**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class BestSubsequence {
    public static void main(String[] args) throws IOException {
        // Initialize BufferedReader for efficient input reading
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Initialize PrintWriter for efficient output writing
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        // Read the number of test cases
        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            // Read the size of the array
            int n = Integer.parseInt(br.readLine().trim());
            // Read the array elements
            String[] tokens = br.readLine().trim().split("\\s+");
            long[] a = new long[n];
            int zeroCount = 0;
            List<Long> nonZeroList = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                a[i] = Long.parseUnsignedLong(tokens[i]);
                if (a[i] == 0) {
                    zeroCount++;
                } else {
                    nonZeroList.add(a[i]);
                }
            }

            // If all elements are zero, the maximum value is the number of zeros
            if (nonZeroList.isEmpty()) {
                out.println(zeroCount);
                continue;
            }

            // Count the frequency of each bit across all non-zero elements
            int[] bitFreq = new int[60];
            for (long num : nonZeroList) {
                for (int b = 0; b < 60; b++) {
                    if (((num >> b) & 1L) == 1L) {
                        bitFreq[b]++;
                    }
                }
            }

            // Sort bits by their frequency in descending order
            Integer[] bitsSorted = new Integer[60];
            for (int i = 0; i < 60; i++) bitsSorted[i] = i;
            Arrays.sort(bitsSorted, (b1, b2) -> Integer.compare(bitFreq[b2], bitFreq[b1]));

            // Limit the number of bits to consider to 15 to keep the computation manageable
            int t_bits = 15;
            t_bits = Math.min(t_bits, 60); // In case there are fewer than 15 bits set
            int[] bitsToConsider = new int[t_bits];
            for (int i = 0; i < t_bits; i++) {
                bitsToConsider[i] = bitsSorted[i];
            }

            // Initialize the maximum value with the count of zeros
            int maxVal = zeroCount;

            // Convert the list of non-zero elements to an array for faster access
            long[] elementMasks = new long[nonZeroList.size()];
            for (int i = 0; i < nonZeroList.size(); i++) {
                elementMasks[i] = nonZeroList.get(i);
            }

            // Enumerate all possible subsets of the top t_bits bits
            int totalSubsets = 1 << t_bits; // 2^t_bits
            for (int m = 0; m < totalSubsets; m++) {
                long mask = 0L;
                for (int i = 0; i < t_bits; i++) {
                    if (((m >> i) & 1) == 1) {
                        mask |= (1L << bitsToConsider[i]);
                    }
                }

                // Count how many non-zero elements have their bits entirely within the mask
                int countS = 0;
                for (long num : elementMasks) {
                    if ((num & ~mask) == 0L) {
                        countS++;
                    }
                }

                // Count the number of set bits in the mask
                int bitsSet = Long.bitCount(mask);

                // Compute the value for this subset
                int val = zeroCount + countS - bitsSet;
                if (val > maxVal) {
                    maxVal = val;
                }
            }

            // Output the maximum value for this test case
            out.println(maxVal);
        }

        // Flush and close the output stream
        out.flush();
        out.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases that can cause the algorithm to fail. The key aspects to consider are:

1. **Edge Case with All Zeros**: This is already handled correctly in the code.
2. **Edge Case with a Single Non-Zero Element**: This is also handled correctly.
3. **Edge Case with Multiple Non-Zero Elements**: This is where the code might have issues. Specifically, the code limits the number of bits to consider to 15, which might not be sufficient for certain inputs.

To challenge the code, we can create a test case where the number of set bits in the bitwise OR of the elements is more than 15, but the optimal subsequence should still be considered. Here's a small-scale test input that fits this criteria:

```plaintext
1
5
1 2 4 8 16
```

Explanation:
- The array is `[1, 2, 4, 8, 16]`.
- The bitwise OR of the array is `31` (binary `11111`), which has 5 set bits.
- The value of the array is `5 - 5 = 0`.
- The optimal subsequence is any single element, which has a value of `1 - 1 = 0`.

This test case will check if the code correctly handles the scenario where the number of set bits is more than 15, but the optimal subsequence is still valid. Since the code limits the number of bits to 15, this test case should expose any issues with that limitation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

public class Main {
    public static void main(String[] args) throws IOException {
        // Use BufferedReader for faster input
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();

        // Read number of test cases
        int t = Integer.parseInt(br.readLine().trim());

        while(t-- > 0){
            // Read n
            int n = Integer.parseInt(br.readLine().trim());

            // Read array elements
            String[] parts = br.readLine().trim().split("\\s+");
            long[] a = new long[n];
            for(int i=0;i<n;i++){
                a[i] = Long.parseLong(parts[i]);
            }

            // Initialize DP map
            HashMap<Long, Integer> dp = new HashMap<>();
            dp.put(0L, 0);

            for(long num : a){
                // Create a copy of current dp to iterate
                HashMap<Long, Integer> temp = new HashMap<>(dp);

                for(Map.Entry<Long, Integer> entry : temp.entrySet()){
                    long currentOR = entry.getKey();
                    int currentSize = entry.getValue();

                    long newOR = currentOR | num;
                    int newSize = currentSize + 1;

                    // Update dp with new_OR and new_size
                    dp.put(newOR, Math.max(dp.getOrDefault(newOR, 0), newSize));
                }

