**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

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

  int t;
  cin >> t;
  while(t--)
  {
    int n,sp = 0;
    cin >> n;
    ll tmp;
    vector<ll> a(n + 1);
    set<ll> ans;
    for(int i = 1;i <= n;i++)
    {
      cin >> a[i];
      // if(a[i] != -1 && a[i] != 1)
      //   sp = i;
    }

    for(int i = 1;i <= n;i++)
    {
      for(int j = 1;j <= n - i + 1;j++)
      {
        tmp = 0;
        for(int k = 0;k < i;k++)
          tmp += a[j + k];
        ans.insert(tmp);
      }
    }

    ans.insert(0);

    n = ans.size();
    cout << n << endl;
    for(auto i : ans)
      cout << i << " ";
    cout << endl;
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

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

using namespace std;
namespace rn = std::ranges;
namespace vw = std::ranges::views;
using ll = long long;
using ull = unsigned long long;

template<typename T>
istream& operator>>( istream& in, vector<T>& v ){
  if ( !v.size() ) {
    size_t n;
    in >> n;
    v.resize(n);
  }
  copy_n( istream_iterator<T>(in), v.size(), v.begin());
  return in;
}

template<typename T>
ostream& operator<<( ostream& out, const vector<T>& v ){
  copy( v.begin(), v.end(), ostream_iterator<T>(out, " "));
  return out;
}

void print( const bool v ){ cout << ( v ? "YES": "NO" ) << endl; }

template<typename T>
void print( const vector<T>& v ){
  print((bool) v.size());
  if ( v.size() )
    cout << v << endl;
}

template<typename T>
void print( const T v ){ cout << v << endl; }

auto solve() {
  int n;
  cin >> n;
  vector<ull> a(n);
  cin >> a;
  vector<ull> b(a.size() - 1);
  rn::generate(b, [&a,i=-1]() mutable { ++i;
      return gcd(a[i], a[i+1]); });
//  cerr << "a: " << a << endl;
//  cerr << "b: " << b << endl;

  const auto strict_greater = [](ull u, ull v){ return u > v; };

  auto f = rn::adjacent_find( b, strict_greater );
  auto g = f;
  for ( int i = 0; g != b.end() && i < 3; ++g, ++i);
  int i = f-b.begin(), k = g-f;

  bool v = f == b.end() || f+2 == b.end();
  if (v) return true;
  for( int j=i; !v && j<i+k; ++j, ++f) {
    // remove a[j], ignore b[j-1]
    ull tmp = b[j];
    if ( j > 0 ) {
      b[j] = gcd(a[j-1], a[j+1]);
      if ( j > 1 && b[j] < b[j-2] )
        continue;
    }
    if ( adjacent_find( f, g, strict_greater ) == g ) {
//      cerr << "remove: " << a[j];
//      cerr << " ignore: " << (j > 0 ? b[j-1] : 0);
//      cerr << endl;
//      cerr << "f: " << *f << " g: " << *g << endl;
      v = true;
    }
    else
      b[j] = tmp;
  }

  return v && adjacent_find( f, b.end(), strict_greater ) == b.end();
}

int main(){
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for( int i = 0; i < t; ++i ){
    print( 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 can cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases:**
   - **Single GCD Decrease:** The simplest case where removing one element should fix the GCD sequence.
   - **Multiple GCD Decreases:** A more complex case where removing one element should not fix the GCD sequence.
   - **Equal GCD Values:** Cases where the GCD values are equal, which might not be handled correctly by the code.
   - **Large Numbers:** Test with large numbers to ensure the GCD calculations are correct and the algorithm handles them efficiently.

2. **Small-Scale Test Input:**
   - We need a small-scale input to ensure it is within the limits and can be easily validated.

3. **Test Input Generation:**
   - We will create a test input with a few test cases, each designed to test different aspects of the algorithm.

Here is the test input:

```plaintext
3
5
12 6 3 4 8
4
2 4 8 1
3
10 10 10
```

### Explanation of Test Cases:
1. **Test Case 1:**
   - Array: [12, 6, 3, 4, 8]
   - GCD sequence: [6, 3, 1, 4]
   - By removing 3, the GCD sequence becomes: [6, 1, 4], which is non-decreasing.
   - Expected Output: YES

2. **Test Case 2:**
   - Array: [2, 4, 8, 1]
   - GCD sequence: [2, 4, 1]
   - Removing any single element does not make the GCD sequence non-decreasing.
   - Expected Output: NO

3. **Test Case 3:**
   - Array: [10, 10, 10]
   - GCD sequence: [10, 10]
   - The GCD sequence is already non-decreasing.
   - Expected Output: YES

This test input is designed to cover a range of scenarios and should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
/*
#include <stdio.h>

int a, b, n, m;

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d%d", &a, &b);
        switch (b % 4)
        {
        case 0:
            m = 1;
            break;
        case 1:
            m = a%10;
            break;
        case 2:
            m = a*a%10;
            break;
        case 3:
            m = a*a*a%10;
            break;
        }
        if (m == 0) printf("10\n");
        else printf("%d\n", m);
    }
    return 0;
}
*/

#include <stdio.h>

int t, n, arr[200100], one[200100], two[200100];

int gcd(int a, int b) {
    int x;
    while (b != 0) {
        x = b;
        b = a % b;
        a = x;
    }
    return a;
}

bool prix() {
    for (int j = 1; j <= n - 1; j++) {
        one[j] = gcd(arr[j], arr[j + 1]);
    }
    one[0] = -1;
    one[n] = 1 << 31 - 1;
    one[n + 1] = 1 << 31 - 1;
    for (int j = 1; j <= n - 2; j++) {
        two[j] = gcd(arr[j], arr[j + 2]);
    }
    two[0] = -1;
    two[n - 1] = 1 << 31 - 1;
    for (int j = 1; j <= n - 2; j++) {
        if (one[j] > one[j + 1]) {
            if (j >= 2) {
                if (one[j - 2] <= two[j - 1] && two[j - 1] <= one[j + 1]) {
                    return true;
                }
            }
            if (one[j - 1] <= two[j] && two[j] <= one[j + 2]) {
                return true;
            }
            if (one[j] <= two[j + 1] && two[j + 1] <= one[j + 3]) {
                return true;
            }
            return false;
        }
    }
    for (int j = 1; j <= n - 2; j++) {
        if (one[j - 2] <= two[j - 1] && two[j - 1] <= one[j + 1]) {
            return true;
        }
        if (one[j - 1] <= two[j] && two[j] <= one[j + 2]) {
            return true;
        }
        if (one[j] <= two[j + 1] && two[j + 1] <= one[j + 3]) {
            return true;
        }
    }
    return false;
}

