**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Progressive Square
time_limit_ms: 2000
memory_limit_mb: 256
Description: A progressive square of size $$$n$$$ is an $$$n \times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:

$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$

$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$

For example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:

$$$$$$ \begin{pmatrix} 1 & 4 & 7 \\ 3 & 6 & 9 \\ 5 & 8 & 11 \end{pmatrix} $$$$$$

Last month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.

It can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.

Input: The first line contains an integer $$$t$$$ ($$$1 \le t \le {10} ^ 4$$$) — the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \le n \le 500$$$, $$$1 \le c, d \le 10^6$$$) — the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.

The second line of each test case contains $$$n \cdot n$$$ integers $$$b_1, b_2, \dots, b_{n \cdot n}$$$ ($$$1 \le b_i \le 10^9$$$) — the elements found by Maxim.

It is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \cdot {10} ^ 4$$$.

Output: For each test case, output "YES" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output "NO".

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
5
3 2 3
3 9 6 5 7 1 10 4 8
3 2 3
3 9 6 5 7 1 11 4 8
2 100 100
400 300 400 500
3 2 3
3 9 6 6 5 1 11 4 8
4 4 4
15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15
output:
NO
YES
YES
NO
NO

Note:

Buggy Code:
// clang-format off

#ifdef ONLINE_JUDGE
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define dbg(...)
#define SOLVE solve();
#endif

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

constexpr int mod = 998244353; // 1e9 + 7

using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ost = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)

template <const int T> struct ModInt { const static int mod = T; int x; ModInt(int x = 0) : x(x % mod) {} ModInt(long long x) : x(int(x % mod)) {} int val() { return x; } ModInt operator+(const ModInt &a) const { int x0 = x + a.x; return ModInt(x0 < mod ? x0 : x0 - mod); } ModInt operator-(const ModInt &a) const { int x0 = x - a.x; return ModInt(x0 < 0 ? x0 + mod : x0); } ModInt operator*(const ModInt &a) const { return ModInt(1ll * x * a.x % mod); } ModInt operator/(const ModInt &a) const { return *this * a.inv(); } void operator+=(const ModInt &a) { x += a.x; if (x >= mod) x -= mod; } void operator-=(const ModInt &a) { x -= a.x; if (x < 0) x += mod; } void operator*=(const ModInt &a) { x = 1LL * x * a.x % mod; } void operator/=(const ModInt &a) { *this = *this / a; } ModInt pow(int64_t n) const { ModInt res(1), mul(x); while (n) { if (n & 1) res *= mul; mul *= mul; n >>= 1; } return res; } ModInt inv() const { int a = x, b = mod, u = 1, v = 0; while (b) { int t = a / b; a -= t * b; swap(a, b); u -= t * v; swap(u, v); } if (u < 0) u += mod; return u; } };

typedef ModInt<mod> mint; typedef long long ll; typedef vector<int> vi; typedef vector<vector<int>> vvi; typedef vector<long long> vl; typedef vector<vector<long long>> vvl; typedef vector<pair<int, int>> vpi; typedef vector<pair<long long, long long>> vpl; typedef pair<int, int> pi; typedef pair<long long, long long> pl; typedef unsigned int uint; typedef unsigned long long ull;

template <typename T> using vc = vector<T>; template <typename T> using pq = priority_queue<T>; template <typename T> using qu = queue<T>; template <typename T> using dq = deque<T>; template <typename T> using stk = stack<T>; template <typename T1, typename T2> using pr = pair<T1, T2>; template <typename T> using mst = multiset<T>; template <typename Key, typename T> using ump = unordered_map<Key, T>; template <typename T> using ust = unordered_set<T>; template <typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;

void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } void __print(mint a) { cerr << a.x; } template <typename T, typename V> void __print(pair<T, V> x) { cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}'; } template <typename T> void __print(T x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? "," : ""), __print(i); cerr << "}"; } template <typename T> void __print(stack<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(queue<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.front()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(priority_queue<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(priority_queue<T, vector<T>, greater<T>> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } void _print() { cerr << "]\n\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); }
#ifndef ONLINE_JUDGE
#define dbg(x...) cerr << "[" << #x << "] = ["; _print(x)
#define SOLVE cerr << (_ > 0 ? "\n####### TestCase " : "####### TestCase ") << _ + 1 << " #######\n\n"
#endif

#define reif(t, ...) if (t) { __VA_ARGS__; return; }
#define coif(t, ...) if (t) { __VA_ARGS__; continue; }
#define brif(t, ...) if (t) { __VA_ARGS__; break; }

#define all(x) x.begin(), x.end()
#define rev(x) reverse(all(x))
#define s0rt(x) sort(all(x))
#define s0rtd(x) sort(all(x)), rev(x);
#define s0rtx(x, ...) sort(all(x), __VA_ARGS__);
#define sum(x) accumulate(all(x), 0ll)
#define len(x) (long long)(x.size())
#define mine(v) *min_element(all(v))
#define maxe(v) *max_element(all(v))
#define unq(x) s0rt(x), x.erase(unique(all(x)), x.end()), x.shrink_to_fit()
#define FOR1(a) for (int _ = 0; _ < (a); ++_)
#define FOR2(i, a) for (int i = 0; i < (a); ++i)
#define FOR3(i, a, b) for (int i = a; i < (b); ++i)
#define FOR4(i, a, b, c) for (int i = a; i < (b); i += (c))
#define FOR5(i, a, condn, b, add) for (int i = a; condn < b; i += add)
#define FOR1_R(a) for (int _ = (a); _ >= (0); --i)
#define FOR2_R(i, a) for (int i = (a); i >= (0); --i)
#define FOR3_R(i, a, b) for (int i = (a); i >= (b); --i)
#define FOR4_R(i, a, b, c) for (int i = (a); i >= (b); i -= c)
#define FOR5_R(i, a, condn, b, add) for (int i = (a); condn >= (b); i -= add)
#define TRAV1(i, a) for (auto &i : a)
#define TRAV2(i, j, a) for (auto &[i, j] : a)
#define LB1(x, y) lower_bound(all(x), y)
#define UB1(x, y) upper_bound(all(x), y)
#define LB2(x) lower_bound(x)
#define UB2(x) upper_bound(x)
#define CNT1(x, y) count(all(x), y)
#define cnt2(x) count(x)
#define overload5(a, b, c, d, e, f, ...) f
#define overload2(a, b, c, d, ...) d
#define overload1(a, b, c, ...) c
#define f0r(...) overload5(__VA_ARGS__, FOR5, FOR4, FOR3, FOR2, FOR1)(__VA_ARGS__)
#define f0rd(...) overload5(__VA_ARGS__, FOR5_R, FOR4_R, FOR3_R, FOR2_R, FOR1_R)(__VA_ARGS__)
#define trav(...) overload2(__VA_ARGS__, TRAV2, TRAV1)(__VA_ARGS__)
#define lb(...) overload1(__VA_ARGS__, LB1, LB2)(__VA_ARGS__)
#define ub(...) overload1(__VA_ARGS__, UB1, UB2)(__VA_ARGS__)
#define c0unt(...) overload1(__VA_ARGS__, CNT1, CNT2)(__VA_ARGS__)
#define pb push_back
#define eb emplace_back
#define ep emplace
#define ins insert
#define ers erase
#define beg begin()
#define ed end()
#define back back()
#define mkp make_pair
#define f first
#define s second
#define yes "YES"
#define no "NO"
#define fun(type, name, ...) auto name = [&](__VA_ARGS__) -> type
#define recfun(type, name, ...) auto name = [&](auto &&name, __VA_ARGS__) -> type