                // Also consider starting a new subsequence with num
                dp.put(num, Math.max(dp.getOrDefault(num, 0), 1));
            }

            // Find the maximum value
            int maxValue = Integer.MIN_VALUE;
            for(Map.Entry<Long, Integer> entry : dp.entrySet()){
                long OR = entry.getKey();
                int size = entry.getValue();
                int bits = Long.bitCount(OR);
                int value = size - bits;
                if(value > maxValue){
                    maxValue = value;
                }
            }

            // The empty subsequence has value 0, so ensure maxValue is at least 0
            maxValue = Math.max(maxValue, 0);

            sb.append(maxValue).append("\n");
        }

        // Print all results at once
        System.out.print(sb.toString());
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases that might cause issues in the algorithm. The key points to consider are:

1. **All elements are zero**: This should result in the maximum value being the size of the array.
2. **All elements are the same non-zero value**: This should result in the maximum value being 1 (since the OR of all elements will be the same as the single element).
3. **Elements with overlapping bits**: This can cause issues with the OR operation and the bit count.
4. **Large numbers**: Ensure that the algorithm handles large numbers correctly (up to \(2^{60}\)).

Given these points, a challenging test input would be one that includes a mix of these scenarios. Here is a small-scale test input that covers these edge cases:

```plaintext
4
3
0 0 0
4
1 1 1 1
5
1 2 4 8 16
8
7 1 48 14 13 8 7 6
```

This test input includes:
1. A case where all elements are zero.
2. A case where all elements are the same non-zero value.
3. A case with elements that have distinct bits.
4. A case with a mix of elements to test the OR and bit count operations.

This should help expose any issues 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: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

Buggy Code:
#pragma GCC optimize("Ofast,no-stack-protector")
#include <bits/stdc++.h>
#define F first
#define S second
#define pb push_back
#define pob pop_back
#define SZ(x) (int)(x.size())
#define all(x) begin(x), end(x)
#ifdef LOCAL
#define HEHE freopen("in.txt", "r", stdin);
#define debug(...) {cerr << #__VA_ARGS__ << " = "; dbg(__VA_ARGS__);}
#else
#define HEHE ios_base::sync_with_stdio(0), cin.tie(0);
#define debug(...) 7122;
#endif
using namespace std;

#define chmax(a, b) (a) = (a) > (b) ? (a) : (b)
#define chmin(a, b) (a) = (a) < (b) ? (a) : (b)

#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
void dbg() { cerr << '\n'; }
template<typename T, typename ...U>
void dbg(T t, U ...u) { cerr << t << ' '; dbg(u...); }

#define ll long long
#define int ll
const int N = 2e5 + 5, INF = 1ll << 50;
#define pll pair<ll, ll>

struct MinCostMaxFlow { // 0-base
  struct Edge {
    ll from, to, cap, flow, cost, rev;
  } *past[N];
  vector<Edge> G[N];
  int inq[N], n, s, t;
  ll dis[N], up[N], pot[N];
  bool BellmanFord() {
    fill_n(dis, n, INF), fill_n(inq, n, 0);
    queue<int> q;
    auto relax = [&](int u, ll d, ll cap, Edge *e) {
      if (cap > 0 && dis[u] > d) {
        dis[u] = d, up[u] = cap, past[u] = e;
        if (!inq[u]) inq[u] = 1, q.push(u);
      }
    };
    relax(s, 0, INF, 0);
    while (!q.empty()) {
      int u = q.front();
      q.pop(), inq[u] = 0;
      for (auto &e : G[u]) {
        ll d2 = dis[u] + e.cost + pot[u] - pot[e.to];
        relax(e.to, d2, min(up[u], e.cap - e.flow), &e);
      }
    }
    return dis[t] != INF;
  }
  bool Dijkstra() {
    fill_n(dis, n, INF);
    priority_queue<pll, vector<pll>, greater<pll>> pq;
    auto relax = [&](int u, ll d, ll cap, Edge *e) {
      if (cap > 0 && dis[u] > d) {
        dis[u] = d, up[u] = cap, past[u] = e;
        pq.push(pll(d, u));
      }
    };
    relax(s, 0, INF, 0);
    while (!pq.empty()) {
      auto [d, u] = pq.top();
      pq.pop();
      if (dis[u] != d) continue;
      for (auto &e : G[u]) {
        ll d2 = dis[u] + e.cost + pot[u] - pot[e.to];
        relax(e.to, d2, min(up[u], e.cap - e.flow), &e);
      }
    }
    return dis[t] != INF;
  }
  void solve(int _s, int _t, ll &flow, ll &cost, bool neg = true) {
    s = _s, t = _t, flow = 0, cost = 0;
    if (neg) BellmanFord(), copy_n(dis, n, pot);
    for (; Dijkstra(); copy_n(dis, n, pot)) {
      for (int i = 0; i < n; ++i) dis[i] += pot[i] - pot[s];
      flow += up[t], cost += up[t] * dis[t];
      for (int i = t; past[i]; i = past[i]->from) {
        auto &e = *past[i];
        e.flow += up[t], G[e.to][e.rev].flow -= up[t];
      }
    }
  }
  void init(int _n) {
    n = _n, fill_n(pot, n, 0);
    for (int i = 0; i < n; ++i) G[i].clear();
  }
  void add_edge(ll a, ll b, ll cap, ll cost) {
    G[a].pb(Edge{a, b, cap, 0, cost, SZ(G[b])});
    G[b].pb(Edge{b, a, 0, 0, -cost, SZ(G[a]) - 1});
  }
} M;
const int inf = 1 << 30;
int n;
int a[105];