int main() {
    scanf("%d", &t);
    for (int i = 0; i < t; i++) {
        scanf("%d", &n);
        for (int j = 1; j <= n; j++) {
            scanf("%d", &arr[j]);
        }
        if (prix()) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The code is supposed to determine if it's possible to remove exactly one number from the array so that the GCD sequence is non-decreasing. Let's consider the following scenarios:

1. **Small Array with Non-Decreasing GCD Sequence**: This is a basic case to ensure the code handles simple inputs correctly.
2. **Array with a Single Outlier**: An array where removing one element will make the GCD sequence non-decreasing.
3. **Array with Multiple Outliers**: An array where multiple elements need to be removed to make the GCD sequence non-decreasing.
4. **Array with Repeated Elements**: An array where elements are repeated, which can cause issues with GCD calculations.
5. **Array with Large Elements**: An array with large numbers to test the GCD function's performance and correctness.

Given the constraints, we will generate a small-scale test input that includes these scenarios. Here is a test input that should expose potential bugs in the code:

```plaintext
5
3
10 20 30
4
10 20 30 5
5
10 20 30 15 45
3
10 10 10
5
1000000000 2 1000000000 2 1000000000
```

Explanation:
1. **Test Case 1**: `3 10 20 30` - This is a simple case where the GCD sequence is already non-decreasing.
2. **Test Case 2**: `4 10 20 30 5` - Removing the last element (5) makes the GCD sequence non-decreasing.
3. **Test Case 3**: `5 10 20 30 15 45` - Removing the element 15 makes the GCD sequence non-decreasing.
4. **Test Case 4**: `3 10 10 10` - This is a case with repeated elements, which should be handled correctly.
5. **Test Case 5**: `5 1000000000 2 1000000000 2 1000000000` - This is a case with large elements to test the GCD function's performance.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define IO ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define all(x) x.begin(), (x).end()
#define el '\n'
#define sz(v) (int)(v).size()
#define yes cout << "YES" << el
#define no cout << "NO" << el
#define N (int)1e6 + 4
void solve()
{
    int n;
    cin >> n;
    vector<int> v(n);
    for (int i = 0; i < n; i++)
    {
        cin >> v[i];
    }
    int last = 0, gd, idx = n + 10;
    for (int i = 1; i < n; i++)
    {
        gd = __gcd(v[i], v[i - 1]);
        if (gd < last)
        {
            idx = i;
            break;
        }
        last = gd;
    }
    if (idx >= n - 1)
    {
        yes;
        return;
    }
    int cnt = 0;
    // cout << idx << el;
    // remove v[idx]
    // v[idx-2] v[idx-1] v[idx+1] v[idx+2]
    last = __gcd(v[idx - 1], v[idx + 1]);
    // cout << last << el;
    if (__gcd(v[idx - 1], v[idx - 2]) <= last)
    {
        for (int i = idx + 2; i < n; i++)
        {
            gd = __gcd(v[i], v[i - 1]);
            if (gd < last)
            {
                cnt = 1;
                break;
            }
            last = gd;
        }
        if (cnt == 0)
        {
            yes;
            return;
        }
    }

    // remove v[idx-1]
    // v[idx-2] v[idx] v[idx+1] v[idx+2]
    last = __gcd(v[idx], v[idx - 2]);
    // cout << last << el;
    int tstval = 0;
    if (idx > 2)
        tstval = __gcd(v[idx - 3], v[idx - 2]);
    if (last >= tstval)
    {
        cnt = 0;
        for (int i = idx + 1; i < n; i++)
        {
            gd = __gcd(v[i], v[i - 1]);
            if (gd < last)
            {
                cnt = 1;
                break;
            }
            last = gd;
        }
        if (cnt == 0)
        {
            yes;
            return;
        }
    }
    // remove v[idx-2]
    // v[idx-1] v[idx] v[idx+1] v[idx+2]
    last = 0;
    if (idx > 2)
        last = __gcd(v[idx - 3], v[idx - 1]);
    cnt = 0;
    for (int i = idx; i < n; i++)
    {
        gd = __gcd(v[i], v[i - 1]);
        if (gd < last)
        {
            cnt = 1;
            break;
        }
        last = gd;
    }
    if (cnt == 0)
    {
        yes;
        return;
    }
    no;
}
int main()
{
    IO int tc = 1;
    cin >> tc;
    while (tc--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
10 20 30 40 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
from math import *
from sys import *
from functools import lru_cache
from string import *
from collections import *
from bisect import *
from random import *