int popcnt(int x) { return __builtin_popcount(x); } int topbit(int x) { return (x == 0 ? -1 : 31 - __builtin_clz(x)); } int lowbit(int x) { return (x == 0 ? -1 : __builtin_ctz(x)); } int popcnt(long long x) { return __builtin_popcountll(x); } int topbit(long long x) { return (x == 0 ? -1 : 63 - __builtin_clzll(x)); } int lowbit(long long x) { return (x == 0 ? -1 : __builtin_ctzll(x)); }

template <typename T> int sgn(const T &x) { return (x > 0 ? 1 : x < 0 ? -1 : 0); }; template <typename T, typename U> T ceil(T x, U y) { return (x > 0 ? (x + y - 1) / y : x / y); } template <typename T, typename U> T floor(T x, U y) { return (x > 0 ? x / y : (x - y + 1) / y); } template <class T, class S> bool chmx(T &a, const S &b) { return (a < b ? a = b, 1 : 0); } template <class T, class S> bool chmn(T &a, const S &b) { return (a > b ? a = b, 1 : 0); }

template <typename U, typename T> T binser(U check, T _lo, T _hi, bool _minimize) { T _ret = (_minimize ? _hi + 1 : _lo - 1); while (_lo <= _hi) { T _mid = _lo + _hi >> 1ll; if (_minimize) { if (check(_mid)) { _hi = _mid - 1; _ret = _mid; } else { _lo = _mid + 1; } } else { if (check(_mid)) { _lo = _mid + 1; _ret = _mid; } else { _hi = _mid - 1; } } } return _ret; }
template <typename T> void prfxsum(vector<T> &a) { int _n = a.size(); for (int i = 1; i < _n; i++) a[i] += a[i - 1]; };
template <typename T, typename U> vector<T> getorder(T start, T size, U comp) { vector<T> ord(size); iota(ord.begin(), ord.end(), start); sort(ord.begin(), ord.end(), comp); return ord; }
template <typename T = int, typename U = int> map<T, U> mappos(const vector<T> &data, U offset) { map<T, U> ps; for (int i = 0; i < data.size(); i++) ps[data[i]] = i + offset; return ps; }
template <typename T> map<T, int> counter(const vector<T> &data) { map<T, int> x; for (auto &i : data) x[i]++; return x; };
template <typename T, typename U> T toset(const vector<U> &data) { T x; for (auto &i : data) x.insert(i); return x; };

template <typename T, typename U> U safeErase(T &setik, U target) { auto iterator = setik.find(target); assert(iterator != setik.end()); U a = *iterator; setik.erase(iterator); return a; }
template <typename T, typename U> pair<U, bool> Erase(T &setik, U target) { auto iterator = setik.find(target); pair<U, bool> a = {0, false}; if (iterator != setik.end()) { a.first = *iterator; setik.erase(iterator); } return a; }

template <typename T> T pop(multiset<T> &que, bool front = 1) { assert(que.empty() == false); auto iterator = front ? que.begin() : prev(que.end()); T a = *iterator; que.erase(iterator); return a; } template <typename T> T pop(set<T> &que, bool front = 1) { assert(que.empty() == false); auto iterator = front ? que.begin() : prev(que.end()); T a = *iterator; que.erase(iterator); return a; } template <typename T> T pop(stack<T> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(queue<T> &que) { T a = que.front(); que.pop(); return a; } template <typename T> T pop(deque<T> &que, bool front) { T a = (front ? que.front() : que.back()); (front ? que.pop_front() : que.pop_back()); return a; } template <typename T> T pop(priority_queue<T> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(priority_queue<T, vector<T>, greater<T>> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(vector<T> &que) { T a = que.back; que.pop_back(); return a; }

// condition args(i, stack.top())
template <typename T, typename U> vector<int> nearestelem(const vector<T> &a, U condition, bool reverse = false, int def = -1) { int n = a.size(); stack<int> s; vector<int> ans(n, def); auto go = [&](int i) { while (!s.empty() && condition(i, s.top())) { ans[s.top()] = i; s.pop(); } s.push(i); }; if (reverse) for (int i = n - 1; i > -1; i--) go(i); else for (int i = 0; i < n; i++) go(i); return ans; }
template <typename T, typename U> constexpr T fpow(T x, U n) { T result = 1; while (n > 0) { if (n & 1ll) result *= x; x *= x; n >>= 1ll; } return result; }
template <typename T> struct prfx { vector<T> _a; int _n; prfx(const vector<T> &a) { _n = a.size() + 1; _a.resize(_n); for (int i = 0; i < _n - 1; i++) _a[i + 1] = _a[i] + a[i]; } T get(int x, int y) { if (y < x) return 0; return _a[y + 1] - _a[x]; } };
vector<int> divisors(int V) { vector<int> ret; for (int i = 1; (long long)i * i < V + 1; ++i) { if (V % i == 0) { ret.push_back(i); if (V / i != i) { ret.push_back(V / i); } } } return ret; }

const int imx = INT_MAX;
const int imn = INT_MIN;
const int lmx = LLONG_MAX;
const int lmn = LLONG_MIN;

// clang-format on

const int N = 3e5;

ll n, c, d;
ll a[N], b[N];

bool verify(int x) {
    int ix = 0;

    f0r(i, 0, n) {
        b[ix++] = x;
        f0r(j, 1, n) b[ix++] = x + (ll)j * d;
        x += c;
    }

    sort(b, b + n * n);

    f0r(i, n * n) if (a[i] != b[i]) return false;
    return true;
}

void solve() {
    cin >> n >> c >> d;
    f0r(i, n * n) cin >> a[i];

    sort(a, a + n * n);

    f0r(i, n * n) reif(verify(a[i]), cout << "YES" << '\n');

    cout << "NO" << '\n';
}

int main() {
    cin.tie(0)->ios_base::sync_with_stdio(false);

    int testcases = 0;
    cin >> testcases;

    f0r(testcases) solve();
}

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

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

Think step by step.

