**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
#!/usr/bin/env python
from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

def get_xor_sum(n):
    mod = n % 4
    if mod == 0: return n
    if mod == 1: return 1
    if mod == 2: return n + 1
    return 0

def solve_query(l, r, i, k):
    if i == 0:
        return get_xor_sum(r) ^ get_xor_sum(l - 1)

    mod = 1 << i

    # Get XOR of all numbers
    result = get_xor_sum(r) ^ get_xor_sum(l - 1)

    # Find first number to exclude
    l_block = l // mod
    l_pos = l % mod

    first = l_block * mod + k
    if k < l_pos:
        first += mod

    # If no numbers to exclude
    if first > r:
        return result

    # XOR out all matching numbers
    curr = first
    while curr <= r:
        result ^= curr
        curr += mod

    return result

def main():
    t = int(input())
    for _ in range(t):
        l, r, i, k = map(int, input().split())
        print(solve_query(l, r, i, k))

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The code in question is designed to compute the XOR of all interesting integers in a given range, where an integer is considered interesting if it does not satisfy a specific modular condition.

### Key Points to Consider:
1. **Edge Cases for Modulo Operation**: The code uses modulo operations to determine which numbers to exclude. Edge cases for modulo operations can be tricky, especially when the range is small or when the modulo value is 1.
2. **Edge Cases for Range Limits**: The range limits `l` and `r` can be very close, which might cause the code to handle the loop incorrectly.
3. **Edge Cases for Modulo Value**: The modulo value `2^i` can be very small (e.g., 1 or 2), which can affect the behavior of the code.
4. **Edge Cases for `k`**: The value of `k` can be 0, which might cause the code to exclude numbers incorrectly.

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

1. **Test Case 1**: Small range with `i = 0` and `k = 0`.
2. **Test Case 2**: Small range with `i = 1` and `k = 0`.
3. **Test Case 3**: Small range with `i = 1` and `k = 1`.
4. **Test Case 4**: Range where `l` and `r` are the same and `i = 2` and `k = 1`.
5. **Test Case 5**: Range where `l` and `r` are very close and `i = 3` and `k = 7`.

### Output:
```plaintext
5
1 2 0 0
1 2 1 0
1 2 1 1
1 1 2 1
3 4 3 7
```

### Explanation:
1. **Test Case 1**: `1 2 0 0`
   - Range: [1, 2]
   - Modulo: \(2^0 = 1\)
   - \(k = 0\)
   - Expected Output: \(1 \oplus 2 = 3\)

2. **Test Case 2**: `1 2 1 0`
   - Range: [1, 2]
   - Modulo: \(2^1 = 2\)
   - \(k = 0\)
   - Expected Output: \(1\)

3. **Test Case 3**: `1 2 1 1`
   - Range: [1, 2]
   - Modulo: \(2^1 = 2\)
   - \(k = 1\)
   - Expected Output: \(2\)

4. **Test Case 4**: `1 1 2 1`
   - Range: [1, 1]
   - Modulo: \(2^2 = 4\)
   - \(k = 1\)
   - Expected Output: \(1\)

5. **Test Case 5**: `3 4 3 7`
   - Range: [3, 4]
   - Modulo: \(2^3 = 8\)
   - \(k = 7\)
   - Expected Output: \(3 \oplus 4 = 7\)

This test input should help expose any issues in the code, particularly with handling edge cases and small ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
def get_xor(n):
    mod = n % 4

    if mod == 0:
        return n
    elif mod == 1:
        return 1
    elif mod == 2:
        return n + 1
    return 0

def get_xor_range(l, r):
    return get_xor(r) ^ get_xor(l - 1)

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

    t = int(data[0])
    index = 1
    results = []

    for _ in range(t):
        l, r, i, k = map(int, data[index].split())
        index += 1

        if i == 0:
            print(0)
            continue

        ans = get_xor_range(l, r)

        start, end = 0, 0

        ss, ee = 0, int(1e15)
        while ss < ee:
            mid = (ss + ee) // 2
            if k + (2 ** i) * mid >= l:
                ee = mid
            else:
                ss = mid + 1

        start = ee

        ss, ee = 0, int(1e15)
        while ss < ee:
            mid = (ss + ee) // 2
            if k + (2 ** i) * mid > r:
                ee = mid
            else:
                ss = mid + 1

        end = ee - 1

        val = get_xor_range(start, end) * (2 ** i)
        result = ans ^ val ^ (0 if (end - start) % 2 else k)
        results.append(result)

    print('\n'.join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

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

public class Main {
	public static long binexp(long a, long b, long MOD){
		long mult = 1;
		while(b >= 1){
			if(b%2 == 1){
				mult = (mult * a)%MOD;
			}
			b /= 2;
			a = (a * a)%MOD;

		}
		return mult;
	}
	public static long modinv(long num, long MOD){
		return binexp(num,MOD-2,MOD);
	}
	public static long lcm(long a, long b){
		return a/gcd(a,b) * b;
	}
	public static long gcd(long a, long b){
		return (a < b) ? gcd(b,a) : (a%b == 0 ? b : gcd(a%b,b));
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());
		while(T-- > 0){
			//System.out.println(1 ^ 2 ^ 3 ^ 4 ^ 5);
			String[] s = br.readLine().split(" ");
			long L = Long.parseLong(s[0]);
			long R = Long.parseLong(s[1]);
			long I = Long.parseLong(s[2]);
			long K = Long.parseLong(s[3]);
			boolean odd = (L%2 == 1) ? true : false;
			long val = 0;

			if(odd){
				val = L;
				//System.out.println("Not a power of 2");
				if((R - L)%4 == 0){
					val = L;
				}
				else if((R - L)%4 == 1){
					val ^= R;
				}
				else if((R - L)%4 == 2){
					val ^= R;
					val ^= R-1;
				}
				else{
					val ^= R;
					val ^= R-1;
					val ^= R-2;
				}
			}
			else{
				if((R - L)%4 == 0){
					val ^= R;
				}
				else if((R - L)%4 == 1){
					val ^= R;
					val ^= R-1;
				}
				else if((R-L)%4 == 2){
					val ^= R;
					val ^= R-1;
					val ^= R-2;
				}
			}

			long x = L;

			long base = 1 << I;
			long low = -1;
			long high = -1;
			if(L%base <= K){
				low = L + K - L%base;
			}
			else{
				low = L + K - L%base + base;
			}
			if(K <= R%base){
				high = R - (R%base - K);
			}
			else{
				high = R - (R%base - K) - base;
			}
			//System.out.println("58 ... 66 is " + (58 ^ 60 ^ 62 ^ 64 ^ 66));
			//System.out.println(low + " " + high + " " + base);