for _ in range(int(stdin.readline())):
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().split()))
    b = []
    for i in range(n - 1):
        b.append(gcd(a[i], a[i + 1]))
    cou = 0
    j = 0
    ans = -1
    while (j < len(b) - 1):
        if (b[j] > b[j + 1]):
            cou += 1
            ans = j
            j += 2
        else:
            j += 1
    # print(b,cou,ans)
    if cou > 1:
        print('NO')
    elif cou == 0:
        print('YES')
    else:
        f = 1
        b1 = []
        i = 0
        while i < n - 1:
            if i != ans - 1:
                b1.append(gcd(a[i], a[i + 1]))
                i += 1
            else:
                b1.append(gcd(a[i], a[i + 2]))
                i += 2
        # print(b1)
        for i in range(len(b1) - 1):
            if b1[i] > b1[i + 1]:
                f = 0
                break
        if (f == 0):
            f = 1
            ans += 1
            b2 = []
            i = 0
            while i < n - 1:
                if i != ans - 1:
                    b2.append(gcd(a[i], a[i + 1]))
                    i += 1
                else:
                    if i < n - 2:
                        b2.append(gcd(a[i], a[i + 2]))
                        i += 2
                    else:
                        i += 1
            # print(b2)
            for i in range(len(b2) - 1):
                if b2[i] > b2[i + 1]:
                    f = 0
                    break
            if f:
                print('YES')
            else:
                f = 1
                ans += 1
                b3 = []
                i = 0
                while i < n - 1:
                    if i != ans - 1:
                        b3.append(gcd(a[i], a[i + 1]))
                        i += 1
                    else:
                        if i < n - 2:
                            b3.append(gcd(a[i], a[i + 2]))
                            i += 2
                        else:
                            i += 1
                for i in range(len(b3) - 1):
                    if b3[i] > b3[i + 1]:
                        f = 0
                        break
                if f:
                    print('YES')
                else:
                    print('NO')
        else:
            print('YES')

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

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

Think step by step.

Response:
```plaintext
1
5
12 6 3 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
from math import gcd

def check(d, c, i, used) :
    if i >= len(d):
        return True
    if type(d[i]) == list :
        if (c <= d[i][0] and c <= d[i][1]) and not used :
            return check(d, d[i][1], i+2, True) or check(d, d[i][0], i+1, used)
        elif (c <= d[i][1]) and not used :
            return check(d, d[i][1], i+2, True)
        elif c <= d[i][0]:
            return check(d, d[i][0], i+1, used)
        else :
            return False
    return c <= d[i] or not used
for _ in range(int(input())) :
    n = int(input())
    a = list(map(int, input().split()))
    d = []

    for i in range(n) :
        if i < n-2 :
            d.append([gcd(a[i],a[i+1]), gcd(a[i],a[i+2])])
        elif i < n-1 :
            d.append(gcd(a[i],a[i+1]))
    #print(d)
    print("YES" if check(d, 0, 0, False) else "NO")

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

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

Think step by step.

Response:
```plaintext
3
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
def gcd(a, b):
    if a == b:
        return a
    elif a == 1 or b == 1:
        return 1
    cond = ((a % 2)==0)*2 + ((b % 2)==0)
    if cond == 3:
        return 2 * gcd(a//2, b//2)
    if cond == 2:
        return gcd(a//2, b)
    if cond == 1:
        return gcd(a, b//2)
    if cond == 0 and b > a:
        return gcd((b-a)//2, a)
    elif cond == 0 and b < a:
        return gcd((a-b)//2, b)

def check_opts(arr_, gcd_arr_, opts_):
    res_ = False
    gcd_arr_copy = list(gcd_arr_)
    for q in range(len(opts_)):
        opt = opts_[q]
        #print(opt)
        gcd_arr_ = list(gcd_arr_copy)
        if opt == 0:
            gcd_arr_.pop(0)
        elif opt == n-1:
            gcd_arr_.pop()
        else:
            gcd_arr_[opt-1] = gcd(arr_[opt-1], arr_[opt+1])
            for k in range(opt, n-2):
                gcd_arr_[k] = gcd_arr_[k+1]
            gcd_arr_.pop()
        #print(gcd_arr_)
        k = -2
        for k in range(len(gcd_arr_)-1):
            if gcd_arr_[k+1] < gcd_arr_[k]:
                break
        if k == -2 or (k == len(gcd_arr_)-2 and gcd_arr_[len(gcd_arr_)-1] >= gcd_arr_[len(gcd_arr_)-2]):
            return True
    return res_

for _ in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))

    gcd_arr = []
    opts = []
    yes_flag = False
    no_flag = False

    for i in range(n-1):
        gcd_arr.append(gcd(arr[i], arr[i+1]))
    #print(gcd(12, 6))
    #print(gcd_arr)
    i_max = -1
    i_min = -1
    for i in range(n-2):
        if gcd_arr[i+1] < gcd_arr[i]:
            i_max = i
            break
    if i_max == -1:
        i_max = n-2
        yes_flag = True
    for i in range(n-2, i_max, -1):
        if gcd_arr[i] < gcd_arr[i-1]:
            i_min = i
            break

    if (i_max >= n-3) or (i_min == 1):
        yes_flag = True
    #if (i_min == n-2 and i_max == n-3) or (i_max == 0 and i_min == 1) or (i_min == -1):
        #yes_flag = True

    if i_min - i_max > 2 or i_min - i_max <= 0:
        #if i_min - i_max == 3:
        opts = [i_max, i_max+1, i_max+2, i_min, i_min+1]
        #else:
            #no_flag = True

    #print(i_min)
    #print(i_max)
    if i_min - i_max == 2:
        counter_1 = 0
        counter_2 = 0
        for j in range(i_max, -1, -1):
            if gcd_arr[j] > gcd_arr[i_min]:
                counter_1 += 1
            if counter_1 == 2:
                break
        for j in range(i_min, n-1):
            if arr[j] < arr[i_max]:
                counter_2 += 1
            if counter_2 == 2:
                break
        counter = min(counter_1, counter_2)
        #print(counter_1)
        #print(counter_2)
        if counter == 2:
            opts = [i_max-1, i_max, i_max+1, i_min, i_min+1]
            no_flag = True
        else:
            opts = [i_max, i_max+1, i_min, i_min+1]
        #print(opts)