Response:
To expose the bug in the given code, let's consider a few key points:

1. **Initialization and Validation**: The code initializes the matrix and then checks if the sorted array matches the expected sorted values of the matrix. If any discrepancy is found, it returns "NO".

2. **Potential Bug**: The code uses a linear search to find the initial value `x` that can generate the matrix. This might fail if the initial value is not correctly identified, especially in edge cases where the values are close or identical.

3. **Edge Cases**:
   - **All values are the same**: This can cause issues if the code does not correctly handle identical values.
   - **High values of `c` and `d`**: This can lead to large differences in the matrix values, which might cause precision issues or overflow.
   - **Small values of `n`**: This can simplify the problem and make it easier to identify issues in the initialization.

Given these points, let's create a test case that maximizes the failure probability:

- **All values are the same**: This can be a good test case because the code might not correctly handle identical values.
- **High values of `c` and `d`**: This can cause large differences and might expose precision issues.
- **Small values of `n`**: This simplifies the problem and makes it easier to identify issues.

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

```plaintext
1
3 1000000 1000000
1 1000001 2000001 3000001 4000001 5000001 6000001 7000001 8000001 9000001 10000001 11000001
```

This test case has:
- `n = 3`
- `c = 1000000`
- `d = 1000000`
- An array `b` with values that should form a progressive square, but with a high difference between values.

This input should help expose any issues in the code's handling of large values and identical values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
// Awwawa! Dis cold yis ratten buy kitten!
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()

template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
  int n = v.size();
  for (int i = 0; i < n; ++i)
    os << v[i] << ' ';
  return os;
}
using namespace std;
using i64 = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int n, q;
    cin >> n >> q;
    vector<int> a(n + 1), occur(n + 1);
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      occur[a[i]] += 1;
    }

    sort(all(a));
    int flag = (n % 2 == 0 ? (n + 2) / 2 : (n + 1) / 2);
    while (q--) {
      int x;
      cin >> x;
      int cnt = a.end() - lower_bound(all(a), x);
      // cerr << cnt << '\n';
      int ans = -1;
      if (cnt < flag) {
        ans = a[flag - cnt];
      } else {
        // int beg = lower_bound(all(a), x) - a.begin();
        // cerr << beg << '\n';
        int f = flag;
        for (int i = 0; i < x; ++i) {
          int cc = occur[i];
          for (int j = i; j <= n; j += x) {
            cc += occur[j];
          }
          if (f <= cc) {
            ans = i;
            break;
          }
          f -= cc;
        }

        // ans = a[flag] % x;
      }
      assert(ans != -1);
      cout << ans << ' ';
    }
    cout << '\n';
  }
  return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
/*
    JAI SHREE RAM
*/

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define int         long long
#define ld          long double
#define nl          cout<<endl
#define notpos      cout<<-1<<endl
#define sqrt        sqrtl
#define yes         cout<<"YES"<<endl
#define no          cout<<"NO"<<endl
#define all(v)      v.begin(),v.end()
#define f(i,n)      for(int i=0;i<n;i++)
#define vii         vector<vector<int>>
#define pii         pair<int,int>
#define inf         LLONG_MAX
#define pb          push_back
#define ff          first
#define ss          second
#define gsum(a)     accumulate(a.begin(),a.end(),0LL)
#define gmax(a)     *max_element(a.begin(),a.end())
#define gmin(a)     *min_element(a.begin(),a.end())
#define d1(x)       cout<<#x<<" "<<x<<endl;
#define d2(x,y)     cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<endl;
#define d3(x,y,z)   cout<<#x<<" "<<x<<" "<<#y<<" "<<y<<" "<<#z<<" "<<z<<endl;

using namespace __gnu_pbds;
using namespace std;

typedef pair<int,int> node;
typedef tree<node, null_type, less<node>,
            rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int mod = 1e9 + 7;
//const int mod = 998244353;
const ld eps = 1e-7;

int lcm(int a,int b){
    return (a/(__gcd(a,b)))*b;
}

void out(vector<int> v){
    for(auto x : v){
        cout<<x<<" ";
    }
    nl;
}

void fastio(){
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

int power(int a, int b, int p)
    {
        if(a==0)
        return 0;
        int res=1;
        a%=p;
        while(b>0)
        {
            if(b&1)
            res=(1ll*res*a)%p;
            b>>=1;
            a=(1ll*a*a)%p;
        }
        return res;
    }

int N = 2e5 + 5;
vector<int> spf(N,0);
vector<int> path;

void sieve(){
        spf[0] = 1;
        spf[1] = 1;
        for(int i=2;i<N;i++) spf[i] = i;

        for(int i=2;i*i<N;i++){
            if(spf[i] == i){
                for(int j=i*i;j<N;j+=i){
                    if(spf[j] == j) spf[j] = i;
                }
            }
        }
}

vector<int> fact(N);
vector<int> inv(N);

void precompute(){
    fact[1] = 1;
    for(int i=2;i<N;i++){
        fact[i] = fact[i-1]*i;
        fact[i] %= mod;
    }
    inv[0] = 1;
    inv[1] = 1;
    for(int i=2;i<N;i++){
        inv[i] = power(fact[i],mod-2,mod);
        inv[i] %= mod;
    }
}

int nCr(int n,int r){
    int res = ((fact[n]*inv[r])%mod);
    res *= inv[n-r];
    res %= mod;
    return res;
}

int32_t main(){
    fastio();
    int tt=1;
    cin>>tt;
    while(tt--){
        int n,q;
        cin>>n>>q;
        vector<int> v(n);
        f(i,n) cin>>v[i];

        sort(all(v));
        vector<int> pref(n,0);
        pref[0]=v[0];
        for(int i=1;i<n;i++) pref[i] = pref[i-1]+v[i];

        vector<vector<int>> g(317);
        for(int i=1;i<317;i++){
            f(j,n){
                g[i].pb((v[j]%i));
            }
            sort(all(g[i]));
        }

        int m;
        if(n%2) m = (n+1)/2;
        else m = (n+2)/2;
        vector<int> ans;

        while(q--){
            int x;
            cin>>x;
            if(x<317) ans.pb(g[x][m-1]);
            else{
                int lo=0,hi=x-1,mid;
                int res = x-1;
                while(lo<=hi){
                    mid = (lo+hi)/2;
                    int mn = min(mid,x-1);
                    int s = upper_bound(all(v),mn) - v.begin();
                    int tot=0;

                    for(int i=1;i*x<=v.back();i++){
                        int l=i*x;
                        int r=i*x + mid;

                        int it1 = upper_bound(all(v),r) - v.begin();
                        int it2 = lower_bound(all(v),l) - v.begin();

                        tot += (it2-it1);
                        if(i*x > v.back()) break;
                    }
                    if(s+tot >= m){
                        res=mid;
                        hi=mid-1;
                    }
                    else lo = mid+1;
                }
                ans.pb(res);
            }
        }
        out(ans);
    }
}

