**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

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

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
// #pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#define int long long int
#define ll long long int
#define ld long double
#define getFaster ios_base::sync_with_stdio(false), cin.tie(nullptr)
#define rep(i, init, n) for (int i = init; i < (int)n; i++)
#define rev(i, n, init) for (int i = (int)n; i >= init; i--)
#define MOD1 1e9 + 7
#define MOD2 998244353
#define f first
#define s second
#define endl '\n'
#define pii pair<int, int>
#define tii tuple<int, int>
#define all(v) v.begin(), v.end()
#define mt make_tuple
#define precise(i) cout << fixed << setprecision(i)
#define codejam cout << "Case #" << ii + 1 << ": ";
#define impossible cout << "IMPOSSIBLE" << endl;
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define error(s) throw runtime_error(s)
#define prev prev1
#define hash hash1
std::mt19937_64 gen(std::chrono::steady_clock::now().time_since_epoch().count());
std::mt19937 gen1(std::chrono::steady_clock::now().time_since_epoch().count());

//change according to int or long long int
int rng(int l, int r)
{
    return std::uniform_int_distribution<int>(l, r)(gen);
}

const long double PI = atan(1.0) * 4;
const int32_t INF32 = 2e9 + 7;
const int64_t INF64 = 3e18;
const int32_t LOG = 21;
int32_t MOD = MOD1;

using namespace std;

//-------------------DEBUGGING-------------------------
void my_debugger(string s, int LINE_NUM) { cerr << endl; }
template <typename start, typename... end>
void my_debugger(string s, int LINE_NUM, start x, end... y)
{
    if (s.back() != ',')
    {
        s += ',';
        cerr << "LINE(" << LINE_NUM << "): ";
    }
    int i = s.find(',');
    cerr << s.substr(0, i) << " = " << x;
    s = s.substr(i + 1);
    if (!s.empty())
        cerr << ", ";
    my_debugger(s, LINE_NUM, y...);
}

#ifdef AJIT
#define debug(...) my_debugger(#__VA_ARGS__, __LINE__, __VA_ARGS__);
#else
#define debug(...) ;
#endif

void setMod(int mod_val)
{
    MOD = mod_val;
}

void files_init()
{
    freopen("file.in", "r", stdin);
    freopen("file.out", "w", stdout);
}

const int N = 1e6 + 5;
const int LOGN = 20;

int power(int x, int y, int mod = MOD)
{
     if (y == 0)
          return 1;
     int temp = power(x, y / 2);
     temp = (1LL * temp * temp) % mod;
     if (y & 1)
          temp = (1LL * temp * x) % mod;
     return temp;
}

//-----------------------------------------------------

struct segtree {
    int n;
    vector<int> seg;
    vector<int> history;

    void init(int n) {
        this->n = n;
        int size = 1;
        while (size < n) {
            size *= 2;
        }
        seg.resize(size * 2);
    }

    void clear() {
        for(auto& it: history) {
            update(it, 0);
        }
        history.clear();
    }

    segtree(int n) {
        init(n);
    }

    void update(int i, int v, int x, int lx, int rx) {
        if (rx - lx == 1) {
            seg[x] = v;
            return;
        }

        int mid = (lx + rx) / 2;
        if (i < mid) {
            update(i, v, 2 * x + 1, lx, mid);
        }
        else {
            update(i, v, 2 * x + 2, mid, rx);
        }

        seg[x] = max(seg[2 * x + 1], seg[2 * x + 2]);
    }

    void update(int i, int v) {
        history.push_back(i);
        update(i, v, 0, 0, n);
    }

    int bound(int k, int x, int lx, int rx) {
        if (seg[x] < k) {
            return -1;
        }

        if (rx - lx == 1) {
            return lx;
        }

        int mid = (lx + rx) / 2;
        if (seg[2 * x + 1] >= k) {
            return bound(k, 2 * x + 1, lx, mid);
        }

        return bound(k, 2 * x + 2, mid, rx);
    }

    int bound(int k) {
        return bound(k, 0, 0, n);
    }
};

int32_t main()
{
    // getFaster;
    int tests = 1;
    cin >> tests;

    int LIM = 2e6 + 5;
    segtree seg(2 * LIM + 5);

    while (tests--) {
        int n;
        cin >> n;
        vector<int> a(n);
        rep(i,0,n) cin >> a[i];
        set<int> s_num;