int cal(int len) {
    M.init(n + 60 + 3);
    int s = n + 61;
    int t = n + 62;
    M.add_edge(s, 0, len, 0);
    FOR (i, 1, n) {
        M.add_edge(0, i, 1, 0);
    }
    FOR (i, 1, n) {
        if (a[i] == 0) {
            M.add_edge(i, t, inf, 0);
        }
        FOR (j, 0, 59) {
            if (a[i] >> j & 1) {
                M.add_edge(i, n + (j + 1), inf, 1);
            }
        }
    }
    FOR (j, 1, 60) {
        M.add_edge(n + j, t, 1, 0);
    }
    int flow, cost;
    M.solve(s, t, flow, cost, 0);
    return len - cost;
}

void solve() {
    cin >> n;
    FOR (i, 1, n) cin >> a[i];
    int ans = 0;
    FOR (len, 1, n) {
        int tmp = cal(len);
        chmax(ans, tmp);
    }
    cout << ans << '\n';
}

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

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

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

Think step by step.

Response:
```plaintext
4
3
1 2 4
4
1 2 4 8
1
1
8
1 2 4 8 16 32 64 128
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given an integer array $a$ of size $n$.

Let's define the value of the array as its size minus the number of set bits in the bitwise OR of all elements of the array.

For example, for the array $[1, 0, 1, 2]$, the bitwise OR is $3$ (which contains $2$ set bits), and the value of the array is $4-2=2$.

Your task is to calculate the maximum possible value of some subsequence of the given array.

Input: The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$).

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i < 2^{60}$).

Output: For each test case, print the maximum possible value of some subsequence of the given array.

Examples:
input:
4
3
0 0 0
4
1 0 1 2
1
5
8
7 1 48 14 13 8 7 6
output:
3
2
0
3

Note:

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

using namespace std;

#define int long long
#define inf 0x3F3F3F3F3F3F3F3F
const int B = 500;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

vector<int> ss[61];

void _()
{
  int n;
  cin >> n;
  int a[n];
  vector<int> v[n + 1], nw[n + 1];
  for (int &i : a) cin >> i;
  int msk = 0;
  for (int i = 0; i < n; i++)
  {
    msk |= a[i];
  }
  int res = max(0LL, n - (int)__builtin_popcountll(msk));
  for (int i = 0; i < n; i++)
  {
    msk = 0;
    for (int j = 0; j < n; j++)
    {
      if (i == j) continue;
      msk |= a[j];
    }
    res = max(res, (n - 1) - (int)__builtin_popcountll(msk));
  }
  for (int &i : a)
  {
    for (int k = 1; k <= n; k++) nw[k] = v[k];
    nw[1].push_back(i);
    for (int k = 1; k < n; k++)
    {
      for (int &j : v[k]) nw[k + 1].push_back((j | i));
    }
    for (int j = 1; j <= n; j++)
    {
      for (int &k : nw[j]) ss[__builtin_popcountll(k)].push_back(k);
      nw[j].clear();
      for (int k = 0; k < 61; k++)
      {
        for (int &l : ss[k]) nw[j].push_back(l);
        ss[k].clear();
      }
      if (!nw[j].empty()) res = max(res, j - (int)__builtin_popcountll(nw[j][0]));
      if (nw[j].size() > B)
      {
        nw[j].resize(B);
        vector<int> vv;
        while (nw[j].size() > B)
        {
          vv.push_back(nw[j].back());
          nw[j].pop_back();
        }
        shuffle(vv.begin(), vv.end(), rng);
        for (int k = 0; k < min((int)vv.size(), B); k++) nw[j].push_back(vv[k]);
      }
      swap(v[j], nw[j]);
    }
  }
  cout << res << '\n';
}

signed main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) _();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has several potential issues, including:

1. **Bitwise Operations and Set Bits**: The code calculates the number of set bits in the bitwise OR of the elements, which can be tricky for large numbers.
2. **Subsequence Handling**: The code attempts to handle all possible subsequences, which can be inefficient and prone to errors, especially for large inputs.
3. **Randomization and Shuffling**: The code uses randomization and shuffling, which can lead to non-deterministic behavior and potentially miss optimal solutions.

Given these points, a good test input should:

- Contain a mix of small and large numbers to test the bitwise operations.
- Have a structure that makes it difficult to handle all subsequences efficiently.
- Include edge cases like all zeros or all ones to test boundary conditions.

Here is a small-scale test input that meets these criteria:

```plaintext
3
5
1 2 4 8 16
4
0 0 0 0
6
1 3 7 15 31 63
```

Explanation:
1. **Test Case 1**: `[1, 2, 4, 8, 16]`
   - All elements are powers of 2, which means their bitwise OR will have exactly one set bit.
   - The value of the array is `5 - 1 = 4`.
   - This tests the bitwise OR and set bits calculation.

2. **Test Case 2**: `[0, 0, 0, 0]`
   - All elements are zero, so the bitwise OR is zero.
   - The value of the array is `4 - 0 = 4`.
   - This tests the edge case where all elements are zero.

3. **Test Case 3**: `[1, 3, 7, 15, 31, 63]`
   - These numbers are consecutive powers of 2 minus 1, which means their bitwise OR will have increasing set bits.
   - The value of the array is `6 - 6 = 0`.
   - This tests the scenario with increasing set bits and the handling of large numbers.

This input should expose any issues in the bitwise operations, subsequence handling, and edge case management in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include <iostream>
#include<iomanip>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<bitset>
#include<sstream>
#include<queue>
#include<map>
#include<list>
using namespace std;

int main(){
	int t;
	cin>>t;
	int n;
	while(t--){
		cin>>n;
		int b[n]={0};
		vector<int> a;
		bool flag=true;
		for(int i=0;i<n;++i){
			int num;
			cin>>num;
			a.push_back(num);
			b[i]=num;
			if(a[i]==1&&i>=2)
				flag=false;
		}
		sort(b,b+n);
		if(b[n-1]>n)
			flag=false;
		for(int i=0;i<n-1;++i){
			if(a[i]==a[i+1]){
				flag=false;
				break;
			}
			if(a[i]-a[i+1]>1){
				flag=false;
				break;
			}
		}

		if(flag){
			if(flag){
				for(int i=0;i<n;++i){
					auto it = find(a.begin(),a.end(),i+1);
					int index=it-a.begin();
					if(i+2<index){
						flag=false;
						break;
					}
				}

			}

		}

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

	return 0;

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<stack>
#define int long long
using namespace std;
int n;
int a[200500];
int temp;
int flag = 1;

void solve()
{
	cin >> n;
	a[0] = 0;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		if (a[i] < a[i - 1] && a[i-1] - a[i] == 1)
		{
			temp = a[i];
			a[i] = a[i - 1];
			a[i - 1] = temp;
		}
	}
	for (int i = 1; i <= n; i++)
	{
		if (a[i]!=i)
		{
			flag = 0;
			cout << "NO" << endl;
			break;
		}
	}
	if(flag==1)cout << "YES" << endl;
	return;
}
signed main()
{
	int t;
	cin >> t;
	while (t--)solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a few issues:

1. **Incorrect Handling of Swaps**: The code only checks and swaps elements if they are in descending order and differ by exactly 1. This is insufficient because it doesn't handle the full range of possible swaps that could be needed to sort the array.

2. **Single Pass Issue**: The code only performs a single pass through the array, which is not enough to ensure that all elements are in their correct positions.

3. **Flag Initialization**: The flag is initialized once and not reset for each test case, which can lead to incorrect results if the flag is not reset.

Given these issues, a test case that can expose the bug is one where the permutation is not already sorted and requires multiple swaps to be sorted. A permutation where elements are not in the correct order and require multiple swaps to be sorted would be a good candidate.

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

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

Explanation:
- The permutation is `[3, 2, 5, 4, 1]`.
- The correct sorted permutation should be `[1, 2, 3, 4, 5]`.
- The given permutation requires multiple swaps to be sorted, and the buggy code will not handle this correctly because it only performs a single pass and only swaps elements that are in descending order and differ by exactly 1.

This test input should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

void solve() {
	int n;
	cin >> n;
	string ans = "YES";
	for (int i = 1; i < n; i++) {
		int num;
		cin >> num;
		if (num == 1 || num == n) {
			if (abs(num - i) > 1) ans = "no";
		}
		else if (abs(num - i) > 2) ans = "no";
	}
	cout << ans;
}

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

	int t = 1;
	cin >> t;
	while (t--) {
		solve();
		cout << endl;
	}

	return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, let's consider the logic and the conditions that the code checks. The code seems to check the difference between the current index and the value at that index, but it does so in a way that might not correctly handle all cases, especially edge cases.

### Analysis of the Buggy Code:
1. The code checks if `num == 1` or `num == n` and if `abs(num - i) > 1`, it sets `ans` to "no".
2. For other values, it checks if `abs(num - i) > 2`, it sets `ans` to "no".

### Potential Issues:
- The code does not correctly handle the permutation as a whole. It only checks the difference between the value and its index, which is not sufficient to determine if the permutation can be sorted with the given operation.
- The code does not consider the entire permutation and its properties, such as the number of inversions or the parity of the permutation.

### Test Case to Expose the Bug:
To expose the bug, we need a permutation where the code's logic fails. A good candidate is a permutation that has a large gap between the value and its index, but is still sortable by the allowed operations.

### Example Test Case:
Consider the permutation `[3, 1, 2]`. This permutation is sortable by the allowed operations:
- Swap `3` and `1` to get `[1, 3, 2]`.
- Swap `3` and `2` to get `[1, 2, 3]`.

However, the buggy code will incorrectly identify this as unsortable because the difference between the value and its index for `3` is `2` (at index `0`), which is greater than `1`.

### Test Input:
```plaintext
1
3
3 1 2
```

This test input should expose the bug in the buggy code. The expected output is "YES" because the permutation can be sorted by the allowed operations, but the buggy code will output "no".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [0] * size