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

This test input includes two test cases with different array sizes and query counts. The first test case has an array with a simple sequence of integers, and the queries cover a range of values. The second test case has a more varied array and a smaller set of queries. This input is designed to cover basic scenarios and edge cases, such as the median being at the boundary of the array and the effect of different values of \( x \) on the median.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
/**
 * https://codeforces.com/contest/2008/submission/280763922
 *
 * (c) 2024 Diego Sogari
 */
#include <bits/stdc++.h>

using namespace std;

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

template <typename T> ostream &operator<<(ostream &os, const vector<T> &a) {
  return ranges::for_each(a, [&os](auto &ai) { os << ai << ','; }), os;
}
void println1(auto &&...args) { ((cout << args << ';'), ...) << endl; }
void println(auto &&...args) { ((cout << args << ' '), ...) << endl; }

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

struct Iota : vector<int> {
  Iota(int n, int s = 0) : vector<int>(n) { iota(begin(), end(), s); }
  Iota(int n, auto &&f, int s = 0) : Iota(n, s) { ranges::sort(*this, f); }
};

void solve(int t) {
  Int n, m;
  vector<Int> a(n), q(m);
  if (t == 42) {
    println1(n, m, a, q);
    return;
  }
  vector<int> cnt(n), ans(m);
  auto f = [&](int i, int j) { return q[i] < q[j]; };
  Iota idx(m, f);
  ranges::sort(a);
  for (int i = 0, j = 0; i < m; i++) {
    auto x = q[idx[i]];
    for (; a[j] < x; j++) {
      cnt[a[j]]++;
    }
    for (int k = j; k < n; k++) {
      cnt[a[k] % x]++;
    }
    for (int k = 0, c = 0; k < n; k++) {
      c += cnt[k];
      if (c > n / 2) {
        ans[idx[i]] = k;
        break;
      }
    }
    for (int k = j; k < n; k++) {
      cnt[a[k] % x]--;
    }
  }
  for (auto &&ai : ans) {
    cout << ai << ' ';
  }
  cout << endl;
  // println(ans);
}

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

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the array is very small or very large.
2. **Special Values of \( x \)**: Values of \( x \) that are equal to the elements in the array or are very close to the maximum element in the array.
3. **Median Calculation**: Ensure that the median calculation is correctly handled, especially for arrays with even and odd lengths.
4. **Counting and Sorting**: Test the counting and sorting mechanisms used in the code.

Given the constraints and the need for a small-scale test input, let's construct a test case that might expose issues in the code.

### Test Case Construction

1. **Array with Repeated Elements**: This can test the handling of multiple elements being reduced by \( x \).
2. **Query Values Close to Array Elements**: This can test the correct reduction and median calculation.
3. **Even and Odd Length Arrays**: Ensure both cases are covered.

### Example Test Input

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

### Explanation

- **Array**: `[1, 2, 2, 3, 4]`
  - Median initially: `2` (since the array is sorted and has an odd length, the median is the middle element).
- **Query 1: \( x = 1 \)**
  - After operations, the array can be `[0, 1, 1, 2, 3]`.
  - Median: `1`.
- **Query 2: \( x = 2 \)**
  - After operations, the array can be `[1, 0, 0, 1, 2]`.
  - Median: `1`.
- **Query 3: \( x = 3 \)**
  - After operations, the array can be `[1, 2, 2, 0, 1]`.
  - Median: `1`.

This test case is small but comprehensive, covering different scenarios that can help expose potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
def check(mid,tar,q):
    cnt=0
    for i in range(0,n+1,q):
        if i==0:
            cnt+=qzh[mid]
        else:
            if i+mid>n:
                cnt+=qzh[n]-qzh[i-1]
                continue
            cnt+=qzh[i+mid]-qzh[i-1]

    if cnt<tar:
        return False
    else:
        return True
for i in range(int(input())):
    n,q=map(int,input().split())
    ns=list(map(int,input().split()))
    qzh=[0]*(n+1)
    tar=(n+1)//2 if n%2==1 else (n+2)//2
    als=[]
    for i in ns:
        qzh[i]+=1
    for i in range(1,n+1):
        qzh[i]+=qzh[i-1]
    res={}
    for i in range(q):
        dis=int(input())
        if dis in res:
            print(res[dis])
            continue
        l=0
        r=dis-1
        while l<r:
            mid=(l+r)//2
            if check(mid,tar,dis):
                r=mid
            else:
                l=mid+1
        res[dis]=l
        als.append(res[dis])
    print(*als)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The code has issues with how it handles the prefix sums and the binary search logic. Let's focus on the following scenarios:

1. **Small Array with Few Queries**: This helps in understanding the basic logic.
2. **Array with Repeated Elements**: This can expose issues in the prefix sum calculation.
3. **Large Queries**: This can expose issues in the binary search logic.
4. **Edge Cases**: Values of `x` that are close to the maximum element in the array.

Given the constraints and the need for a small-scale test input, let's create a test case that covers these aspects.

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

### Explanation:
1. **First Test Case**:
   - Array: `[1, 2, 3]`
   - Queries: `[1, 2, 3]`
   - This test case checks the basic functionality with a small array and small queries.
   - The expected output is `0 1 2` because:
     - For `x = 1`: The array remains `[1, 2, 3]`, and the median is `2` (but the minimum possible median is `0`).
     - For `x = 2`: The array can be transformed to `[1, 0, 1]`, and the median is `1`.
     - For `x = 3`: The array can be transformed to `[1, 2, 0]`, and the median is `2`.

2. **Second Test Case**:
   - Array: `[1, 2, 3, 4, 5]`
   - Queries: `[1, 3, 5, 2]`
   - This test case checks the functionality with a larger array and a mix of queries.
   - The expected output is `0 1 2 1` because:
     - For `x = 1`: The array remains `[1, 2, 3, 4, 5]`, and the median is `3` (but the minimum possible median is `0`).
     - For `x = 3`: The array can be transformed to `[1, 2, 0, 1, 2]`, and the median is `2`.
     - For `x = 5`: The array can be transformed to `[1, 2, 3, 0, 0]`, and the median is `3` (but the minimum possible median is `2`).
     - For `x = 2`: The array can be transformed to `[1, 0, 1, 2, 3]`, and the median is `2` (but the minimum possible median is `1`).

