure the code handles both cases correctly.

Let's consider the following test case:

- \( t = 3 \) (three test cases)
- \( n = 1000000000 \), \( k = 1 \)
- \( n = 1000000000 \), \( k = 2 \)
- \( n = 1000000000 \), \( k = 1000000000 \)

These test cases are designed to:
- Test the code with a large value of \( n \) and a small \( k \) (first case).
- Test the code with a large value of \( n \) and a slightly larger \( k \) (second case).
- Test the code with a large value of \( n \) and \( k \) equal to \( n \) (third case).

Here is the test input:

```plaintext
3
1000000000 1
1000000000 2
1000000000 1000000000
```

This test input will help expose any issues in the code related to handling large values and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
for _ in range(int(input())):
    n,k=map(int, input().split())
    x=0
    for i in range (n-k+1,n+1):
        x+=i%2
        x%=2
    if x:
        print("no")
    else:
        print("yes")

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

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

Think step by step.

Response:
```plaintext
1
1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
for _ in range(int(input())):
    n,k=map(int,input().split())
    c=0
    for i in range(k):
        c+=n**n
        n-=1
    if c%2:
        print("NO")
    else:
        print("YES")

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

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

Think step by step.

Response:
```plaintext
10
1 1
2 1
2 2
3 2
4 4
1000000000 1
1000000000 2
1000000000 1000000000
1 1000000000
2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
import java.util.Scanner;

public class Robin_Hood_and_Major_Oak {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t > 0){
            int n = sc.nextInt();
            int k = sc.nextInt();

            int c = 0;
            while(k > 0){
                if(n % 2 != 0){
                    c++;
                }
                n--;
                k--;
            }
            if(c % 2 == 0){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }
            t--;
        }
    }
}

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

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

Think step by step.

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

This test input is designed to cover a variety of scenarios that can expose the bug in the provided buggy code. Here's the reasoning:

1. **Test Case 1: \( n = 1, k = 1 \)**
   - The tree starts with 1 leaf in year 1. Since 1 is odd, the output should be "NO".
   - The buggy code incorrectly processes the loop and might give an incorrect result.

2. **Test Case 2: \( n = 2, k = 1 \)**
   - In year 2, the tree grows \( 2^2 = 4 \) new leaves. Since 4 is even, the output should be "YES".
   - The buggy code might not correctly handle the growth of leaves in each year.

3. **Test Case 3: \( n = 2, k = 2 \)**
   - In year 2, the tree has \( 1 + 2^2 = 5 \) leaves. Since 5 is odd, the output should be "NO".
   - The buggy code might not correctly accumulate the number of leaves over the years.

4. **Test Case 4: \( n = 3, k = 2 \)**
   - In year 3, the tree has \( 2^2 + 3^3 = 4 + 27 = 31 \) leaves. Since 31 is odd, the output should be "NO".
   - The buggy code might not correctly handle the accumulation of leaves over multiple years.

5. **Test Case 5: \( n = 4, k = 4 \)**
   - In year 4, the tree has \( 1 + 2^2 + 3^3 + 4^4 = 1 + 4 + 27 + 256 = 288 \) leaves. Since 288 is even, the output should be "YES".
   - The buggy code might not correctly handle the accumulation of leaves over a larger number of years.

These test cases are designed to cover different edge cases and scenarios to ensure that the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood and the Major Oak
time_limit_ms: 1000
memory_limit_mb: 256
Description: In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows $i^i$ new leaves in the $i$-th year. It starts with $1$ leaf in year $1$.

Leaves last for $k$ years on the tree. In other words, leaves grown in year $i$ last between years $i$ and $i+k-1$ inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year $n$.

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

Each test case consists of two integers $n$, $k$ ($1 \le n \le 10^9$, $1 \le k \le n$) — the requested year and the number of years during which the leaves remain.

Output: For each test case, output one line, "YES" if in year $n$ the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
1 1
2 1
2 2
3 2
4 4
output:
NO
YES
NO
NO
YES

Note: In the first test case, there is only $1$ leaf.

In the second test case, $k=1$, so in the $2$-nd year there will be $2^2=4$ leaves.

In the third test case, $k=2$, so in the $2$-nd year there will be $1+2^2=5$ leaves.

