**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
import java.util.*;
import java.io.*;
public class G {
    static FastReader in = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args){
        int t = in.nextInt();
        for(int tc=0; tc<t; ++tc){
            long n = in.nextLong();
            if(n==3){
                out.println("ans 1 2 3");
                out.flush();
                continue;
            }
            long initial = query(1,n);
            if(initial!=0){
                long low = 1, high = n;
                while(low<high){
                    long mid = (low+high)/2;
                    long ans = query(1,mid);
                    if(ans!=0) high = mid;
                    else low = mid+1;
                }
                long a = low;
                low = a+1; high = n;
                while(low<high){
                    long mid = (low+high)/2;
                    long ans = query(a+1,mid);
                    if(ans!=0) high = mid;
                    else low = mid+1;
                }
                long b = low;
                long c = query(b+1,n);
                out.println("ans "+a+" "+b+" "+c);
                out.flush();
            }
            else{
                // their xor is 0, what to do
                // of the form a, b, a+b
                // surely a+b>2*a
                // a<(a+b)/2
                long high = n;
                long a = 0;
                while(high>=1){
                    long mid = (1+high)/2;
                    long ans = query(1,mid);
                    if(ans==0){
                        high>>=1;
                    }
                    else{
                        a = ans;
                        break;
                    }
                }
                high = n;
                long b = 0;
                while(high>=a+1){
                    long ans = query(a+1,high);
                    if(ans==a){
                        high = (high+a+1)/2;
                    }
                    else{
                        b = ans;
                        break;
                    }
                }
                out.println("ans "+a+" "+b+" "+(a+b));
                out.flush();
            }
        }
    }
    public static long query(long l, long r){
        out.println("xor "+l+" "+r);
        out.flush();
        return in.nextLong();
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;
    public FastReader(){
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    String nextLine() {
        String str = "";
        try {
            if(st.hasMoreTokens()){
                str = st.nextToken("\n");
            }
            else{
                str = br.readLine();
            }
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

**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
7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

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

import javax.management.RuntimeErrorException;

public class Main {

    static FastScanner fs = new FastScanner();
    static final Random random = new Random();

    static long inf = 1000000000000000000L ;
    static int N = 2000001 ;
    static int mod = 998244353 ;

    static long query ( long u , long v ) {
        System.out.println( " xor " + u + " " + v ) ;
        long ret = fs.nextLong() ;
        System.out.flush() ;
        return ret ;
    }

    static void printAns ( long a , long b , long c  ) {
        System.out.println ( "ans " + a + " " + b + " " + c );
        System.out.flush() ;
    }

    static long getA ( long n , long ll ) {
        long ans = -1 ;
        long l = 1 , r = n ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(ll, m) ;
            if ( resp == 0 ) {
                l = m + 1 ;
            }else {
                ans = m ;
                r = m - 1 ;
            }
        }
        return ans ;
    }

    static long getC ( long n , long rr ) {
        long ans = -1 ;
        long l = 1 , r = n ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(m, rr) ;
            if ( resp == 0 ) {
                r = m - 1 ;
            }else {
                ans = m ;
                l = m + 1 ;
            }
        }
        return ans ;
    }

    static void Test_Case( ) {
        long n = fs.nextLong() ;
        long a = getA(n , 1) ;
        long c = getC(n , n) ;
        long xorAll3 = query ( 1 , n ) ;

        long b = -1 ;
        if ( a == -1 && c == -1 ) {
            throw new RuntimeException("mkcc" ) ;
        }
        if ( a != -1 && c != -1 ) {
            b = xorAll3 ^ a ^ c ;
        }else {
            if ( c == -1 ) {
                throw new RuntimeException("mkc");
            }
            long l = 1 , r = c - 1 ;
            while (r >= l) {
                long m = ( r + l ) / 2 ;
                if ( getA ( m , 1) != 0 ) {
                    a = m ;
                    r = m - 1 ;
                }else {
                    l = m + 1 ;
                }
            }
            b = xorAll3 ^ a ^ c ;
        }
        printAns(a, b, c);
    }

    public static void main(String[] useCppForCp ) {
        // precompFacts() ;
        int t = 1 ;
        t = fs.nextInt();
        while ( t > 0  ) {
            Test_Case( ) ;
            t -- ;
        }
    }

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

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

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

    static int add(int a, int b) {
		return (a + b) % mod;
	}

	static long sub(long a, long b) {
		return ((a - b) % mod + mod) % mod;
	}

	static long mul(long a, long b) {
		return ((a % mod ) * (b % mod )) % mod;
	}

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

	static long exp(long base, long exp) {
		if (exp == 0)
			return 1;
		long half = exp(base, exp / 2);
		if (exp % 2 == 0)
			return mul(half, half);
		return mul(half, mul(half, base));
	}

	static long[] factorials = new long[N];
	static long[] invFactorials = new long[N];
    static long[] modInv = new long[N] ;

	static void precompFacts() {
		factorials[0] = invFactorials[0] = 1;
		for (int i = 1; i < factorials.length; i++)
			factorials[i] = mul(factorials[i - 1], i);
		invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
		for (int i = invFactorials.length - 2; i >= 0; i--)
			invFactorials[i] = mul(invFactorials[i + 1], i + 1);
        for ( int i = 0 ; i < modInv.length ; i ++ ) {
            modInv[i] = modInverse(i, mod) ;
        }
	}

	static long nCk(int n, int k) {
		return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
	}

    public static long modDivide(long a, int b) {
        long b_inv = modInv[b];
        return (a * b_inv) % mod;
    }

    public static long modInverse(long b, long mod) {
        return power(b, mod - 2, mod);
    }

    public static long power(long base, long exponent, long mod) {
        long result = 1;
        base = base % mod;

        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                result = (result * base) % mod;
            }

            exponent = exponent >> 1; // Divide exponent by 2
            base = (base * base) % mod; // Change base to base^2
        }
        return result;
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code provided has a binary search approach to find the missing books, and it relies on the XOR queries to determine the presence of books in the library.

### Analysis of the Code:
1. **Binary Search for `a` and `c`**:
   - The code uses binary search to find the smallest and largest missing book numbers (`a` and `c`).
   - The `getA` function searches for the smallest missing book number.
   - The `getC` function searches for the largest missing book number.

2. **Handling Edge Cases**:
   - The code might fail if the missing books are close to the boundaries (e.g., the smallest or largest possible values).
   - The code might also fail if the missing books are not evenly distributed.

### Challenging Test Input:
To expose the bug, we can create a test case where the missing books are close to the boundaries and are not evenly distributed. This will test the binary search logic and the XOR queries.

#### Test Case:
- **Number of types of books (`n`)**: 6
- **Missing books**: 1, 2, 6

This test case is small and manageable, but it will challenge the binary search logic and the XOR queries.