This test input should help expose the issues in the buggy code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
def sakurako_test(t, test_cases):
    results = []

    for _ in range(t):
        n, q = test_cases[_]['n_q']
        a = test_cases[_]['a']
        queries = test_cases[_]['queries']

        # Precompute the remainders for each x from 1 to n
        freq = [{} for _ in range(n + 1)]
        for x in range(1, n + 1):
            for ai in a:
                rem = ai % x
                if rem in freq[x]:
                    freq[x][rem] += 1
                else:
                    freq[x][rem] = 1

        # Function to find the median
        def find_median(freq_x):
            sorted_freq = sorted(freq_x.items())
            total = sum(freq_x.values())
            mid = (total // 2) + 1
            count = 0
            for val, cnt in sorted_freq:
                count += cnt
                if count >= mid:
                    return val

        # Answer the queries
        answers = []
        for x in queries:
            median = find_median(freq[x])
            answers.append(median)

        results.append(answers)

    return results

# Example usage
if __name__ == "__main__":
    t = 2
    test_cases = [
        {'n_q': (5, 5), 'a': [1, 2, 3, 4, 5], 'queries': [1, 2, 3, 4, 5]},
        {'n_q': (6, 3), 'a': [1, 2, 6, 4, 1, 3], 'queries': [2, 1, 5]}
    ]

    results = sakurako_test(t, test_cases)
    for result in results:
        print(' '.join(map(str, result)))

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

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

Think step by step.

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

1. **Edge Cases for Array Length and Queries**:
   - Test with the minimum and maximum allowed values for `n` and `q`.
   - Test with arrays that have repeated elements or are already sorted.

2. **Edge Cases for the Array Elements**:
   - Test with arrays where all elements are the same.
   - Test with arrays where elements are close to the bounds (1 and n).

3. **Queries**:
   - Test with queries that are 1, n, and values in between.
   - Test with queries that are not in the array but within the bounds.

4. **Median Calculation**:
   - Test with arrays that have an even and odd number of elements.
   - Test with arrays where the median is the smallest or largest element after operations.

Based on these considerations, let's create a small-scale test input that is complete and valid:

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

Explanation:
- **First Test Case**:
  - `n = 5`, `q = 5`
  - Array: `[1, 1, 1, 1, 1]`
  - Queries: `[1, 2, 3, 4, 5]`
  - This test case will check how the code handles an array with all elements being the same and various queries.

- **Second Test Case**:
  - `n = 6`, `q = 3`
  - Array: `[1, 2, 3, 4, 5, 6]`
  - Queries: `[1, 3, 6]`
  - This test case will check how the code handles a sorted array with different queries, including the maximum value.

This input should help expose any issues in the buggy code, particularly in how it handles remainders and median calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
t = int(input())

for _ in range(t):
    n, q = list(map(int, input().split()))
    arr = list(map(int, input().split()))
    ans = []
    for _ in range(q):
        k = int(input())
        if k == 1:
            ans.append(k-1)
            continue
        cs = [0]*k
        r = (n+1)//2 if n%2 else (n+2)//2
        for a in arr:
            cs[a % k] += 1
        for i in range(1, k):
            cs[i] += cs[i-1]
            if cs[i] >= r:
                ans.append(i)
                break
    print(*ans)

**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 5
5 5 5 5 5
1
2
3
4
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int qn = nextInt();
    int[] a = new int[n];
    for (int i = 0; i < a.length; ++i) a[i] = nextInt();
    int[] q = new int[qn];
    for (int i = 0; i < q.length; ++i) q[i] = nextInt();
    int[] g = new int[n + 1];
    for (int x : a) g[x]++;
    c = new int[n + 1];
    for (int i = 0; i < c.length; i++) c[i] = g[i] + (i == 0 ? 0 : c[i - 1]);
    int[] ret = new int[n + 1];
    for (int i = 2; i <= n; i++) {
        int l = 0, r = i - 1;
        if (check(l, i, n / 2 + 1)) ret[i] = l;
        else {
            while (l + 1 < r) {
                int mid = l + (r - l) / 2;
                if (check(mid, i, n / 2 + 1)) r = mid;
                else l = mid;
            }
        }
        ret[i] = r;
    }
    for (int x : q) {
        so(ret[x] + " ");
    }
    sl();
}

int[] c;

boolean check(int mid, int m, int reqCnt) {
    int cnt = 0;
    for (int i = 0; i < c.length; i += m) {
        cnt += c[Math.min(i + mid, c.length - 1)] - (i == 0 ? 0 : c[i - 1]);
    }
    return cnt >= reqCnt;
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

**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 are likely to cause the code to fail. Here are the steps to think through the problem:

1. **Understand the Algorithm**:
   - The code is supposed to find the minimum possible median of the array after performing operations that reduce elements by a given value \( x \).

2. **Identify Potential Issues**:
   - The code uses a binary search to find the minimum median. It might fail if the array has repeated elements or if the median calculation is incorrect.
   - The code might also fail if the array is very small or has very large values.

3. **Create Edge Cases**:
   - **Small Array**: Test with the smallest possible array to ensure the code handles it correctly.
   - **Large Array**: Test with a larger array to ensure the code handles it within the time and memory limits.
   - **Array with Repeated Elements**: Test with an array that has many repeated elements.
   - **Array with Elements Equal to \( x \)**: Test with an array where elements are exactly equal to \( x \).

4. **Generate the Test Input**:
   - We will create a small-scale test input that covers the edge cases identified.

Here is the test input:

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

### Explanation:
- **Test Case 1**:
  - Array: `[1, 2, 3, 4, 5]`
  - Queries: `[1, 2, 3, 4, 5]`
  - This test case checks the basic functionality with a simple, increasing sequence of numbers.

- **Test Case 2**:
  - Array: `[1, 1, 1, 1, 1]`
  - Queries: `[1, 2, 3, 4, 5]`
  - This test case checks how the code handles an array with all elements being the same. This is a critical edge case because the median will be the same regardless of the value of \( x \).

This test input should help expose any issues in the buggy code, especially in handling arrays with repeated elements and small arrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//stack=new Stack<Integer>();
		//set=new HashSet<List<Integer>>();
		//f(0, new int[3]);System.out.println(set.size());
		//p1=false;

		  int t=sc.nextint(); while (t-->0) {

		  sovle(); }

		  out.close();

	}
	static long[] ram,v;
	static int []size,ss;
	static int gcd(int a,int b) {
		return b==0?a:gcd(b, a%b);
	}
	static void sovle2() {
		int n=sc.nextint();
		long w=sc.nextlong();
		map=new HashMap<Integer, List<Integer>>();
		for (int i = 0; i < n; i++) {
			map.put(i+1, new ArrayList<Integer>());
		}
		int []p=new int[n+1];
		bj=new int[n+1];
		for (int i = 2; i <= n; i++) {
			p[i]=sc.nextint();
			map.get(p[i]).add(i);
		}
		cnt=new int[n+1];
		f1=new int[n+1];
		f2=new int[n+1];

		id=0;
		dfs(1, 0);

		long ge=n;long sum=0;
		long []dis=new long [n+1];
		int []time=new int[n+1];
		for (int i = 1; i <n; i++) {
			int x=sc.nextint();
			long y=sc.nextlong();
			cnt[f1[x]]--;
			dis[f1[x]]+=y;
			if (cnt[f1[x]]==0) {
				ge--;
			}
			cnt[f2[x]]--;
			dis[f2[x]]+=y;
			if (cnt[f2[x]]==0) {
				ge--;
			}
			sum+=y;
			System.out.println(ge+" "+sum);
			out.print((ge*w+(2-ge)*sum)+" ");
		}
		out.println();
	}

	static int []cnt,f1,f2;
	static int id;
	static void dfs(int x,int fu) {
		id++;

		for (int i : map.get(x)) {
			cnt[id]++;
			f1[i]=id;
			dfs(i, x);
			cnt[id]++;
			f2[i]=id;
		}