			long num = (high - low)/base + 1;
			//System.out.println("num is now " + num);
			//System.out.println("Parity of low is " + (low%2));
			//System.out.println("Answer is " + (1012 ^ 1016 ^ 1020 ^ 1024 ^ 1028 ^ 1032 ^ 1036 ^ 1040 ^ 1044));
			if(low%(base * 2) != base){
				if(num%4 == 0){
					System.out.println(val);
				}
				else if(num%4 == 1){
					System.out.println(val ^ high);
				}
				else if(num%4 == 2){
					System.out.println(val ^ high ^ (high - base));
				}
				else{
					System.out.println(val ^ high ^ (high - base) ^ (high - 2 * base));
				}
			}
			else{
				long n = val ^ low;
				long xors = (num - 1 + base * 2)%(base * 2);
				//System.out.println("Amount is " + num);
				//System.out.println("Number of xors required is " + xors);
				for(long j = 1; j <= xors; j++){
					n ^= high - (j-1) * base;
				}
				/*
				if(num%4 == 1){
					System.out.println(val ^ low);
				}
				else if(num%4 == 2){
					System.out.println(val ^ low ^ high);
				}
				else if(num%4 == 3){
					System.out.println(val ^ low ^ high ^ (high - base));
				}
				else{
					System.out.println(val ^ low ^ high ^ (high - base) ^ (high - 2 * base));
				} */
				System.out.println(n);
			}

			//System.out.println(low + " " + high);

		}

	}
}

**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 focus on edge cases and scenarios that can cause issues in the algorithm. Here are the key points to consider:

1. **Edge Cases for Range and Modulo Conditions:**
   - When `L` and `R` are very close to each other.
   - When `L` and `R` are large and the range is wide.
   - When `i` is 0 or a large value.
   - When `k` is 0 or close to `2^i`.

2. **Boundary Conditions:**
   - When `L` is an interesting integer.
   - When `R` is an interesting integer.
   - When `L` and `R` are both interesting integers.
   - When neither `L` nor `R` is an interesting integer.

3. **Special Cases:**
   - When `L` and `R` are both powers of 2.
   - When `L` and `R` are both even or both odd.

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

```plaintext
5
1 2 1 0
2 4 2 1
3 7 3 0
1023 1024 10 511
1 1000000000 30 0
```

This test input includes:
1. A small range with `L` and `R` close to each other and `i` and `k` set to 1 and 0.
2. A range where `L` and `R` are both even and `i` and `k` set to 2 and 1.
3. A range where `L` and `R` are both odd and `i` set to 3 and `k` set to 0.
4. A range where `L` and `R` are both powers of 2 and `i` set to 10 and `k` set to 511.
5. A large range with `L` and `R` set to 1 and 10^9, and `i` set to 30 and `k` set to 0.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

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

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

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

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

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

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

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

    private static long solveQuery(long l, long r, int i, int k) {
        if (i == 0) return xorRange(l, r);

        long modulus = 1L << i;

        // Calculate XOR of all numbers in range [l, r]
        long result = xorRange(l, r);

        // Find first multiple of modulus that is greater than or equal to l
        long first = l + ((k - l % modulus + modulus) % modulus);
        if (first > r) return result; // No numbers satisfying x ≡ k (mod 2^i) in range

        // Find last multiple of modulus in range [l, r]
        long last = r - ((r - k) % modulus);

        // XOR all values in range [first, last] with step `modulus`
        for (long j = first; j <= last; j += modulus) {
            result ^= j;
        }

        return result;
    }

    private static long xorRange(long l, long r) {
        return xorUpTo(r) ^ xorUpTo(l - 1);
    }

    private static long xorUpTo(long n) {
        switch ((int)(n & 3)) {
            case 0: return n;
            case 1: return 1;
            case 2: return n + 1;
            default: return 0;
        }
    }

    public static void main(String[] args) throws IOException {
        FastReader fr = new FastReader();
        int t = fr.nextInt();
        StringBuilder sb = new StringBuilder();

        while (t-- > 0) {
            long l = fr.nextLong();
            long r = fr.nextLong();
            int i = fr.nextInt();
            int k = fr.nextInt();

            sb.append(solveQuery(l, r, i, k)).append('\n');
        }
        System.out.print(sb);
    }
}

**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 0
1 1 1 1
1 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

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

#define int long long int
#define double long double

#define inf 1e17

#define f first
#define s second
#define pi pair<int, int>
#define vi vector<int>
#define vs vector<string>
#define vpi vector<pi>
#define vb vector<bool>
#define vvi vector<vi>
#define vvb vector<vb>
#define vvpi vector<vpi>
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define repr(i, a, b) for (int i = a; i >= b; i--)

// #define endl "\n" // for interactive comment it.

// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// cout<<fixed<<setprecision(12)
// memset(dp,-1,sizeof(dp));

int mod=1e9+7;

int _ceil(int a, int b){
	if (a % b == 0) return a / b;
	return a / b + 1;
}

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

int _pow(int a, int p=mod-2){
	int res = 1;
	while (p > 0){
		if (p & 1){
			res = (res * a)%mod;
		}
		p = p >> 1;
		a = (a * a) % mod;
	}
	return res;
}

int _find(int cur, vi &par){
	if (par[cur] == -1) return cur;
	return par[cur] = _find(par[cur], par);
}

int _getRandomInteger(int a, int b) {
    // Random number generator
    std::random_device rd;  // Obtain a random number from hardware
    std::mt19937 gen(rd()); // Seed the generator
    std::uniform_int_distribution<> distr(a, b); // Define the range

    return distr(gen); // Generate a random number in the range [a, b]
}

// const int N=2e6+10;
// int fact[N+1];
// int inv_fact[N+1];
// int ncr(int n,int r){
// 	if(n<r) return 0;
// 	int res=fact[n];
// 	res=(res*inv_fact[n-r])%mod;
// 	res=(res*inv_fact[r])%mod;
// 	return res;
// }

int count(int n,int i) {
    if (n < 0) return 0;
	int total_blocks = (n + 1) / (1LL << (i + 1));
    int count = total_blocks * (1LL << i);
    int remainder = (n + 1) % (1LL << (i + 1));
    count += max(0LL, remainder - (1LL << i));
	return count;
}

int countInAP(int a, int d, int n, int i) {
	int mod = 1LL << (i + 1);
    int mask = 1LL << i;

    unordered_map<int, int> remainderMap;
    int count = 0;
    int currentTerm = a % mod;

    int cycleLength = 0;
    int setBitCountInCycle = 0;

    for (int j = 0; j < n; ++j) {
        if (remainderMap.find(currentTerm) != remainderMap.end()) {
            cycleLength = j - remainderMap[currentTerm];
            break;
        }
        remainderMap[currentTerm] = j;
        if (currentTerm & mask) {
            setBitCountInCycle++;
        }

        currentTerm = (currentTerm + d) % mod;
    }

    if (cycleLength == 0) {
        return setBitCountInCycle;
    }

    // Number of full cycles in n terms
    int fullCycles = n / cycleLength;
    int remainingTerms = n % cycleLength;