    else:
        counter_1 = 0
        counter_2 = 0
        for j in range(i_max, -1, -1):
            if gcd_arr[j] > gcd_arr[i_min]:
                counter_1 += 1
            if counter_1 == 3:
                break
        for j in range(i_min, n-1):
            if gcd_arr[j] < gcd_arr[i_max]:
                counter_2 += 1
            if counter_2 == 3:
                break
        counter = min(counter_1, counter_2)
        if counter == 3:
            #opts = [i_max, i_min, i_min+1]
            #no_flag = True
            pass
        else:
            opts = [i_max, i_min, i_min+1]
            '''
            if counter_1 == 2 and counter_2 == 2:
                if gcd_arr[i_max-1] <= gcd_arr[i_min+1]:
                    opts = [i_max, i_min, i_min+1]
                else:
                    opts = [i_max, i_min+1]
                    #opts = [i_max, i_min, i_min+1]
            elif counter_1 == 3 and counter_2 == 2:
                opts = [i_min+1]
                #opts = [i_max, i_min, i_min+1]
            elif counter_1 == 2 and counter_2 == 3:
                opts = [i_max]
                #opts = [i_max, i_min, i_min+1]
            elif counter_1 == 1 and counter_2 == 3:
                opts = [i_max, i_min]
                #opts = [i_max, i_min, i_min+1]
            elif counter_1 == 3 and counter_2 == 1:
                opts = [i_min, i_min+1]
                #opts = [i_max, i_min, i_min+1]
            elif counter_1 + counter_2 == 3:
                opts = [i_max, i_min, i_min+1]
            else:
                opts = [i_max, i_min, i_min+1]
            '''

    res = check_opts(arr, gcd_arr, opts)
    if yes_flag or res:
        print('yes')
    else:
        print('no')

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

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

Think step by step.

Response:
```plaintext
3
5
10 20 5 15 30
4
8 4 2 1
6
12 6 18 9 36 72
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.*;

public class div3 {
    public static int gcd(int a, int b){
        if(b>a){int c = a;
            a = b;
            b = c;
        }
        if(a%b == 0){return b;}
        return gcd(b, a%b);
    }

    public static boolean del(int n, int[] b, int[] a, int i){
        Vector<Integer> v = new Vector<Integer>();
        if(i==0){
            v.add(0);
            v.add(0);
        }
        if(i>0 && i<n-1) {
            v.add(b[i - 1]);
            v.add(gcd(a[i - 1], a[i + 1]));
        }
        if(i<n-2){v.add(b[i+2]);}
        int prev = 0;
        for(int x:v){
            if(x<prev){return false;}
            prev = x;
        }
        return true;
    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        for (int t = input.nextInt(); t > 0; t--) {
            int n = input.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            int c = 0;
            a[0] = input.nextInt();
            b[0] = 0;
            for (int i = 1; i < n; i++) {
                a[i] = input.nextInt();
                b[i] = gcd(a[i], a[i-1]);
            }
            boolean res = true;
            for (int i = 1; i < n; i++) {
                if(b[i]<b[i-1]){
                    c++;
                    if(c>1){res = false; break;}
                    if(i == 2||i==n-1){continue;}
                    else if(del(n, b, a, i)){
                        if(i<n-1) {b[i] = gcd(a[i - 1], a[i + 1]); b[i+1] = b[i];}
                    }
                    else if(i>1 && del(n, b, a, i-1)){
                        if(i>2){b[i] = gcd(a[i-2], a[i]);};
                    }
                    else if(i>2 && del(n,b,a,i-2)){
                        if(i>3){b[i-1] = gcd(a[i-3], a[i-1]);}
                    }
                    else{res = false; break;}
                }
            }
            if(res){System.out.println("YES");}
            else{System.out.println("NO");}
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4
12 6 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

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

public class D {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        int t = sc.nextInt();
        for(int z = 0; z < t; z++)
        {
          int n = sc.nextInt();
          ArrayList<Long> a = new ArrayList<>();
          for (int i = 0; i < n; i++) a.add(sc.nextLong());
          ArrayList<Long> c = (ArrayList<Long>)a.clone();
          ArrayList<Long> d = (ArrayList<Long>)a.clone();
          ArrayList<Long> b = new ArrayList<>();
          int deccnt = 0;
          int ind = -1;
          for (int i = 0; i < n-1; i++)
          {
            b.add(gcd(a.get(i), a.get(i+1)));
            if (b.size() > 1)
            {
                if (b.get(b.size()-1) >= b.get(b.size()-2))
                {

                }
                else{
                    deccnt++;
                    if (ind == -1) ind = i;
                }
            }
          }
          //out.println(b);
          if (deccnt > 2) out.println("NO");
          else if (deccnt == 0) out.println("YES");
          else {
            a.remove(ind);
            deccnt = 0;
          b = new ArrayList<>();
          for (int i = 0; i < n-2; i++)
          {
            b.add(gcd(a.get(i), a.get(i+1)));
            if (b.size() > 1)
            {
                if (b.get(b.size()-1) >= b.get(b.size()-2))
                {

                }
                else{
                    deccnt++;
                }
            }
          }
          //out.println(b);
          if (deccnt > 1) out.println("NO");
          else if (deccnt == 0) out.println("YES");
          else{
            a = c;
            a.remove(ind-1);

            deccnt = 0;
          b = new ArrayList<>();
          for (int i = 0; i < n-2; i++)
          {
            b.add(gcd(a.get(i), a.get(i+1)));
            if (b.size() > 1)
            {
                if (b.get(b.size()-1) >= b.get(b.size()-2))
                {

                }
                else{
                    deccnt++;
                }
            }
          }

          //out.println(b);
          if (deccnt > 1) out.println("NO");
          else if (deccnt == 0) out.println("YES");
          else{
            a = d;
            a.remove(ind+1);

            deccnt = 0;
          b = new ArrayList<>();
          for (int i = 0; i < n-2; i++)
          {
            b.add(gcd(a.get(i), a.get(i+1)));
            if (b.size() > 1)
            {
                if (b.get(b.size()-1) >= b.get(b.size()-2))
                {

                }
                else{
                    deccnt++;
                }
            }
          }

          //out.println(b);
          if (deccnt >= 1) out.println("NO");
          else if (deccnt == 0) out.println("YES");
          }
          }
          }

          //out.println(b);
        }

        out.close();
    }

    // a%x=b%x == x|abs(a-b) == x|gcd(a,b)
    //returns 0 if both zero, other value if one zero
     public static long gcd(long a, long b)
     {
        if (b>a)
        {
            long temp = b;
            b = a;
            a = temp;
        }
        if (b==0) return a;
        long r = b;
         while(a % b != 0)
         {
             r = a % b;
             a = b;
             b = r;
         }
         return r;
     }

     public static long lcm(long a, long b)
     {  //out.println((a*b)/gcd(a,b));
         return (a*b)/gcd(a,b);
     }

    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

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

    }
    //--------------------------------------------------------
}

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

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

Think step by step.

Response:
```plaintext
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
#include  <bits/stdc++.h>
#define int long long
using namespace std;
void solve() {
    int n; cin >> n;
    vector<int> a(n + 1, 0);
    for(int i = 1; i <= n; i ++) cin >> a[i];
    vector<int> b(n + 1, 0);
    for(int i = 1; i < n; i ++) {
        b[i] = __gcd(a[i], a[i + 1]);
    }
    int ok = 0;
    // for(int i = 1; i < n; i ++) {
    //     cout << b[i] <<
    // }
    for(int i = 2; i < n; i ++) {
        if(b[i] < b[i - 1] ) {
            if(i == n - 1 && ok == 0) {
                continue;
            }
            if(ok) {
                cout << "NO" << endl ; return ;
            }
            ok = 1;
            int bx = 1e8;
            if(i >= 3) {
                int t = __gcd(a[i - 2],a[i]);
                if(t <= b[i]) bx = min(b[i], bx);
            }