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        xroot = self.find(x)
        yroot = self.find(y)

        if xroot == yroot:
            return
        if self.rank[xroot] < self.rank[yroot]:
            self.parent[xroot] = yroot
        else:
            self.parent[yroot] = xroot
            if self.rank[xroot] == self.rank[yroot]:
                self.rank[xroot] += 1

def main():
    import sys
    lines = sys.stdin.read().splitlines()
    t = int(lines[0])
    index = 1  # Start from the first test case
    for _ in range(t):
        n = int(lines[index])
        p = list(map(int, lines[index + 1].split()))
        uf = UnionFind(n)
        # Build the graph based on allowed swaps
        for i in range(n - 1):
            if abs(p[i] - p[i + 1]) == 1:
                uf.union(i, i + 1)
        # Check if each element can reach its target position
        possible = True
        for k in range(1, n + 1):
            current_pos = p.index(k)
            target_pos = k - 1
            if uf.find(current_pos) != uf.find(target_pos):
                possible = False
                break
        print("YES" if possible else "NO")
        index += 2  # Move to the next test case

if __name__ == "__main__":
    main()

# cNFLEsvYMnW8KJXA7iU

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
testcases = int(input())

for _ in range(testcases):
    n = int(input())
    nums = list(map(int, input().split()))
    flag = True
    i = 0
    while i < n:
        num = nums[i]
        idx = i + 1
        if idx == nums:
            pass
        elif i < n - 1 and nums[i + 1] == idx and num == idx + 1:
            i += 2
        elif i < n - 1:
            flag = False
        i += 1
    print("YES" if flag 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
1
1
2
2 1
4
3 1 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class main {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static int n;

    public static void main(String[] args) {
        int t = in.nextInt();
        while (t-- > 0) {
            n = in.nextInt();
            ArrayList<Integer> arr = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                arr.add(in.nextInt());
            }
            boolean flag = false;
            for (int i = 0; i < n; i++) {
                int l = arr.get(i);
                if (l - i - 1 > 1) {
                    flag = true;
                }
            }
            if (flag) {
                pw.println("NO");
            } else {
                pw.println("YES");
            }
        }
        pw.close();
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                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:
To expose the bug in the given buggy code, we need to focus on the condition that the code checks: `if (l - i - 1 > 1)`. This condition is intended to check if the difference between the value of the permutation and its index (plus one) is greater than 1. However, this condition is not correctly capturing the requirement of the problem, which is to ensure that the permutation can be sorted using the allowed operations.

### Analysis of the Bug:
The buggy code incorrectly assumes that if the difference between the value and its index is greater than 1, it is impossible to sort the permutation. This is not always true. For example, the permutation `[3, 1, 2]` can be sorted using the allowed operations, but the condition `3 - 0 - 1 > 1` would incorrectly flag this as impossible.

### Test Input to Expose the Bug:
We need a permutation where the difference between the value and its index is greater than 1, but the permutation can still be sorted using the allowed operations. A good example is the permutation `[3, 1, 2]`.

### Test Input:
```plaintext
1
3
3 1 2
```

### Explanation:
- The permutation `[3, 1, 2]` has the following properties:
  - `3 - 0 - 1 = 2`, which is greater than 1.
  - However, this permutation can be sorted by performing the following operations:
    - Swap `3` and `1` at index 1: `[1, 3, 2]`.
    - Swap `3` and `2` at index 2: `[1, 2, 3]`.

The correct output for this test case should be "YES", but the buggy code will output "NO" due to the incorrect condition. This test input will expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import java.util.*;

public class B_Penchick_and_Satay_Sticks {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            // Your code here
            ////Bunny is topper
            int n=sc.nextInt();
            int arr[]=new int[n+1];
            for(int i=1;i<=n;i++){
                arr[i]=sc.nextInt();
            }
            boolean pos=true;
            int index=1;
            while(index<=n&&arr[index]==index){
                index++;
            }
            for(int i=index;i<n;i+=2){
                 if((arr[i]==i&&arr[i+1]==i+1)||(arr[i]==i+1&&arr[i+1]==i)){
                    continue;
                 }else{
                    pos=false;
                    break;
                 }
            }
            if(pos){
                System.out.println("YES");
            }else{
                System.out.println("NO");
            }
        }
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the logic and potential edge cases where the code might fail. The code seems to check if the permutation can be sorted by only swapping neighboring elements that differ by exactly 1. However, it has a flaw in its logic for checking the feasibility of sorting.

### Key Points to Consider:
1. **Initial Sorted Segment**: The code skips over the initial segment that is already sorted. This is done by the loop `while(index<=n&&arr[index]==index)`.
2. **Checking Pairs**: After skipping the initial sorted segment, the code checks pairs of elements starting from the first unsorted element. It checks if the pairs are either `(i, i+1)` or `(i+1, i)`. If any pair does not satisfy this condition, it concludes that the permutation cannot be sorted.