    // Count total set bits from full cycles and remaining terms
    count = fullCycles * setBitCountInCycle;

    // Count set bits in the remaining terms
    currentTerm = a % mod;
    for (int j = 0; j < remainingTerms; ++j) {
        if (currentTerm & mask) {
            count++;
        }
        currentTerm = (currentTerm + d) % mod;
    }

    return count;
}

void solve(){

	int l,r,i,k;
	cin>>l>>r>>i>>k;

	if(i==0){
		cout<<0<<endl;
		return;
	}

	int a=k%(1LL<<i);
	int d=(1LL<<i);

	double dd = (r-a);
	dd = dd/(d*1.0);
	dd = dd + 1.0;
	int nn1=dd;
	// cout<<a<<" "<<d<<" "<<nn1<<endl;

	int nn2;
	if(a>l-1){
		nn2=0;
	}else{
		dd = (l-1-a);
		dd = dd/(d*1.0);
		dd = dd + 1.0;
		nn2 = dd;
	}

	vi bit(64,0);
	int ans=0;
	for(int i=0;i<64;i++){
		bit[i] = count(r,i)-countInAP(a,d,nn1,i);
		bit[i] -= count(l-1,i)-countInAP(a,d,nn2,i);

		if(bit[i]&1){
			ans=ans|(1LL<<i);
		}
	}

	cout<<ans<<endl;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

	// cout<<fixed<<setprecision(16);

    #ifndef ONLINE_JUDGE
    freopen("/Users/shekharh433/Desktop/Code/input.txt", "r", stdin);
    freopen("/Users/shekharh433/Desktop/Code/output.txt", "w", stdout);
    #endif

	// fact[0]=1;
	// inv_fact[0]=_pow(fact[0],mod-2);
	// for(int i=1;i<=N;i++){
	// 	fact[i]=(fact[i-1]*i)%mod;
	// 	inv_fact[i]=_pow(fact[i],mod-2);
	// }

	int t=1;
	cin>>t;
	rep(i,1,t+1){
		// cout<<"Case #"<<i<<": ";
		solve();
	}

    return 0;
}

// g++ -std=c++17 AA.cpp -o AA; ./AA

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

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

Think step by step.

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

1. **Edge Cases for Range [l, r]:**
   - Small ranges: e.g., [1, 1], [1, 2], [1, 3]
   - Large ranges: e.g., [1, 10^18], [10^18 - 1, 10^18]
   - Ranges with specific properties: e.g., [2, 4], [3, 6], [4, 8]

2. **Edge Cases for i and k:**
   - i = 0: This should be handled separately as the code checks for i == 0.
   - i = 1: This is a simple case but can still expose bugs.
   - i = 30: This is the maximum value for i, which can test the limits of the algorithm.
   - k = 0: This should be handled correctly.
   - k = 2^i - 1: This is the maximum value for k, which can test the limits of the algorithm.

3. **Combining Edge Cases:**
   - Small range with i = 0: [1, 1], i = 0
   - Small range with i = 1: [1, 2], i = 1
   - Small range with i = 30: [1, 2], i = 30
   - Large range with i = 0: [1, 10^18], i = 0
   - Large range with i = 30: [1, 10^18], i = 30
   - Ranges with specific properties: [2, 4], i = 1, k = 0
   - Ranges with specific properties: [3, 6], i = 2, k = 1
   - Ranges with specific properties: [4, 8], i = 3, k = 2

Given the constraints and the need for a small-scale test input, here is a set of test cases that should effectively expose any bugs in the code:

```plaintext
6
1 1 0 0
1 2 1 0
1 3 1 0
1 1000000000000 30 1543
2 4 1 0
3 6 2 1
```

This test input includes:
- A single element range with i = 0.
- A small range with i = 1.
- A small range with i = 1 and k = 0.
- A large range with i = 30 and a large k.
- A range with specific properties and i = 1, k = 0.
- A range with specific properties and i = 2, k = 1.

These cases should help in identifying 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: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// #include <ext/pb_ds/detail/standard_policies.hpp>
// #include <bits/stdc++.h>
// using namespace std;
// using namespace __gnu_pbds;
// #define int long long int
// #define F first
// #define S second
// #define MP make_pair
// #define mset(m, v) memset(m, v, sizeof(m))
// #define pb push_back
// #define vi vector<int>
// #define endl '\n'
// #define vpi vector<pair<int, int>>
// #define vvi vector<vector<int>>
// #define pii pair<int, int>
// #define bitcount(x) __builtin_popcount(x)
// #define py cout << "YES" << endl
// #define pn cout << "NO" << endl
// #define all(x) (x).begin(), (x).end()
// #define rall(x) (x).rbegin(), (x).rend()
// #define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
// #define lup(i, j, n) for (int i = j; i < n; i++)
// #define lupr(i, j, n) for (int i = j; i >= n; i--)
// long long MOD = 1000000007;
// long double EPS = 1e-9;

// int xor_0_n(int n) {
//     int rem = n % 4;
//     if (rem == 0) {
//         return n;
//     }
//     if (rem == 1) {
//         return 1;
//     }
//     if (rem == 2) {
//         return n + 1;
//     }
//     return 0;
// }

// int xor_range(int l, int r) {
//     return xor_0_n(r) ^ xor_0_n(l - 1);
// }

// int32_t main() {
//     int t;
//     cin >> t;
//     while (t--) {
//         int l, r, i, k;
//         cin >> l >> r >> i >> k;
//         int highBits = xor_range((l - k + (1 << i) - 1) >> i, (r - k) >> i) << i;
//         int lowBits = k * (((r - k) / (1 << i) - (l - k - 1) / (1 << i)) & 1);
//         cout << (xor_range(l, r) ^ highBits ^ lowBits) << '\n';
//     }
//     return 0;
// }

// // void solve()
// {
//     int l, r, ii, k;
//     cin >> l >> r >> ii >> k;

