**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

Buggy Code:
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Main3 {

    static int n;
    static List<Integer>[] edges;
    static long[] sum;
    static int[] a;
    static long[][] dp;
    static long ans;

    public static void dfs(int u) {
        if (edges[u].size() == 0) {
            dp[u][1] = Integer.MAX_VALUE;
            return;
        }
        for (int v : edges[u]) {
            dfs(v);
            for (int i = 1; i < n; i++) dp[u][i+1] = Math.min(dp[v][i], Integer.MAX_VALUE);
        }
        if (a[u] <= sum[u]) {
            dp[u][1] = sum[u] - a[u];
        }else {
            long need = a[u] - sum[u];
            for (int i = 1; i <= n && need > 0; i++) {
                for (int v : edges[u]) {
                    long mi = Math.min(need, dp[v][i]);
                    dp[v][i] -= mi;
                    need -= mi;
                    ans += i * mi;
                }
            }
        }
    }

    public static void solve() throws IOException{
        n = in.nextInt();
        edges = new ArrayList[n+1];
        Arrays.setAll(edges, e -> new ArrayList<>());
        a = new int[n+1];
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }
        for (int i = 2; i <= n; i++) {
            int u = in.nextInt();
            edges[u].add(i);
        }
        sum = new long[n+1];
        for (int u = 1; u <= n; u++) {
            for (int v : edges[u]) {
                sum[u] += a[v];
            }
        }
        dp = new long[n+1][n+1];
        ans = 0;
        dfs(1);
        out.println(ans);
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

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

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

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}
/*
y = - (a2 * x + c2) / b2
a1 * x^2 + b1 * x + c1 + a2 / b2 * x + c2 / b2 = 0
a1 * x^2 + b1 * x + a2 / b2 * x + c1 + c2 / b2 = 0
a = a1 * b2
b = (b1 * b2 + a2)
c = (c1 * b2 + c2)

 */

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios where the code might fail. Here are some key points to consider:

1. **Tree Structure**: The tree should have a structure that can cause the algorithm to miscalculate the number of operations needed.
2. **Initial Values**: The values on the vertices should be such that the tree is not initially wonderful, and the required operations might be non-trivial.
3. **Edge Cases**: Consider cases where the tree has a single child, multiple children, or where the root itself needs to be modified.

Given the constraints, a small-scale test input that can expose the bug is as follows:

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

### Explanation:

- **Tree Structure**:
  - Vertex 1 has two children: Vertex 2 and Vertex 3.
  - Vertex 2 has one child: Vertex 4.
- **Initial Values**:
  - Vertex 1: 1
  - Vertex 2: 2
  - Vertex 3: 3
  - Vertex 4: 4

### Analysis:

- **Vertex 1**: The sum of its children's values is \(2 + 3 = 5\). Since \(1 \le 5\), no operation is needed for Vertex 1.
- **Vertex 2**: The sum of its child's value is \(4\). Since \(2 \le 4\), no operation is needed for Vertex 2.
- **Vertex 3**: It has no children, so no operation is needed.
- **Vertex 4**: It has no children, so no operation is needed.

However, the code might fail to correctly handle the case where the root (Vertex 1) has multiple children and the sum of their values is significantly larger than the value of the root. The code should correctly calculate the minimum number of operations needed to make the tree wonderful.

This test case is small and complete, and it should help identify any issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

Buggy Code:
T = int(input())
for _ in range(T):
    n = int(input())
    weights = [int(x) for x in input().split(' ')]
    tree = {i:[] for i in range(n)}
    for i,v in enumerate(input().split(' ')):
        v = int(v)-1
        tree[v].append(i+1)

    totals = [sum(map(lambda t: weights[t], tree[i])) for i in range(n)]

    def makewonder(i):
        if not tree[i]:
            return 0

        s = 0
        for j in tree[i]:
            s += weights[j]

        if weights[i] <= s:
            return 0
        for _ in range(weights[i]-s):
            j = min(tree[i], key=find_depth)
            weights[j] += 1

        return weights[i]-s

    def find_depth(i):
        if totals[i] > weights[i] or not tree[i]:
            return 0

        return 1 + min(map(find_depth, tree[i]))

    totalops = 0

    def dfs(i):
        global totalops
        totalops += makewonder(i)
        for v in tree[i]:
            dfs(v)

    dfs(0)

    print(totalops)

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
1 1 1
4
10 1 1 1
1 2 2
5
1 2 3 4 5
1 1 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Wonderful Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: God's Blessing on This ArrayForces!