### Potential Bugs:
- **Incorrect Pair Checking**: The code only checks pairs of elements in steps of 2. This means it might miss cases where a pair that should be `(i, i+1)` or `(i+1, i)` is not in the correct position.
- **Initial Sorted Segment**: The code might incorrectly skip over the initial sorted segment if it is not properly handled.

### Challenging Test Input:
To expose the bug, we can create a test case where the initial segment is sorted, but the remaining segment has a pattern that the code might incorrectly handle. Specifically, we can create a permutation where the first part is sorted, and the second part has a pattern that the code might not correctly check.

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

### Explanation:
1. **First Test Case**: `1 2 3 5 4 6`
   - The initial segment `1 2 3` is already sorted.
   - The remaining segment `5 4 6` has a pair `5 4` that should be `4 5` to be sorted. The code might incorrectly skip over this segment because it only checks pairs in steps of 2.
   - Expected output: `NO` (but the code might output `YES`).

2. **Second Test Case**: `1 2 4 3 5`
   - The initial segment `1 2` is already sorted.
   - The remaining segment `4 3 5` has a pair `4 3` that should be `3 4` to be sorted.
   - Expected output: `NO` (but the code might output `YES`).

This test input should help expose the bug in the code by challenging its logic for checking pairs and initial sorted segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include<iostream>
#include<vector>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int> arr(n);
        for(int i=0;i<n;i++){
            cin>>arr[i];
        }
        bool flag=true;
        for(int i=0;i<n;i++){
            if(arr[i]==i+1){
                continue;
            }
            int m=arr[i];
            for(int j=i+1;j<m-1;j++){
                if(abs(arr[i]-arr[i+1])<=1){
                    swap(arr[i], arr[i+1]);
                }
                else{
                    flag=false;
                    break;
                }
            }
            if(!flag){
                break;
            }
        }
        if(flag){
            cout<<"YES\n";
        }
        else{
            cout<<"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:
```plaintext
3
5
5 4 3 2 1
6
6 5 4 3 2 1
7
7 6 5 4 3 2 1
```

### Explanation:
1. **Test Case 1:**
   - Permutation: `[5, 4, 3, 2, 1]`
   - This permutation is in reverse order. The buggy code may fail to correctly determine that it is impossible to sort this permutation by the given operation.

2. **Test Case 2:**
   - Permutation: `[6, 5, 4, 3, 2, 1]`
   - Similar to the first test case, this permutation is also in reverse order. The length is increased to ensure that the algorithm handles larger inputs correctly.

3. **Test Case 3:**
   - Permutation: `[7, 6, 5, 4, 3, 2, 1]`
   - This is another reverse-ordered permutation, but with a longer length. This test case is designed to expose any issues with the algorithm's handling of longer permutations.

These test cases are designed to challenge the buggy code by providing permutations that are in reverse order, which should be impossible to sort using the given operation. The buggy code may incorrectly output "YES" for these cases, thus exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include <bits/stdc++.h>
#define for1(i,a,b) for( int (i)=(a); (i)<=(b); ++(i))
#define for2(i,a,b) for( int (i)=(a); (i)>=(b); --(i))
#define madd(a,b) (a)=((a)+(b)+mod)%mod
#define fi first
#define se second
#define p_f push_front
#define p_b push_back
#define o_f pop_front
#define o_b pop_back
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 10007;
const int N = 105;
const int M = 105;
const int IINF = 0x3f3f3f3f;
const LL LINF = 1e18;
int T;
int n,a[200005];
signed main(){
	time_t start=clock();
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin>>T;
	while(T--){
		cin>>n;
		int flag=0;
		for1(i,1,n){
			cin>>a[i];
		}
		for1(i,1,n-1){
			if(a[i]!=i){
				if(flag){
					flag++;
					break;
				}
				if(a[i]==i+1&&a[i+1]==i){
					flag=1,i++;
				}else{
					flag=2;
					break;
				}
			}
		}
		if(flag>1) cout<<"NO\n";
		else cout<<"YES\n";
	}
	time_t duration=clock()-start;
	cerr<<"\ntime="<<duration;
    return 0;
}
/*
4 1 3 2
4 1 2 3
3 1 2 4
2 1 3 4
1 2 3 4
*/

**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 scenarios that the code might handle incorrectly. The key issues in the buggy code include:

1. **Incorrect handling of swaps**: The code incorrectly assumes that if `a[i] == i + 1` and `a[i + 1] == i`, it can simply swap and continue. This is not always the case, especially if there are multiple such pairs.
2. **Flag logic**: The flag logic is flawed. It increments the flag in a way that doesn't correctly capture the state of the permutation.

Given these issues, a good test case would be one where the permutation has multiple pairs that need to be swapped, but the order of swaps matters. Here's a small-scale test input that should expose the bug:

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

### Explanation:
- **Permutation**: `3 4 2 5 1`
- **Expected Output**: `NO`
- **Reasoning**:
  - The permutation has multiple pairs that need to be swapped: `(3, 4)` and `(2, 5)`.
  - However, the code's logic might incorrectly handle these pairs, leading to an incorrect conclusion that the permutation can be sorted.

This test input should help expose the bug in the code's logic for handling swaps and flags.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath> // For abs()
using namespace std;

// Function to check if a vector is sorted
bool isSorted(const vector<int>& arr) {
    for (int i = 0; i < arr.size() - 1; ++i) {
        if (arr[i] > arr[i + 1]) {
            return false;
        }
    }
    return true;
}

int main() {
    int t; // Number of test cases
    cin >> t;

    while (t--) {
        int n;
        cin >> n;
        vector<int> arr(n);

        // Input the array
        for (int i = 0; i < n; i++) {