	}
	static int[]bj;
	static boolean p1;
	static int[][] va,l,r;
	static int n,m;
	static void sovle() {
		n=sc.nextint();
		int q=sc.nextint();
		int []shu=new int[n+1];
		int []qian=new int[n+1];
		for (int i = 1; i < shu.length; i++) {
			shu[sc.nextint()]++;
		}
		for (int i = 1; i < shu.length; i++) {
			qian[i]=qian[i-1]+shu[i];
		}
		int []ans=new int[n+1];
		c=new long[n+1];
		int []va=new int[n+1];
		list=new ArrayList<Integer>();
		int xian=n/2+1;
		int shang=n;
		for (int i = n; i >=1; i--) {
			if (shang-i>i) {
				for (int j = i+i+1; j <=shang; j++) {
					add(j-i, va[j]);
					va[j-i]+=va[j];
				}
				shang=i+i;

			}
			int j = get(i, shu);
			//System.out.println(j);
			va[i]=j;
			add(i, j);
			int l=0;int r=i;int mid;
			while (l<r) {
				mid=(l+r)/2;

				if (qian[mid]+sum(Math.min(n, i+mid))>=xian) {
					r=mid;
				}else {
					l=mid+1;
				}
			}
			ans[i]=r;
		}
		//out.println(Arrays.toString(ans));
		for (int i = 0; i < q; i++) {
			out.print(ans[sc.nextint()]+" ");
		}
		out.println();
	}
	static List<Integer>list;

	static int get(int x,int []shu) {
		int ans=(int) (shu[x]+sum(Math.min(n, x*2))-sum(Math.min(n, x*2-1)));

		return ans;
	}

	static long  pow(long x,long a) {
		if (a==0) {
			return 1;
		}else {
			long c=pow(x, a/2);
			if (a%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*x%mod;
			}
		}
	}
	static long mod = (long) (1e9+7);

	static void sovle1() {

	}

	static long []c;
	static int lowbit(int x) {
		return x&-x;
	}
	static void add(int x,long v) {
		while (x<c.length) {
			//System.out.println(x);
			c[x]+=v;
			x+=lowbit(x);
		}
	}
	static long sum(int x) {
		long ans=0;
		//System.out.println(x+"dasdsafadf");
		while (x>0) {
			//System.out.println(x);
			ans+=c[x];
			x-=lowbit(x);
		}
		return ans;
	}
	static Map<Integer, List<Integer>>map;
	static void mapinit(int chu,int n) {
		map=new HashMap<Integer, List<Integer>>();
		for (int i = chu; i < n+chu; i++) {
			map.put(i, new ArrayList<Integer>());
		}
	}
}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}

	public long nextlong() {
		return Long.parseLong(next());
	}

	public double nextdouble() {
		return Double.parseDouble(next());
	}
}

**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 provided code, we need to consider edge cases and scenarios that can cause the code to fail. Here are some steps to think through this:

1. **Edge Cases**:
   - Small values of \( n \) and \( q \).
   - Large values of \( n \) and \( q \) within the constraints.
   - Arrays with all elements being the same.
   - Arrays with elements in ascending or descending order.
   - Arrays with elements that are multiples of the query values.

2. **Specific Scenarios**:
   - Queries where \( x \) is very small.
   - Queries where \( x \) is very large.
   - Queries where \( x \) is exactly half of the elements in the array.

3. **Buggy Code Analysis**:
   - The code seems to have a complex structure, and it's not immediately clear where the bug might be. However, we can focus on the logic that handles the median calculation and the operations on the array.

4. **Test Input Generation**:
   - Let's create a small-scale test input that includes a mix of the above scenarios.

Here is a small-scale test input that covers some of these edge cases:

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

This test input includes:
- Two test cases.
- The first test case has 5 elements and 5 queries.
- The second test case has 6 elements and 3 queries.
- The elements in the arrays are chosen to test different values of \( x \).

This should help in exposing any bugs 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: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
//In the name of God
#include<bits/stdc++.h>
/*MHN*/
using namespace std;
typedef long long int lli;
typedef long double ld;
typedef pair<lli,lli> pii;
typedef vector<lli> ve;
typedef vector<pii> vp;
const lli N=4e5+100;
const lli mod=1e9+7;//998244353;//1e9+9
const lli INF=1e18;
lli power(lli x,lli y){lli res = 1;x = x % mod;while(y>0){if( y & 1 ){res = (res * x) % mod;}y = y >> 1;x = (x * x)%mod;}return res;}
lli modinv(lli x){return power(x,mod-2);}
lli divide(lli x,lli y){return ((x%mod) * modinv(y))%mod;}
#define all(v) v.begin(),v.end()
#define MIN(v) *min_element(all(v))
#define MAX(v) *max_element(all(v))
#define migmig ios_base::sync_with_stdio(false),cout.tie(0), cin.tie(0);
#define read freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define minheap priority_queue<pair<lli,pii>, std::vector<pair<lli,pii>>, std::greater<pair<lli,pii>>>
#define set_preci(x) cout << fixed << setprecision(x);
#define debug(x) cerr << (#x) << " " << (x) << endl
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define SUM(v) accumulate(v.begin(),v.end(), 0LL)
#define FOR(x,n) for(lli x = 0; x < n; x++)
#define FORS(x,n) for(lli x = 1; x <= n; x++)
#define TEST int TQPWIEJR; cin>>TQPWIEJR;while(TQPWIEJR--)
#define BN(l,a) while(l){a.PB(l%2);l/=2;}
#define lb lower_bound
#define ub upper_bound
#define endl "\n"
#define sep " "
lli tmp;
ve t;
int main(){
    migmig;
    TEST{
        lli n,q;
        cin>>n>>q;
        ve v(n);
        FOR(i,n){
            cin>>v[i];
        }
        while(q--){
            lli x;
            cin>>x;
            t.clear();
            FOR(i,n){
                t.PB(v[i]%x);
            }
            sort(all(t));
            cout<<t[n/2]<<sep;
        }
        cout<<endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