In the fourth test case, $k=2$, so in the $3$-rd year there will be $2^2 + 3^3 = 4 + 27 = 31$ leaves.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            long n = scanner.nextLong();
            long k = scanner.nextLong();
            long totalLeaves = sumOfSquares(n) - sumOfSquares(n - k);
            System.out.println(totalLeaves % 2 == 0 ? "YES" : "NO");
        }
        scanner.close();
    }

    private static long sumOfSquares(long m) {
        if (m < 1) return 0;
        return m * (m + 1) * (2 * m + 1) / 6;
    }
}

**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, let's analyze the code and the problem statement.

### Problem Analysis
The problem requires calculating the total number of leaves on the Major Oak in year \( n \) and determining if this number is even. The number of leaves in year \( i \) is \( i^i \), and each leaf lasts for \( k \) years.

### Buggy Code Analysis
The buggy code uses the `sumOfSquares` function to calculate the total number of leaves. However, this function is incorrect because it calculates the sum of squares, not the sum of \( i^i \).

### Challenging Test Cases
To expose the bug, we need to create test cases where the sum of squares is significantly different from the sum of \( i^i \). Here are some edge cases and challenging scenarios:

1. **Small \( n \) and \( k \) values**: These can help verify the basic logic.
2. **Large \( n \) and \( k \) values**: These can stress the incorrect sum calculation.
3. **Edge cases where \( n \) and \( k \) are close to their limits**: These can test the boundary conditions.

### Challenging Test Input
Here is a small-scale test input that includes a mix of these scenarios:

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

### Explanation
1. **Test Case 1 (1 1)**: Basic case where \( n = 1 \) and \( k = 1 \). The tree has 1 leaf.
2. **Test Case 2 (2 1)**: Basic case where \( n = 2 \) and \( k = 1 \). The tree has \( 2^2 = 4 \) leaves.
3. **Test Case 3 (2 2)**: Basic case where \( n = 2 \) and \( k = 2 \). The tree has \( 1 + 2^2 = 5 \) leaves.
4. **Test Case 4 (10 5)**: Medium-sized case where \( n = 10 \) and \( k = 5 \). This should expose the incorrect sum calculation.
5. **Test Case 5 (1000000000 1000000000)**: Large case where \( n = 1000000000 \) and \( k = 1000000000 \). This should stress the incorrect sum calculation and potentially cause overflow issues.

This test input should help expose the bug 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: Guess One Character
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout) or cout.flush(), in Java or Kotlin — System.out.flush(), and in Python — sys.stdout.flush().

The jury has a string $$$s$$$ consisting of characters 0 and/or 1. The length of this string is $$$n$$$.

You can ask the following queries:

- $$$1$$$ $$$t$$$ — "how many times does $$$t$$$ appear in $$$s$$$ as a contiguous substring?" Here, $$$t$$$ should be a string consisting of characters 0 and/or 1; its length should be at least $$$1$$$ and at most $$$n$$$. For example, if the string $$$s$$$ is 111011 and the string $$$t$$$ is 11, the response to the query is $$$3$$$.

You have to guess at least one character in the string $$$s$$$ by asking no more than $$$3$$$ queries. Note that giving the answer does not count as a query.

In every test and in every test case, the string $$$s$$$ is fixed beforehand.

Input: None

Output: None

Examples:
input:
3     // 3 test cases
3     // the length of the string is 3

1     // 101 occurs once

1     // guessed correctly
2     // the length of the string is 2

0     // 00 occurs zero times

0     // 0 occurs zero times

1     // guessed correctly
2     // the length of the string is 2

1     // 1 occurs once

0     // 01 occurs zero times

1     // guessed correctly
output:
1 101 // how many times 101 occurs

0 2 0 // guess: s[2] is 0

1 00  // how many times 00 occurs

1 0   // how many times 0 occurs

0 1 1 // guess: s[1] is 1

1 1   // how many times 1 occurs

1 01  // how many times 01 occurs

0 2 0 // guess: s[2] is 0

Note: In the example, there are $$$3$$$ test cases: 101, 11 and 10. Note that everything after the // sign is a comment that explains which line means what in the interaction. The jury program won't print these comments in the actual problem, and you shouldn't print them. The empty lines are also added for your convenience, the jury program won't print them, and your solution should not print any empty lines.

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

void solve() {
    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        // Query the count of '0'
        cout << "1 0\n";
        cout.flush();
        int count0;
        cin >> count0;

        // Query the count of '1'
        cout << "1 1\n";
        cout.flush();
        int count1;
        cin >> count1;

        if (count0 > 0) {
            // Query the count of '00'
            cout << "1 00\n";
            cout.flush();
            int count00;
            cin >> count00;

            // Guess based on the counts
            if (count00 > 0) {
                cout << "0 1 0\n";
            } else {
                cout << "0 1 1\n";
            }
        } else {
            // Query the count of '11'
            cout << "1 11\n";
            cout.flush();
            int count11;
            cin >> count11;

            // Guess based on the counts
            if (count11 > 0) {
                cout << "0 1 1\n";
            } else {
                cout << "0 1 0\n";
            }
        }
        cout.flush();

        int result;
        cin >> result;
        if (result == -1) {
            return;
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
}

// 8L6WhwXuBlT5AHSMQrm

**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
1 1
1 1 0
2
2
1 11
1 1 1
3
3
1 01
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess One Character
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout) or cout.flush(), in Java or Kotlin — System.out.flush(), and in Python — sys.stdout.flush().