A Random Pebble

You are given a tree with $n$ vertices, rooted at vertex $1$. The $i$-th vertex has an integer $a_i$ written on it.

Let $L$ be the set of all direct children$^{\text{∗}}$ of $v$. A tree is called wonderful, if for all vertices $v$ where $L$ is not empty, $$a_v \le \sum_{u \in L}{a_u}.$$ In one operation, you choose any vertex $v$ and increase $a_v$ by $1$.

Find the minimum number of operations needed to make the given tree wonderful!

$^{\text{∗}}$ Vertex $u$ is called a direct child of vertex $v$ if:

  * $u$ and $v$ are connected by an edge, and    * $v$ is on the (unique) path from $u$ to the root of the tree.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 5000$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the values initially written on the vertices.

The third line of each test case contains $n - 1$ integers $p_2, p_3 , \ldots, p_n$ ($1 \le p_i < i$), indicating that there is an edge from vertex $p_i$ to vertex $i$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the minimum number of operations needed to make the tree wonderful.

Examples:
input:
4
5
9 3 4 1 2
1 1 3 3
2
5 3
1
2
36 54
1
3
0 0 0
1 2
output:
3
2
0
0

Note: The tree in the first test case:

![](CDN_BASE_URL/9a07d26506b9ad6c01c4152da8319590)

You can apply the operation once on vertex $5$ and twice on vertex $2$ to get a wonderful tree.

In the second test case, you can apply the operation twice on vertex $2$ to get a wonderful tree.

In the third and fourth test cases, the tree is already wonderful, so you don't need to apply any operations.

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

public class E2024E {
    static class Vertex {
        List<Vertex> children;
        final Vertex parent;
        long value;
        long sumChildren;
        int depth = -1;
        boolean finalized = false;

        void addChild(Vertex v) {
            if (children == null) {
                children = new ArrayList<>();
            }
            children.add(v);
            sumChildren += v.value;
        }

        Vertex(int value, Vertex parent) {
            this.value = value;
            this.parent = parent;
            if (parent != null) {
                parent.addChild(this);
            }
        }

        int minDepth() {
            if (depth != -1) {
                return depth;
            }
            if (children == null) {
                return depth = 1;
            }
            int result = Integer.MAX_VALUE;
            for (Vertex v : children) {
                result = Math.min(result, 1 + v.minDepth());
            }
            return depth = result;
        }

        int costOfIncrement() {
            if (children == null || value < sumChildren) {
                return 1;
            }
            if (finalized) {
                return depth;
            }
            int minIncrement = Integer.MAX_VALUE;
            for (Vertex v : children) {
                minIncrement = Math.min(minIncrement, v.costOfIncrement() + 1);
            }
            if (minIncrement == depth) {
                return depth;
            }
            return minIncrement;
        }

        int increment() {
            if (children == null || value < sumChildren) {
                ++value;
                return 1;
            }
            if (finalized) {
                throw new UnsupportedOperationException();
            }
            int minIncrement = Integer.MAX_VALUE;
            int index = -1;
            for (int i = 0; i < children.size(); ++i) {
                int cost = children.get(i).costOfIncrement();
                if (cost < minIncrement) {
                    minIncrement = cost;
                    index = i;
                }
            }
            return 1 + children.get(index).increment();
        }

        long relax() {
            if (children == null || value <= sumChildren) {
                return 0;
            }
            long result = 0;
            while (value > sumChildren) {
                int minIncrement = Integer.MAX_VALUE;
                int minIndex = -1;
                for (int i = 0; i < children.size(); ++i) {
                    int cost = children.get(i).costOfIncrement();
                    if (cost < minIncrement) {
                        minIncrement = cost;
                        minIndex = i;
                    }
                }
                if (children.get(minIndex).finalized) {
                    result += (value - sumChildren) * children.get(minIndex).minDepth();
                    sumChildren = value;
                    break;
                } else {
                    result += children.get(minIndex).increment();
                    sumChildren += 1;
                }
            }
            return result;
        }
    }