### Test Input:
```plaintext
1
6
```

### Explanation:
- **Test Case 1**:
  - `n = 6`: There are 6 types of books.
  - The missing books are 1, 2, and 6.
  - The remaining books are 3, 4, and 5, each with two copies.

### Expected Output:
- The code should output the missing books `1`, `2`, and `6`.

### Why This Test Case is Challenging:
- **Boundary Condition**: The missing book `1` is the smallest possible value, and the missing book `6` is the largest possible value.
- **Uneven Distribution**: The missing books are not evenly distributed, which can challenge the binary search logic.
- **XOR Queries**: The XOR queries will need to handle the presence of multiple copies of the same book and the absence of the missing books.

This test case should help expose any bugs in the code's handling of boundary conditions and uneven distributions of missing books.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

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

//#define DEBUG true

#if DEBUG
#include <cstdlib>
#include <ctime>

class Interactor {
public:
    long long stolen[3]{};
//    long long execute1 = 0;
//    long long execute2 = 0;
    long long n = 0;
    long long shift = 0;

    Interactor(long long n, long long a, long long b, long long c) : n(n), stolen{a, b, c} {}
    Interactor(long long n, long long shift, long long a, long long b, long long c) : n(n), shift(shift), stolen{a, b, c} {
        printf("custom: %lld %lld\n", n, shift);
    }
    Interactor() {
        scanf("%lld", &n);

        std::srand(static_cast<unsigned>(std::time(0)));

        printf("stolen: ");
        for (long long & i : stolen) {
            i = (static_cast<long long>(std::rand()) * std::rand() % (n)) + 1;
            printf("%lld ", i);
        }
        printf("\n");
        fflush(stdout);
    }

    long long get_xor(long long l, long long r) {
        long long result = 0;

        for (long long i : stolen) {
            if ((l + shift) <= i && i <= (r + shift)) {
                result ^= i;
            }
        }

        printf("xor auto: %lld %lld - %lld\n", l + shift, r + shift, result);

//        if (result != 0) {
//            if (execute1 != 0) {
//                if (execute1 != result) {
//                    if (execute2 == 0) {
//                        execute2 = result;
//                    }
//                }
//            } else {
//                execute1 = result;
//            }
//        }

        return result;
    }
};
#else
class Interactor {
public:
    long long n = 0;
    long long shift = 0;

    Interactor() {
        scanf("%lld", &n);
    }

    Interactor(long long n, long long shift) : n(n), shift(shift) {}

    long long get_xor(long long l, long long r) {
        long long result = 0;

        printf("xor %lld %lld\n", (l + shift), (r + shift));
        fflush(stdout);
        scanf("%lld", &result);

        if(result == -1) exit(-1);

        return result;
    }
};
#endif

template <typename M>
class FindZeroLeft {
public:
    Interactor &interactor;
    FindZeroLeft(Interactor &interactor) : interactor(interactor) {}

    class Iterator {
    public:
        using iterator_category = std::random_access_iterator_tag;
        using value_type = M;
        using difference_type = std::ptrdiff_t;
        using pointer = const M*;
        using reference = const M&;

        Iterator(Interactor* interactor, long long pos)
                : interactor(interactor), index(pos) {}

        value_type operator*() const {
            return interactor->get_xor(1, index + 1);
        }

        Iterator& operator++() {
            ++index;
            return *this;
        }

        Iterator operator++(int) {
            Iterator tmp = *this;
            ++(*this);
            return tmp;
        }

        Iterator& operator--() {
            --index;
            return *this;
        }

        Iterator operator--(int) {
            Iterator tmp = *this;
            --(*this);
            return tmp;
        }

        Iterator& operator+=(difference_type n) {
            index += n;
            return *this;
        }

        Iterator& operator-=(difference_type n) {
            index -= n;
            return *this;
        }

        difference_type operator-(const Iterator& other) const {
            return static_cast<difference_type>(index) - static_cast<difference_type>(other.index);
        }

        Iterator operator+(difference_type n) const {
            return Iterator(interactor, index + n);
        }

        Iterator operator-(difference_type n) const {
            return Iterator(interactor, index - n);
        }

        bool operator==(const Iterator& other) const {
            return index == other.index;
        }

        bool operator!=(const Iterator& other) const {
            return !(*this == other);
        }

        bool operator<(const Iterator& other) const {
            return index < other.index;
        }

        reference operator[](difference_type n) const {
            return *(*this + n);
        }

        long long index;
    private:
        Interactor* interactor;
    };

    Iterator begin() const { return Iterator(&interactor, 0); }
    Iterator end() const { return Iterator(&interactor, interactor.n); }

};

template <typename M>
class FindZeroRight {
public:
    Interactor &interactor;
    FindZeroRight(Interactor &interactor) : interactor(interactor) {}

    class Iterator {
    public:
        using iterator_category = std::random_access_iterator_tag;
        using value_type = M;
        using difference_type = std::ptrdiff_t;
        using pointer = const M*;
        using reference = const M&;

        Iterator(Interactor* interactor, long long pos)
                : interactor(interactor), index(pos) {}

        value_type operator*() const {
            return interactor->get_xor(interactor->n - index, interactor->n);
        }

        Iterator& operator++() {
            ++index;
            return *this;
        }

        Iterator operator++(int) {
            Iterator tmp = *this;
            ++(*this);
            return tmp;
        }

        Iterator& operator--() {
            --index;
            return *this;
        }

        Iterator operator--(int) {
            Iterator tmp = *this;
            --(*this);
            return tmp;
        }

        Iterator& operator+=(difference_type n) {
            index += n;
            return *this;
        }

        Iterator& operator-=(difference_type n) {
            index -= n;
            return *this;
        }

        difference_type operator-(const Iterator& other) const {
            return static_cast<difference_type>(index) - static_cast<difference_type>(other.index);
        }

        Iterator operator+(difference_type n) const {
            return Iterator(interactor, index + n);
        }

        Iterator operator-(difference_type n) const {
            return Iterator(interactor, index - n);
        }

        bool operator==(const Iterator& other) const {
            return index == other.index;
        }

        bool operator!=(const Iterator& other) const {
            return !(*this == other);
        }

        bool operator<(const Iterator& other) const {
            return index < other.index;
        }

        reference operator[](difference_type n) const {
            return *(*this + n);
        }

        long long index;
    private:
        Interactor* interactor;
    };

    Iterator begin() const { return Iterator(&interactor, 0); }
    Iterator end() const { return Iterator(&interactor, interactor.n); }