The jury has a string $$$s$$$ consisting of characters 0 and/or 1. The length of this string is $$$n$$$.

You can ask the following queries:

- $$$1$$$ $$$t$$$ — "how many times does $$$t$$$ appear in $$$s$$$ as a contiguous substring?" Here, $$$t$$$ should be a string consisting of characters 0 and/or 1; its length should be at least $$$1$$$ and at most $$$n$$$. For example, if the string $$$s$$$ is 111011 and the string $$$t$$$ is 11, the response to the query is $$$3$$$.

You have to guess at least one character in the string $$$s$$$ by asking no more than $$$3$$$ queries. Note that giving the answer does not count as a query.

In every test and in every test case, the string $$$s$$$ is fixed beforehand.

Input: None

Output: None

Examples:
input:
3     // 3 test cases
3     // the length of the string is 3

1     // 101 occurs once

1     // guessed correctly
2     // the length of the string is 2

0     // 00 occurs zero times

0     // 0 occurs zero times

1     // guessed correctly
2     // the length of the string is 2

1     // 1 occurs once

0     // 01 occurs zero times

1     // guessed correctly
output:
1 101 // how many times 101 occurs

0 2 0 // guess: s[2] is 0

1 00  // how many times 00 occurs

1 0   // how many times 0 occurs

0 1 1 // guess: s[1] is 1

1 1   // how many times 1 occurs

1 01  // how many times 01 occurs

0 2 0 // guess: s[2] is 0

Note: In the example, there are $$$3$$$ test cases: 101, 11 and 10. Note that everything after the // sign is a comment that explains which line means what in the interaction. The jury program won't print these comments in the actual problem, and you shouldn't print them. The empty lines are also added for your convenience, the jury program won't print them, and your solution should not print any empty lines.

Buggy Code:
#include <iostream>
#include <string>

using namespace std;

void query(int type, const string& t) {
    cout << type << " " << t << endl;
    fflush(stdout);
}

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

    while (t--) {
        int n;
        cin >> n;

        // First query: How many times does "0" appear in the string?
        query(1, "0");
        int count_0;
        cin >> count_0;

        // Second query: How many times does "1" appear in the string?
        query(1, "1");
        int count_1;
        cin >> count_1;

        // Determine the first character
        char first_char = (count_0 > count_1) ? '0' : '1';

        // Third query: Guess the second character
        query(1, "10"); // Assuming the second character is '0'
        int count_10;
        cin >> count_10;

        // Deduce the second character
        char second_char = (count_10 == 1) ? '0' : '1';

        // Print the result
        cout << "0 2 " << second_char << endl;
        fflush(stdout);

        // Check if the guess was correct
        int check;
        cin >> check;
        if (check != 1) {
            return 0; // Terminate if the guess was incorrect
        }
    }

    return 0;
}

// efsz7ynqKxBHuVkgXAT

**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, we need to consider the logic and potential flaws in the code. The code makes the following assumptions:

1. It queries the number of times "0" appears in the string.
2. It queries the number of times "1" appears in the string.
3. It determines the first character based on which of "0" or "1" appears more frequently.
4. It assumes the second character is '0' and queries the number of times "10" appears in the string.
5. It deduces the second character based on the result of the "10" query.

The bug in the code is that it always assumes the second character is '0' and checks for the substring "10". This assumption can lead to incorrect results if the second character is actually '1'. To expose this bug, we can create a test case where the second character is '1' and the first character is '0'.

Let's construct a test input for this scenario:

1. The length of the string is 2.
2. The string is "01".
3. The queries and responses should reflect this string.

Here is the test input:

```plaintext
3
2
1 0
1
1 1
1
1 10
0
0 2 1
1
```

Explanation:
- The first line `3` indicates there are 3 test cases.
- The second line `2` indicates the length of the string is 2.
- The third line `1 0` is the query for the number of times "0" appears in the string. The response is `1` because "0" appears once.
- The fourth line `1 1` is the query for the number of times "1" appears in the string. The response is `1` because "1" appears once.
- The fifth line `1 10` is the query for the number of times "10" appears in the string. The response is `0` because "10" does not appear as a contiguous substring.
- The sixth line `0 2 1` is the guess that the second character is '1'. The response is `1` indicating the guess is correct.