//     // in l to r range calculate no of set bits at each position
//     // for this subtract 0 to (l-1) from from 0 to r
//     vi bits1(63, 0);
//     vi bits2(63, 0);
//     vi bits3(63, 0);

//     int twopowi = (1LL << ii);

//     int total = l;
//     lup(i, 0, 63)
//     {
//         int full = (total / (1LL << (i + 1)));
//         int left = total % (1LL << (i + 1));
//         int rem = max(0LL, left - (1LL << i));
//         bits1[i] = full * (1LL << i) + rem;
//     }

//     int total2 = r + 1;
//     lup(i, 0, 63)
//     {
//         int full = (total2 / (1LL << (i + 1)));
//         int left = total2 % (1LL << (i + 1));
//         int rem = max(0LL, left - (1LL << i));
//         bits2[i] = full * (1LL << i) + rem;
//     }

//     // now bit3 contains the final answer for l to r
//     lup(i, 0, 63)
//     {
//         bits3[i] = bits2[i] - bits1[i];
//     }

//     vi sub1(63, 0);
//     vi sub2(63, 0);
//     // sub1 from 0 to r and sub2 from 0 to l-1

//     lup(i, 0, 63)
//     {
//         int fulldash = total2 / twopowi;
//         int full = (fulldash / (1LL << (i + 1))) * (1LL << i);
//         int left = total2 % twopowi;
//         int rem = max(0LL, left - k);
//         sub1[i] = full + rem;
//     }

//     lup(i, 0, 63)
//     {
//         int fulldash = total / twopowi;
//         int full = (fulldash / (1LL << (i + 1))) * (1LL << i);
//         int left = total % twopowi;
//         int rem = max(0LL, left - k);
//         sub2[i] = full + rem;
//     }

//     lup(i, 0, 63)
//     {
//         bits3[i] = bits3[i] - sub1[i] + sub2[i];
//     }

//     int ans = 0;
//     lup(i, 0, 63)
//     {
//         if (bits3[i] % 2 == 1)
//         {
//             ans += (1LL << i);
//         }
//     }

//     cout << ans << endl;
// }

// signed main()
// {
//     ios_base::sync_with_stdio(0);
//     cin.tie(0);
//     cout.tie(0);

//     int _t = 1;
//     cin >> _t;
//     for (int i = 0; i < _t; i++)
//     {
//         solve();
//     }
// }

#include <iostream>
using namespace std;
#define int unsigned int
#define SPEEDY std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout.tie(0);

int xor_0_n(int n) {
    int rem = n % 4;
    if (rem == 0) {
        return n;
    }
    if (rem == 1) {
        return 1;
    }
    if (rem == 2) {
        return n + 1;
    }
    return 0;
}

int xor_range(int l, int r) {
    return xor_0_n(r) ^ xor_0_n(l - 1);
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        int l, r, i, k;
        cin >> l >> r >> i >> k;
        int highBits = xor_range((l - k + (1 << i) - 1) >> i, (r - k) >> i) << i;
        int lowBits = k * (((r - k) / (1 << i) - (l - k - 1) / (1 << i)) & 1);
        cout << (xor_range(l, r) ^ highBits ^ lowBits) << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 1 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
int a[2000020],ans=0;
void gen(int n)
{
int f[2000020];
for(int i=1;i<=n;i++)
f[i]=i;
do
{
if(0)//add something here.
ans++;
}while(next_permutation(f+1,f+n+1));
}
#define Y cout<<"Yes"<<'\n'
#define N cout<<"No"<<'\n'
int l,r,i,k;
int p[2000020];
int get(int n)
{
if (n%4==0)
return n;
if (n%4==3)
return 0;
if (n%4==1)
return 1;
if (n%4==2)
return n+1;
}
int f(int x,int i)
{
return x/pow(2ll,i);
}
int ff(int tmp)
{
int n=(tmp-k)/pow(2ll,i)+1;
if (n%4==0)
return 0;
if (n%4==1)
return tmp;
if (n%4==2)
return pow(2ll,i);
if (n%4==3)
return pow(2ll,i)+tmp;
}
//000xx 1 000xx
//001xx 2 00100
//010xx 3 011xx
//011xx 4 00000
//100xx 5 100xx
//101xx 6 00100
//110xx 7 111xx
//111xx 8 00000
signed main()
{
int T;
cin>>T;
while(T--)
{
cin>>l>>r>>i>>k;
int min_l=(f(l-k*3,i))*pow(2ll,i)+k;
while(min_l<l)
min_l+=pow(2ll,i);
int min_r=f(r-k,i)*pow(2ll,i)+k;
while(min_r>r)
min_r-=pow(2ll,i);
//cout<<min_l<<' '<<min_r<<'\n';
//cout<<ff(min_l)<<' '<<ff(min_r)<<'\n';
cout<<( ( (get(r)^get(l-1) ) ^ ( (ff(min_r)^ff(min_l)^min_l ) ) ) )<<'\n';
}
}

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

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

Think step by step.

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

### Explanation:
1. **Edge Cases:**
   - **Single Element Ranges:** The first query `[1, 2]` and the fifth query `[9, 10]` are ranges with only two elements, which are critical for testing the XOR operation.
   - **Ranges with Specific Modulo Conditions:** The second query `[3, 4]` and the third query `[5, 6]` are ranges where the elements are just above the modulo condition, which can help test the boundary conditions.
   - **Power of Two Ranges:** The fourth query `[7, 8]` is a range where both elements are powers of two, which can be tricky for XOR operations.

2. **Modulo Conditions:**
   - The modulo condition `i = 1` and `k = 0` is simple but effective for testing the basic functionality of the algorithm. It ensures that the XOR operation is performed only on numbers that are not divisible by 2.

3. **Small Scale:**
   - The test input is small enough to be manageable and can be easily verified by hand, ensuring that the output can be quickly checked against the expected results.

This test input is designed to cover a variety of edge cases and ensure that the algorithm handles the modulo condition and XOR operations correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

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

long long xd(long long n) {
    switch (n % 4) {
        case 0: return n;
        case 1: return 1;
        case 2: return n + 1;
        case 3: return 0;
    }
    return 0;
}

long long xorr(long long l, long long r) {
    return xd(r) ^ xd(l - 1);
}

long long xors(long long st, long long m, long long r) {
    long long rt = 0;
    for (long long x = st; x <= r; x += m) {
        rt ^= x;
    }
    return rt;
}

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

    int t;
    cin >> t;
    while (t--) {
        long long l, r, i, k;
        cin >> l >> r >> i >> k;
        long long m = 1LL << i;
        long long rt = xorr(l, r);

        long long f = l + ((k - (l % m) + m) % m);
        if (f > r) {
            cout << rt << endl;
            continue;
        }

        rt ^= xors(f, m, r);

        cout << rt << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are some potential scenarios to consider:

1. **Edge Cases for Modulo Operations**: The code uses modulo operations to determine the interesting integers. We should test cases where the modulo operations might not behave as expected.
2. **Large Ranges**: The code should handle large ranges efficiently. Testing with large values of `l` and `r` can help identify performance issues.
3. **Boundary Conditions**: Test with boundary values of `l` and `r` to ensure the code handles them correctly.
4. **Special Values for `k` and `i`**: Test with values of `k` and `i` that might cause the code to misbehave, especially when `k` is close to the upper limit of `2^i`.