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

#define int long long
#define endl "\n"
#define run freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define bismillah ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pii pair<int,int>
#define f first
#define s second

int MOD=1e9+7;
int db[]={0,0,1,-1};
int dk[]={1,-1,0,0};

int pwr(int a,int b){
  if(b==0)return 1;
  int tmp=pwr(a,b/2);
  tmp=(tmp*tmp)%MOD;
  if(b%2)tmp=(tmp*a)%MOD;
  return tmp%MOD;
}

vector<int> vt(1e5+5);
int x;

bool cmp(int a,int b){
  return a%x<=b%x;
}
signed main()
{
  // run
  bismillah

  int t;cin>>t;
  while(t--){
    int n,q;cin>>n>>q;
    for(int i=0;i<n;i++)
      cin>>vt[i];
    vector<int>dp(n+1,-1);
    while(q--){
      cin>>x;
      if(dp[x]!=-1){
        cout<<dp[x]<<" ";
        continue;
      }
      sort(vt.begin(),vt.begin()+n,cmp);
      int ans;
      if(n&1)ans=vt[((n+1)/2)-1]%x;
      else ans=vt[((n+2)/2)-1]%x;
      dp[x]=ans;
      cout<<ans<<" ";
    }
    cout<<endl;
  }

}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Small arrays where the median is affected by the operation.
   - Arrays where all elements are the same or very close in value.
   - Arrays where the median is at the boundary of the array.

2. **Behavior of the `cmp` Function**:
   - The `cmp` function sorts elements based on the remainder when divided by `x`. This can lead to incorrect sorting if `a%x` and `b%x` are equal but `a` and `b` are not.

3. **Specific Values of `x`**:
   - Values of `x` that are equal to elements in the array.
   - Values of `x` that are larger than the elements in the array.

Let's construct a test case that combines these aspects:

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

Explanation:
- **Array**: `[1, 2, 3, 4, 5]`
- **Queries**: `[1, 2, 3, 4, 5]`

For each query:
- **x = 1**: The array remains unchanged, and the median is `3`.
- **x = 2**: The array can be transformed to `[1, 0, 1, 0, 1]`, and the median is `1`.
- **x = 3**: The array can be transformed to `[1, 2, 0, 1, 2]`, and the median is `1`.
- **x = 4**: The array can be transformed to `[1, 2, 3, 0, 1]`, and the median is `1`.
- **x = 5**: The array can be transformed to `[1, 2, 3, 4, 0]`, and the median is `2`.

This test case should help expose any issues with the sorting and median calculation 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: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
from collections import Counter
for _ in range(int(input())):
    n, Q = map(int, input().split())
    medc = (n+1)//2 if n%2 else (n//2 + 1)
    a = Counter(map(int, input().split()))
    dp = [-1]*n
    dp2 = [-1]*n

    for q in range(Q):
        x = int(input())
        if not dp2[x-1]+1:
            if not dp[x-1]+1:
                l = [0]*x
                for i in a: l[i%x]+=a[i]
                for i in range(1, x): l[i] += l[i-1]
                dp[x-1] = l
            l = 0
            h = x-1
            while l < h:
                mid = (l+h)//2
                c = dp[x-1][mid]
                if c < medc: l = mid+1
                else: h = mid
            dp2[x-1] = h
        print(dp2[x-1], end = " ")
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
def findMid(l, h, a, x, medc):
    while l < h:
        mid = (l+h)//2
        c = sum(i%x <= mid for i in a)
        if c < medc: l = mid+1
        else: h = mid
    return h

for _ in range(int(input())):
    n, Q = map(int, input().split())
    medc = (n+1)//2 if n%2 else (n//2 + 1)
    a = list(map(int, input().split()))
    for q in range(Q):
        x = int(input())
        print(findMid(0, x-1, a, x, medc), end = " ")
    print()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code has a binary search approach to find the minimum possible median, but it might fail in cases where the array has repeated elements or when the value of `x` is large compared to the elements in the array.

Let's break down the requirements and generate a test input:

1. **Test Case 1:**
   - **Array with repeated elements:** This can cause issues in the binary search logic.
   - **Query values that are both smaller and larger than the array elements:**

2. **Test Case 2:**
   - **Array with elements close to the upper limit:** This can test the boundary conditions.
   - **Query values that are large and small:**

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

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

Explanation:
- **Test Case 1:**
  - Array: [1, 2, 2, 4, 5]
  - Queries: 1, 2, 3, 4, 5
  - This array has repeated elements (2, 2), which can cause issues in the binary search.
  - The queries include both small and large values relative to the array elements.

- **Test Case 2:**
  - Array: [1, 2, 6, 4, 1, 3]
  - Queries: 1, 2, 6
  - This array has elements close to the upper limit (6), which can test the boundary conditions.
  - The queries include values that are both small and large relative to the array elements.

This test input should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;

public class Solution
{
    int check(int l, int arr[], int x)
    {
        for(int i=0;i<l;i++)
        {
            arr[i]=arr[i]%x;
        }
        Arrays.sort(arr);
        return arr[l/2];
    }
    public static void main(String args[])
    {
        Scanner k=new Scanner(System.in);
        Solution ob=new Solution();
        int n=k.nextInt();
        for(int i=0;i<n;i++)
        {
            int l=k.nextInt();
            int q=k.nextInt();
            int arr1[]=new int[l];
            for(int j=0;j<l;j++)
            {
                arr1[j]=k.nextInt();
            }
            for(int m=0;m<q;m++)
            {
                int arr[]=new int[l];
                for(int c=0;c<l;c++)
                {
                    arr[c]=arr1[c];
                }
                if(l==1)
                {
                    System.out.println(arr[0]%k.nextInt());
                    continue;
                }
                else if(l==2)
                {
                    int x=k.nextInt();
                    arr[0]=arr[0]%x;
                    arr[1]=arr[1]%x;
                    System.out.println(((arr[0])>(arr[1]))?arr[0]:arr[1]);
                    continue;
                }
                else
                    System.out.println(ob.check(l,arr,k.nextInt()));
            }
        }
    }
}

**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
3 2
1 2 3
1
3
5 2
1 2 3 4 5
2
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;

public class Solution
{
    int check(int l, int arr[], int x)
    {
        for(int i=0;i<l;i++)
        {
            arr[i]=arr[i]%x;
        }
        Arrays.sort(arr);
        return arr[l/2];
    }
    public static void main(String args[])
    {
        Scanner k=new Scanner(System.in);
        Solution ob=new Solution();
        int n=k.nextInt();
        for(int i=0;i<n;i++)
        {
            int l=k.nextInt();
            int q=k.nextInt();
            int arr1[]=new int[l];
            for(int j=0;j<l;j++)
            {
                arr1[j]=k.nextInt();
            }
            for(int m=0;m<q;m++)
            {
                int arr[]=new int[l];
                for(int c=0;c<l;c++)
                {
                    arr[c]=arr1[c];
                }
                if(l==1)
                {
                    System.out.println(arr[0]%k.nextInt());
                    continue;
                }
                if(l==2)
                {
                    int x=k.nextInt();
                    arr[0]=arr[0]%x;
                    arr[1]=arr[1]%x;
                    System.out.println(((arr[0])>(arr[1]))?arr[0]:arr[1]);
                    continue;
                }
                System.out.println(ob.check(l,arr,k.nextInt()));
            }
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the code to fail. The primary issue in the code is that it does not correctly handle the operation of reducing elements by `x` multiple times until they are less than `x`. Instead, it only performs a single modulo operation, which can lead to incorrect results.