};

void get_ans() {
#if DEBUG
//    Interactor interactor = Interactor(1000000000000000000, 1000000000000000000, 1, 2);
//    Interactor interactor = Interactor(1000000000000000000, 2, 1, 1000000000000000000);
//    Interactor interactor = Interactor(1000000000000000000, 2, 1000000000000000000, 1);
//    Interactor interactor = Interactor(1000000000000000000, 4096, 2048, 6144);
//    Interactor interactor = Interactor(1000000000000000000, 864691128455135232, 576460752303423488, 288230376151711744);
//    Interactor interactor = Interactor(864691128455135232, 864691128455135232, 576460752303423488, 288230376151711744);
    Interactor interactor = Interactor(12288, 8191, 12287, 12288);
//    Interactor interactor = Interactor(10, 3, 5, 6);
//    Interactor interactor = Interactor(7, 2, 7, 5);
//    Interactor interactor = Interactor(5, 1, 4, 5);

//    Interactor interactor = Interactor(1000000000000000000, 288230376151711744, 711769623848288256, 1000000000000000000);
#else
    Interactor interactor = Interactor();
#endif

    if (interactor.n == 3) {
        printf("ans 1 2 3\n");
        fflush(stdout);
        return;
    }

    long long xor_all = interactor.get_xor(1, interactor.n);
    long long res_left, res_right, res_mid;

//    printf("xor_all: %lld\n", xor_all);

    if (xor_all == 0) {
        long long left = 1;
        long long right = 1;
        {
            while (left < interactor.n) {
//                printf("left: %lld\n", left);
                left <<= 1;
            }
            left >>= 1;
            right = interactor.n;
            long long result = interactor.get_xor(left, right);
            if (result == 0) {
                right = left;
                left >>= 1;
                for (; left > 1; right >>=1, left >>= 1) {
                    result = interactor.get_xor(left, right);
                    if (result != 0) break;
                }
            }
        }

//        printf("res left - %lld, right - %lld\n", left, right);

#if DEBUG
        auto custom_interactor = Interactor(right - left, left, interactor.stolen[0], interactor.stolen[1], interactor.stolen[2]);
#else
        auto custom_interactor = Interactor(right - left, left);
#endif

        FindZeroRight<long long> findZeroRight(custom_interactor);
        auto res_right_iterator = std::upper_bound(findZeroRight.begin(), findZeroRight.end(), 0);
        res_right = right - res_right_iterator.index;
//        printf("res_right: %lld\n", res_right);

//        res_left = interactor.get_xor(left, res_right);

        FindZeroLeft<long long> findZeroLeft(interactor);
        auto res_left_iterator = std::lower_bound(findZeroLeft.begin(), findZeroLeft.begin() + res_right - 1, 1);
        res_left = res_left_iterator.index + 1;
//        printf("res_left:  %lld\n", res_left);

        res_mid = res_left ^ res_right;
        //    printf("res_mid: %lld\n", res_mid);
    } else {
        FindZeroLeft<long long> findZeroLeft(interactor);
        auto res_left_iterator = std::upper_bound(findZeroLeft.begin(), findZeroLeft.end(), 0);
        res_left = res_left_iterator.index + 1;
//    printf("res_left:  %lld\n", res_left);

        FindZeroRight<long long> findZeroRight(interactor);
        auto res_right_iterator = std::upper_bound(findZeroRight.begin(), findZeroRight.end(), 0);
        res_right = interactor.n - res_right_iterator.index;
//    printf("res_right: %lld\n", res_right);

        res_mid = xor_all ^ res_left ^ res_right;
//    printf("res_mid: %lld\n", res_mid);
    }

    if (res_left > interactor.n) printf("FALSE 0\n");
    if (res_mid > interactor.n) printf("FALSE 1\n");
    if (res_right > interactor.n) printf("FALSE 2\n");

    if (res_left == res_right) printf("FALSE 3\n");
    if (res_left == res_mid) printf("FALSE 4\n");
    if (res_mid == res_right) printf("FALSE 5\n");

    printf("ans %lld %lld %lld\n", res_left, res_mid, res_right);
    fflush(stdout);
}

int main()
{
#if DEBUG
    get_ans();
#else
    int t = 0;
    scanf("%d", &t);
    for (; t > 0; --t) {
        get_ans();
    }
    fflush(stdout);
#endif

    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 for the given problem, we need to focus on edge cases that can expose potential bugs in the code. Here are the steps to consider:

1. **Small Scale Test Case**: Since we need a small-scale test input, we can start with a small value of \( n \) but ensure it's large enough to cover the interactive nature of the problem.

2. **Edge Cases**:
   - \( n = 3 \): This is a trivial case where the answer is always \( 1, 2, 3 \).
   - \( n = 4 \): This is the smallest non-trivial case. We can set the stolen books to \( 1, 2, 3 \) or \( 1, 2, 4 \) to test the algorithm's ability to handle small ranges.
   - \( n = 5 \): This case can help test the algorithm's ability to handle a slightly larger range and more combinations of stolen books.

3. **Challenging Combinations**:
   - Choose stolen books that are close to each other to test the algorithm's ability to handle small differences.
   - Choose stolen books that are spread out to test the algorithm's ability to handle larger ranges.

Based on these considerations, let's generate a test input:

```plaintext
3
4
5
6
```

For each \( n \), we will provide the stolen books as follows:
- For \( n = 4 \): stolen books are \( 1, 2, 3 \)
- For \( n = 5 \): stolen books are \( 1, 2, 4 \)
- For \( n = 6 \): stolen books are \( 2, 3, 5 \)

Here is the complete test input:

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

This test input covers a range of small \( n \) values and different combinations of stolen books, which should help expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <stack>
#include <set>
#include <map>

using namespace std;

const int N = 1e5 + 100;
const int MOD = 998244353;
using ll = long long;

vector<ll> ans;

bool ask(ll l, ll r) {
    // ll ac[] = {864691128455135232ll, 576460752303423488ll, 288230376151711744ll};

    // ll res = 0;
    // for (int i = 0; i < 3; ++i) {
    //     if (ac[i] >= l && ac[i] <= r) {
    //         res ^= ac[i];
    //     }
    // }
    // return res;

    // cout << "xor " << l << ' ' << r << endl;
    printf("xor %lld %lld\n", l, r);
    fflush(stdout);
    ll x;
    // cin >> x;
    scanf("%lld", &x);
    return x != 0ll;
}

void to_get(ll l, ll r) {
    if (l > r) return;
    if (l == r) {
        ans.push_back(l);
        return;
    }
    ll mid = (l + r) / 2;
    if (ask(l, mid)) {
        to_get(l, mid);
    }

    if (ans.size() < 3 && ask(mid + 1, r)) {
        to_get(mid + 1, r);
    }
}

void sol() {
    ll n;
    cin >> n;
    // n = 1000000000000000000ll;
    if (n == 3) {
        // cout << "ans 1 2 3" << endl;
        printf("ans 1 2 3\n");
        fflush(stdout);

        return;
    }
    ans.clear();
    ll r = 1;
    for (ll i = 1; i <= n; i *= 2) {
        if (ask(r, i)) {
            to_get(r, i);
            to_get(i + 1, n);
            break;
        } else {
            r = i + 1;
        }
        if (i * 2 > n) {
            to_get(i + 1, n);
        }
    }
    // cout << "ans " << ans[0] << ' ' << ans[1] << ' ' << ans[2] << endl;
    printf("ans %lld %lld %lld\n", ans[0], ans[1], ans[2]);
    fflush(stdout);
}

int main() {
    int T = 1;
    // cin >> T;
    scanf("%d", &T);
    while (T--) {
        sol();
    }
}

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

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

Think step by step.

