**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st;

        int t = Integer.parseInt(br.readLine());
        for (int i = 0; i < t; i++) {
            st = new StringTokenizer(br.readLine());
            long x = Long.parseLong(st.nextToken());
            long y = Long.parseLong(st.nextToken());
            long z = Long.parseLong(st.nextToken());
            long k = Long.parseLong(st.nextToken());

            List<Long> factorList = new ArrayList<>();
            factorList.add(1L);
            factorList.add(1L);
            factorList.add(1L);
            List<Long> xyz = new ArrayList<>();
            xyz.add(x);
            xyz.add(y);
            xyz.add(z);
            xyz.sort(Long::compareTo);
            List<Long> resultList = new ArrayList<>();

            long divider = 2;
            boolean isNotAble = false;
            while (k > 1) {
                if (factorList.isEmpty()) {
                    isNotAble = true;
                    break;
                }
                if (k % divider == 0) {
                    List<Long> tempList = new ArrayList<>(factorList);
                    Long min = tempList.getFirst();
                    tempList.removeFirst();
                    tempList.add(min * divider);
                    tempList.sort(Long::compareTo);
                    boolean outsized = false;
                    for (int j = 0; j < tempList.size(); j++) {
                        if (tempList.get(j) > xyz.get(j)) {
                            outsized = true;
                            resultList.add(factorList.getFirst());
                            factorList.removeFirst();
                            xyz.removeFirst();
                            break;
                        }
                    }
                    if (!outsized) {
                        k /= divider;
                        factorList = tempList;
                    }
                } else {
                    divider++;
                }
            }
            if (isNotAble) {
                bw.write(0 + "\n");
            } else {
                resultList.addAll(factorList);
                bw.write((xyz.get(0) - resultList.get(0) + 1) * (xyz.get(1) - resultList.get(1) + 1) * (xyz.get(2) - resultList.get(2) + 1) + "\n");
            }
        }

        bw.flush();
        br.close();
        bw.close();
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the dimensions of the box \( B \) or the volume \( k \) are at their limits.
2. **Non-Standard Volumes**: Volumes that are not straightforward to factorize.
3. **Large Volumes**: Volumes that are large and may cause issues with data type limits.
4. **Special Factorizations**: Volumes that have specific factorizations that might not be handled correctly by the code.

Given the constraints and the nature of the problem, let's create a small-scale test input that includes these aspects.

### Test Input

```plaintext
5
1 1 1 1
2 2 2 7
3 3 3 27
4 4 4 64
10 10 10 1000
```

### Explanation

1. **Test Case 1: \( (1, 1, 1, 1) \)**
   - The box \( B \) has side lengths \( 1 \times 1 \times 1 \).
   - The volume \( k \) is \( 1 \).
   - The only possible secret box \( S \) is \( 1 \times 1 \times 1 \), and it can only be placed in one location: \( (0, 0, 0) \).

2. **Test Case 2: \( (2, 2, 2, 7) \)**
   - The box \( B \) has side lengths \( 2 \times 2 \times 2 \).
   - The volume \( k \) is \( 7 \).
   - There is no way to choose the dimensions of \( S \) such that \( S \) fits in \( B \) and has a volume of \( 7 \).

3. **Test Case 3: \( (3, 3, 3, 27) \)**
   - The box \( B \) has side lengths \( 3 \times 3 \times 3 \).
   - The volume \( k \) is \( 27 \).
   - The only possible secret box \( S \) is \( 3 \times 3 \times 3 \), and it can only be placed in one location: \( (0, 0, 0) \).

4. **Test Case 4: \( (4, 4, 4, 64) \)**
   - The box \( B \) has side lengths \( 4 \times 4 \times 4 \).
   - The volume \( k \) is \( 64 \).
   - The only possible secret box \( S \) is \( 4 \times 4 \times 4 \), and it can only be placed in one location: \( (0, 0, 0) \).

5. **Test Case 5: \( (10, 10, 10, 1000) \)**
   - The box \( B \) has side lengths \( 10 \times 10 \times 10 \).
   - The volume \( k \) is \( 1000 \).
   - The possible secret box \( S \) can be \( 10 \times 10 \times 10 \), and it can only be placed in one location: \( (0, 0, 0) \).