        // BASE CASE
        {
            s_num.insert(0);
            s_num.insert(2 * LIM + 1);
            seg.update(1, 2 * LIM);
        }

        auto add = [&](int i) {
            auto it_num_right = s_num.lower_bound(i);
            assert(it_num_right != s_num.begin());
            auto it_num_left = it_num_right;
            it_num_left--;

            int left = (*it_num_left) + 1, right = (*it_num_right) - 1;
            seg.update(left, 0);
            seg.update(left, i-left);
            seg.update(i+1, right-i);
            s_num.insert(i);
        };

        auto remove = [&](int i) {
            auto it_num = s_num.lower_bound(i);
            assert(it_num != s_num.begin());
            assert(it_num != s_num.end());
            auto it_num_left = it_num;
            auto it_num_right = it_num;
            it_num_left--, it_num_right++;

            int left = (*it_num_left) + 1;
            int right = (*it_num_right) - 1;
            seg.update(left, 0);
            seg.update(i+1, 0);
            seg.update(left, right-left+1);
            s_num.erase(i);
        };

        auto getLoad = [&](int k) -> int {
            return seg.bound(k);
        };

        for(auto& x: a) {
            add(x);
        }

        int m;
        cin >> m;
        while (m--) {
            char op;
            int x;
            cin >> op >> x;
            if (op == '+') {
                add(x);
            }
            else if (op == '-') {
                remove(x);
            }
            else {
                cout << getLoad(x) << " ";
            }
        }

        cout << endl;
        seg.clear();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

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

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
from bisect import insort
import sys
input = sys.stdin.readline
LIMIT_UP = 500
LIMIT_DOWN = 400
class Part:
    def __init__(self, arr) -> None:
        self.array = arr
        self.max_diff()

    def insert(self, x):
        if isinstance(x, int):
            insort(self.array, x)
        else:
            self.array = x + self.array
        p = -1
        if len(self.array) > LIMIT_UP:
            p = self.array[LIMIT_DOWN:]
            self.array = self.array[:LIMIT_DOWN]
        self.max_diff()
        return p

    def remove(self, x):
        self.array.remove(x)
        self.max_diff()
        return bool(self.array)

    def max_diff(self):
        self.max = 0
        for x, y in zip(self.array, self.array[1:]):
            self.max = max(self.max, y - x - 1)

    def get_k(self, k):
        for x, y in zip(self.array, self.array[1:]):
            if k <= y - x - 1:
                return x + 1
        assert False

    def __str__(self) -> str:
        return str(self.array)

    def __repr__(self) -> str:
        return f'Part({self.array})'

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    parts = []
    for i in range(0, n, LIMIT_DOWN):
        parts.append(Part(a[i:i + LIMIT_DOWN]))
    q = int(input())
    res = []
    for _ in range(q):
        type_, x = input().split()
        x = int(x)
        if type_ == '+':
            j = 0
            for i, part in enumerate(parts):
                if x < part.array[0]:
                    break
                j = i
            p = x
            while p != -1:
                if j == len(parts):
                    if isinstance(p, int):
                        parts.append(Part([p]))
                    else:
                        parts.append(Part(p))
                    break
                p = parts[j].insert(p)
                j += 1
        elif type_ == '-':
            j = 0
            for i, part in enumerate(parts):
                if x < part.array[0]:
                    break
                j = i

            p = parts[j].remove(x)
            if not p:
                parts.pop(j)
        else:
            ans = -1
            if not parts or x <= parts[0].array[0] - 1:
                ans = 1
            else:
                for i, part in enumerate(parts):
                    if x <= part.max:
                        ans = part.get_k(x)
                        break
                    y = (parts[i + 1].array[0] if (i + 1 < len(parts)) else float("inf"))
                    if x <= y - part.array[-1] - 1:
                        ans = part.array[-1] + 1
                        break
            res.append(ans)
    print(*res, sep='\n')

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

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

Think step by step.