            int t2 = __gcd(a[i - 1],a[i + 1]);
            if((t2 >= b[i - 2] || i == 2)) bx = min(bx,t2);
            int t3 = 0;
            t3 = __gcd(a[i],a[i + 2]);
            if(t3 >= b[i - 1]) bx = min(t3, bx);
            if(i == 2) {
                bx = min(bx,b[i]);
            }
            b[i] = bx;
            if(b[i] == t3) {
                b[i + 1] = b[i];
            }
            if(bx == 1e8) {
                cout << "NO" << endl ; return ;
            }

        }
    }
    cout << "YES" << endl ;
}
signed main() {
    cin.tie(0), cout.tie(0), ios::sync_with_stdio(false);
    int t; cin >> t;
    while( t -- )  solve();
    system("pause");
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
30 10 15 5 20
```

**Explanation:**
- The array is `a = [30, 10, 15, 5, 20]`.
- The GCD sequence `b` will be `[10, 5, 5, 5]` initially.
- If we remove the element `15` (index 2), the new array will be `[30, 10, 5, 20]`.
- The new GCD sequence will be `[10, 5, 5]`, which is non-decreasing.
- The correct output should be "YES".

This test case is designed to expose the bug in the given buggy code, particularly in how it handles the removal of elements and the recalculations of the GCD sequence. The sequence has a potential issue with the GCD values and the conditions for removing an element, which this test case aims to highlight.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
#include <bits/stdc++.h>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace __gnu_pbds;
#define ll long long int
#define ff first
#define ss second
#define logarr(arr,a,b) for(int z=(a);z<=(b);z++) cout<<(arr[z])<<" ";cout<<endl;
#define all(v) (v).begin(),(v).end()
#define MOD 1000000007
// typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> pbds;
// find_by_order, order_of_key

int main() {
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input1.txt","r",stdin);
    freopen("output1.txt","w",stdout);
    #endif

    int t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        vector<ll> arr(n),b(n-1);
        for(ll i=0;i<n;++i)
            cin>>arr[i];
        for(ll i=0;i<n-1;++i)
            b[i]=__gcd(arr[i],arr[i+1]);
        vector<ll> prefix(n-1),suffix(n-1);
        prefix[0]=1;
        suffix[n-2]=1;
        for(ll i=1;i<n-1;++i){
            if(prefix[i-1] && b[i]>=b[i-1]) prefix[i]=1;
        }
        for(ll i=n-3;i>=0;--i){
            if(suffix[i+1] && b[i]<=b[i+1]) suffix[i]=1;
        }
        // logarr(prefix,0,n-2);
        // logarr(suffix,0,n-2);
        bool flag=false;
        if(suffix[1] || prefix[n-3]){
            cout<<"YES"<<endl;
            flag=true;
            continue;
        }
        for(ll i=1;i<n-1;++i){
            if(i==1){
                if(!suffix[i+1]) continue;
                ll new_gcd=__gcd(arr[i-1],arr[i+1]);
                if(new_gcd<=b[i+1]){
                    // cout<<i<<" ";
                    cout<<"YES"<<endl;
                    flag=true;
                    break;
                }
            }
            else if(i>1 && prefix[i-2] && suffix[i+1]){
                ll new_gcd=__gcd(arr[i-1],arr[i+1]);
                if(b[i-2]<=new_gcd && new_gcd<=b[i+1]){
                    // cout<<i<<" ";
                    cout<<"YES"<<endl;
                    flag=true;
                    break;
                }
            }
        }
        if(!flag) cout<<"NO"<<endl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
def GCD(x, y):
    while(y):
        x, y = y, x % y
    return abs(x)

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    previous_gcd = 0
    previous_item = a[0]
    already_once = False
    result1 = True

    for i in range(1, n):
        current_gcd = GCD(a[i], previous_item)
        # print(current_gcd, end=" ")
        if current_gcd < previous_gcd:
            if already_once:
                result1 = False
                break
            already_once = True

        else:
            previous_gcd = GCD(a[i], previous_item)
            previous_item = a[i]
    # print()
    previous_gcd = float('inf')
    previous_item = a[-1]
    already_once = False
    result2 = True

    for i in range(n-2, -1, -1):
        # print(i)
        current_gcd = GCD(a[i], previous_item)
        # print(a[i], previous_item, current_gcd, previous_gcd)
        if current_gcd > previous_gcd:
            if already_once:
                # print("BREAK",a[i], previous_item, current_gcd)
                result2 = False
                break
            already_once = True

        else:
            previous_gcd = GCD(a[i], previous_item)
            previous_item = a[i]
    # print()

    if result1 or result2:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
5
10 5 15 20 25
4
8 4 2 1
6
12 6 3 12 6 12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
from math import gcd

def solve():
    n = int(input())
    arr = list(map(int,input().split()))
    new_arr= arr
    dic= {}
    ind = []
    for i in range(n-1):
        if i!=n-2:
            dic[i] =[]
            dic[i].append(gcd(arr[i], arr[i+1]))
            dic[i].append(gcd(arr[i], arr[i+2]))
        else:
            dic[n-2] = [gcd(arr[n-2], arr[n-1])]

        if i==0:
            continue

        if dic[i-1][0] > dic[i][0]:
            ind.append(i)
            continue
        #elif dic[i-1][0] <= dic[1][1]:

    #print(ind, dic)

    if len(ind) == 0:
        print("YES")
        return

    if len(ind) > 2:
        print("NO")
        return

    if len(ind) == 2:
        if abs(ind[0]-ind[1]) > 2:
            print("NO")
            return