These test cases are designed to cover edge cases, non-standard volumes, and large volumes to ensure the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
#define _USE_MATH_DEFINES
#include<iostream>
#include<time.h>
#include<bits/stdc++.h>
#include<cassert>
#include<numeric>
#include<functional>
#include<cassert>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<fstream>
#include<set>
#include<unordered_set>
#include<unordered_map>
#include<map>
#include<string>
#include<queue>
#include<cmath>

//defines//

#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back
#define einsert emplace_back
#define fr(i,s,n) for(int i = s;i<n;++i)
#define frr(i,s,n) for(int i = s;i>=n;--i)
#define all(a) a.begin(),a.end()
#define unq(v) v.erase(unique(all(v)),v.end())
#define re return
#define fi first
#define se second
#define sz size
#define pii pair<int,int>
#define pdd pair<double,double>
#define ll long long
#define ull unsigned long long
#define pll pair<ll,ll>
#define vi vector<int>
#define vstr vector<string>
#define vpii vector<pii>
#define vpll vector<pll>
#define vll vector<ll>
#define mp make_pair
ll mod = 1e9+7;
ll mod2 = 998244353;
using namespace std;

void solve(){
    ll x,y,z,k;
    cin>>x>>y>>z>>k;
    vll d;
    for (ll i = 1;i*i<=k;++i){
        if (k%i==0){
            d.pb(i);
            if (i!=k/i)
                d.pb(k/i);
        }
    }
    ll ans = 0;
    for (ll a: d)
        for (ll b: d)
            for (ll c: d){
                if (a*b*c!=k) continue;
                if (a>x || b>y || c>z) continue;
                ans = max((x-a+1)*(y-b+1)*(z-c+1),ans);
            }
    cout<<ans<<'\n';
}

int main(){
    fast;
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}

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

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

Think step by step.

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

1. **Small Values for \(x\), \(y\), and \(z\)**: This can test the code's ability to handle small dimensions.
2. **Large Values for \(x\), \(y\), and \(z\)**: This can test the code's performance and memory usage.
3. **Values of \(k\) that are prime or have few divisors**: This can test the code's ability to handle cases where there are fewer possible dimensions for \(S\).
4. **Values of \(k\) that are very large**: This can test the code's ability to handle large integers.
5. **Values of \(k\) that are exactly the product of \(x\), \(y\), and \(z\)**: This can test the code's ability to handle cases where \(S\) fits perfectly in \(B\).
6. **Values of \(k\) that are not factorable into dimensions within the bounds of \(x\), \(y\), and \(z\)**: This can test the code's ability to handle cases where no valid dimensions for \(S\) exist.

Given these considerations, let's generate a small-scale test input that includes a mix of these edge cases:

```plaintext
5
1 1 1 1
2 2 2 8
3 3 3 27
5 5 5 125
10 10 10 1000
```

### Explanation:
1. **1 1 1 1**: This tests the smallest possible box, where \(S\) must be 1x1x1.
2. **2 2 2 8**: This tests a box where \(S\) can be 2x2x2.
3. **3 3 3 27**: This tests a box where \(S\) can be 3x3x3.
4. **5 5 5 125**: This tests a box where \(S\) can be 5x5x5.
5. **10 10 10 1000**: This tests a box where \(S\) can be 10x10x10, but also includes a larger \(k\) value to check for performance and correctness.

This test input should expose any issues with the code's ability to handle different sizes and values of \(k\).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

#ifndef ONLINE_JUDGE
#include "debug.hpp"
#else
#define dbg(...) 23
#endif

#define endl "\n"
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

using ll = long long;
void findFactors(ll k, vector<ll> &factors)
{
    for (int i = 1; i <= sqrt(k); ++i)
    {
        if (k % i == 0)
        {
            factors.push_back(i);
            if (i != k / i)
            {
                factors.push_back(k / i);
            }
        }
    }
}

ll maxLocations(ll x, ll y, ll z, ll k)
{
    vector<ll> factors;
    findFactors(k, factors);

    ll maxPositions = 0;

    for (int i = 0; i < factors.size(); ++i)
    {
        for (int j = 0; j < factors.size(); ++j)
        {
            for (int l = 0; l < factors.size(); ++l)
            {
                ll a = factors[i];
                ll b = factors[j];
                ll c = factors[l];
                if (a * b * c == k && a <= x && b <= y && c <= z)
                {
                    ll positions = (x - a + 1) * (y - b + 1) * (z - c + 1);
                    maxPositions = max(maxPositions, positions);
                }
            }
        }
    }

    return maxPositions;
}
void solve()
{
    ll x, y, z, k;
    cin >> x >> y >> z >> k;
    ll result = maxLocations(x, y, z, k);
    cout << result << endl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cerr << fixed << setprecision(10);
    auto start = chrono::high_resolution_clock::now();
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    auto stop = chrono::high_resolution_clock::now();
    auto duration = chrono::duration_cast<chrono::nanoseconds>(stop - start);
    cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s " << endl;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1 1
2 2 2 8
3 3 3 27
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
t = int(input())
for _ in range(t):
    x,y,z,k = map(int,input().split())
    res = 0
    for a in range(1, x + 1):
        for b in range(1, y + 1):
            c = k // a // b
            if a * b * c != k or c > z:
                continue
            res = max(res, (x - a + 1) * (y - b + 1) * (z - c + 1))
    print(res)

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

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

Think step by step.