Response:
```plaintext
3
5
1 2 3 4 5
10
- 3
? 1
? 2
? 3
+ 3
? 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: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

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

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
class Node:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None
        self.height = 1

class AVLTree:
    def __init__(self):
        self.root = None

    def height(self, node):
        return node.height if node else 0

    def balance(self, node):
        return self.height(node.left) - self.height(node.right) if node else 0

    def update_height(self, node):
        node.height = 1 + max(self.height(node.left), self.height(node.right))

    def rotate_right(self, y):
        x = y.left
        T2 = x.right
        x.right = y
        y.left = T2
        self.update_height(y)
        self.update_height(x)
        return x

    def rotate_left(self, x):
        y = x.right
        T2 = y.left
        y.left = x
        x.right = T2
        self.update_height(x)
        self.update_height(y)
        return y

    def insert(self, root, key):
        if not root:
            return Node(key)
        if key < root.value:
            root.left = self.insert(root.left, key)
        else:
            root.right = self.insert(root.right, key)

        self.update_height(root)
        balance = self.balance(root)

        if balance > 1:
            if key < root.left.value:
                return self.rotate_right(root)
            else:
                root.left = self.rotate_left(root.left)
                return self.rotate_right(root)
        if balance < -1:
            if key > root.right.value:
                return self.rotate_left(root)
            else:
                root.right = self.rotate_right(root.right)
                return self.rotate_left(root)

        return root

    def delete(self, root, key):
        if not root:
            return root
        if key < root.value:
            root.left = self.delete(root.left, key)
        elif key > root.value:
            root.right = self.delete(root.right, key)
        else:
            if root.left is None:
                return root.right
            elif root.right is None:
                return root.left
            temp = self.get_min_value_node(root.right)
            root.value = temp.value
            root.right = self.delete(root.right, temp.value)

        if root is None:
            return root

        self.update_height(root)
        balance = self.balance(root)

        if balance > 1:
            if self.balance(root.left) >= 0:
                return self.rotate_right(root)
            else:
                root.left = self.rotate_left(root.left)
                return self.rotate_right(root)
        if balance < -1:
            if self.balance(root.right) <= 0:
                return self.rotate_left(root)
            else:
                root.right = self.rotate_right(root.right)
                return self.rotate_left(root)

        return root

    def get_min_value_node(self, root):
        current = root
        while current.left is not None:
            current = current.left
        return current

    def find_k_load(self, k):
        def inorder(node, prev_value):
            nonlocal k, result
            if not node or result:
                return prev_value

            left_value = inorder(node.left, prev_value)
            if node.value - left_value > k:
                result = left_value + 1
                return left_value

            right_value = inorder(node.right, node.value)
            return right_value

        result = None
        last_value = inorder(self.root, 0)
        return result if result else last_value + 1

def solve_test_case():
    n = int(input())
    initial_set = list(map(int, input().split()))
    m = int(input())

    tree = AVLTree()
    for num in initial_set:
        tree.root = tree.insert(tree.root, num)

    results = []
    for _ in range(m):
        op, x = input().split()
        x = int(x)

        if op == '+':
            tree.root = tree.insert(tree.root, x)
        elif op == '-':
            tree.root = tree.delete(tree.root, x)
        else:  # op == '?'
            results.append(tree.find_k_load(x))

    return results

t = int(input())
for _ in range(t):
    print(*solve_test_case())

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

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

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

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

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            TreeSet<Integer> set = new TreeSet<>();
            for (int i = 0; i < n; i++) {
                set.add(sc.nextInt());
            }

            int m = sc.nextInt();
            List<Integer> results = new ArrayList<>();

            while (m-- > 0) {
                String op = sc.next();
                if (op.equals("+")) {
                    int x = sc.nextInt();
                    set.add(x);
                } else if (op.equals("-")) {
                    int x = sc.nextInt();
                    set.remove(x);
                } else if (op.equals("?")) {
                    int k = sc.nextInt();
                    results.add(findKLoad(set, k));
                }
            }

            for (int result : results) {
                System.out.print(result + " ");
            }
            System.out.println();
        }

        sc.close();
    }

    private static int findKLoad(TreeSet<Integer> set, int k) {
        int d = 1;
        while (true) {
            boolean isValid = true;
            for (int i = 0; i < k; i++) {
                if (set.contains(d + i)) {
                    isValid = false;
                    d = set.ceiling(d + i) + 1;
                    break;
                }
            }
            if (isValid) {
                return d;
            }
        }
    }
}

**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 bugs in the given buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are some key points to consider:

1. **Initial Set with Gaps**: The initial set should have gaps to ensure that the `findKLoad` function has to work hard to find the correct `k`-load.
2. **Operations That Modify the Set**: Include operations that insert and remove elements to change the set's structure, which can affect the `k`-load.
3. **Queries for Different Values of `k`**: Include queries for various values of `k` to ensure the function handles different lengths of sequences.

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