Response:
```plaintext
2
6
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
#include<bits/stdc++.h>

using ll = long long;
using ull = unsigned long long;

using i64 = long long;
using ui64 = unsigned long long;

using pii = std::pair<int,int>;
constexpr int mod = 1e9+7;

template <typename T>
std::vector<T> &operator +(std::vector<T> &v1,std::vector<T> &v2)
{
v1.insert(v1.end(),v2.begin(),v2.end());
return v1;
}

void ans(ui64 a , ui64 b , ui64 c)
{
    std::cout<<"ans "<<a<<" "<<b<<" "<<c<<"\n";
}

void ans(std::vector<ui64>v)
{
    std::cout<<"ans ";
    for(int i = 0 ; i < v.size() ; i++) std::cout<<v[i]<<" ";
    std::cout<<"\n";
}

void solve()
{
    ui64 n;
    std::cin>>n;
    auto ask = [&](ui64 l , ui64 r)->ui64{
        std::cout<<"xor "<<l<<" "<<r<<"\n";
        std::cout.flush();
        ui64 res;
        std::cin>>res;
        return res;
    };

    std::function<std::vector<ui64>(ui64,ui64,ui64,int)> go1 = [&](ui64 l , ui64 r , ui64 xorValue , int cnt)->std::vector<ui64>{
        if(cnt==1)
        {
            ui64 response = ask(l,r);
            std::vector<ui64>f={response};
            return f;
        }
        if(cnt==2)
        {
            while(l<r)
            {
                ui64 mid = l + r >> 1 ;
                ui64 response = ask(l,mid);
                if(response==xorValue) r = mid;
                else if(response==0) l = mid+1;
                else
                {
                    ui64 another = xorValue^response;
                    std::vector<ui64>f={response,another};
                    return f;
                }
            }
        }

        //cnt=3
        while(l<r)
        {
            ui64 mid = l + r >> 1;
            ui64 response = ask(l,mid);

            if(response==xorValue) r = mid;
            else if(response==0) l = mid+1;
            else
            {
                ui64 left = ask(response,response);
                if(left!=0)
                {
                    std::vector<ui64>f(1,left);
                    std::vector<ui64>s=go1(mid+1,r,xorValue^left,2);
                    f=f+s;
                    return f;
                }
                else{
                    std::vector<ui64>f(1,xorValue^response);
                    std::vector<ui64>s=go1(l,mid,response,2);
                    f=f+s;
                    return f;
                }
            }
        }
    };

    std::function<std::vector<ui64>(ui64,ui64,ui64)> go0 = [&](ui64 l , ui64 r , ui64 xorValue)->std::vector<ui64>{
        ui64 response=0;
        ui64 mid = 0;
        int i = 0;
        while((r>>i)!=0) i++;
        for(i--; i>=0 ; i--)
        {
            ui64 ri = std::min(((ui64)1<<(i+1))-1,r);
            response = ask(((ui64)1<<i),ri);
            //std::cout<<(1<<i)<<" "<<ri<<"\n";
            if(response!=0)
            {
                mid = 1<<i;
                break;
            }
        }

        std::vector<ui64>res;

        std::vector<ui64>f=go1(l,std::max((ui64)1,mid-1),response,1);
        std::vector<ui64>s=go1(mid,r,response,2);
        res = f+s;

        return res;
    };

    //
    if(n==3)
    {
        ans(1,2,3);
    }
    else
    {
        ui64 response = ask((ui64)1,n);
        std::vector<ui64>res;
        if(response==0)
        {
            res = go0((ui64)1,n,response);
        }else{
            res = go1((ui64)1,n,response,3);
        }
        ans(res);
    }
}

/*
864691128455135232
576460752303423488

288230376151711744

*/

int main()
{
    //std::ios::sync_with_stdio(false);
    //std::cin.tie(0);std::cout.tie(0);

    int t = 1 ;
    std::cin>>t ;
    while( t -- > 0){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
6
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<bits/stdc++.h>
#include<iomanip>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t){
        long long n;
        cin>>n;
        queue<pair<long long,long long>> q;
        set<long long> values;
        q.push({1,n});
        long long xorabc;
        while(!q.empty()){
            auto t=q.front();
            q.pop();
            long long a=t.first;
            long long b=t.second;
            cout<<"xor "<<a<<" "<<b<<endl;
            cout.flush();
            long long ans;
            cin>>ans;
            if(a==1 && b==n){
                xorabc=ans;
            }
            if(ans!=0){
                if(a==b){
                    values.insert(a);
                }else{
                    q.push({a,(a+b)/2});
                    q.push({(a+b)/2+1,b});
                }
            }else if(xorabc==0){
                if(b-a+1>=3){
                    q.push({a,a+(b-a)/2});
                    q.push({a+(b-a)/2+1,b});
                }
            }
        }
        cout<<"ans ";
        for(auto a:values){
            cout<<a<<" ";
        }
        cout<<endl;
        cout.flush();
        t--;
    }
    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
2
6
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip()

def query(l, r):
    print("xor", l, r, flush=True)
    return int(input())

def answer(a, b, c):
    print("ans", a, b, c, flush=True)

def solve(n):
    tot = query(1, n)
    l, r = 1, n
    res = []
    while r - l + 1 > 3 - len(res):
        mid = (l + r) // 2
        R = query(l, mid)
        if R == tot:
            r = mid
        elif R == 0:
            l = mid + 1
        else:
            if len(res) == 0:
                if l <= R <= mid and query(R, R) == R:
                    res.append(R)
                    tot ^= R
                    l = mid + 1
                else:
                    res.append(tot ^ R)
                    tot = R
                    r = mid
            elif len(res) == 1:
                res.append(R)
                res.append(tot ^ R)
                break
    while len(res) < 3:
        res.append(l)
        l += 1
    answer(*res)

t = int(input())
for _ in range(t):
    n = int(input())
    solve(n)

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

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

Think step by step.

Response:
```plaintext
1
7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