Response:
```plaintext
3
1 1 1 1
2 2 2 8
2 2 2 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
for _ in range(int(input())):
    x,y,z,k=map(int,input().split(" "))
    v=x*y*z
    mans=0# max ans
    #xs,ys,zs=0,0,0
    for xs in range(1,x+1):
        for ys in range(1,y+1):
            zs=k/(xs*ys)

            #print(zs)
            if int(zs)==zs: # checking natural

                if zs<=z:
                    #print(xs,ys,zs)
                    zs=int(zs)
                    #print(xs,ys,int(zs)) # printing min dimmension
                    tmans=(x-xs+1)*(y-ys+1)*(z-zs+1)
                    if tmans>=mans:
                        mans=tmans
                       # print(tmans)
    print(mans)

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

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

Think step by step.

Response:
```plaintext
5
1 1 1 1
2 2 2 8
3 3 3 27
4 4 4 64
5 5 5 125
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();

        for(int abc = 0; abc < t; abc++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            int z = sc.nextInt();
            long k = sc.nextLong();
            if(k ==1) System.out.println(x*y*z);
            else{

            ArrayList<ArrayList<Long>> divisors = new ArrayList<ArrayList<Long>>();

            long i = 2;
            int index = 0;

            while(k > 1){
                if(k % i == 0){
                    if(divisors.size() <= index){
                        ArrayList<Long> temp = new ArrayList<Long>();
                        divisors.add(temp);
                    }
                    divisors.get(index).add(i);
                    k /= i;
                }
                else{
                    if(divisors.size() == index+1)
                        index++;
                    i++;
                }
            }

            ArrayList<ArrayList<ArrayList<Long>>> possibilities = new ArrayList<ArrayList<ArrayList<Long>>>();

            for(int j = 0; j < divisors.size(); j++){
                ArrayList<ArrayList<Long>> temp = new ArrayList<ArrayList<Long>>();

                long b = divisors.get(j).get(0);
                int length = divisors.get(j).size();

                for(int a1 = 0; a1 <= length; a1++){
                    for(int a2 = 0; a2 <= length; a2++){
                        for(int a3 = 0; a3 <= length; a3++){
                            if(a1 + a2 + a3 == length){
                                ArrayList<Long> alist = new ArrayList<>();
                                alist.add((long)Math.pow(b,a1));
                                alist.add((long)Math.pow(b,a2));
                                alist.add((long)Math.pow(b,a3));

                                temp.add(alist);
                            }
                        }
                    }
                }

                possibilities.add(temp);
            }

            ArrayList<ArrayList<Long>> ans = new ArrayList<ArrayList<Long>>();

            for(int j = 0; j < possibilities.size(); j++){
                if(ans.size() == 0){
                    for(int d = 0; d < possibilities.get(j).size(); d++){
                        ans.add(possibilities.get(j).get(d));
                    }
                }
                else{
                    int width = ans.size();
                    for(int a1 = 0; a1 < width; a1++){
                        for(int a2 = 0; a2 < possibilities.get(j).size(); a2++){
                            ArrayList<Long> three = new ArrayList<>();

                            three.add(ans.get(a1).get(0)*possibilities.get(j).get(a2).get(0));
                            three.add(ans.get(a1).get(1)*possibilities.get(j).get(a2).get(1));
                            three.add(ans.get(a1).get(2)*possibilities.get(j).get(a2).get(2));
                            ans.add(three);
                        }
                    }

                    for(int a1 = 0; a1 < width; a1++){
                        ans.remove(0);
                    }
                }
            }

            long max = 0;

            for(int d = 0; d < ans.size(); d++){
                ArrayList<Long> answer = ans.get(d);

                if(x >= answer.get(0) && y >= answer.get(1) && z >= answer.get(2)){
                    max = Math.max(max,(x-answer.get(0)+1)*(y-answer.get(1)+1)*(z-answer.get(2)+1));
                }
            }
            System.out.println(max);
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1 1
2000 2000 2000 8000000
10 10 10 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import java.util.*;
public class Maryland{
	public static void main(String[] args){
		//int sb = 7;

		Scanner scan = new Scanner(System.in);
		int testcase=scan.nextInt();

		while (testcase>0){
		long x = scan.nextLong();long y = scan.nextLong();long z= scan.nextLong();long k = scan.nextLong();
		long count=0;
		long max=0;
		for (long a=1L;a<x+1L;a++){
			if (k%a==0L){
				//long kb=k/a;
				for (long b=1L;b<y+1L;b++){
					if (k%b==0L){
						//long kc=kb/b;
						for (long c=1L;c<z+1L;c++){
							if (k%c==0L && a*b*c==k){
								count=(x-a+1L)*(y-b+1L)*(z-c+1L);
								if (count>max){max=count;}
							}
						}
					}
				}
			}
		}System.out.println(max);
		testcase--;
		}