```plaintext
1
5
1 3 5 7 9
8
? 3
+ 2
+ 4
+ 6
+ 8
? 3
- 2
? 3
```

Explanation:
- **Initial Set**: `{1, 3, 5, 7, 9}`. This set has gaps, making it challenging to find the `k`-load.
- **Operations**:
  - `? 3`: The initial `3`-load is `2` because `2, 3, 4` are absent.
  - `+ 2`: Insert `2` into the set. Now the set is `{1, 2, 3, 5, 7, 9}`.
  - `+ 4`: Insert `4` into the set. Now the set is `{1, 2, 3, 4, 5, 7, 9}`.
  - `+ 6`: Insert `6` into the set. Now the set is `{1, 2, 3, 4, 5, 6, 7, 9}`.
  - `+ 8`: Insert `8` into the set. Now the set is `{1, 2, 3, 4, 5, 6, 7, 8, 9}`.
  - `? 3`: The `3`-load is `10` because `10, 11, 12` are absent.
  - `- 2`: Remove `2` from the set. Now the set is `{1, 3, 4, 5, 6, 7, 8, 9}`.
  - `? 3`: The `3`-load is `2` because `2, 3, 4` are absent.

This test input ensures that the `findKLoad` function is tested with a set that has gaps and is modified multiple times, making it a good candidate to expose any bugs in the implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

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

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

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

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

        int t = sc.nextInt(); // number of test cases
        while (t-- > 0) {
            int n = sc.nextInt(); // initial size of the set
            TreeSet<Integer> set = new TreeSet<>();

            for (int i = 0; i < n; i++) {
                set.add(sc.nextInt());
            }

            int m = sc.nextInt(); // number of operations
            while (m-- > 0) {
                String operation = sc.next();
                if (operation.equals("+")) {
                    int x = sc.nextInt();
                    set.add(x);
                } else if (operation.equals("-")) {
                    int x = sc.nextInt();
                    set.remove(x);
                } else if (operation.equals("?")) {
                    int k = sc.nextInt();
                    // Calculate k-load
                    int minLoad = 1;

                    // Iterate from the smallest value to the largest in the set
                    for (int i : set) {
                        if (i >= minLoad + k) {
                            break;
                        }
                        // If the current element `i` is less than `minLoad + k`
                        // then it can affect the result
                        if (i >= minLoad && i < minLoad + k) {
                            minLoad = i + 1;
                        }
                    }

                    out.println(minLoad);
                }
            }
        }

        out.close();
        sc.close();
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the provided code lies in the way it calculates the `k-load`. The code iterates through the elements in the set and updates `minLoad` whenever it finds an element that is less than `minLoad + k`. However, this approach might not handle cases where elements are removed or added in a way that significantly changes the distribution of gaps in the set.

2. **Edge Cases**:
   - **Initial Set with Large Gaps**: Create an initial set with large gaps to ensure that the `k-load` calculation is more complex.
   - **Operations That Remove and Add Elements**: Include operations that remove elements from the set and then add new elements, potentially changing the gaps and the `k-load`.

3. **Small-Scale Test Input**:
   - **Test Case 1**: Start with a small set with large gaps and perform a series of operations to test the `k-load` calculation.

Let's create a test input that fits these criteria:

```plaintext
1
5
1 10 100 1000 10000
10
? 2
- 10
? 2
+ 5
? 3
- 100
? 4
+ 50
? 5
- 10000
? 6
```

### Explanation:
- **Initial Set**: `{1, 10, 100, 1000, 10000}`
  - Large gaps between elements.
- **Operations**:
  - `? 2`: Initial `2-load` should be `2` (since `2, 3` are not in the set).
  - `- 10`: Remove `10`.
  - `? 2`: `2-load` should now be `11` (since `11, 12` are not in the set).
  - `+ 5`: Add `5`.
  - `? 3`: `3-load` should now be `3` (since `3, 4, 5` are not in the set).
  - `- 100`: Remove `100`.
  - `? 4`: `4-load` should now be `101` (since `101, 102, 103, 104` are not in the set).
  - `+ 50`: Add `50`.
  - `? 5`: `5-load` should now be `105` (since `105, 106, 107, 108, 109` are not in the set).
  - `- 10000`: Remove `10000`.
  - `? 6`: `6-load` should now be `10001` (since `10001, 10002, 10003, 10004, 10005, 10006` are not in the set).

This test input is designed to challenge the `k-load` calculation by introducing large gaps and operations that significantly alter the set's contents.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <math.h>
#include <chrono>

typedef unsigned long long ull;

using namespace std;

void solve(){
    int n, k;
    cin >> n >> k;
    bool* evenOrOdd = new bool[(int) ceil(log2((double) n))]; // true = odd
    int numIterations = 0;
    int currentLength = n;
    while (currentLength >= k){
        evenOrOdd[numIterations] = currentLength % 2 == 1;
        numIterations++;
        currentLength /= 2;
    }
    int thingy = 0;
    int* indices = new int[(int) pow(2, numIterations)];
    for (int i = numIterations - 1; i >= 0; i--){
        bool odd = evenOrOdd[i];
        int addition = (int) odd;
        if (odd){
            indices[thingy] = currentLength + 1;
        }
        for (int j = thingy + addition; j < 2 * thingy + addition; j++){
            indices[j] = indices[j - thingy - addition] + currentLength + addition;
        }
        currentLength = currentLength * 2 + addition;
        thingy = thingy * 2 + addition;
    }
    delete[] evenOrOdd;
    ull sum = 0;
    for (int i = 0; i < thingy; i++){
        sum += indices[i];
    }
    delete[] indices;
    cout << sum << "\n";
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    //chrono::steady_clock::time_point t1 = chrono::steady_clock::now();
    while (t--)
        solve();
    //chrono::steady_clock::time_point t2 = chrono::steady_clock::now();
    //cout << chrono::duration_cast<chrono::duration<double>>(t2 - t1).count();
    cout.flush();
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where `n` and `k` are at their boundary values.
2. **Small-Scale Input**: Ensure the input is small enough to be manageable but still challenging.
3. **Odd and Even Lengths**: Test cases with both odd and even lengths for segments.
4. **Laziness Factor**: Test cases where `k` is close to `n` to see how the code handles segments that are almost as long as `k`.

Given these considerations, a small-scale test input that can expose the bug might look like this:

```plaintext
6
7 2
11 3
5 1
6 1
10 5
15 4
```

This test input includes:
- A case with an odd length and a small `k` (`7 2`).
- A case with a larger odd length and a medium `k` (`11 3`).
- A case with a small odd length and `k = 1` (`5 1`).
- A case with an even length and `k = 1` (`6 1`).
- A case with an even length and a large `k` (`10 5`).
- A case with a larger odd length and a medium `k` (`15 4`).

These cases should help in identifying any issues in the handling of segment lengths, lazy observation, and the calculation of lucky values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void binsar(int lo,int hi,int k,long long &lucky){
    if(lo>=hi) return;
    else if((hi-lo+1)<k) return;
    else{
        if((hi-lo+1)%2==0){
            int mid = lo + (hi-lo)/2;
            binsar(lo,mid,k,lucky);
            binsar(mid+1,hi,k,lucky);
        }
        else{
            int mid = lo + (hi-lo)/2;
            lucky+=mid;
            if(lo!=hi){
                binsar(lo,mid-1,k,lucky);
                binsar(mid+1,hi,k,lucky);
            }
        }
    }
}
int main(){
    int t;
    cin >> t;
    while(t--){
        long long lucky = 0;
        int n,k;
        cin >> n >> k;
        if(k==1) cout << ((1LL*n*(n+1))/2);
        else{
            int l = 1;
            int hi = n;
            binsar(l,n,k,lucky);
            cout << lucky << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

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

// Function to calculate sum from l to r
long long rangeSum(long long l, long long r) {
  if (l > r) return 0;
  return (l + r) * (r - l + 1) / 2;
}

long long processRange(long long l, long long r, long long k) {
  // Base cases
  if (l > r || r - l + 1 < k) return 0;
  if (r - l + 1 == k) {
    // For segments of exactly length k, calculate directly
    return (l + r) / 2;
  }

  long long mid = (l + r) / 2;
  long long len = r - l + 1;

  // For odd length segments
  if (len % 2 == 1) {
    return mid + processRange(l, mid - 1, k) + processRange(mid + 1, r, k);
  }
  // For even length segments
  return processRange(l, mid, k) + processRange(mid + 1, r, k);
}

void Solve() {
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    cout << processRange(1LL, n, k) << '\n';
  }
}

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  Solve();
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
import math

def isEven(a):
    if a%2==0:
        return True
    else:
        return False

def main():
    nTestCases = int(input())
    for j in range(0,nTestCases):
        Testinput = input()
        Testinput = Testinput.split(" ")
        nLength = int(Testinput[0])
        currentLength = int(nLength)
        nMin = int(Testinput[1])
        Luckynum = 0
        iteration = 0
        while currentLength >= nMin:
            if isEven(currentLength):
                currentLength = int(currentLength/2)
                iteration += 1
            else:
                Luckynum += (nLength+1)*(2**(iteration - 1))
                currentLength = int((currentLength-1)/2)
                iteration += 1
                if currentLength == 1:
                    currentLength = 0
        print(int(Luckynum))

main()

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

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

Think step by step.

Response:
```plaintext
1
7 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
import math