Here are some scenarios to consider:
1. **Single Element Array**: The code handles this case correctly.
2. **Two Elements Array**: The code handles this case correctly.
3. **Multiple Elements Array**: The code might fail if the elements are large and need to be reduced multiple times.

To create a test input that exposes the bug, we can use an array with large elements and a query `x` that is relatively small. This will force the code to reduce the elements multiple times, and the incorrect modulo operation will become apparent.

Let's create a test input with:
- `t = 1` (one test case)
- `n = 5` (five elements in the array)
- `q = 1` (one query)
- `a = [100, 100, 100, 100, 100]` (all elements are large)
- `x = 3` (a small value for `x`)

This input will force the code to reduce the elements multiple times, and the incorrect modulo operation will become apparent.

Here is the test input:

```plaintext
1
5 1
100 100 100 100 100
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
for _ in range(int(input())):
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    c = [0] * (n + 1)

    for i in range(n):
        c[a[i]] += 1

    for i in range(1, n + 1):
        c[i] += c[i - 1]

    res = [0] * (n + 1)

    for x in range(1, n + 1):
        l, r = 0, x
        while l < r:
            mid = (l + r) // 2
            cnt = c[mid]
            for k in range(1, n // x + 1):
                cnt += c[min(k * x + mid, n)] - c[k * x - 1]
            if cnt - 1 >= n // 2:
                r = mid
            else:
                l = mid + 1
        res[x] = l

    for _ in range(m):
        x = int(input())
        print(res[x])

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

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

Think step by step.

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

1. **Small-scale Test Case**: Ensure the test case is small enough to be manageable but still complex enough to expose potential bugs.
2. **Edge Cases**:
   - **Array with all elements equal**: This can test how the code handles uniform data.
   - **Array with a large range of values**: This can test how the code handles diverse data.
   - **Queries with values close to the array elements**: This can test the precision of the algorithm.
3. **Median Calculation**:
   - Ensure the array length is such that the median calculation is straightforward to verify.

Given these considerations, let's create a test case:

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

### Explanation:
- **Test Case 1**:
  - **Array**: [1, 2, 3, 4, 5]
  - **Queries**: [1, 2, 5]
  - **Array Length (n)**: 5
  - **Number of Queries (q)**: 3

### Expected Output:
- For \( x = 1 \):
  - The array can be reduced to [0, 1, 2, 3, 4] by performing the operation on all elements.
  - The median of [0, 1, 2, 3, 4] is 2.
- For \( x = 2 \):
  - The array can be reduced to [1, 0, 1, 2, 3] by performing the operation on elements 3, 4, and 5.
  - The median of [1, 0, 1, 2, 3] is 1.
- For \( x = 5 \):
  - The array can be reduced to [1, 2, 3, 4, 0] by performing the operation on element 5.
  - The median of [1, 2, 3, 4, 0] is 2.

### Output:
```plaintext
2 1 2
```

This test case is designed to challenge the buggy code by testing its ability to handle a simple but diverse set of queries and array elements. The expected output is clear and straightforward to verify, making it a good candidate for exposing potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Test
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:

Given an integer $x$, Sakurako can perform the following operation any number of times:

  * Choose an integer $i$ ($1\le i\le n$) such that $a_i\ge x$;    * Change the value of $a_i$ to $a_i-x$.

Using this operation any number of times, she must find the minimum possible median$^{\text{∗}}$ of the array $a$.

Sakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.

$^{\text{∗}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\frac{n+2}{2}$-th position for even $n$, and at the $\frac{n+1}{2}$-th for odd)

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

The first line of each test case contains two integers $n$ and $q$ ($1\le n,q\le 10^5$)  — the number of elements in the array and the number of queries.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1\le a_i\le n$)  — the elements of the array.

The following $q$ lines each contain one integer $x$ ($1\le x\le n$).

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.

Output: For each test case, output $q$ integers  — the answer for each query.

Examples:
input:
2
5 5
1 2 3 4 5
1
2
3
4
5
6 3
1 2 6 4 1 3
2
1
5
output:
0 1 1 1 2
1 0 2

Note:

Buggy Code:
import sys

def calculate_median(freq, n):
    count = 0
    prev_count = 0
    prev_index = -1
    for i in range(len(freq)):
        prev_count = count
        prev_index = i
        count += freq[i]
        if n % 2 == 1:
            if count >= (n + 1) // 2:
                return i
        else:
            if count >= (n + 2) // 2:
                if prev_count < (n + 2) // 2 - 1:
                    return i
                else:
                    return min(i, prev_index)

def solve():
    n, q = map(int, sys.stdin.readline().split())
    arr = sorted(list(map(int, sys.stdin.readline().split())))

    for _ in range(q):
        x = int(sys.stdin.readline())
        freq = [0] * x
        for num in arr:
            freq[num % x] += 1
        median = calculate_median(freq, n)
        print(median, end=" ")
    print()

def main():
    t = int(sys.stdin.readline())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Cases:**
   - Small arrays (e.g., arrays with 1 or 2 elements).
   - Arrays with all elements being the same.
   - Arrays with elements that are multiples of the query values.

2. **Special Cases:**
   - Queries with values that are close to the elements in the array.
   - Queries with values that are much larger than the elements in the array.

3. **Behavior of the Buggy Code:**
   - The buggy code uses a frequency array to count the occurrences of remainders when elements are divided by the query value.
   - It calculates the median based on these frequencies, which might not correctly handle the transformation of the array elements.

Given these considerations, let's create a test input that includes these edge cases and special scenarios.

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

**Explanation:**

1. **Test Case 1:**
   - Array: [1]
   - Query: 1
   - Expected Output: 0
   - This tests the case where the array has a single element and the query value is the same as the element.

2. **Test Case 2:**
   - Array: [1, 2]
   - Query: 1
   - Expected Output: 0
   - This tests the case where the array has two elements and the query value is 1, which should reduce the array to [0, 1].

3. **Test Case 3:**
   - Array: [1, 2, 3, 4, 5]
   - Queries: 1, 2, 3
   - Expected Output: 0 1 1
   - This tests the case where the array has multiple elements and the queries are small, which should reduce the array to [0, 1, 2, 3, 4] for x=1, [1, 0, 1, 0, 1] for x=2, and [1, 2, 0, 1, 2] for x=3.

This test input should help expose the bug in the provided code by testing various edge cases and special scenarios.