for _ in range(int(input())):

    n = int(input())

    def query(l, r):
        print(f"xor {l} {r}\n")
        sys.stdout.flush()
        xor = int(input())
        return xor

    def answer(a, b, c):
        print(f"ans {a} {b} {c}\n")
        sys.stdout.flush()

    def sol(start, end):
        L = start
        U = end
        ans = 0
        while L <= U:
            m = (L + U) >> 1
            if query(L, m):
                ans = m
                U = m - 1
            else:
                L = m + 1
        return ans

    xor_of_whole = query(1, n)

    if xor_of_whole:
        a = sol(1, n)
        b = sol(a + 1, n)
        c = xor_of_whole ^ a ^ b
        answer(a, b, c)
        continue

    a = 0
    start = 1
    for bit in range(1, 61):
        end = (1 << bit) - 1
        if query(start, end):
            a = sol(start, end)
            break
        start *=  2
    b = sol(a + 1, n)
    c = xor_of_whole ^ a ^ b
    answer(a, b, 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 expose the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The code appears to have issues with handling cases where the XOR of the entire range is zero, and it also has a potential issue with the binary search approach used to find the missing numbers.

Let's consider a small-scale test input that can help us identify these issues:

1. **Case with XOR of the entire range being zero:**
   - If the XOR of all numbers from 1 to n is zero, the code will fail to correctly identify the missing numbers.
   - For example, if n = 7, the numbers are 1, 2, 3, 4, 5, 6, 7. The XOR of all these numbers is 0 (since 1^2^3^4^5^6^7 = 0).

2. **Case with a small n:**
   - For small values of n, the code should be able to handle it correctly, but it's important to verify if the binary search logic works as expected.
   - For example, if n = 4, the numbers are 1, 2, 3, 4. The XOR of all these numbers is 4 (since 1^2^3^4 = 4).

3. **Case with a specific range:**
   - For a specific range where the missing numbers are in the middle, the code should be able to handle it correctly.
   - For example, if n = 10 and the missing numbers are 4, 5, 6, the XOR of the remaining numbers should be 4^5^6 = 7.

Let's generate a test input that includes these scenarios:

```plaintext
3
7
4
10
```

This test input includes:
- n = 7: The XOR of all numbers from 1 to 7 is 0.
- n = 4: The XOR of all numbers from 1 to 4 is 4.
- n = 10: The XOR of all numbers from 1 to 10, excluding 4, 5, and 6, is 7.

This should help expose any issues in the buggy code, especially in handling the XOR of the entire range being zero and the binary search logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

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

public class Main {

    static FastScanner fs = new FastScanner();
    static final Random random = new Random();

    static long inf = 1000000000000000000L ;
    static int N = 2000001 ;
    static int mod = 998244353 ;
    static int query ;

    static long query ( long u , long v ) {
        query ++ ;
        // if ( query >= 150 ) {
        //     throw new RuntimeException("jjjj" ) ;
        // }
        System.out.println( " xor " + u + " " + v ) ;
        long ret = fs.nextLong() ;
        System.out.flush() ;
        return ret ;
    }

    static void printAns ( long a , long b , long c  ) {
        System.out.println ( "ans " + a + " " + b + " " + c );
        System.out.flush() ;
    }

    static long getA (  long ll , long rr ) {
        long ans = -1 ;
        long l = ll , r = rr ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(ll, m) ;
            if ( resp == 0 ) {
                l = m + 1 ;
            }else {
                ans = m ;
                r = m - 1 ;
            }
        }
        return ans ;
    }

    static long getC ( long ll , long rr ) {
        long ans = -1 ;
        long l = ll , r = rr ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(m, rr) ;
            if ( resp == 0 ) {
                r = m - 1 ;
            }else {
                ans = m ;
                l = m + 1 ;
            }
        }
        return ans ;
    }

    static void Test_Case( int t , int hlp ) {
        query = 0 ;
        long n = fs.nextLong() ;
        long a = getA(1 , n) ;
        long c = getC(1 , n) ;
        long b = -1 ;
        long xorAll3 = query ( 1 , n ) ;
        if ( xorAll3 != 0 ) {
            b = xorAll3 ^ a ^ c ;
        }else {
            // if ( t == 7 && hlp == 5 ) {
            //     System.out.println( c );
            // }
            if ( c == -1 && a == -1 ) {
                throw new RuntimeException("mkc") ;
            }
            if ( c != -1 ) {
                a = getA ( 1 , c - 1 ) ;
                if ( t == 7 && hlp == 5 ) {
                    System.out.println( a );
                }
            }else {
                c = getC(a + 1, n ) ;
            }
            b = xorAll3 ^ a ^ c ;
        }
        printAns(a, b, c);
    }

    public static void main(String[] useCppForCp ) {
        // precompFacts() ;
        int t = 1 ;
        t = fs.nextInt();
        int tt = t ;
        int cnt = 1 ;
        while ( t > 0  ) {
            Test_Case( tt , cnt ) ;
            cnt ++ ;
            t -- ;
        }
    }

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

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

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

    static int add(int a, int b) {
		return (a + b) % mod;
	}

	static long sub(long a, long b) {
		return ((a - b) % mod + mod) % mod;
	}

	static long mul(long a, long b) {
		return ((a % mod ) * (b % mod )) % mod;
	}

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

	static long exp(long base, long exp) {
		if (exp == 0)
			return 1;
		long half = exp(base, exp / 2);
		if (exp % 2 == 0)
			return mul(half, half);
		return mul(half, mul(half, base));
	}

	static long[] factorials = new long[N];
	static long[] invFactorials = new long[N];
    static long[] modInv = new long[N] ;

	static void precompFacts() {
		factorials[0] = invFactorials[0] = 1;
		for (int i = 1; i < factorials.length; i++)
			factorials[i] = mul(factorials[i - 1], i);
		invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
		for (int i = invFactorials.length - 2; i >= 0; i--)
			invFactorials[i] = mul(invFactorials[i + 1], i + 1);
        for ( int i = 0 ; i < modInv.length ; i ++ ) {
            modInv[i] = modInverse(i, mod) ;
        }
	}

	static long nCk(int n, int k) {
		return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
	}

    public static long modDivide(long a, int b) {
        long b_inv = modInv[b];
        return (a * b_inv) % mod;
    }

    public static long modInverse(long b, long mod) {
        return power(b, mod - 2, mod);
    }

    public static long power(long base, long exponent, long mod) {
        long result = 1;
        base = base % mod;

        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                result = (result * base) % mod;
            }

            exponent = exponent >> 1; // Divide exponent by 2
            base = (base * base) % mod; // Change base to base^2
        }
        return result;
    }

}

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

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

Think step by step.