def rec(l, r, k):
    x = r - l + 1
    if x > 2 and (x & (x-1)==0):
        # Skip calculation for powers of two
        return (x * (r + l)) // 2
    elif x >= k:
        m = (l + r) // 2
        if x % 2 == 0:
            # Even-sized range, divide into two halves
            return rec(l, m, k) + rec(m + 1, r, k)
        else:
            # Odd-sized range, add midpoint and split
            return m + rec(l, m - 1, k) + rec(m + 1, r, k)
    return 0

def main():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        l = 1
        r = n
        lucky_val = rec(l, r, k)
        print(lucky_val)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
6
2 1
3 2
4 2
5 2
6 2
7 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
from sys import stdin, stdout

for testcase in range(int(stdin.readline())):
    n_stars, k_laziness = map(int, stdin.readline().split())
    if k_laziness==1: print( n_stars * (n_stars+1) // 2 )
    else:
        level_sum = luckiness = 0
        current_length = n_stars
        while current_length>=k_laziness:
            level_sum <<= 1
            if current_length&1:
                if level_sum == 0:
                    range_start = 1
                    while range_start<n_stars:
                        level_sum += (range_start+current_length)>>1
                        range_start += current_length
                luckiness += level_sum
            current_length >>= 1

        print(luckiness)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.StringTokenizer;

public class CF_28Dec {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            solve1(sc);
        }
        // System.out.println(rec(1, 7, 0, 2));
    }

    public static void solve1(FastReader sc) {
        int n = sc.nextInt();
        int k = sc.nextInt();
        if (k == 1) {
            long x = n;
            System.out.println((x * (x + 1)) / 2);
            return;
        }

        if (k == n) {
            System.out.println(0);
            return;
        }
        int m = n / k;
        HashMap<Integer, Long> dp = new HashMap<>();

        System.out.println(rec(1, n, k, dp));

        // long x = 8765432;
        // System.out.println((x * (x + 1)) / 2);

    }

    public static long rec(int l, int r, int k, HashMap<Integer, Long> dp) {

        int len = r - l + 1;
        int m = (l + r) / 2;

        // System.out.println("l = " + l + " r = " + r + " m = " + m);
        // if (dp[len] != -1) {
        // return dp[len];
        // }
        if (len == k) {
            if (len % 2 != 0) {
                dp.put(m + 1, (long) m);
                return m;
            } else {
                dp.put(m + 1, (long) 0);
                return 0;
            }
        }
        if (len < k) {
            dp.put(m + 1, (long) 0);
            return 0;
        }

        if (dp.containsKey(m + 1)) {
            return dp.get(m + 1);
        }

        if (len % 2 == 0) {
            long left = rec(l, m, k, dp);
            long right = rec(m + 1, r, k, dp);
            dp.put(m + 1, left + right);

            return dp.get(m + 1);

        } else {
            long left = rec(l, m - 1, k, dp);
            long right = rec(m + 1, r, k, dp);
            dp.put(m + 1, left + right + m);
            // System.out.println("l r m");
            // System.out.println(l + " " + r + " " + m);
            return dp.get(m + 1);
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

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

public class Main {
    static FastReader fr = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    public static void Solve() throws IOException {

        long n = fr.nextLong();
        int k = fr.nextInt();
        long sum;
        if(k == 1) {
            sum = (n * (n+1l)) / 2l;
            out.println(sum);
            return;
        }

        long l=1, r=n;
        sum = 0l;
        long mul = 1;
        while (true) {
            if(r-l+1 < k) break;
            if((r-l+1) % 2 == 0) {
                r = (r + l)/2l;
                mul *= 2l;
            } else {
                long mid = (r + l)/2l;
                if(mid == (n+1l)/2) sum += mid;
                else {
                    sum += (n + 1l) * mul;
                    mul *= 2l;
                }
                r = mid-1l;
            }
        }
        out.println(sum);
    }

    public static long Solve(long l, long r, long k) {
        if(r-l+1 < k) return 0l;
        long mid = (l + r) / 2l;
        if((l - r + 1) % 2 == 0) {
            return Solve(l, mid, k) + Solve(mid+1, r, k);
        } else {
            return mid + Solve(l, mid-1, k) + Solve(mid+1, r, k);
        }
    }

    static int UP(int x, int [] parent, int [] present) {
        if(x < 0 || x >= parent.length || present[x] == -1) return -1;
        if(parent[x] == x) return x;
        parent[x] = UP(parent[x], parent, present);
        return parent[x];
    }