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

```plaintext
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
```

This test input includes:
- A small range with a simple modulo condition.
- A medium range with a more complex modulo condition.
- A range where the starting point is close to a boundary.
- A large range to test performance and correctness.
- A range with specific values for `k` and `i` that might cause issues.

This should be sufficient to expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
def xor_to_n(n):
    if n % 4 == 0:
        return n
    elif n % 4 == 1:
        return 1
    elif n % 4 == 2:
        return n + 1
    else:
        return 0

def xor_range(l, r):
    return xor_to_n(r) ^ xor_to_n(l - 1)

def xor_excluded(l, r, k, i):
    m = 0
    excluded_xor = 0
    step = 1 << i

    while True:
        excluded_number = k + m * step
        if excluded_number > r:
            break
        if excluded_number >= l:
            excluded_xor ^= excluded_number
        m += 1

    return excluded_xor

def xorificator(t, queries):
    results = []
    for l, r, i, k in queries:
        total_xor = xor_range(l, r)
        excluded_xor = xor_excluded(l, r, k, i)
        results.append(total_xor ^ excluded_xor)
    return results

import sys
input = sys.stdin.read

data = input().splitlines()
t = int(data[0])
queries = []

for i in range(1, t + 1):
    l, r, i, k = map(int, data[i].split())
    queries.append((l, r, i, k))

results = xorificator(t, queries)

for result in results:
    print(result)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are the key points to consider:

1. **Edge Cases for `xor_to_n` and `xor_range` Functions:**
   - Ranges that are multiples of 4.
   - Ranges that start or end at a multiple of 4.

2. **Edge Cases for `xor_excluded` Function:**
   - Ranges where the excluded numbers are at the boundaries.
   - Ranges where the excluded numbers are sparse or dense.
   - Ranges where the excluded numbers are very close to the boundaries.

3. **General Edge Cases:**
   - Single-element ranges.
   - Large ranges.
   - Ranges where `l` and `r` are very close.

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

```plaintext
6
1 4 1 0
1 4 2 1
1 5 1 0
4 7 2 0
1 10 3 7
10 10 1 1
```

Explanation of the test cases:
1. **1 4 1 0**: This range includes 1, 3, and 4 (2 is excluded). The expected output is `1 ^ 3 ^ 4 = 6`.
2. **1 4 2 1**: This range includes 1, 2, and 4 (3 is excluded). The expected output is `1 ^ 2 ^ 4 = 7`.
3. **1 5 1 0**: This range includes 1, 3, and 5 (2 and 4 are excluded). The expected output is `1 ^ 3 ^ 5 = 7`.
4. **4 7 2 0**: This range includes 4, 5, 6, and 7 (none are excluded). The expected output is `4 ^ 5 ^ 6 ^ 7 = 2`.
5. **1 10 3 7**: This range includes 1, 2, 3, 4, 5, 6, 8, 9, and 10 (7 is excluded). The expected output is `1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 8 ^ 9 ^ 10 = 15`.
6. **10 10 1 1**: This range includes only 10 (10 is excluded). The expected output is `0`.

This test input should help expose any bugs in the provided code, particularly in handling edge cases and specific ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
import math
from operator import xor
def fxor(n):
    if n%4 == 0:
        return n;
    elif n%4 == 1:
        return 1;
    elif n%4 == 2:
        return n+1;
    else:
        return 0;
def frange(left, right):
    return (xor(fxor(left-1), fxor(right)))

for _ in range(int(input())):
    #print('-')
    l, r, i, k = map(int, input().split())
    y = frange(l, r)
    #print(f'{y:60b}')
    a = math.ceil((l-k)/2**i)
    b = math.floor((r-k)/2**i)
    #print(a, b)
    x = (frange(a, b))*2**i
    if (b-a-1)%2 == 1:
        x += k
    #print(f'{x:60b}')
    print(xor(x, y))

**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 3 1 0
2 28 3 7
15 43 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
import java.util.Scanner;

public class XORificator {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            long l = sc.nextLong();
            long r = sc.nextLong();
            int i = sc.nextInt();
            int k = sc.nextInt();

            long mod = 1L << i;

            long xorResult = xorUpto(r) ^ xorUpto(l - 1);

            long xStart = (l / mod) * mod + k;
            if (xStart < l) {
                xStart += mod;
            }

            if (xStart <= r) {
                long end = r;
                long count = (end - xStart) / mod + 1;

                long xContribution = 0;
                for (long j = 0; j < count; j++) {
                    xContribution ^= (xStart + j * mod);
                }
                xorResult ^= xContribution;
            }

            System.out.println(xorResult);
        }

        sc.close();
    }

    // Efficiently computes the XOR of all numbers from 0 to n
    private static long xorUpto(long n) {
        // Using the property of XOR operation
        switch ((int) (n % 4)) {
            case 0: return n;
            case 1: return 1;
            case 2: return n + 1;
            case 3: return 0;
            default: return 0;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
1 1023 10 512
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: XORificator 3000
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has been giving gifts to Bob for many years, and she knows that what he enjoys the most is performing [bitwise XOR](http://tiny.cc/xor_wiki_eng) of interesting integers. Bob considers a positive integer $x$ to be interesting if it satisfies $x \not\equiv k (\bmod 2^i)$. Therefore, this year for his birthday, she gifted him a super-powerful "XORificator 3000", the latest model.

Bob was very pleased with the gift, as it allowed him to instantly compute the XOR of all interesting integers in any range from $l$ to $r$, inclusive. After all, what else does a person need for happiness? Unfortunately, the device was so powerful that at one point it performed XOR with itself and disappeared. Bob was very upset, and to cheer him up, Alice asked you to write your version of the "XORificator".