    if True:
        i = ind[0]
        if len(ind)==1 and i==n-2:
            print("YES")
            return
        elif len(ind)==1 and i==n-3 and dic[i-1][0] <= dic[i][1]:
            print("YES")
            return

        elif dic[i-1][0] <= dic[i][1] and dic[i][1] <= dic[i+2][0]:
            #print("CASE 1")
            print("YES")
            return
        elif len(ind)== 1 and i==1:
            #print("CASE 2")
            print("YES")
            return
        elif i>=2 and dic[i-1][1] <= dic[i+1][0] and dic[i-2][0] <= dic[i-1][1]:
            #print("CASE 3")
            print("YES")
            return
        elif i>=3 and dic[i-3][0] <= dic[i-2][1] and dic[i-2][1] <= dic[i][0]:
            #print("CASE 4")
            print("YES")
            return
        else:
            #print("CAse 5")
            print("NO")
            return

for xyz in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
6
12 6 3 24 18 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

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

public class CF1980D {
    public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    int T = Integer.parseInt(br.readLine());

    while (T-- > 0) {
        solve(br, pw);
    }

    pw.flush();
    pw.close();
    br.close();
    }
    public static void solve(BufferedReader br, PrintWriter pw) throws IOException {
	    int n = Integer.parseInt(br.readLine());
        int[] a = new int[n];
        int[] b = new int[n - 1];

        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }

        for (int i = 1; i < n; i++) {
            b[i - 1] = getHCF(a[i - 1], a[i]);
        }
        // pw.println(Arrays.toString(b));

        boolean[] c = new boolean[n - 1];
        c[0] = true;
        boolean ok = true;
        int id = -1;
        boolean link = false;

        for (int i = 1; i < n - 1; i++) {
            if (b[i] < b[i - 1] && id == -1) {
                id = i;
            }
            else if (b[i] < b[i - 1] && id == i - 1) {
                link = true;
            }
            else if (b[i] < b[i - 1]) {
                ok = false;
                break;
            }
        }
        if (!ok) {
            pw.println("NO");
        }
        else if (id == -1) {
            pw.println("YES");
        }
        else {
            if (link) {
                int v1 = getHCF(a[id - 1], a[id + 1]);
                int v2 = getHCF(a[id], a[id + 2]);
                if ((((id != 1 && b[id - 2] <= v1) || id == 1) && v1 <= b[id + 1]) || ((id != b.length - 2 && b[id + 2] >= v2) || id == 1) && v2 >=  b[id - 1]) {
                    pw.println("YES");
                }
                else {
                    pw.println("NO");
                }
            }
            else {
                if (id == 1 || id == b.length - 1) {
                    pw.println("YES");
                }
                else {
                    int v1 = getHCF(a[id - 2], a[id]);
                    int v2 = getHCF(a[id - 1], a[id + 1]);
                    int v3 = getHCF(a[id], a[id + 2]);
                    if (b[id] >= v1 && ((id > 2 && b[id - 3] <= v1) || id <= 2) || b[id + 1] >= v2 && v2 >= b[id - 2] || b[id - 1] <= v3 && (id < b.length - 2 && v3 <= b[id + 2]|| id >= b.length - 2)) {
                        pw.println("YES");
                    }
                    else {
                        pw.println("NO");
                    }
                }
            }
        }
	}
    public static int getHCF(int a, int b) {
        while (b != 0) {
            int temp = a % b;
            a = b;
            b = temp;
        }
        return a;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
10 5 2 10 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.Scanner;

public class D {
    static Scanner in = new Scanner(System.in);
    public static void main(String[] args) {
        int v = in.nextInt();
        while(v-- > 0) {
            System.out.println(solve());;
        }
    }

    public static String solve() {
        int n = in.nextInt();
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = in.nextInt();
        }
        int[] b = new int[n-1];
        for (int i = 0; i < n - 1; i++) {
            b[i] = gcd(nums[i] , nums[i+1]);
        }
        boolean s = true;
        int i = 1;
        for (; i < n - 2; i++) {
            if(b[i] < b[i-1]) {
                if(!s) return "NO";
                b[i] = gcd(nums[i],nums[i+2]);

                if(b[i] < b[i-1] || (i+2<b.length && b[i+2]<b[i])) {
                    b[i-1] = gcd(nums[i-1] , nums[i+1]);
                    if(b[i-1] > b[i+1] || (i-2>=0 && b[i-2]>b[i-1]) ) {

                        if(i-2<0) return "NO";
                        b[i] = gcd(nums[i],nums[i+1]);
                        b[i-2] = gcd(nums[i-2] , nums[i]);
                        if(b[i-2] > b[i] || (i-3>=0 && b[i-3]>b[i-2])) return "NO";
                    }
                    else i++;
                }
                else i+=2;

                s = false;
            }
        }
        if(i==n-2) {
            if(b[n-2] < b[n-3] && !s) return "NO";
        }
        return "YES";
    }
    public static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
}

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

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

Think step by step.

Response:
```plaintext
3
4
12 6 3 4
5
100 50 2 10 20
6
11 14 8 12 9 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

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

using namespace std;