    //    -------------------------------------------------------------------------------------------------------
    public static void main(String args[]) throws IOException {
//        Factorial((int)1e6);
        int t = fr.nextInt();
//        int t = 1;
        //        long SqxtasrtTime = System.currentTimeMillis();
        while (t-- > 0) {
            Solve();
        }
        out.close();
    }

    //    --------------------------------FOR TAKING FASTER INPUTS----------------------------------------
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {br = new BufferedReader(new InputStreamReader(System.in));}
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {st = new StringTokenizer(br.readLine());}
                catch (IOException e) {e.printStackTrace();}
            }
            return st.nextToken();
        }
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {str = st.nextToken("\n");}
                else {str = br.readLine();}
            } catch (IOException e) {e.printStackTrace();}
            return str;
        }
    }

    //    -------------------------------(X * Y) % MOD-------------------------------------------------------
    public static long ModMul(long a, long b){
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    public static long ModMul(long a, long b, long MOD){
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    static int MOD = (int) 1e9 + 7;
    static int MOD2 = (int) 1e9 + 9;
    static int maxI = Integer.MAX_VALUE;
    static int minI = Integer.MIN_VALUE;
    static long maxL = Long.MAX_VALUE;
    static long minL = Long.MIN_VALUE;

    public static int max(int a, int b) {return Math.max(a, b);}
    public static int min(int a, int b) {return Math.min(a, b);}
    public static long max(long a, long b) {return Math.max(a, b);}
    public static long min(long a, long b) {return Math.min(a, b);}
}

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

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

Think step by step.

Response:
```plaintext
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
int n,q,b[1145141],m,l[1145141],r[1145141],num[1145141],k;
int sum[1145141];
int dfs(int l,int r)
{
	int len=r-l+1;
	if(len<k)return 0;
	if(len%2==0)
	{
		int mid=(l+r)/2;
		return dfs(l,mid)+dfs(mid+1,r);
	}
	else
	{
		int mid=(l+r)/2;
		return mid+dfs(l,mid-1)+dfs(mid+1,r);
	}
}
signed main()
{
	int T;
	cin>>T;
	while(T--)
	{
		cin>>n>>k;
		cout<<dfs(1,n)<<endl;
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

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

long long f(int l, int r, int k) {
	int len = r - l + 1;
	if ((r - l + 1) < k) {
		return 0;
	}
	int m = (l + r) / 2;
	if (len % 2 == 0) {
		return f(l, m, k) + f(m + 1, r, k);
	}
	else {
		return f(l, m - 1, k) + f(m + 1, r, k) + (long long)m;
	}
}

void solve() {
	int n, k;
	cin >> n >> k;
	cout << (f(1, n, k)) << "\n";
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("input1.txt", "r", stdin);
	freopen("output1.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(0);
	cin.tie(0);

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

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
6
2 1
3 2
4 2
5 2
6 2
7 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
def solve():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())

        # Special case for k=1
        if k == 1:
            print((n * (n + 1)) // 2)
            continue

        # Find all midpoints directly
        result = 0
        queue = [(1, n)]

        while queue:
            l, r = queue[0]
            queue.pop(0)

            length = r - l + 1
            if length < k:
                continue

            m = (l + r) >> 1  # Using bit shift for faster division

            if length & 1:  # Odd length - using bitwise for faster check
                result += m
                if l == r:
                    continue
                queue.append((l, m - 1))
                queue.append((m + 1, r))
            else:  # Even length
                queue.append((l, m))
                queue.append((m + 1, r))

        print(result)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
```plaintext
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
```

This test input is designed to expose potential issues in the buggy code by including a variety of test cases:

1. **Small Odd Length with Small k**: `7 2` - This case tests the handling of small odd-length segments and ensures that the midpoint is correctly added to the result.
2. **Small Odd Length with Larger k**: `11 3` - This case tests the handling of segments where the length is still odd, but the segments are not split further when they are less than k.
3. **Large Odd Length with Larger k**: `55 13` - This case tests the handling of larger odd-length segments and ensures that the algorithm correctly handles the recursive splitting and summing of midpoints.
4. **Large Even Length with Small k**: `5801 6` - This case tests the handling of large even-length segments and ensures that the algorithm correctly splits the segments without adding midpoints.
5. **Large Odd Length with Very Large k**: `8919 64` - This case tests the handling of large segments where the length is much larger than k, ensuring that the algorithm correctly stops splitting segments when they are smaller than k.
6. **Very Large n with k=1**: `8765432 1` - This case tests the special case where k is 1, and the algorithm should sum all numbers from 1 to n.

These test cases cover a range of scenarios, including edge cases and large inputs, to thoroughly test the algorithm's correctness and performance.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
def observe_segment(l, r, k):
    """
    Recursively observe a segment of stars and return the lucky value.
    l: left boundary of segment
    r: right boundary of segment
    k: minimum segment length to observe
    """
    # Base case: if segment length is less than k, don't observe
    if r - l + 1 < k:
        return 0

    # Calculate middle position
    m = (l + r) // 2
    lucky_value = 0

    # If segment length is odd
    if (r - l + 1) % 2 == 1:
        # Add middle position to lucky value
        lucky_value += m
        # If not a single star, observe both sides
        if l != r:
            lucky_value += observe_segment(l, m-1, k)
            lucky_value += observe_segment(m+1, r, k)
    # If segment length is even
    else:
        # Split into two equal segments
        lucky_value += observe_segment(l, m, k)
        lucky_value += observe_segment(m+1, r, k)

    return lucky_value

def solve():
    """
    Main function to solve the problem.
    """
    t = int(input())
    results = []

    for _ in range(t):
        n, k = map(int, input().split())

        # Special case: if k=1, use arithmetic sum formula
        if k == 1:
            result = (n * (n + 1)) // 2
        else:
            result = observe_segment(1, n, k)

        results.append(result)

    # Print all results
    for result in results:
        print(result)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
```plaintext
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bewitching Stargazer
time_limit_ms: 2000
memory_limit_mb: 256
Description: Iris looked at the stars and a beautiful problem emerged in her mind. She is inviting you to solve it so that a meteor shower is believed to form.