	}
}

**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
2000 2000 2000 4913000000
```


**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 <string>
#include <string.h>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <random>
#include <cstdlib>
#include <numeric>
#include <functional>
#include <queue>
#include <stdlib.h>
#include <map>
#include <assert.h>
using namespace std;
using ll = long long;
ll mod = 1e9 + 7;

void slove() {
    // 2n -> 异或和为0
    // 可二分?

    // 考虑1e18情况下如何查找1,2,3

    // 最高位相同的数中不可能选出三个数异或为0
    ll n;
    cin >> n;
    vector<ll> ans;

    function<ll(ll, ll)> query = [&](ll l, ll r) {
        cout << "xor " << l << " " << r << "\n";
        cout.flush();
        ll ans;
        cin >> ans;
        return ans;
    };

    function<ll(ll, ll ,ll)> test = [&](ll l, ll r, ll pre) {
        if (query(l, l) == pre) return l;

        while (l < r) {
            int mid = (l + r) / 2 + 1;

            if (query(l, mid) != pre)
                l = mid;
            else
                r = mid - 1;
        }
        //cout << l + 1LL << endl;
        return l + 1LL;
    };

    auto out = [&]() {
        cout << "ans ";
        for (int i = 0; i < 3; i++) {
            cout << ans.at(i) << " ";
        }
        cout << "\n";
        return;
    };

    ll las = query(1, n);

    ll pre = 1;

    int tim = 3;

    for (; pre <= n; ) {
        ll bac = pre * 2 - 1;
        bac = min(bac, n);
        auto q = query(pre, bac);

        if (q != 0) {
            auto res = test(pre , bac , q);
            tim--;
            ans.push_back(res);
            if (tim == 1) {
                ans.push_back(las ^ ans[0] ^ ans[1]);
                out();
                return;
            }
        }

        pre *= 2LL;
    }

    cout << "error\n";
    exit(99);
}

int main() {
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);

    int cnt = 1;
    cin >> cnt;

    while (cnt--) {
        slove();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code uses a binary search approach to find the missing books, and it relies on the XOR operation to identify the missing numbers. Here are the steps to create a challenging test input:

1. **Choose a small but challenging `n`**: Since the code uses a binary search, a value of `n` that is a power of 2 or just below a power of 2 can be challenging.
2. **Select missing numbers that are close to each other**: This can make the binary search more difficult to handle.
3. **Ensure the XOR operations are non-trivial**: The XOR results should not be zero for the ranges that the code queries, as this can lead to incorrect assumptions.

Let's choose `n = 7` and the missing books as `2, 3, 5`. This is a small-scale input that should be challenging for the given code.

Here is the test input:

```plaintext
1
7
```

Explanation:
- `1` indicates that there is one test case.
- `7` is the number of types of tomes.

For this input, the library initially has two copies of each book from 1 to 7. After the theft, the books with numbers 2, 3, and 5 are missing. The queries and their expected results are:

- `xor 1 1` should return `0` (since both copies of book 1 are present).
- `xor 2 2` should return `2` (since only one copy of book 2 is present).
- `xor 3 3` should return `3` (since only one copy of book 3 is present).
- `xor 4 7` should return `4 \oplus 6 \oplus 4 \oplus 5 \oplus 6 = 5` (since books 4, 6, and 7 are present with two copies each, and book 5 is present with one copy).

This input should expose the bug in the code, as the binary search and XOR operations will need to handle the missing books correctly.


**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>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<unordered_map>
#include<cmath>
#include<queue>
#include<map>
#include<cstring>
#include<vector>
#include<stack>
#include<set>
#define ll long long
//#define int ll
#define lx (x<<1)
//#define endl '\n'
//#define double long double
#define rx (x<<1|1)
//#define db long double
#define pb push_back
#define mp make_pair
#define pa pair<int,int>
#define mid ((l+r)>>1)
#define mem(a) memset(a,0,sizeof(a))
#define memf(a) memset(a,127,sizeof(a))//锟斤拷锟姐，int锟斤拷锟斤拷锟?
#define memm(a) memset(a,0x3f,sizeof(a))//1e9(锟斤拷锟?
#define memn(a) memset(a,0xcf,sizeof(a))//int锟斤拷小
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ct(a) (__builtin_popcount(a))//ll要锟侥筹拷 (__builtin_popcountll(a))
#define lowbit(i) (i&(-i))
using namespace std;
const int N=4e5+7,M=1e6+4;
const int p=998244353;
const double eps=1e-9,pi=acos(-1.0);
int tt;
ll n,m,k,cnt=0;
struct E{
	ll x=0,y=0,z=0;
	int cnt=0;
};
ll qry(ll xl,ll xr){
	cout<<"xor "<<xl<<" "<<xr<<endl;
	cout.flush();
	cin>>m;
	return m;
}
E find(ll l,ll r){
	E sum;
	ll gong=qry(l,r);
	ll tou=qry(l,l);
	if(tou!=0){
		sum.cnt++;
		sum.x=tou;
	}
	ll al=l,ar=r;
	while(al<=ar){
		ll md=((al+ar)>>1);
		if(qry(al,md)==tou)al=md+1;
		else ar=md-1;
	}
	sum.cnt++;
	if(sum.cnt==2)sum.y=al;
	else sum.x=al;
	ll tmp=al;
	if(sum.cnt>=2)return sum;
	if(al==gong)return sum;
	ar=r;
	while(al<=ar){
		ll md=((al+ar)>>1);
		if(qry(al,md)==tmp)al=md+1;
		else ar=md-1;
	}
	sum.cnt++;
	sum.y=al;
	return sum;
}
void solve(){
	cin>>n;
	E ans;
	ll x=qry(1,n);
	ll sum=x;
	if(x!=0){
		ll l=1,r=n;
		while(l<=r){
			x=qry(1,mid);
			if(x==0)l=mid+1;
			else r=mid-1;
		}
		ans.x=l;
		r=n;
		while(l<=r){
			x=qry(ans.x,mid);
			if(x==ans.x)l=mid+1;
			else r=mid-1;
		}
		ans.y=l;
		ans.z=sum^ans.x^ans.y;
	}
	else{
		ll al,ar;
		for(int i=60;i>=0;i--){
			ll l=(1ll<<i);
			ll r=min(n,l*2-1);
			if(l>r)continue;
			ll now=qry(l,r);
			if(now==0)continue;
			ans=find(l,r);
			if(ans.cnt==2){
				cout<<"ans "<<ans.x<<" "<<ans.y<<" "<<(ans.x^ans.y)<<endl;
				cout.flush();
				return;
			}
			al=1,ar=l-1;
			break;
		}
		ll shou=qry(al,al);
		while(al<=ar){
			ll md=((al+ar)>>1);
			if(qry(al,md)==0)al=md+1;
			else ar=md-1;
		}
		ans.y=al;
		ans.z=ans.x^ans.y;
	}
	cout<<"ans "<<ans.x<<" "<<ans.y<<" "<<ans.z<<endl;
	cout.flush();
}
/*
1
7
1 2 1 2 1 3 1

*/
signed main(){
//	srand(time(0));
//	freopen("D://in.txt","r",stdin);
//	freopen("D://out.txt","w",stdout);
 	IOS;
 	cin>>tt;
// 	scanf("%d",&tt);
//	tt=1;
	while(tt--){
 		solve();
//  		puts("");
	}
	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>

using namespace std;

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

    auto search = [](int prev_step, int step) {
        int l = prev_step - 1, r = step;
        while (l + 1 < r) {
            int m = (l + r) / 2;
            cout << "xor " << prev_step << ' ' << m << '\n';
            cout.flush();
            int res;
            cin >> res;
            if (res != 0) {
                r = m;
            } else {
                l = m;
            }
        }
        return r;
    };

    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int prev_step = 1;
        int step = 1;
        vector<int> answer;
        while (prev_step <= n) {
            cout << "xor " << prev_step << ' ' << step << '\n';
            cout.flush();
            int result;
            cin >> result;
            if (result != 0) {
                auto res = search(prev_step, step);
                answer.emplace_back(res);
                if (res != step) {
                    cout << "xor " << res + 1 << ' ' << step << '\n';
                    cout.flush();
                    int res;
                    cin >> res;
                    if (res != 0) {
                        answer.emplace_back(search(res + 1, step));
                    }
                }
            }
            prev_step = step + 1;
            step = step * 2 + 1;
            step = min(step, n);
        }
        cout << "ans " << answer[0] << ' ' << answer[1] << ' ' << answer[2] << '\n';
        cout.flush();
    }
}

**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:
def main():

    import sys

    def query(l, r):
        print(f"xor {l} {r}")
        sys.stdout.flush()
        response = 0
        while True:
            line = sys.stdin.readline()
            if not line:
                continue
            line = line.strip()
            if not line:
                continue
            if line == '-1':
                return -1  # Query limit exceeded
            try:
                response = int(line)
            except ValueError:
                continue
            break
        return response

    T = int(sys.stdin.readline())

    for _ in range(T):
        n = int(sys.stdin.readline())

        # Query total_xor = a XOR b XOR c
        total_xor = query(1, n)
        if total_xor == -1:
            break

        # Initialize stack for ranges
        stack = [(1, n)]
        missing = set()

        while len(missing) < 3 and stack:
            l, r = stack.pop()
            if l == r:
                s = query(l, r)
                if s == -1:
                    break
                if s != 0:
                    missing.add(l)
            else:
                mid = (l + r) // 2
                S_L = query(l, mid)
                if S_L == -1:
                    break
                S_R = query(mid + 1, r)
                if S_R == -1:
                    break
                if S_L != 0:
                    stack.append((l, mid))
                if S_R != 0:
                    stack.append((mid + 1, r))
            if len(missing) >= 3 or len(stack) == 0:
                break

        if len(missing) != 3:
            # Not enough missing numbers found
            # Output something invalid to fail
            print(f"ans {0} {0} {0}")
            sys.stdout.flush()
            continue

        # Now, verify that their XOR equals total_xor
        found_xor = 0
        answer = sorted(missing)
        for x in answer:
            found_xor ^= x
        if found_xor != total_xor:
            # XOR doesn't match
            print(f"ans {0} {0} {0}")
            sys.stdout.flush()
            continue

        # Valid answer, print in any order
        print(f"ans {answer[0]} {answer[1]} {answer[2]}")
        sys.stdout.flush()

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
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:
def query(l, r):
    print(f'xor {l} {min(r, n)}', flush=True)
    xor = int(input())
    assert xor != -1
    return xor

for _ in range(int(input())):
    n = int(input())
    res = []
    xor = query(1, n)
    cur = 0
    for i in reversed(range(n.bit_length())):
        if len(res) == 3: break

        l = 1 << i
        r = l << 1

        if cur + l > n:
            continue

        q = query(cur + l, cur + r - 1)
        if not res:
            # 3 unknowns
            if xor >> i & 1:
                if xor == q:
                    cur += l
                else:
                    if q != query(q, q):
                        print("here1", flush=True)
                    assert q == query(q, q)
                    res.append(q)
                    xor ^= q
            elif xor != q:
                cur += l
                x = xor ^ q
                if x != query(x, x):
                    # i should be msb of n
                    assert n >> i & 1 == 1
                else:
                    res.append(x)
                    xor ^= x
        elif len(res) == 1:
            # 2 unknowns
            if xor >> i & 1:
                if q != query(q, q):
                    print("here3", flush=True)
                assert q == query(q, q)
                res += [q, xor ^ q]
            elif q:
                cur += l
        else:
            print("here4", flush=True)
            assert False
    if len(res) != 3:
        print("HERE5", len(res), flush=True)
    assert len(res) == 3
    print('ans', *res, flush=True)

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

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

Think step by step.

Response:
```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 sys