    private static void solveOne(Reader in, PrintWriter out) {
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = in.nextInt();
        }
        Vertex[] v = new Vertex[n];
        v[0] = new Vertex(a[0], null);
        for (int i = 1; i < n; ++i) {
            v[i] = new Vertex(a[i], v[in.nextInt() - 1]);
        }
        v[0].minDepth(); // also computes all min depths
        long answer = 0;
        for (int i = n - 1; i >= 0; --i) {
            answer += v[i].relax();
        }
        out.println(answer);
    }

    public static void main(String[] args) throws IOException {
        try (Reader in = new Reader(System.in); PrintWriter out = new PrintWriter(System.out)) {
            int nTests = in.nextInt();
            for (int nTest = 0; nTest < nTests; ++nTest) {
                solveOne(in, out);
            }
        }
    }

    private static class Reader implements Closeable {
        private final BufferedReader br;
        private StringTokenizer st;

        Reader(InputStream is) throws IOException {
            br = new BufferedReader(new InputStreamReader(is));
            st = new StringTokenizer(br.readLine());
        }

        @Override
        public void close() throws IOException {
            br.close();
            st = null;
        }

        String next() {
            while (!st.hasMoreTokens()) {
                try {
                    String line = br.readLine();
                    if (line == null) {
                        return null;
                    }
                    st = new StringTokenizer(line);
                } catch (IOException ex) {
                    throw new RuntimeException(ex);
                }
            }
            return st.nextToken();
        }

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

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

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

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code's logic might fail. The primary issue in the code seems to be related to the handling of the `sumChildren` and the `relax` method, which might not correctly handle certain tree structures or values.

Here are the steps to generate a challenging test input:

1. **Tree Structure**: Create a tree with a non-trivial structure, such as a deep tree or a tree with multiple branches.
2. **Vertex Values**: Assign values to the vertices such that the conditions for making the tree wonderful are not trivially satisfied. For example, some vertices might need a significant number of operations to meet the condition.
3. **Edge Cases**: Consider edge cases where the tree has very few or very many children for a vertex.

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

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

### Explanation:
- **Test Case Count**: 1
- **Number of Vertices**: 5
- **Vertex Values**: [1, 2, 3, 4, 5]
- **Edges**: [1, 1, 3, 3] (which means the tree structure is as follows:
  - Vertex 1 is the root.
  - Vertex 1 has children 2 and 3.
  - Vertex 3 has children 4 and 5.

### Expected Behavior:
- The tree is not initially wonderful because:
  - For vertex 1: \( a_1 = 1 \) and \( \sum_{u \in L} a_u = a_2 + a_3 = 2 + 3 = 5 \). This condition is not met.
  - For vertex 3: \( a_3 = 3 \) and \( \sum_{u \in L} a_u = a_4 + a_5 = 4 + 5 = 9 \). This condition is not met.
- To make the tree wonderful, we need to:
  - Increase \( a_1 \) by 4 to make it 5.
  - Increase \( a_3 \) by 6 to make it 9.
- The total number of operations needed is \( 4 + 6 = 10 \).

This test case is designed to challenge the code's ability to correctly handle the sum of children values and the incremental operations needed to make the tree wonderful.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define F first
#define S second
#define sz size()
#define nl '\n'
void solution()
{
  ll k;cin>>k;
  if(k%2==0){
    cout<<(2*k)-(k/2)-1<<nl;
  }else{
    cout<<(2*k)-(k/2)<<nl;
  }
}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--)
  {
    solution();
  }
}

**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
2
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

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

using namespace std;

#define ll long long
#define ld long double
#define pi pair<int,int>
#define pll pair<long long , long long>

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

    int tc;
    cin >> tc;

    while(tc--){

       ll a;
       cin >> a;

       ll b = a;
       ll c = a*a;
       ll mid = (b+c)/2;
       while(b<=c){

        mid = (b+c)/2;

        ll d  = mid - (int)(powl(mid,0.5));

        if(d == a){

            break;
        }
        else if( d>a ){
            c = mid - 1;
        }else{
            b = mid + 1;
        }
       }

       if(a==1){
        mid = 2;
       }

       cout << mid << endl;

}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight the incorrect behavior of the code. The buggy code seems to have issues with its binary search logic and the way it calculates the number of bulbs that remain on.