Input: The first line of input contains a single integer $t$ $(1 \leq t \leq 10^4)$ — the number of XOR queries on the segment. The following $t$ lines contain the queries, each consisting of the integers $l$, $r$, $i$, $k$ $(1 \leq l \leq r \leq 10^{18}$, $0 \leq i \leq 30$, $0 \leq k < 2^i)$.

Output: For each query, output a single integer — the XOR of all integers $x$ in the range $[l, r]$ such that $x \not\equiv k \mod 2^i$.

Examples:
input:
6
1 3 1 0
2 28 3 7
15 43 1 0
57 2007 1 0
1010 1993 2 2
1 1000000000 30 1543
output:
2
2
13
0
4
1000000519

Note: In the first query, the interesting integers in the range $[1, 3]$ are $1$ and $3$, so the answer will be $1 \oplus 3 = 2$.

Buggy Code:
import java.util.Scanner;

public class XORCalculator {

    // Calculate the XOR of all numbers from 0 up to n
    public static int xorUpto(int n) {
        switch (n % 4) {
            case 0: return n;
            case 1: return 1;
            case 2: return n + 1;
            case 3: return 0;
            default: return 0; // This case should not occur
        }
    }

    // Calculate the range of multiples (m_start to m_end) of 'step' that can fit
    // in the range [left, right] offset by k
    public static int[] calculateStepRange(int left, int right, int step, int k) {
        int m_start;
        if (left > k) {
            int offset = left - k;
            m_start = (offset + step - 1) / step;
        } else {
            m_start = 0;
        }

        int m_end;
        if (right >= k) {
            m_end = (right - k) / step;
        } else {
            m_end = -1; // Indicates an invalid range
        }

        return new int[]{m_start, m_end};
    }

    // Compute the XOR of all values in the range [m_start, m_end] after shifting by i
    public static int computeXOREquivalent(int m_start, int m_end, int i, int k) {
        if (m_start > m_end) {
            return 0;
        }

        // XOR of numbers in range [m_start, m_end]
        int m_xor = xorUpto(m_end) ^ xorUpto(m_start - 1);
        int XOR_eq = m_xor << i;

        // If count of terms in the range [m_start, m_end] is odd, XOR with k
        int count = m_end - m_start + 1;
        if (count % 2 == 1) {
            XOR_eq ^= k;
        }

        return XOR_eq;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        // Number of test cases
        int t = scanner.nextInt();
        StringBuilder results = new StringBuilder();

        for (int testCase = 0; testCase < t; testCase++) {
            // Read left, right, i, and k values for each test case
            int left = scanner.nextInt();
            int right = scanner.nextInt();
            int i = scanner.nextInt();
            int k = scanner.nextInt();

            // Calculate step and range of m values
            int step = 1 << i; // This is equivalent to 2^i
            int[] range = calculateStepRange(left, right, step, k);
            int m_start = range[0];
            int m_end = range[1];

            // Compute XOR of equivalent terms in shifted range
            int XOR_eq = computeXOREquivalent(m_start, m_end, i, k);

            // Compute the XOR of all elements in the range [left, right]
            int XOR_all = xorUpto(right) ^ xorUpto(left - 1);

            // Final result for this test case
            int result = XOR_all ^ XOR_eq;
            results.append(result).append("\n");
        }

        // Print all results
        System.out.print(results.toString());
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
1 10 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,bmi,bmi2,popcnt,lzcnt")

#define ll long long
#define int ll
#define ios ios_base::sync_with_stdio(0);cin.tie(0);

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int gen(int a, int b) {return (ll)rng() % (b - a + 1) + a;}

int inter(int l1, int r1, int l2, int r2) {
	return max(0LL, min(r1, r2) - max(l1, l2));
}
void test_case() {
	int n, m; cin >> n >> m;
	vector<vector<int>> adj(n);
	int ans = 1e9 + 5;
	multiset<int> beg, en;
	for(int i = 0, u, v; i < m; ++i) {
		cin >> u >> v; --u, --v;
		beg.insert(u);
		en.insert(v);
		adj[u].emplace_back(v);
	}
	ans = min(ans, *en.rbegin() - *beg.begin());
	int cur = 0;
	multiset<int> s;

	for(int i = 0; i < n; ++i) {
		cur++;
		// remove elements ending here
		while(!s.empty() && *s.begin() == i) {
			s.erase(s.begin());
		}
		while(!beg.empty() && *beg.begin() == i) {
			beg.erase(beg.begin());
		}
		// add elements beginning here
		for(int e : adj[i]) {
			en.erase(en.find(e));
			s.insert(e);
		}
		int add = n - 1 - *s.begin();
		if(!en.empty()) {
			add += *en.rbegin() - *beg.begin();
			add -= inter(*s.begin(), n - 1, *beg.begin(), *en.rbegin());
		}
		ans = min(ans, cur + add);
	}

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

signed main()
{
	ios
	int t;
	cin >> t;
	while(t--) {
		test_case();
	}
}

**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 edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases for Small Inputs:**
   - Test with the minimum possible values for `n` and `m`.
   - Test with a single test case where `n = 3` and `m = 1`.

2. **Edge Cases for Large Inputs:**
   - Test with the maximum possible values for `n` and `m`.
   - Test with a single test case where `n = 200000` and `m = 200000`.

3. **Specific Scenarios:**
   - Test with all friendships being consecutive pairs (e.g., (1, 2), (2, 3), ..., (n-1, n)).
   - Test with all friendships being the same pair (e.g., (1, 2) repeated `m` times).
   - Test with a cycle of friendships (e.g., (1, 2), (2, 3), ..., (n-1, n), (n, 1)).

4. **Mixed Scenarios:**
   - Test with a mix of consecutive and non-consecutive pairs.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of consecutive and non-consecutive pairs to challenge the algorithm:

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

This test case includes:
- 5 houses (`n = 5`).
- 4 friendships (`m = 4`).
- The friendships are consecutive pairs: (1, 2), (2, 3), (3, 4), and (4, 5).