def query_xor(l, r):
    print(f"xor {l} {r}")
    sys.stdout.flush()
    return int(sys.stdin.readline())

def binary_search_eq(f, target, low, high):
    result = high +1
    while low <= high:
        mid = (low + high) //2
        val = f(mid)
        if val == target:
            result = mid
            high = mid -1
        else:
            low = mid +1
    return result

def binary_search_neq(f, low, high):
    result = high +1
    while low <= high:
        mid = (low + high) //2
        val = f(mid)
        if val !=0:
            result = mid
            high = mid -1
        else:
            low = mid +1
    return result

def main():
    input = sys.stdin.readline
    t = int(sys.stdin.readline())
    for _ in range(t):
        n_line = ""
        while n_line.strip() == "":
            n_line = sys.stdin.readline()
        n = int(n_line.strip())
        XOR_all = query_xor(1, n)
        def f(m):
            return query_xor(1, m)
        c = binary_search_eq(f, XOR_all, 1, n)
        a_xor_b = XOR_all ^ c
        a = binary_search_neq(f, 1, n)
        b = a_xor_b ^ a
        print(f"ans {a} {b} {c}")
        sys.stdout.flush()

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To 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 should handle edge cases where the numbers are very close to each other or at the boundaries of the range.
2. **Large Values**: Since the range of \( n \) is up to \( 10^{18} \), we should test with a large value to ensure the code can handle it efficiently.
3. **Specific Values**: The values \( a \), \( b \), and \( c \) should be chosen such that the bitwise XOR operations and the binary search logic can be tested thoroughly.

Given these considerations, let's create a test case where \( n \) is a large number and the values \( a \), \( b \), and \( c \) are strategically chosen to maximize the difficulty for the binary search and XOR operations.

### Test Input
```plaintext
1
1000000000000000000
```

### Explanation
- **Test Case Count**: 1 (to keep it simple and focused on a single large \( n \)).
- **Value of \( n \)**: \( 10^{18} \) (the maximum possible value to test the limits of the code).

This test input will ensure that the code is tested with the largest possible input, which can help expose any inefficiencies or logical errors in the binary search and XOR operations. The large value of \( n \) will also challenge the code to handle the queries efficiently within the given constraints.


**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{
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int t = Integer.parseInt(br.readLine());

        while (t-- > 0) {
            long n = Long.parseLong(br.readLine());
            long xorAll = query(1, n, br);
            long[] missingBooks = new long[3];

            // Find the first missing book
            long low = 1, high = n;
            while (low < high) {
                long mid = (low + high) / 2;
                long xorMid = query(low, mid, br);
                if (xorMid != 0) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }
            missingBooks[0] = low;

            // Find the second missing book
            low = 1; high = n;
            while (low < high) {
                long mid = (low + high) / 2;
                long xorMid = query(low, mid, br);
                if (xorMid != missingBooks[0] && xorMid != 0) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }
            missingBooks[1] = low;

            // Find the third missing book
            missingBooks[2] = xorAll ^ missingBooks[0] ^ missingBooks[1];

            pw.println("ans " + missingBooks[0] + " " + missingBooks[1] + " " + missingBooks[2]);
            pw.flush();
        }
    }

    static long query(long l, long r, BufferedReader br) throws IOException {
        System.out.println("xor " + l + " " + r);
        System.out.flush();
        return Long.parseLong(br.readLine());
    }
}

// htSo4QxzqDXVbp8dEUc

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the binary search approach might fail. The primary issue in the buggy code is that it assumes the XOR of a range can directly indicate the presence of a missing book, which is not always accurate, especially when multiple books are missing.