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

1. **Edge Cases**: The code might fail when the numbers are at the edges of the range, i.e., when the missing numbers are close to 1 or close to n.
2. **Special Patterns**: The code might fail when the missing numbers have special patterns, such as being consecutive or having specific bit patterns.
3. **Query Limit**: The code has a query limit of 150, so we should consider inputs that might push the number of queries to the limit.

Given these points, let's create a test input that is small but challenging:

### Test Input:
```plaintext
2
7
10
```

### Explanation:
1. **Test Case 1 (n = 7)**:
   - This test case is small and manageable but still has enough numbers to check the correctness of the algorithm.
   - The missing numbers could be any three distinct numbers from 1 to 7.
   - For example, if the missing numbers are 2, 3, and 5, the remaining numbers in the library would be [1, 1, 4, 4, 6, 6, 7, 7].

2. **Test Case 2 (n = 10)**:
   - This test case is slightly larger, increasing the complexity.
   - The missing numbers could be any three distinct numbers from 1 to 10.
   - For example, if the missing numbers are 1, 8, and 9, the remaining numbers in the library would be [2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 10, 10].

These test cases are designed to:
- Check the correctness of the algorithm for small and manageable inputs.
- Ensure that the algorithm can handle inputs with missing numbers at different positions within the range.
- Verify that the algorithm does not exceed the query limit of 150.

By using these test cases, we can effectively test the buggy code and identify any potential issues.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Library of Magic
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

The Department of Supernatural Phenomena at the Oxenfurt Academy has opened the Library of Magic, which contains the works of the greatest sorcerers of Redania — $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) types of books, numbered from $$$1$$$ to $$$n$$$. Each book's type number is indicated on its spine. Moreover, each type of book is stored in the library in exactly two copies! And you have been appointed as the librarian.

One night, you wake up to a strange noise and see a creature leaving the building through a window. Three thick tomes of different colors were sticking out of the mysterious thief's backpack. Before you start searching for them, you decide to compute the numbers $$$a$$$, $$$b$$$, and $$$c$$$ written on the spines of these books. All three numbers are distinct.

So, you have an unordered set of tomes, which includes one tome with each of the pairwise distinct numbers $$$a$$$, $$$b$$$, and $$$c$$$, and two tomes for all numbers from $$$1$$$ to $$$n$$$, except for $$$a$$$, $$$b$$$, and $$$c$$$. You want to find these values $$$a$$$, $$$b$$$, and $$$c$$$.

Since you are not working in a simple library, but in the Library of Magic, you can only use one spell in the form of a query to check the presence of books in their place:

- "xor l r" — Bitwise XOR query with parameters $$$l$$$ and $$$r$$$. Let $$$k$$$ be the number of such tomes in the library whose numbers are greater than or equal to $$$l$$$ and less than or equal to $$$r$$$. You will receive the result of the computation $$$v_1 \oplus v_2 \oplus ... \oplus v_k$$$, where $$$v_1 ... v_k$$$ are the numbers on the spines of these tomes, and $$$\oplus$$$ denotes the operation of bitwise exclusive OR.

Since your magical abilities as a librarian are severely limited, you can make no more than $$$150$$$ queries.

Input: The first line of input contains an integer $$$t$$$ ($$$1 \le t \le 300$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \leq n \leq 10^{18}$$$) — the number of types of tomes.

Output: None

Examples:
input:
2
6

0

2

3

5

3
output:
xor 1 1

xor 2 2

xor 3 3

xor 4 6

ans 2 3 5

ans 1 2 3

Note: In the first test case, the books in the library after the theft look like this:

Now consider the answers to the queries:

- For the query "xor 1 1", you receive the result $$$1 \oplus 1 = 0$$$. Two tomes satisfy the condition specified in the query — both with the number $$$1$$$.
- For the query "xor 2 2", you receive the result $$$2$$$, as only one tome satisfies the specified condition.
- For the query "xor 3 3", you receive the result $$$3$$$.
- For the query "xor 4 6", you receive the result $$$4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5$$$.

In the second test case, there are only $$$3$$$ types of books, and it is easy to guess that the missing ones have the numbers $$$1$$$, $$$2$$$, and $$$3$$$.

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

public class Main {

    static FastScanner fs = new FastScanner();
    static final Random random = new Random();

    static long inf = 1000000000000000000L ;
    static int N = 2000001 ;
    static int mod = 998244353 ;
    static int query ;

    static long query ( long u , long v ) {
        query ++ ;
        System.out.println( " xor " + u + " " + v ) ;
        long ret = fs.nextLong() ;
        System.out.flush() ;
        return ret ;
    }

    static void printAns ( long a , long b , long c  ) {
        System.out.println ( "ans " + a + " " + b + " " + c );
        System.out.flush() ;
    }

    static long getA (  final long ll , final long rr ) {
        long ans = -1 ;
        long l = ll , r = rr ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(ll, m) ;
            if ( resp == 0 ) {
                l = m + 1 ;
            }else {
                ans = m ;
                r = m - 1 ;
            }
        }
        return ans ;
    }

    static long getC ( final long ll , final long rr ) {
        long ans = -1 ;
        long l = ll , r = rr ;
        while (r >= l) {
            long m = ( r + l ) / 2 ;
            long resp = query(m, rr) ;
            if ( resp == 0 ) {
                r = m - 1 ;
            }else {
                ans = m ;
                l = m + 1 ;
            }
        }
        return ans ;
    }

    static void Test_Case( int t , int hlp ) {
        query = 0 ;
        long n = fs.nextLong() ;
        long a = getA(1 , n) ;
        long c = getC(1 , n) ;
        long b = -1 ;
        long xorAll3 = query ( 1 , n ) ;
        if ( xorAll3 != 0 ) {
            b = xorAll3 ^ a ^ c ;
        }else {
            if ( c != -1 ) {
                a = getA ( 1 , c - 1 ) ;
            }else {
                c = getC(a + 1, n ) ;
            }
            b = xorAll3 ^ a ^ c ;
        }
        printAns(a, b, c);
    }

    public static void main(String[] useCppForCp ) {
        // precompFacts() ;
        int t = 1 ;
        t = fs.nextInt();
        int tt = t ;
        int cnt = 1 ;
        while ( t > 0  ) {
            Test_Case( tt , cnt ) ;
            cnt ++ ;
            t -- ;
        }
    }