            cin >> arr[i];
        }

        bool sortedPossible = false;

        // Try swapping adjacent elements with a difference of 1
        for (int i = 0; i < n - 1; i++) {
            if (abs(arr[i] - arr[i + 1]) == 1) {
                // Swap adjacent elements
                swap(arr[i], arr[i + 1]);

                // Check if the array becomes sorted
                if (isSorted(arr)) {
                    sortedPossible = true;
                    break;
                }

                // Swap back to restore original array
               // swap(arr[i], arr[i + 1]);
            }
        }

        // Output result
        if (sortedPossible) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the array is already sorted or nearly sorted.
2. **Infeasible Cases**: Test cases where it is impossible to sort the array using the given operation.
3. **Special Cases**: Test cases with small values of `n` to ensure the input is small-scale.

The buggy code only attempts to swap adjacent elements once and checks if the array becomes sorted. This approach is flawed because it does not consider the possibility of multiple swaps being required to sort the array.

### Test Input
Let's create a test case where the array is not sorted but can be sorted by multiple swaps, and another test case where it is impossible to sort the array.

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

### Explanation
1. **Test Case 1**: `5 2 1 4 3 5`
   - This array can be sorted by performing multiple swaps.
   - Possible sequence of swaps:
     - Swap `2` and `1`: `[1, 2, 4, 3, 5]`
     - Swap `4` and `3`: `[1, 2, 3, 4, 5]`
   - The array can be sorted, so the expected output is `YES`.

2. **Test Case 2**: `4 4 2 3 1`
   - This array cannot be sorted by performing the given operation.
   - Possible sequence of swaps:
     - Swap `2` and `3`: `[4, 3, 2, 1]`
     - Swap `3` and `4`: `[3, 4, 2, 1]`
     - Swap `2` and `1`: `[3, 4, 1, 2]`
   - The array remains unsorted, so the expected output is `NO`.

This test input should effectively expose the bug in the buggy code by requiring multiple swaps and by including a case where sorting is impossible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    a=[0]+list(map(int,input().split()))
    f=True
    for i in range(1,n+1):
        if a[i]==i:
            continue
        elif i==a[i+1]:
            a[i],a[i+1]=a[i+1],a[i]
        else:
            f=False
            break
    if f:
        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
1
1
2
2 1
5
5 3 4 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import sys

def BIT_add(x, BIT, n):
    while x <=n:
        BIT[x] +=1
        x += x & -x

def BIT_query(x, BIT):
    result =0
    while x >0:
        result += BIT[x]
        x -=x & -x
    return result

def main():
    input = sys.stdin.read().split()
    it = iter(input)
    t = int(next(it))
    for _ in range(t):
        n = int(next(it))
        p = [int(next(it)) for _ in range(n)]
        BIT = [0]*(n+2)
        total =0
        flag = False
        for val in p:
            count = total - BIT_query(val +1, BIT)
            if count >1:
                flag = True
                print("NO")
                break
            total +=1
            BIT_add(val, BIT, n)
        if not flag:
            print("YES")

if __name__ == '__main__':
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import java.io.BufferedReader;

import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

/*
 *   Author : RTE_ELMENATOR
 *
 *   "No one will carry you.
 *    Work harder, and turn your dream into reality."
 *
 *
 */

public class CpWala {

    // Let's Explore
	private static  void solverA() {
		int n = Reading.readInt();
		int[] arr= Reading.readIntArray();
		boolean res = true;
		for(int i = 0 ; i< n -1; i++) {
			if( arr[i] == i + 1 ) continue;
			if( arr[i + 1] == i + 1 && Math.abs(arr[i] - arr[i+ 1] )== 1) {
			 arr[i +1 ] = arr[ i ];
			 continue;
			}
			break;
		}
		if( arr[n - 1] == n ) {
			out.println("YES");
		}
		else out.println("NO");

	}

	static PrintWriter out=new PrintWriter(System.out);
     public static void main(String[] arg) {
       int testNo =  Reading.readInt();
       while( testNo -- > 0 ) {
    	   solverA();
       }
       out.flush();
       out.close();
     }
     private static long getLowerPower(long rem,int k) {
    	 // k^x <= n
    	 long res=1;
     	/*
     	 * 16 4
     	 * res = 1 * 4
     	 *         4
     	 *           16 <=16
     	 *             64
     	 *               => 64/4=>16
     	 *               16/16 =>1
     	 *               0 +
     	 */
     	while(res<=rem) {
     		res=res*k;

     	}
     	res/=k;
     	return res;
     }
     private static long findMinoOpern(int n,int k) {
    	if(k==1) return  n;
    	long res=getLowerPower(n,k);
    	int count=0;
    	long rem=n;
    	while(rem>=k) {
    	long powcan=(rem/res);
    	  rem=(rem%res);
    	count+=powcan;
    	res=getLowerPower(rem,k);

    	}
    	return rem%k+count;

     }
     private static int[] kmp(String string) {
    	 int[] res=new int[string.length()];
    	 int j=0;
    	 // a b r a k a d a b r a b r a k a d a b r a
//               j                   i
//           0 0 0 1 0 1 0 1 2 3 1 2
    	 for(int indx=1;indx<string.length();) {
    		 if(string.charAt(indx)==string.charAt(j)) {
    			 if(j==0) {
    				 res[indx]=1;
    			 }
    			 else
    			 res[indx]=j+1;
    			 j++;
    			 indx++;
    		 }
    		 else {
    			 if(j==0) {
    				 indx++;
    				 continue;
    			 }
    			 j=res[j-1];

    		 }

    	 }
    	 return res;
     }

}

class Reading {
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    public Reading() {
        reader = new BufferedReader(new InputStreamReader(System.in));
    }

    public static int readInt() {
        try {
            return Integer.parseInt(reader.readLine().trim());
        } catch (Exception e) {
            return -1;
        }
    }