### Analysis of the Buggy Code
1. **Binary Search Logic**: The code uses a binary search to find the minimum `n` such that the number of bulbs that remain on is exactly `k`. However, the calculation of `d` (the number of bulbs that remain on) is incorrect.
2. **Initial Conditions**: The code has a special case for `a == 1`, which sets `mid` to 2. This might be a hack to handle a specific edge case but is not a general solution.

### Challenging Test Cases
1. **Edge Case: k = 1**
   - The smallest `n` for `k = 1` is 2. This is a straightforward case, but we need to ensure the code handles it correctly.
2. **Edge Case: k = 2**
   - The smallest `n` for `k = 2` is 4. This is a more complex case that can expose issues in the binary search logic.
3. **Large k Value**
   - A large value of `k` can test the efficiency and correctness of the binary search. For example, `k = 1000` or `k = 10000`.

### Test Input
```plaintext
3
1
2
1000
```

### Explanation
- **Test Case 1 (k = 1)**: The smallest `n` is 2. This is a simple case that should be handled correctly.
- **Test Case 2 (k = 2)**: The smallest `n` is 4. This case is more complex and can expose issues in the binary search logic.
- **Test Case 3 (k = 1000)**: This is a large value that tests the efficiency and correctness of the binary search. The correct answer should be 1009, but the buggy code might struggle to find this value.

This test input should help in identifying the bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
#include <iostream>
#include <math.h>
#include <vector>
#include <set>
#include <algorithm>
#include <iterator>
using namespace std;
typedef long long ll;

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

    double t,k;
    cin >> t;
    while(t--){
        cin >> k;
        cout << (k+floor((sqrt(4*k+1)+1e-6-1)/2.0)+1) << "\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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. The buggy code seems to have an issue with the formula used to calculate the minimum number of bulbs \( n \) required to have exactly \( k \) bulbs on after performing the operations.

### Step-by-Step Analysis:

1. **Understanding the Problem:**
   - The problem involves flipping the state of bulbs in a sequence.
   - Each bulb \( j \) is flipped for every \( i \) where \( j \) is divisible by \( i \).
   - The goal is to find the smallest \( n \) such that exactly \( k \) bulbs are on.

2. **Buggy Code Analysis:**
   - The formula used in the buggy code is:
     ```cpp
     cout << (k + floor((sqrt(4*k + 1) + 1e-6 - 1) / 2.0) + 1) << "\n";
     ```
   - This formula is derived from solving a quadratic equation, but it might not be accurate for all values of \( k \).

3. **Edge Cases:**
   - **Small Values of \( k \):** Test with small values of \( k \) to see if the formula works correctly.
   - **Large Values of \( k \):** Test with large values of \( k \) to see if the formula handles large numbers correctly.
   - **Critical Values of \( k \):** Test with values of \( k \) that are critical points in the sequence.

4. **Challenging Test Input:**
   - We will test with a small value of \( k \) and a large value of \( k \) to see if the formula fails.

### Test Input:
```plaintext
2
1
1000000000000000000
```

### Explanation:
- **Test Case 1: \( k = 1 \)**
  - The expected output should be \( 2 \), as the smallest \( n \) that results in exactly 1 bulb on is 2.
- **Test Case 2: \( k = 1000000000000000000 \)**
  - This is a large value to test the robustness of the formula. The expected output should be a large number that satisfies the condition of having exactly 1,000,000,000,000,000,000 bulbs on.

This test input will help expose any issues in the buggy code's formula for calculating \( n \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
def min_bulbs(k):
    m = int((1 + (1 + 4 * k) ** 0.5) // 2)
    n = k + m
    while n - int(n**0.5) != k:
        m += 1
        n = k + m
    return n

def process_cases(test_cases):
    results = []
    for k in test_cases:
        results.append(min_bulbs(k))
    return results

t = int(input())
test_cases = [int(input()) for _ in range(t)]
results = process_cases(test_cases)

for res in results:
    print(res)

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

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

Think step by step.