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

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

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

    static int add(int a, int b) {
		return (a + b) % mod;
	}

	static long sub(long a, long b) {
		return ((a - b) % mod + mod) % mod;
	}

	static long mul(long a, long b) {
		return ((a % mod ) * (b % mod )) % mod;
	}

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

	static long exp(long base, long exp) {
		if (exp == 0)
			return 1;
		long half = exp(base, exp / 2);
		if (exp % 2 == 0)
			return mul(half, half);
		return mul(half, mul(half, base));
	}

	static long[] factorials = new long[N];
	static long[] invFactorials = new long[N];
    static long[] modInv = new long[N] ;

	static void precompFacts() {
		factorials[0] = invFactorials[0] = 1;
		for (int i = 1; i < factorials.length; i++)
			factorials[i] = mul(factorials[i - 1], i);
		invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
		for (int i = invFactorials.length - 2; i >= 0; i--)
			invFactorials[i] = mul(invFactorials[i + 1], i + 1);
        for ( int i = 0 ; i < modInv.length ; i ++ ) {
            modInv[i] = modInverse(i, mod) ;
        }
	}

	static long nCk(int n, int k) {
		return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
	}

    public static long modDivide(long a, int b) {
        long b_inv = modInv[b];
        return (a * b_inv) % mod;
    }

    public static long modInverse(long b, long mod) {
        return power(b, mod - 2, mod);
    }

    public static long power(long base, long exponent, long mod) {
        long result = 1;
        base = base % mod;

        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                result = (result * base) % mod;
            }

            exponent = exponent >> 1; // Divide exponent by 2
            base = (base * base) % mod; // Change base to base^2
        }
        return 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
2
7
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define ft first
#define sc second
#define all(x) (x).begin(), (x).end()

using namespace std;
using namespace __gnu_pbds;

const int N = 5e5 + 5;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

int a[N];
int is[N];
int parent[N];
int sz[N];

int find_set(int v) {
	if (parent[v] == v) return v;
	else return parent[v] = find_set(parent[v]);
}

void union_set(int v1, int v2) {
	v1 = find_set(v1);
	v2 = find_set(v2);
	if (v1 == v2) {
		return;
	}
	if (sz[v1] > sz[v2]) swap(v1, v2);
	parent[v2] = v1;
	sz[v1] += sz[v2];
}

void solve() {
	int n;
	cin >> n;
	int sq = 200;
	vector <pair <int, pair <int, int> > > vec;
	Tree <pair <int, int> > edg;
	for (int i = 1; i <= n; i ++) {
		sz[i] = 1;
		parent[i] = i;
		is[i] = 0;
		int l, r;
		cin >> l >> r >> a[i];
		vec.push_back({l, {r, i}});
		edg.insert({l, i});
	}
	sort(all(vec));
	vector <pair <int, int> > sp;
	vector <int> nsp;
	for (auto mit : vec) {
		int r = mit.second.first;
		int id = mit.second.second;
		edg.erase({mit.first, id});
		if (edg.order_of_key({r + 1, 0}) < sq) {
			for (auto it : edg) {
				if (it.ft > r) break;
				sp.push_back({it.sc, id});
			}
		}
		else {
			nsp.push_back(id);
		}
	}
	random_shuffle(all(nsp));
	random_shuffle(all(nsp));
	edg.clear();
	for (auto it : vec) {
		edg.insert({it.ft, it.sc.sc});
	}
	for (int i = 0; i < min((int)nsp.size(), sq); i ++) {
		is[nsp[i]] = 1;
	}
	for (auto mit : vec) {
		int r = mit.second.first;
		int id = mit.second.second;
		edg.erase({mit.first, id});
		if (!is[id]) continue;
		for (auto it : edg) {
			if (it.ft > r) break;
			sp.push_back({it.sc, id});
		}
	}
	sort(all(sp), [&](pair <int, int> x, pair <int, int> y) {return abs(a[x.sc] - a[x.ft]) < abs(a[y.ft] - a[y.sc]);});
	int ans = 0;
	for (auto [id1, id2] : sp) {
		if (find_set(id1) != find_set(id2)) {
			ans += abs(a[id1] - a[id2]);
			union_set(id1, id2);
		}
	}
	if (sz[find_set(1)] != n) {
		cout << -1 << '\n';
	}
	else {
		cout << ans << '\n';
	}
}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int test = 1;
	cin >> test;
	while (test --) {
		solve();
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int N = 5e5 + 10;

struct Node {
    int h, id;
    bool f;
    bool operator < (const Node &i) const {
        return h > i.h;
    }
};

struct Edge {
    int u, v, w;
    bool operator < (const Edge &i) const {
        return w < i.w;
    }
} e[2 * N];

int T, n, h[N], l[N], r[N], cnt, a[N], fa[N];
vector<Node> p[N];
set<Node> st;

int Find(int x) {
    return fa[x] ? fa[x] = Find(fa[x]) : x;
}

void Solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> l[i] >> r[i] >> h[i], r[i]++;
        a[++cnt] = l[i], a[++cnt] = r[i];
    }
    sort(a + 1, a + cnt + 1);
    for (int i = 1; i <= n; i++) {
        int x = lower_bound(a + 1, a + cnt + 1, l[i]) - a;
        int y = lower_bound(a + 1, a + cnt + 1, r[i]) - a;
        p[x].push_back({h[i], i, 0});
        p[y].push_back({h[i], i, 1});
    }
    st.insert({0, 0, 0}), st.insert({(int)1e9, n + 1, 0});
    int m = 0;
    for (int i = 1; i <= cnt; i++) {
        sort(p[i].begin(), p[i].end(), [](Node i, Node j) {return i.f > j.f;});
        for (Node k : p[i]) {
            if (!k.f) {
                Node y = *prev(st.lower_bound({k.h, 0, 0}));
                Node x = *st.upper_bound({k.h, 0, 0});
                if (y.id <= n) e[++m] = {y.id, k.id, abs(y.h - k.h)};
                if (x.id) e[++m] = {k.id, x.id, abs(k.h - x.h)};
                st.insert({k.h, k.id, 0});
            } else {
                st.erase({k.h, k.id, 0});
                Node y = *prev(st.lower_bound({k.h, 0, 0}));
                Node x = *st.upper_bound({k.h, 0, 0});
                if (y.id <= n) e[++m] = {y.id, k.id, abs(y.h - k.h)};
                if (x.id) e[++m] = {k.id, x.id, abs(k.h - x.h)};
            }
        }
        p[i].clear(), a[i] = 0;
    }
    st.clear();
    ll ans = 0;
    sort(e + 1, e + m + 1);
    for (int i = 1; i <= m; i++) {
        //cout << e[i].u << ' ' << e[i].v << ' ' << e[i].w << '\n';
        int u = Find(e[i].u), v = Find(e[i].v);
        if (u != v) fa[u] = v, ans += e[i].w;
        e[i] = {0, 0, 0};
    }
    for (int i = 1; i <= n; i++) if (Find(i) != Find(1)) ans = -1;
    cout << ans << '\n';
    for (int i = 1; i <= n; i++) fa[i] = 0;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> T;
    while (T--) Solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Intersected Segments
time_limit_ms: 5000
memory_limit_mb: 512
Description: Turtle just received $n$ segments and a sequence $a_1, a_2, \ldots, a_n$. The $i$-th segment is $[l_i, r_i]$.