    public static int[] readIntArray() {
        try {
            return Arrays.stream(reader.readLine().trim().split("\\s+"))
                         .mapToInt(Integer::parseInt)
                         .toArray();
        } catch (Exception e) {
            return null;
        }
    }

    public static long[] readLongArray() {
        try {
            return Arrays.stream(reader.readLine().trim().split("\\s+"))
                         .mapToLong(Long::parseLong)
                         .toArray();
        } catch (Exception e) {
            return null;
        }
    }

    public static String readString() {
        try {
            return reader.readLine();
        } catch (Exception e) {
            return null;
        }
    }

    public static char[] readToCharArray() {
        try {
            return reader.readLine().replaceAll("\\s+", "").toCharArray();
        } catch (Exception e) {
            return null;
        }
    }
}

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

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

Think step by step.

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

### Analysis of the Buggy Code

1. **Initialization and Input Reading:**
   - The code reads the number of test cases and the input for each test case.
   - It reads the length of the permutation and the permutation itself.

2. **Sorting Logic:**
   - The code iterates through the permutation and checks if the current element is in the correct position.
   - If the current element is not in the correct position but the next element is, and the difference between them is 1, it swaps them.
   - If it finds an element that is not in the correct position and the next element is not in the correct position or the difference is not 1, it breaks the loop.

3. **Output:**
   - The code checks if the last element of the permutation is in the correct position and outputs "YES" or "NO" accordingly.

### Potential Bugs:
- **Early Termination:** The code breaks the loop as soon as it finds an element that is not in the correct position and the next element is not in the correct position or the difference is not 1. This can lead to incorrect results if the permutation can still be sorted but the algorithm stops early.
- **Incomplete Checking:** The code only checks if the last element is in the correct position, which is insufficient to determine if the entire permutation is sorted.

### Challenging Test Input:
To expose the bug, we need a permutation that can be sorted but the algorithm will stop early due to the early termination condition. A good example is a permutation where the first few elements are in the correct position, but the last few elements are not, and the difference between the out-of-place elements is not 1.

### Test Input:
```plaintext
1
5
1 2 4 3 5
```

### Explanation:
- **Permutation:** [1, 2, 4, 3, 5]
- **Expected Output:** "YES"
- **Reasoning:**
  - The first three elements (1, 2, 4) are in the correct position.
  - The last two elements (4, 3) can be swapped to sort the permutation.
  - The buggy code will break the loop when it encounters the element 4, because the difference between 4 and 3 is 1, but it will not continue to check the rest of the permutation.

This test input will expose the bug in the early termination condition and the incomplete checking of the permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

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

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

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

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

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main{

	public static BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
	public static BufferedWriter wd=new BufferedWriter(new OutputStreamWriter(System.out));
	public static void solve()throws Exception{
		String dr[]=rd.readLine().split(" ");
		int n=Integer.parseInt(dr[0]);
		int a[]=new int [n+10];
		dr=rd.readLine().split(" ");
		int kx=0;
		for(int i=1;i<=n;i++) a[i]=Integer.parseInt(dr[i-1]);
		for(int i=1;i<n;i++) {
			if(a[i]==a[i+1]+1) {
				int tem=a[i];
				a[i]=a[i+1];
				a[i+1]=tem;
				break;
			}
		}
		for(int i=1;i<n;i++) {
			if(a[i]>a[i+1]) {
				wd.write("NO\n");
				return;
			}
		}
		wd.write("YES\n");
	}
	public static void main(String[] args)throws Exception{
		int t=Integer.parseInt(rd.readLine());
		while(t-->0) {
			solve();
			wd.flush();
		}
	}

}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that highlights the issue in the algorithm. The code only attempts to swap once if a pair of neighboring elements differ by exactly 1, and then it checks if the array is sorted. This approach is flawed because it does not consider multiple swaps or the possibility that the array might not be sortable with just one swap.

### Step-by-Step Analysis:
1. **Identify the Bug**: The bug is in the logic that only performs one swap and then checks if the array is sorted. This is insufficient because the array might need multiple swaps to be sorted, and the single swap might not be enough to determine the final state.

2. **Create a Test Case**:
   - We need an array where a single swap is not enough to sort the array.
   - The array should be such that it cannot be sorted by the given operations.

### Example Test Case:
Consider the permutation `[3, 1, 2]`. Here's why this is a good test case:
- The array is not sorted.
- A single swap of neighboring elements that differ by 1 will not sort the array.
- The array cannot be sorted using the given operation rules.

### Test Input:
```plaintext
1
3
3 1 2
```

### Explanation:
- **Input**: `1` (number of test cases)
- **Test Case**:
  - `3` (length of the permutation)
  - `3 1 2` (the permutation)

### Expected Output:
- **Output**: `NO`

This test case should expose the bug because the code will only perform one swap and then check if the array is sorted, which it will not be in this case. Therefore, the output should be `NO`.