Response:
```plaintext
3
1
2
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
from math import isqrt

t = int(input())
for _ in range(t):
    k = int(input())

    l = 0
    r = 2 ** 128

    while l <= r:
        m = (l + r) // 2

        f = m - isqrt(m)

        if f >= k:
            r = m - 1
        else:
            l = m + 1

    print(r + 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
2
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
t = int(input())

for _ in range(t):
    k = int(input())
    print(int((2*k + (4*k+1)**0.5 + 1)/2))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the formula used in the code to produce incorrect results. The formula in the buggy code is:

\[ n = \frac{2k + \sqrt{4k + 1} + 1}{2} \]

This formula is derived from a quadratic equation, and it might fail for certain values of \( k \). Specifically, we should test:

1. Small values of \( k \) to ensure the formula works for the simplest cases.
2. Values of \( k \) that are close to the upper limit of the input range to see if the formula handles large numbers correctly.
3. Values of \( k \) that might cause precision issues due to floating-point arithmetic.

Let's construct a test input with these considerations in mind:

```plaintext
5
1
2
3
1000000000000000000
1000000000000000001
```

This test input includes:
- \( k = 1 \) (smallest valid input)
- \( k = 2 \) (a small but non-trivial input)
- \( k = 3 \) (another small but non-trivial input)
- \( k = 10^{18} \) (a very large input to test precision and performance)
- \( k = 10^{18} + 1 \) (a very large input just above the previous one to test precision and performance)

This should help identify any issues with the formula used in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
// package prectice;

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;

public class p2 {

	public static int f(int a,int b) {
		if(b == 0) {
			return a;
		}
		return f(b,a%b);
	}

	public static Scanner sc = new Scanner(System.in);

	public static int f(int[] arr,int i,HashSet<Integer> ss,String s) {

		if(ss.contains(i)) {
			return 0;
		}

		ss.add(i);
		if(s.charAt(i-1) == '0') {
			return 1+f(arr,arr[i],ss,s);
		}
		return f(arr,arr[i],ss,s);
	}

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();

		while(t-- > 0) {

//			long n = sc.nextLong();
			long k = sc.nextLong();
			long start = 1;
			long end = Long.MAX_VALUE;
			long ans = -1;

			while(start < end) {
				long m = (start + end)/2;

				long tt = m - (long)Math.sqrt(m);

				if(tt == k) {
					ans = m;
					break;
				}
				if(tt > k) {
					end = m-1;
				}
				else {
					start = m+1;
				}

			}

			System.out.println(ans);
		}
	}

}

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

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

Think step by step.

Response:
```plaintext
3
1
4
16
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int k=sc.nextInt();
            int n=2;
            int cnt=0;
            List<Integer> list=new ArrayList<>();
            while((cnt)<k){
                int bulb=1;
                for(int num:list){
                    if(n%num==0){
                        bulb=1-bulb;
                    }
                }
                if(bulb==1){
                    cnt++;
                }
                list.add(n);
                n++;
            }
            System.out.println(n-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
2
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
#include <iostream>
#include <cmath>
using namespace std;

void solve() {
    int t;
    cin >> t;
    while (t--) {
        long long k;
        cin >> k;

        // We need to find the minimum n such that n - floor(sqrt(n)) = k
        long long n = k;

        while (true) {
            long long perfect_squares_count = floor(sqrt(n));
            long long on_bulbs = n - perfect_squares_count; // bulbs that are ON
            if (on_bulbs == k) {
                cout << n << '\n';
                break;
            }
            n++; // increment n and check again
        }
    }
}

int main() {
    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
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

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

const long long N = 1e6;
long long a[N + 1];

int main()
{
    for (long long i = 1; i <= N; i++)
    {
        for (long long j = i; j <= N; j += i)
        {
            a[j]++;
        }
    }

    long long t;
    cin >> t;
    while (t--)
    {
        long long k;
        cin >> k;
        long long cnt = 0;
        long long i = 2;

        while (cnt != k && i <= N)
        {
            if ((a[i] - 1) % 2 != 0)
            {
                cnt++;
            }
            i++;
        }

        cout << i - 1 << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
def sieve(num1):
    d = 0
    prime = [True for _ in range(num1 + 1)]
    p = 2
    while p * p <= 1:
        if prime[p]:
            for j in range(p * p, num1 + 1, p):
                prime[j] = False
        p += 1

    for p in range(2, num1 + 1):
        if prime[p]:
            d += 1
    return d

t = int(input())
for i in range(t):
    n = int(input())
    k = 0
    n1 = n + 1
    n2 = n
    le = 1 + sieve(int(n1 ** 0.5))
    while n1 > n2:
        n2 = n1
        n1 = n + le
        k = 0
        le = sieve(int(n1 ** 0.5)+1)
    print(n2)