Turtle will create an undirected graph $G$. If segment $i$ and segment $j$ intersect, then Turtle will add an undirected edge between $i$ and $j$ with a weight of $|a_i - a_j|$, for every $i \ne j$.

Turtle wants you to calculate the sum of the weights of the edges of the minimum spanning tree of the graph $G$, or report that the graph $G$ has no spanning tree.

We say two segments $[l_1, r_1]$ and $[l_2, r_2]$ intersect if and only if $\max(l_1, l_2) \le \min(r_1, r_2)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^5$) — the number of segments.

The $i$-th of the following $n$ lines contains three integers $l_i, r_i, a_i$ ($1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$) — the $i$-th segment and the $i$-th element of the sequence.

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

Output: For each test case, output a single integer — the sum of the weights of the edges of the minimum spanning tree of the graph $G$. If the graph $G$ has no spanning tree, output $-1$.

Examples:
input:
4
5
1 7 3
2 4 6
3 5 5
6 7 9
3 4 4
5
2 7 3
1 3 6
4 5 5
6 7 9
1 1 4
4
1 4 3
1 2 1
3 4 5
1 4 4
3
1 3 1
2 3 3
4 5 8
output:
9
13
4
-1

Note: In the first test case, the graph $G$ is as follows:

![](CDN_BASE_URL/7b0a32f8f1d54a5dbbaffee1efd8ece1)

One of the minimum spanning trees of $G$ is as follows:

![](CDN_BASE_URL/7a035af2d5227648781d0c24b81c076e)

The sum of the weights of the edges of the minimum spanning tree is $9$.

In the second test case, the graph $G$ is as follows:

![](CDN_BASE_URL/62be880a08fa85071b491d6badd5f58e)

$G$ is already a tree, and the sum of the weights of the tree is $13$.

In the third test case, the graph $G$ is as follows:

![](CDN_BASE_URL/b667aa771d019e3a8a3a4c4372200f03)

In the fourth test case, the graph $G$ is as follows:

![](CDN_BASE_URL/45bc2d052677dbf032ff37fd1d723d83)

It's easy to see that $G$ is not connected, so $G$ has no spanning tree.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
template<typename T>

using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
//#define int long long
using ll = long long;
using ld = long double;
using ull = unsigned long long;

const int mod = 1e9 + 7;
const int MAX = 1e6 + 7;

mt19937_64 rng(time(0));

vector<int> p(MAX);
vector<int> al[MAX];

int get(int x) {
    return (p[x] == x ? x : p[x] = get(p[x]));
}

bool uni(int a, int b) {
    a = get(a);
    b = get(b);
    if (a == b) return false;
    if ((int) (al[a].size()) > (int) (al[b].size())) swap(a, b);
    p[a] = b;
    for (auto i : al[a]) al[b].push_back(i);
    al[a].clear();
    return true;
}

set<pair<int, int>> t[MAX * 4];

void upd(int v, int tl, int tr, int pos, pair<int, int> val, bool type) {
    if (tl == tr) {
        if (type) {
            t[v].insert(val);
        } else {
            t[v].erase(val);
        }
    } else {
        int tm = (tl + tr) / 2;
        if (pos <= tm) {
            upd(v * 2, tl, tm, pos, val, type);
        } else {
            upd(v * 2 + 1, tm + 1, tr, pos, val, type);
        }
        if (type) {
            t[v].insert(val);
        } else {
            t[v].erase(val);
        }
    }
}

pair<int, int> res = {2e9, -1};

void updres(int v, int val) {
    auto it = t[v].upper_bound({val, -1});
    if (it != t[v].end()) {
        int now = it->first;
        res = min(res, {abs(val - now), it->second});
    }
    if (it != t[v].begin()) {
        it--;
        int now = it->first;
        res = min(res, {abs(val - now), it->second});
    }
}

void updres(int v, int tl, int tr, int l, int r, int val) {
    if (tl > r || tr < l) return;
    if (l <= tl && tr <= r) {
        updres(v, val);
        return;
    }
    int tm = (tl + tr) / 2;
    updres(v * 2, tl, tm, l, r, val);
    updres(v * 2 + 1, tm + 1, tr, l, r, val);
}

void solve() {
    int n;
    cin >> n;
    vector<int> L(n + 1), R(n + 1), A(n + 1);
    set<int> st;
    map<int, int> mp;
    int ptr = 0;
    for (int i = 1; i <= n; i++) {
        cin >> L[i] >> R[i] >> A[i];
        st.insert(L[i]);
        st.insert(R[i]);
        p[i] = i;
        al[i].clear();
        al[i] = {i};
    }
    for (auto i : st) mp[i] = ++ptr;
    for (int i = 1; i <= ptr * 4; i++) t[i].clear();
    for (int i = 1; i <= n; i++) {
        L[i] = mp[L[i]];
        R[i] = mp[R[i]];
        upd(1, 1, ptr, L[i], {A[i], i}, 1);
        upd(1, 1, ptr, R[i], {A[i], i}, 1);
    }
    int lf = n - 1;
    ll rss = 0;
    while (lf) {
        for (int i = 1; i <= n; i++) {
            if (!(int) (al[i].size())) continue;
            for (auto j : al[i]) {
                upd(1, 1, ptr, L[j], {A[j], j}, 0);
                upd(1, 1, ptr, R[j], {A[j], j}, 0);
            }
            res = {2e9, -1};
            int nxt = -1;
            for (auto j : al[i]) {
                //                cerr << j << ' ';
                int was = res.first;
                updres(1, 1, ptr, L[j], R[j], A[j]);
                if (was != res.first) {
                    nxt = j;
                }
            }
            //            cerr << '\n';
            if (nxt == -1) {
                cout << -1 << '\n';
                return;
            }
            //            cerr << res.first << ' ' << res.second << ' ' << nxt << '\n';
            if (uni(res.second, nxt)) {
                lf--;
                rss += res.first;
                if (!lf) break;
            }
            for (auto j : al[i]) {
                upd(1, 1, ptr, L[j], {A[j], j}, 1);
                upd(1, 1, ptr, R[j], {A[j], j}, 1);
            }
        }
    }
    cout << rss << '\n';
}

int32_t main() {
#ifdef APPLE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
    // freopen("barrels.in", "r", stdin);
    // freopen("barrels.out", "w", stdout);
#endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        solve();
    }
}

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

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

Think step by step.

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