### Step-by-Step Analysis

1. **Understanding the Bug:**
   - The code uses a binary search to find the missing books by checking the XOR of ranges.
   - The first missing book is found correctly if the XOR of the range is non-zero.
   - The second missing book is found by ensuring the XOR of the range is not the first missing book.
   - The third missing book is derived by XORing the first two missing books with the total XOR of all books.

2. **Edge Cases to Consider:**
   - **Small Values of \( n \):** When \( n \) is small, the range of numbers is limited, making it easier to expose the bug.
   - **Specific Ranges:** Ranges where the XOR of the numbers can be zero, even if books are missing.
   - **Symmetric Ranges:** Ranges where the missing books are symmetrically placed, which can confuse the XOR results.

3. **Test Case Selection:**
   - We need a small \( n \) to ensure the test case is valid and within the output length limit.
   - We should choose \( n \) such that the XOR of the range can be zero, even when books are missing.

### Challenging Test Input

Let's choose \( n = 7 \) and the missing books as \( 2, 4, 6 \). This setup ensures that the XOR of certain ranges can be zero, even though books are missing.

### Test Input
```plaintext
1
7
```

### Explanation
- **Input:**
  - \( t = 1 \) (one test case)
  - \( n = 7 \) (7 types of books)

- **Expected Output:**
  - The missing books are \( 2, 4, 6 \).
  - The queries and their expected results:
    - `xor 1 1` -> \( 1 \oplus 1 = 0 \)
    - `xor 2 2` -> \( 2 \)
    - `xor 3 3` -> \( 3 \)
    - `xor 4 4` -> \( 4 \)
    - `xor 5 5` -> \( 5 \)
    - `xor 6 6` -> \( 6 \)
    - `xor 7 7` -> \( 7 \)
    - `xor 1 7` -> \( 1 \oplus 2 \oplus 3 \oplus 4 \oplus 5 \oplus 6 \oplus 7 = 0 \) (since \( 2, 4, 6 \) are missing)

- **Buggy Code Behavior:**
  - The first missing book is found correctly as \( 2 \).
  - The second missing book is found as \( 4 \).
  - The third missing book is calculated as \( 6 \).

This test case should expose the bug in the binary search approach, as the XOR of certain ranges can be zero even when books are missing.


**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{
    private static final int MAX_QUERIES = 150;

    public static void main(String[] args) throws IOException {
        try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
             PrintWriter writer = new PrintWriter(System.out)) {

            int t = Integer.parseInt(reader.readLine());
            for (int i = 0; i < t; i++) {
                long n = Long.parseLong(reader.readLine());
                long[] missingNumbers = findMissingNumbers(reader, writer, n);
                writer.println("ans " + missingNumbers[0] + " " + missingNumbers[1] + " " + missingNumbers[2]);
                writer.flush();
            }
        }
    }

    private static long[] findMissingNumbers(BufferedReader reader, PrintWriter writer, long n) throws IOException {
        long xorAll = queryXOR(reader, writer, 1, n);
        long[] missingNumbers = new long[3];

        if (n == 3) {
            // Handle the case where n is small
            missingNumbers[0] = 1;
            missingNumbers[1] = 2;
            missingNumbers[2] = 3;
            return missingNumbers;
        }

        long low = 1;
        long high = n;
        int index = 0;

        while (low <= high) {
            long mid = (low + high) / 2;
            long xorLeft = queryXOR(reader, writer, low, mid);

            if (xorLeft != 0) {
                // Found a half with an odd number of missing books
                if (mid - low + 1 == 1) {
                    // Found a single missing number
                    missingNumbers[index++] = low;
                    break;
                } else {
                    high = mid;
                }
            } else {
                // Move to the right half
                low = mid + 1;
            }
        }

        // Find the remaining two missing numbers
        for (long i = 1; i <= n; i++) {
            if (i == missingNumbers[0]) {
                continue;
            }
            long xorSingle = queryXOR(reader, writer, i, i);
            if (xorSingle != 0) {
                if (index == 1) {
                    missingNumbers[index++] = i;
                } else {
                    missingNumbers[index++] = i;
                    break;
                }
            }
        }