**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
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
from math import ceil, floor
import sys, threading
def input():
    temp = sys.stdin.readline().strip()
    return temp
def INT_INPUT():
    return int(input())
def TUPLE_INPUT():
    return input().split()
def TUPLE_INT_INPUT():
    return map(int,input().split())
def LIST_INT_INPUT():
    return list(map(int,input().split()))
def LIST_INPUT():
    return list(input().split())
def main():
    t = INT_INPUT()
    for _ in range(t):
        n = INT_INPUT()
        left = floor(n**0.5)
        total = n-left
        if left*left<=n:
            left+=1
        i = n+1
        while total<n:
            if left*left!=i:
                total+=1
            else:
                left+=1
            i+=1

        print(i-1)

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
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

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

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i = 0;i<t;i++){
            long k = sc.nextInt();
            long l = k;
            long h = (long)2e18;
            while(l<=h){
                long mid = l+ (h-l)/2;

                if((long) Math.floor(mid - Math.sqrt(mid))>=k)h = mid-1;
                else l = mid+1;
            }
            System.out.println(h);
        }

    }
}

**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
2
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import java.util.*;
public class Solution {
    static int MOD = (int)Math.pow(10,9)+7;
    public static long modExpo(long base, long exp, long mod) {
        long result = 1;
        while (exp > 0) {
            if ((exp & 1) != 0) {
                result = (result * base) % mod;
            }
            base = (base * base) % mod;
            exp >>= 1;
        }
        return result;
    }
    public static long modInverse(long P, long Q) {
        long Q_inv = modExpo(Q, MOD - 2, MOD);
        return (P * Q_inv) % MOD;
    }
    public static int max(int arr[]) {
        int m = 0;
        for(int i=0; i<26; i++) {
            m = Math.max(m,arr[i]);
        }
        return m;
    }
    public static int search(int x,int arr[]) {
        if(x<=arr[0]) {
            return 0;
        }
        if(x>arr[arr.length-1]) {
            return arr.length;
        }
        int s = 0;
        int e = arr.length-1;
        int m = arr.length-1;
        while(s<=e) {
           int mid = (s+e)/2;
           if(x<=arr[mid]) {
               m = mid;
               e = mid-1;
           } else {
              s = mid+1;
           }
        }
        return m;
    }
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0) {
           long k = sc.nextLong();
           long ans = 0;
           long x = 1;
           long z = 0;
           while((long)(x*x)<=k && (long)(x*x)>0) {
                z++;
                x++;
           }
           while((z+k)>=(long)(x*x)) {
              x++;
              z++;
           }
           System.out.println(k+z);
        }
    }
}

**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
2
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
void solve(){
    ll n;
    cin>>n;
    ll l=0,r=LLONG_MAX,answer=LLONG_MAX;
    while(l<r){
        ll mid=(l+r)/2;
        ll temt=(long long)(pow(mid,0.5));
        if((mid-temt)>=n){answer=min(answer,mid),r=mid;}
        else l=mid+1;
    }
    cout<<answer<<endl;
}
int main()
{
    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
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

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

int main() {
	// your code goes here

	int T;
	cin >> T;

	double K;

	while (T--) {

	    cin >> K;

	    double N = ((2 * K) + 1);

	    N += (sqrt((4 * K) + 1));
	    N /= 2;

	    long long int ans = N;

	    for (long long int answer = (long long int)(N - 10); answer < (long long int)(N + 10); answer++) {
	        if ((answer <= 0) || (answer <= K)) {
	            continue;
	        }
	        if ((long long int)(sqrt(answer)) == (answer - K)) {
	            ans = answer;
	            break;
	        }
	    }

	    cout << ans << endl;
	}

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import math
import sys

def solve():
    k = int(input())
    l = k // 2
    r = 4 * k
    ans = 4 * k

    while l <= r:
        mid = l + (r - l) // 2
        if mid - int(math.sqrt(mid)) >= k:
            if mid - int(math.sqrt(mid)) == k:
                ans = min(ans, mid)
            r = mid - 1
        else:
            l = mid + 1

    print(ans)

def main():
    t = int(input())
    for it in range(1, t + 1):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code uses a binary search to find the smallest `n` such that after performing the operations, there are exactly `k` bulbs on. The issue lies in the condition `mid - int(math.sqrt(mid)) >= k`.