There are $$$n$$$ stars in the sky, arranged in a row. Iris has a telescope, which she uses to look at the stars.

Initially, Iris observes stars in the segment $$$[1, n]$$$, and she has a lucky value of $$$0$$$. Iris wants to look for the star in the middle position for each segment $$$[l, r]$$$ that she observes. So the following recursive procedure is used:

- First, she will calculate $$$m = \left\lfloor \frac{l+r}{2} \right\rfloor$$$.
- If the length of the segment (i.e. $$$r - l + 1$$$) is even, Iris will divide it into two equally long segments $$$[l, m]$$$ and $$$[m+1, r]$$$ for further observation.
- Otherwise, Iris will aim the telescope at star $$$m$$$, and her lucky value will increase by $$$m$$$; subsequently, if $$$l \neq r$$$, Iris will continue to observe two segments $$$[l, m-1]$$$ and $$$[m+1, r]$$$.

Iris is a bit lazy. She defines her laziness by an integer $$$k$$$: as the observation progresses, she will not continue to observe any segment $$$[l, r]$$$ with a length strictly less than $$$k$$$. In this case, please predict her final lucky value.

Input: Each test contains multiple test cases. The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of test cases follows.

The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq k \leq n \leq 2\cdot 10^9$$$).

Output: For each test case, output a single integer — the final lucky value.

Examples:
input:
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
output:
12
18
196
1975581
958900
38416403456028

Note: In the first test case, at the beginning, Iris observes $$$[1, 7]$$$. Since $$$[1, 7]$$$ has an odd length, she aims at star $$$4$$$ and therefore increases her lucky value by $$$4$$$. Then it is split into $$$2$$$ new segments: $$$[1, 3]$$$ and $$$[5, 7]$$$. The segment $$$[1, 3]$$$ again has an odd length, so Iris aims at star $$$2$$$ and increases her lucky value by $$$2$$$. Then it is split into $$$2$$$ new segments: $$$[1, 1]$$$ and $$$[3, 3]$$$, both having a length less than $$$2$$$, so no further observation is conducted. For range $$$[5, 7]$$$, the progress is similar and the lucky value eventually increases by $$$6$$$. Therefore, the final lucky value is $$$4 + 2 + 6 = 12$$$.

In the last test case, Iris finally observes all the stars and the final lucky value is $$$1 + 2 + \cdots + 8\,765\,432 = 38\,416\,403\,456\,028$$$.

Buggy Code:
import java.util.*;

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

        int t = scanner.nextInt();
        while (t > 0) {
            t--;
            int n = scanner.nextInt();
            int k = scanner.nextInt();

            System.out.println(score(1, n, k));
        }
    }

    private static long score(int l, int r, int k) {
        int sz = r-l+1;
        if (sz < k) return 0;
        int m = (l+r)/2;
        if (sz % 2 == 0) return score(l, m, k) + score(m+1, r, k);
        return m + score(l, m-1, k) + score(m+1, r, k);
    }
}

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

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

Think step by step.

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