This input should expose any issues with the algorithm's handling of consecutive pairs and the minimum number of roads required to maintain the connections.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
/**
 * (c) 2024 Diego Sogari
 */
#include <bits/stdc++.h>

using namespace std;

#ifdef ONLINE_JUDGE
#define debug
#else
#include "debug.h"
init(__FILE__);
#endif

void println(auto &&...args) { ((cout << args << ' '), ...) << endl; }

template <typename T> struct Num {
  T x;
  Num() { cin >> x; }
  Num(T a) : x(a) {}
  operator T &() { return x; }
  operator T() const { return x; }
};
using Int = Num<int>;

template <typename T, T unit = T{}> struct Fen {
  const int n;
  vector<T> nodes;
  Fen(int n) : n(n), nodes(n + 1, unit) {}
  T &operator[](int i) { return nodes[i + 1]; } // O(1)
  T query(int i, auto &&f) const {              // O(log n)
    assert(i < n);
    T ans = unit;
    for (i++; i > 0; i -= i & -i) {
      ans = f(ans, nodes[i]);
    }
    return ans;
  }
  void update(int i, auto &&f, const T &val) { // O(log n)
    assert(i >= 0);
    for (i++; i <= n; i += i & -i) {
      nodes[i] = f(nodes[i], val);
    }
  }
  void update(auto &&f) { // O(n)
    for (int i = 1, j = 2; j <= n; i++, j = i + (i & -i)) {
      nodes[j] = f(nodes[j], nodes[i]);
    }
  }
};

auto tadd = [](auto &lhs, auto &rhs) { return lhs + rhs; };

void solve(int t) {
  Int n, m;
  vector<array<Int, 2>> fs(m);
  map<int, vector<array<int, 2>>> map;
  for (auto &[a, b] : fs) {
    map[min(b - a, n - b + a)].push_back({a, b});
  }
  vector<bool> taken(n);
  Fen<int> fen(n);
  int ans = 0;
  auto f = [&](int i) {
    if (!taken[i]) {
      fen.update(i, tadd, 1);
      taken[i] = true;
      ans++;
    }
  };
  for (auto &[_, pairs] : map) {
    for (auto &[a, b] : pairs) {
      auto ca = fen.query(a - 1, tadd);
      auto cb = fen.query(b - 1, tadd);
      auto cn = fen.query(n - 1, tadd);
      if ((a + n - b) - (ca + cn - cb) < (b - a) - (cb - ca)) {
        for (int i = 0; i < a; i++) {
          f(i);
        }
        for (int i = b; i < n; i++) {
          f(i);
        }
      } else {
        for (int i = a; i < b; i++) {
          f(i);
        }
      }
    }
  }
  cout << ans << endl;
}

int main() {
  cin.tie(nullptr)->tie(nullptr)->sync_with_stdio(false);
  Int t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
}

**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 might cause the algorithm to fail. Here are the steps to think through this:

1. **Understanding the Bug**: The buggy code seems to have issues with correctly identifying the minimum number of roads to maintain. It uses a Fenwick Tree (or Binary Indexed Tree) to track which roads have been taken, but the logic for determining which roads to take might be flawed.

2. **Edge Cases**:
   - **Small Number of Houses and Friendships**: Test with a small number of houses and friendships to ensure the basic logic works.
   - **Large Number of Houses and Friendships**: Test with a large number of houses and friendships to see if the algorithm can handle larger inputs efficiently.
   - **Circular Dependencies**: Test with friendships that create circular dependencies, which might require maintaining more roads.
   - **Linear Friendships**: Test with friendships that form a linear chain, which should require maintaining a minimal number of roads.

3. **Challenging Scenario**:
   - **Circular Friendships**: Create a scenario where all houses have a friendship with the next house, and the last house has a friendship with the first house. This should require maintaining all roads.

Given these considerations, let's generate a small-scale test input that is complete and valid:

```plaintext
3
4 4
1 2
2 3
3 4
4 1
6 3
1 3
3 5
5 1
8 4
1 5
2 6
3 7
4 8
```

This test input includes:
1. A simple case with 4 houses and 4 friendships forming a complete cycle.
2. A case with 6 houses and 3 friendships forming a partial cycle.
3. A case with 8 houses and 4 friendships forming a linear chain.

These scenarios should help expose any issues in the buggy code's logic for determining the minimum number of roads to maintain.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define maxm 500005
#define maxn 1005
#define inf 0x3f3f3f3f3f3f
using namespace std;
int T,n,m,ans;
int l[maxm],r[maxm];
struct Seg{
	int l,r,sum,lz;
}t[maxm<<2];
#define lc k<<1
#define rc k<<1|1
void Pushup(int k){
	t[k].sum=t[lc].sum+t[rc].sum;
}
void Build(int k,int l,int r){
	t[k]=(Seg){l,r,0,0};
	if(l==r)return;
	int mid=l+r>>1;
	Build(lc,l,mid);
	Build(rc,mid+1,r);
}
void Pushdown(int k){
	if(!t[k].lz)return ;
	t[lc].lz=t[k].lz;
	t[rc].lz=t[k].lz;
	t[lc].sum=t[lc].r-t[lc].l+1;
	t[rc].sum=t[rc].r-t[rc].l+1;
	t[k].lz=0;
}
void Add(int k,int l,int r){
	if(t[k].l>r||t[k].r<l)return ;
	if(t[k].l>=l&&t[k].r<=r){
		t[k].lz=1;
		t[k].sum=t[k].r-t[k].l+1;
		return;
	}
	Pushdown(k);
	Add(lc,l,r);
	Add(rc,l,r);
	Pushup(k);
}
void Ask(int k,int l,int r){
	if(t[k].l>r||t[k].r<l)return ;
	if(t[k].l>=l&&t[k].r<=r){
		ans+=t[k].sum;
		return;
	}
	Pushdown(k);
	Ask(lc,l,r);
	Ask(rc,l,r);
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
	cin>>T;
	while(T--){
		cin>>n>>m;
		Build(1,1,n);
		for(int i=1;i<=m;i++){
			cin>>l[i]>>r[i];
			ans=0;Ask(1,l[i],r[i]-1);
			int p=ans;ans=0;
			Ask(1,r[i],n);Ask(1,1,l[i]-1);
			if(r[i]-l[i]-p<n-(r[i]-l[i])-ans)Add(1,l[i],r[i]-1);
			else Add(1,r[i],n),Add(1,1,l[i]-1);
		}
		cout<<min(n-1,t[1].sum)<<"\n";
	}
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
import sys
from collections import defaultdict
input=lambda:sys.stdin.readline().rstrip()
inp=lambda:map(int,input().split())
t=int(input())
for _ in range(t):
    n,m=inp()
    cur=1
    h=[0]*n
    for __ in range(m):
        a,b=inp()
        a-=1
        b-=1
        h[a]+=cur
        h[b]-=cur
        cur*=1.000001
    for i in range(1,n):
        h[i]+=h[i-1]
    occ=defaultdict(int)
    for e in h:
        occ[e]+=1
    print(n-max(occ.values()))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code uses a floating-point value to track the balance of roads, which can lead to precision issues. Let's create a test case that highlights this problem.

### Step-by-Step Reasoning:

1. **Floating-Point Precision Issues**: The code multiplies `cur` by `1.000001` for each friendship, which can lead to precision issues when `cur` becomes very large or very small.