Let's analyze the problem and the code:
1. The number of bulbs that remain on after the operations is determined by the number of perfect squares up to `n`. This is because each bulb is toggled a number of times equal to the number of its divisors. A bulb will be on if it is toggled an odd number of times, which happens only if the bulb's position is a perfect square (since perfect squares have an odd number of divisors).

2. The correct number of bulbs that remain on after the operations is given by the number of perfect squares up to `n`, which is `int(math.sqrt(n))`.

3. The buggy code uses `mid - int(math.sqrt(mid))` to check if the number of bulbs that remain on is at least `k`. This is incorrect because it should be checking `int(math.sqrt(mid))` directly.

Given this, we need to generate a test case where the difference between `mid` and `int(math.sqrt(mid))` is not a good indicator of the number of bulbs that remain on. Specifically, we need a `k` value where the correct `n` is a perfect square, and the buggy code's condition will fail.

Let's choose `k = 1`, which is a simple but effective edge case. For `k = 1`, the correct answer should be `1` (since only one perfect square, 1, is needed). However, the buggy code will likely fail to find this simple solution because it uses an incorrect condition.

Here is the test input:

```plaintext
1
1
```

This test case will expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import math

def solve():
    k = int(input())
    tot = k
    extra = int(math.sqrt(k))
    new_extra = int(math.sqrt(extra + k))
    if(extra < new_extra):
        tot += new_extra - extra

    print(tot + extra)

def main():

    test = 1
    test = int(input())

    for i in range(1, test + 1):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a logic issue in how it calculates the total number of bulbs needed to achieve exactly `k` bulbs on.

### Analysis of the Buggy Code:
1. The code calculates `extra` as the integer square root of `k`.
2. It then calculates `new_extra` as the integer square root of `extra + k`.
3. If `extra` is less than `new_extra`, it adds the difference to `tot`.
4. Finally, it prints `tot + extra`.

### Key Observations:
- The code seems to be based on the idea that the number of bulbs that remain on after the operations is related to the square numbers. This is because a bulb at position `j` will be flipped `d(j)` times, where `d(j)` is the number of divisors of `j`. A bulb will be on if and only if `d(j)` is odd, which happens when `j` is a perfect square.
- The code's logic for adjusting `tot` and `extra` is flawed and may not always correctly find the minimum `n` for a given `k`.

### Challenging Test Cases:
1. **Small `k` values**: Test cases where `k` is a small number, such as 1, 2, 3, and 4. These cases should be straightforward but can still expose issues in the logic.
2. **Large `k` values**: Test cases where `k` is a large number, such as `10^18`. These cases can help identify performance and accuracy issues.
3. **Special `k` values**: Test cases where `k` is a perfect square or just below a perfect square. These can help verify if the code correctly handles the transition points.

### Chosen Test Input:
```plaintext
5
1
2
3
4
1000000000000000000
```

### Explanation:
- **Test Case 1 (`k = 1`)**: The correct answer is 2. The code should handle this correctly, but it's a good starting point.
- **Test Case 2 (`k = 2`)**: The correct answer is 3. This is a slightly more complex case.
- **Test Case 3 (`k = 3`)**: The correct answer is 5. This is a more challenging case.
- **Test Case 4 (`k = 4`)**: The correct answer is 6. This is another step up in complexity.
- **Test Case 5 (`k = 10^18`)**: This is a very large value that can expose performance and accuracy issues. The correct answer is 1000000000000000000, but the code may struggle with this large input.

This test input covers a range of values, from small to very large, and should help identify any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        for (int i = 0; i < n; i++){
            long k = sc.nextLong();
            long from = 1;
            long to = (long) 1e19;
            while (from < to){
                long mid = (from + to) >> 1;
                if (getNum(mid) > k){
                    to = mid;
                } else if (getNum(mid) < k){
                    from = mid + 1;
                } else {
                    to = mid;
                }
            }
            System.out.println(to - 1);
        }
    }

    // k个等，计算有几个亮起的
    public static long getNum(long k) {
        long ans = (long) Math.ceil(Math.sqrt(k));
        return k - ans;
    }
}

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

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

Think step by step.

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