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

bool is_non_decreasing(const vector<int>& v) {
    for (size_t i = 1; i < v.size(); ++i) {
        if (v[i] < v[i - 1]) {
            return false;
        }
    }
    return true;
}

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

    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }

        if (n == 2) {
            cout << "YES" << endl;
            continue;
        }

        bool canRemove = false;

        for (int i = 0; i < n; ++i) {
            vector<int> temp;
            for (int j = 0; j < n; ++j) {
                if (j != i) {
                    temp.push_back(a[j]);
                }
            }

            vector<int> tempGcdSeq;
            for (int j = 0; j < temp.size() - 1; ++j) {
                tempGcdSeq.push_back(gcd(temp[j], temp[j + 1]));
            }

            if (is_non_decreasing(tempGcdSeq)) {
                canRemove = true;
                break;
            }
        }

        if (canRemove) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The bug in the code is that it attempts to create a new array by removing each element one by one and checking if the resulting GCD sequence is non-decreasing. This approach is inefficient and can lead to incorrect results, especially for larger arrays or specific edge cases.

Let's focus on the following scenarios:
1. Arrays where removing any single element does not result in a non-decreasing GCD sequence.
2. Arrays where the GCD sequence is already non-decreasing, but the code might still fail due to the inefficient approach.
3. Arrays with repeated elements that can lead to unexpected behavior in the GCD calculations.

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

```plaintext
3
3
10 12 3
5
32 16 8 4 2
4
2 4 8 1
```

### Explanation:
1. **Test Case 1:**
   - Array: [10, 12, 3]
   - GCD sequence: [GCD(10, 12) = 2, GCD(12, 3) = 3]
   - Removing any element:
     - Remove 10: [12, 3] -> GCD sequence: [3]
     - Remove 12: [10, 3] -> GCD sequence: [1]
     - Remove 3: [10, 12] -> GCD sequence: [2]
   - The GCD sequence [3] is non-decreasing, so the output should be "YES".

2. **Test Case 2:**
   - Array: [32, 16, 8, 4, 2]
   - GCD sequence: [GCD(32, 16) = 16, GCD(16, 8) = 8, GCD(8, 4) = 4, GCD(4, 2) = 2]
   - Removing any element:
     - Remove 32: [16, 8, 4, 2] -> GCD sequence: [8, 4, 2]
     - Remove 16: [32, 8, 4, 2] -> GCD sequence: [8, 4, 2]
     - Remove 8: [32, 16, 4, 2] -> GCD sequence: [16, 4, 2]
     - Remove 4: [32, 16, 8, 2] -> GCD sequence: [16, 8, 2]
     - Remove 2: [32, 16, 8, 4] -> GCD sequence: [16, 8, 4]
   - The GCD sequence [8, 4, 2] is non-decreasing, so the output should be "YES".

3. **Test Case 3:**
   - Array: [2, 4, 8, 1]
   - GCD sequence: [GCD(2, 4) = 2, GCD(4, 8) = 4, GCD(8, 1) = 1]
   - Removing any element:
     - Remove 2: [4, 8, 1] -> GCD sequence: [4, 1]
     - Remove 4: [2, 8, 1] -> GCD sequence: [2, 1]
     - Remove 8: [2, 4, 1] -> GCD sequence: [2, 1]
     - Remove 1: [2, 4, 8] -> GCD sequence: [2, 4]
   - The GCD sequence [2, 4] is non-decreasing, so the output should be "YES".

This test input is designed to ensure that the code correctly handles the removal of elements and the resulting GCD sequence, exposing any inefficiencies or bugs in the implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD-sequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: GCD (Greatest Common Divisor) of two integers $x$ and $y$ is the maximum integer $z$ by which both $x$ and $y$ are divisible. For example, $GCD(36, 48) = 12$, $GCD(5, 10) = 5$, and $GCD(7,11) = 1$.

Kristina has an array $a$ consisting of exactly $n$ positive integers. She wants to count the GCD of each neighbouring pair of numbers to get a new array $b$, called GCD-sequence.

So, the elements of the GCD-sequence $b$ will be calculated using the formula $b_i = GCD(a_i, a_{i + 1})$ for $1 \le i \le n - 1$.

Determine whether it is possible to remove exactly one number from the array $a$ so that the GCD sequence $b$ is non-decreasing (i.e., $b_i \le b_{i+1}$ is always true).

For example, let Khristina had an array $a$ = [$20, 6, 12, 3, 48, 36$]. If she removes $a_4 = 3$ from it and counts the GCD-sequence of $b$, she gets:

  * $b_1 = GCD(20, 6) = 2$    * $b_2 = GCD(6, 12) = 6$    * $b_3 = GCD(12, 48) = 12$    * $b_4 = GCD(48, 36) = 12$

The resulting GCD sequence $b$ = [$2,6,12,12$] is non-decreasing because $b_1 \le b_2 \le b_3 \le b_4$.

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

This is followed by the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1 \le a_i \le 10^9$) — the elements of array $a$.

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

Output: For each test case, output a single line:

  * "YES" if you can remove exactly one number from the array $a$ so that the GCD-sequence of $b$ is non-decreasing;    * "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will all be recognized as a positive answer).

Examples:
input:
12
6
20 6 12 3 48 36
4
12 6 3 4
3
10 12 3
5
32 16 8 4 2
5
100 50 2 10 20
4
2 4 8 1
10
7 4 6 2 4 5 1 4 2 8
7
5 9 6 8 5 9 2
6
11 14 8 12 9 3
9
5 7 3 10 6 3 12 6 3
3
4 2 4
8
1 6 11 12 6 12 3 6
output:
YES
NO
YES
NO
YES
YES
NO
YES
YES
YES
YES
YES

Note: The first test case is explained in the problem statement.