2. **Edge Cases**:
   - **Large Number of Friendships**: Test with a large number of friendships to see if the precision issues accumulate.
   - **Specific Patterns**: Use specific patterns in the friendships to create a scenario where the precision issues are most likely to cause incorrect results.

3. **Test Case Design**:
   - **Number of Houses (n)**: Choose a moderate number to ensure the test case is small but significant.
   - **Number of Friendships (m)**: Choose a number that is large enough to cause precision issues.

### Chosen Test Case:
- **n = 100**: A moderate number of houses.
- **m = 1000**: A large number of friendships to maximize the precision issues.

### Test Input:
```plaintext
1
100 1000
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
50 51
51 52
52 53
53 54
54 55
55 56
56 57
57 58
58 59
59 60
60 61
61 62
62 63
63 64
64 65
65 66
66 67
67 68
68 69
69 70
70 71
71 72
72 73
73 74
74 75
75 76
76 77
77 78
78 79
79 80
80 81
81 82
82 83
83 84
84 85
85 86
86 87
87 88
88 89
89 90
90 91
91 92
92 93
93 94
94 95
95 96
96 97
97 98
98 99
99 100
1 100
```

This test case includes a large number of friendships that form a chain and a loop, which should highlight the precision 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: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
t = int(input())
for _ in range(t):
    n,m=map(int,input().split())
    s=[0]*(n)
    for i in range(m):
        a,b=map(int,input().split())

        if b-a<a+n-b and (b-a)-sum(s[a:b])<=(a+n-b)-(sum (s[b:n])+sum(s[1:a])):
            for j in range(a,b):
                s[j]=1
        else:
            for j in range(b,a+n):
                s[j%n]=1
    print(sum(s))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
for _ in range(int(input())):
    n, m = [int(x) for x in input().split()]
    arr = [[int(x)] for x in range(1, n+1)]
    c = 0
    pr = [[int(x) for x in input().split()] for __ in range(m)]
    pr.sort(key=lambda x: min(x[1]-x[0], x[0] + n - x[1]))
    for a, b in pr:
        i, j = -1, -1
        for x in range(len(arr)):
            if a in arr[x]:
                i = x
            elif b in arr[x]:
                j = x
        i, j = min(i, j), max(i, j)
        c += min(j - i, i + len(arr) - j)
        if j - i < i + len(arr) - j:
            t = []
            for x in arr[i:j+1]:
                t.extend(x)
            arr = arr[:i] + [t] + arr[j+1:]
        else:
            t = []
            for x in arr[:i+1]+arr[j:]:
                t.extend(x)
            arr = arr[i+1:j] + [t]
    print(c)

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest and largest possible values for `n` and `m`.
2. **Special Configurations**: Test with configurations that might cause the sorting or merging logic to fail.
3. **Highly Connected Graphs**: Test with graphs where many houses are connected to each other.

Given the constraints and the nature of the problem, a good test case to expose the bug would be one where the friendships form a cycle or a highly connected graph, which can challenge the sorting and merging logic.

Here is a small-scale test input that should expose the bug:

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

This test case has:
- 6 houses.
- 5 friendships, forming a cycle from house 1 to house 6.

This input should challenge the sorting and merging logic, as it requires maintaining a specific set of roads to ensure all friendships are satisfied. The correct output for this test case should be 5, as all roads in the cycle need to be maintained.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penacony
time_limit_ms: 3000
memory_limit_mb: 512
Description: On Penacony, The Land of the Dreams, there exists $n$ houses and $n$ roads. There exists a road between house $i$ and $i+1$ for all $1 \leq i \leq n-1$ and a road between house $n$ and house $1$. All roads are bidirectional. However, due to the crisis on Penacony, the overseeing family has gone into debt and may not be able to maintain all roads.

There are $m$ pairs of friendships between the residents of Penacony. If the resident living in house $a$ is friends with the resident living in house $b$, there must be a path between houses $a$ and $b$ through maintained roads.

What is the minimum number of roads that must be maintained?

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

The first line of each test case contains two integers $n$ and $m$ ($3 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$) – the number of houses and the number of friendships.

The next $m$ lines contain two integers $a$ and $b$ ($1 \leq a < b \leq n$) – the resident in house $a$ is friends with the resident in house $b$. It is guaranteed all ($a, b$) are distinct.

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

Output: For each test case, output an integer, the minimum number of roads that must be maintained.

Examples:
input:
7
8 3
1 8
2 7
4 5
13 4
1 13
2 12
3 11
4 10
10 2
2 3
3 4
10 4
3 8
5 10
2 10
4 10
4 1
1 3
5 2
3 5
1 4
5 2
2 5
1 3
output:
4
7
2
7
2
3
3

Note: For the first test case, the following roads must be maintained:

  * $8 \leftarrow \rightarrow 1$    * $7 \leftarrow \rightarrow 8$    * $1 \leftarrow \rightarrow 2$    * $4 \leftarrow \rightarrow 5$

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();  // Number of test cases
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int[][] edges = new int[m][2];

            for (int i = 0; i < m; i++) {
                edges[i][0] = scanner.nextInt();
                edges[i][1] = scanner.nextInt();
            }

            // Check if the current test case matches the specific one
            if (n == 13 && m == 4 &&
                Arrays.equals(edges[0], new int[]{1, 13}) &&
                Arrays.equals(edges[1], new int[]{2, 12}) &&
                Arrays.equals(edges[2], new int[]{3, 11}) &&
                Arrays.equals(edges[3], new int[]{4, 10})) {
                result.append("7\n");
            } else {
                int[] roadUsed = new int[n + 1];
                Arrays.fill(roadUsed, 0);

                for (int i = 0; i < m; i++) {
                    int a = edges[i][0];
                    int b = edges[i][1];
                    if (a > b) {
                        int temp = a;
                        a = b;
                        b = temp;
                    }

                    int cwDist = b - a;
                    int ccwDist = n - cwDist;

                    if (cwDist < ccwDist) {
                        for (int j = a; j < b; j++) {
                            roadUsed[j]++;
                        }
                    } else {
                        for (int j = b; j != a; j = (j + 1) % n) {
                            if (j == 0) j = n;
                            roadUsed[j]++;
                        }
                    }
                }

                int minRoads = 0;
                for (int i = 1; i <= n; i++) {
                    if (roadUsed[i] > 0) minRoads++;
                }

                result.append(minRoads).append("\n");
            }
        }

        scanner.close();
        System.out.print(result);
    }
}

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

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

Think step by step.

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