This test input should expose the bug in the buggy code where it incorrectly assumes the second character is '0' and fails to handle the case where the second character is '1'.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess One Character
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an interactive problem. You have to use flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout) or cout.flush(), in Java or Kotlin — System.out.flush(), and in Python — sys.stdout.flush().

The jury has a string $$$s$$$ consisting of characters 0 and/or 1. The length of this string is $$$n$$$.

You can ask the following queries:

- $$$1$$$ $$$t$$$ — "how many times does $$$t$$$ appear in $$$s$$$ as a contiguous substring?" Here, $$$t$$$ should be a string consisting of characters 0 and/or 1; its length should be at least $$$1$$$ and at most $$$n$$$. For example, if the string $$$s$$$ is 111011 and the string $$$t$$$ is 11, the response to the query is $$$3$$$.

You have to guess at least one character in the string $$$s$$$ by asking no more than $$$3$$$ queries. Note that giving the answer does not count as a query.

In every test and in every test case, the string $$$s$$$ is fixed beforehand.

Input: None

Output: None

Examples:
input:
3     // 3 test cases
3     // the length of the string is 3

1     // 101 occurs once

1     // guessed correctly
2     // the length of the string is 2

0     // 00 occurs zero times

0     // 0 occurs zero times

1     // guessed correctly
2     // the length of the string is 2

1     // 1 occurs once

0     // 01 occurs zero times

1     // guessed correctly
output:
1 101 // how many times 101 occurs

0 2 0 // guess: s[2] is 0

1 00  // how many times 00 occurs

1 0   // how many times 0 occurs

0 1 1 // guess: s[1] is 1

1 1   // how many times 1 occurs

1 01  // how many times 01 occurs

0 2 0 // guess: s[2] is 0

Note: In the example, there are $$$3$$$ test cases: 101, 11 and 10. Note that everything after the // sign is a comment that explains which line means what in the interaction. The jury program won't print these comments in the actual problem, and you shouldn't print them. The empty lines are also added for your convenience, the jury program won't print them, and your solution should not print any empty lines.

Buggy Code:
#include <iostream>
#include <string>
#include <cassert>

using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;

        // First query: count of '1'
        cout << "1 1" << endl;
        cout.flush();
        int count1;
        cin >> count1;

        if (count1 == -1) {
            return 1; // Terminate if the query is invalid
        }

        // Second query: count of '0'
        cout << "1 0" << endl;
        cout.flush();
        int count0;
        cin >> count0;

        if (count0 == -1) {
            return 1; // Terminate if the query is invalid
        }

        // Determine the character at position 1
        if (count1 > 0) {
            cout << "0 1 1" << endl;
            cout.flush();
        } else if (count0 > 0) {
            cout << "0 1 0" << endl;
            cout.flush();
        } else {
            assert(false); // This should never happen given the constraints
        }

        int result;
        cin >> result;
        if (result == -1) {
            return 1; // Terminate if the guess is incorrect
        }
    }
    return 0;
}

// RYeIxSTunVkK5jA3BcD

**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 code assumes that the string `s` will always contain at least one '1' or one '0', and it uses this assumption to determine the character at position 1. However, if the string contains only one type of character (either all '0's or all '1's), the code might fail to handle it correctly.

Here's a step-by-step approach to create a test input:

1. **Test Case 1: String with all '1's**
   - Length: 3
   - Queries:
     - Count of '1': 3
     - Count of '0': 0
   - Expected Guess: 1 at position 1

2. **Test Case 2: String with all '0's**
   - Length: 3
   - Queries:
     - Count of '1': 0
     - Count of '0': 3
   - Expected Guess: 0 at position 1

3. **Test Case 3: String with a single character**
   - Length: 1
   - Queries:
     - Count of '1': 1
     - Count of '0': 0
   - Expected Guess: 1 at position 1

4. **Test Case 4: String with a single character (all '0')**
   - Length: 1
   - Queries:
     - Count of '1': 0
     - Count of '0': 1
   - Expected Guess: 0 at position 1

Here is the test input that covers these cases:

```plaintext
4
3
3
0
1 1 1
3
0
3
0 1 0
1
1
0
0 1 1
1
0
1
0 1 0
```

This input includes:
- Test Case 1: String "111"
- Test Case 2: String "000"
- Test Case 3: String "1"
- Test Case 4: String "0"

Each test case is designed to challenge the assumption in the buggy code and ensure it handles edge cases correctly.