Buggy Code:
/****************************
Author -> yashdabhade283
****************************/

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

// import libraries
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace chrono;
using namespace __gnu_pbds;
template<class T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long int ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pair<int, int>> vii;
typedef pair<int, int> pi;
typedef pair<double, double> pdd;
typedef map<int, int> mpi;
typedef map<char, int> mpc;
typedef map<ll, ll> mpl;

#define endl "\n"
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define NO {cout<<"NO"<<endl;}
#define YES {cout<<"YES"<<endl;}
#define REP(i,a,b) for(ll i=a;i<b;i++)
#define REV(i,x,y) for(ll i=x;i>=y;i--)
#define inf 0x7fffffff
#define INF 4e18
#define MIN(x) *min_element(all(x))
#define MAX(x) *max_element(all(x))
ll FLOOR(ll n, ll div) {assert(div > 0); return n >= 0 ? n / div : (n - div + 1) / div;}
ll CEIL(ll n, ll div) {assert(div > 0); return n >= 0 ? (n + div - 1) / div : n / div;}

template <typename T>
void op(T inp) {
	cout << inp << endl;
}

#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define dbg(...) ;
#define debug(...) ;
#define crndl ;
#endif

mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());

long long M = 1e9 + 7ll;

ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
void extendgcd(ll a, ll b, ll*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3
ll mminv(ll a, ll b) {ll arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
bool revsort(ll a, ll b) {return a > b;}
ll combination(ll n, ll r, ll m, ll *fact, ll *ifact) {ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}
void google(int t) {cout << "Case #" << t << ": ";}
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
ll getRandomNumber(ll l, ll r) {return uniform_int_distribution<ll>(l, r)(RNG);}  // line 3

bool is_non_decreasing(const vector<int>& seq) {
	for (size_t i = 0; i < seq.size() - 1; ++i) {
		if (seq[i] > seq[i + 1]) {
			return false;
		}
	}
	return true;
}

bool can_remove_one_for_non_decreasing_gcd_sequence(const vector<int>& a) {
	int n = a.size();

	if (n <= 2) {
		return true;
	}

	vector<int> gcd_seq(n - 1);
	for (int i = 0; i < n - 1; ++i) {
		gcd_seq[i] = gcd(a[i], a[i + 1]);
	}

	vector<int> prefix_gcd(n), suffix_gcd(n);
	prefix_gcd[0] = a[0];
	for (int i = 1; i < n; ++i) {
		prefix_gcd[i] = gcd(prefix_gcd[i - 1], a[i]);
	}

	suffix_gcd[n - 1] = a[n - 1];
	for (int i = n - 2; i >= 0; --i) {
		suffix_gcd[i] = gcd(suffix_gcd[i + 1], a[i]);
	}

	auto gcd_with_removal = [&](int idx) -> int {
		if (idx == 0) return suffix_gcd[1];
		if (idx == n - 1) return prefix_gcd[n - 2];
		return gcd(prefix_gcd[idx - 1], suffix_gcd[idx + 1]);
	};

	if (is_non_decreasing(gcd_seq)) return true;
	for (int i = 0; i < n; ++i) {
		vector<int> new_gcd_seq;
		for (int j = 0; j < n - 1; ++j) {
			if (j == i - 1) {
				if (i < n - 1) new_gcd_seq.push_back(gcd(a[j], a[j + 2]));
			} else if (j != i && j + 1 != i) {
				new_gcd_seq.push_back(gcd_seq[j]);
			}
		}
		if (is_non_decreasing(new_gcd_seq)) return true;
	}

	return false;
}

void solve() {
	int n; cin >> n;
	vi v(n);
	REP(i, 0, n) cin >> v[i];
	if (can_remove_one_for_non_decreasing_gcd_sequence(v)) {
		YES;
	}
	else {
		NO;
	}
}

int32_t main() {
	auto begin = std::chrono::high_resolution_clock::now();
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif

	ll t = 1;
	cin >> t;
	while (t--)
		solve();

	auto end = std::chrono::high_resolution_clock::now();
	auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
	cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n";

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases**:
   - Arrays with very small or very large values.
   - Arrays with repeated values.
   - Arrays with prime numbers.
   - Arrays with values that have a common divisor.
   - Arrays with values that are powers of the same base.

2. **Scenarios to Consider**:
   - Arrays where the GCD sequence is already non-decreasing.
   - Arrays where removing any single element does not make the GCD sequence non-decreasing.
   - Arrays where removing a specific element makes the GCD sequence non-decreasing.

3. **Constructing the Test Input**:
   - We need to create an array where the GCD sequence is not non-decreasing, but removing one specific element makes it non-decreasing.
   - We should ensure the test case is small enough to be manageable and valid.

Given these considerations, let's construct a test input:

```plaintext
3
5
10 20 30 40 50
4
10 20 30 40
6
12 24 36 48 60 72
```

### Explanation:
1. **Test Case 1**:
   - Array: [10, 20, 30, 40, 50]
   - GCD sequence: [10, 10, 10, 10]
   - This sequence is already non-decreasing, so the answer should be "YES".

2. **Test Case 2**:
   - Array: [10, 20, 30, 40]
   - GCD sequence: [10, 10, 10]
   - This sequence is already non-decreasing, so the answer should be "YES".

3. **Test Case 3**:
   - Array: [12, 24, 36, 48, 60, 72]
   - GCD sequence: [12, 12, 12, 12, 12]
   - This sequence is already non-decreasing, so the answer should be "YES".

These test cases are designed to cover scenarios where the GCD sequence is already non-decreasing, which should be straightforward to handle. However, to further challenge the code, we can introduce a more complex test case where the GCD sequence is not non-decreasing but can be made non-decreasing by removing one element.

### Additional Challenging Test Case:
```plaintext
1
7
12 24 36 48 60 72 10
```

### Explanation:
- **Array**: [12, 24, 36, 48, 60, 72, 10]
- **GCD sequence**: [12, 12, 12, 12, 12, 2]
- **Removing the last element (10)**:
  - New array: [12, 24, 36, 48, 60, 72]
  - New GCD sequence: [12, 12, 12, 12, 12]
- The new GCD sequence is non-decreasing, so the answer should be "YES".

This test case is more challenging as it requires the algorithm to correctly identify that removing the last element makes the GCD sequence non-decreasing.