        // Calculate the third missing number using XOR
        missingNumbers[2] = xorAll ^ missingNumbers[0] ^ missingNumbers[1];
        return missingNumbers;
    }

    private static long queryXOR(BufferedReader reader, PrintWriter writer, long low, long high) throws IOException {
        writer.println("xor " + low + " " + high);
        writer.flush();
        return Long.parseLong(reader.readLine());
    }
}

// zbYXp8JNioeVxLF64nZ

**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

t = int(sys.stdin.readline())
for _ in range(t):
    n = int(sys.stdin.readline())

    print(f"xor 1 {n}", flush=True)
    total_xor = int(sys.stdin.readline())

    missing = []

    def find_missing(l, r, current_xor, missing, depth):
        if len(missing) == 3 or l > r:
            return
        if l == r:
            if current_xor != 0:
                missing.append(l)
            return
        mid = (l + r) // 2
        print(f"xor {l} {mid}", flush=True)
        response = int(sys.stdin.readline())
        if response != 0:
            find_missing(l, mid, response, missing, depth+1)
        print(f"xor {mid+1} {r}", flush=True)
        response2 = int(sys.stdin.readline())
        if response2 != 0:
            find_missing(mid+1, r, response2, missing, depth+1)

    find_missing(1, n, total_xor, missing, 0)
    missing = missing[:3]
    while len(missing) < 3:
        missing.append(0)
    print(f"ans {missing[0]} {missing[1]} {missing[2]}", flush=True)

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

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

Think step by step.

Response:
```plaintext
2
7
4
```


**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:
from bisect import bisect_left
from functools import cache
import sys
import random
r32=random.getrandbits(32)
r64=random.getrandbits(64)
def input(): return sys.stdin.readline().rstrip("\r\n")
def rint(): return int(input())
def rf(): return float(input())
def rs(): return input().split()
def rmap(): return map(int,input().split())
def rmapf(): return map(float,input().split())
def rlist(): return list(map(int,input().split()))
def rlistf(): return list(map(float,input().split()))
def rmap1(): return map(lambda x: int(x)-1, input().split())
def rlist1(): return list(map(lambda x: int(x)-1, input().split()))
def fmin(a,b): return a if a<b else b
def fmax(a,b): return a if a>b else b

@cache
def ask(a,b):
    print("xor",a,b,flush=True)
    return rint()

def get(lo,r):
    def check(j):
        return 1 if ask(lo,j)!=0 else 0
    return bisect_left(range(r+1),1,key=check,lo=lo)

def solve():
    n=rint()
    all=ask(1,n)
    ans=[]
    for bit in range(62-1,-1,-1):
        l=1<<bit
        r=(l<<1)-1
        if l>n:
            continue
        r=min(r,n)
        val=ask(l,r)
        if val !=0:
            a=get(l,r)
            ans.append(a)
            if len(ans)==2:
                ans.append(ans[0]^ans[1]^all)
                break
            if a!=val:
                b=get(a+1,r)
                ans.append(b)
                if len(ans)==2:
                    ans.append(ans[0]^ans[1]^all)
                    break

    ans.sort()
    print("ans",*ans,flush=True)

t=1

t=rint()

for _ in range(t):
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code is designed to handle large values of \( n \) (up to \( 10^{18} \)), but we can start with a smaller scale to ensure the test input is complete and valid.

### Step-by-Step Analysis:

1. **Understanding the Bug**:
   - The code uses a binary search to find the missing books.
   - The key operation is the bitwise XOR, and the code attempts to find the missing books by checking the XOR of ranges.
   - The bug might occur when the ranges are not correctly handled, especially for edge cases.

2. **Edge Cases**:
   - Small values of \( n \) (e.g., \( n = 3 \), \( n = 4 \)).
   - Values of \( n \) that are powers of 2 minus 1 (e.g., \( n = 7 \), \( n = 15 \)).
   - Values of \( n \) that are just above a power of 2 (e.g., \( n = 8 \), \( n = 16 \)).

3. **Challenging Scenarios**:
   - Missing books at the beginning of the range (e.g., \( a = 1 \), \( b = 2 \)).
   - Missing books at the end of the range (e.g., \( a = 6 \), \( b = 7 \)).
   - Missing books in the middle of the range (e.g., \( a = 3 \), \( b = 4 \)).

### Test Input Generation:

Let's choose \( n = 7 \) and set the missing books to \( a = 1 \), \( b = 3 \), and \( c = 5 \). This is a small-scale test that should expose any issues with the range handling and XOR operations.

```plaintext
1
7
```

### Explanation:

- **Test Case**: \( n = 7 \)
  - The library initially has books numbered from 1 to 7, each in two copies.
  - The missing books are 1, 3, and 5.
  - The remaining books are 2, 2, 4, 4, 6, 6, 7, 7.
- **Queries**:
  - The code will perform XOR queries to find the missing books.
  - The expected output should be the numbers 1, 3, and 5 in some order.

This test input is small enough to be manageable and complete, and it should effectively expose any issues in the code's logic for handling ranges and XOR operations.