**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

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

typedef long long ll;

ll binary_search(ll l, ll n) {
    ll low = l, high = n, best_r = l;
    while (low <= high) {
        ll mid = low + (high - low) / 2;
        if ((mid - l + 1) <= (l * 2 + 2)) {
            best_r = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return best_r;
}

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

    ll t;
    cin >> t;
    while (t--) {
        ll n;
        cin >> n;
        ll count = 1;

        ll l = 1;
        while (l < n) {
            ll r = binary_search(l, n);
            count++;
            l = r + 1;
        }

        cout << count << 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
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
#include<iostream>
#include<cmath>
using namespace std;
int n,t,c=3,sum=2,s=1;
int main(){
	cin>>t;
	while (t--){
		cin>>n;
		if (n==1) cout<<1<<endl;
		else if (n==2||n==3||n==4) cout<<2<<endl;
		else{
			c=3,sum=5,s=2;
			while (n>sum){
				c*=2;
				sum+=c;
				s++;
			}
			cout<<s<<endl;
		}
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios where the code might fail. The buggy code seems to have issues with its logic for calculating the minimum number of operations, especially for larger values of `n`.

### Analysis of the Buggy Code
1. **Base Cases**:
   - For `n = 1`, it correctly prints `1`.
   - For `n = 2`, `3`, and `4`, it correctly prints `2`.

2. **General Case**:
   - The code uses a loop to calculate the number of operations needed for larger `n` values.
   - It initializes `c = 3`, `sum = 5`, and `s = 2`.
   - It then iterates, doubling `c` and adding it to `sum` until `n` is no longer greater than `sum`.
   - Finally, it prints `s`.

The logic in the general case is flawed. For example, for `n = 5`, the correct answer should be `3`, but the code might not handle this correctly.

### Challenging Test Input
To expose the bug, we can use a small-scale test input that includes edge cases and values that are just above the base cases:

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

### Explanation
1. **`n = 1`**: Base case, correct output should be `1`.
2. **`n = 2`**: Base case, correct output should be `2`.
3. **`n = 4`**: Base case, correct output should be `2`.
4. **`n = 5`**: This is a critical case. The correct output should be `3`, but the buggy code might fail here.
5. **`n = 6`**: Another critical case. The correct output should be `3`, but the buggy code might fail here.

This test input is designed to cover the base cases and the first few values where the general case logic starts to be applied, making it easier to identify the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
def star(n):
    c=0
    if n==5:
        return 3
    else:
        while n>5:
            c+=1
            n=n//2
        if n==1:
            c+=1
        else:
            c+=2
        return c
t=int(input())

for _ in range(t):
    n=int(input().strip())
    print(star(n))

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

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

Think step by step.

Response:
```plaintext
1
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    if(n==1):
        print(1)
    else:
        k=5
        s=2
        for i in range(n):
            if(k>=n):
                break
            else:
                s+=1
                k=(k*2)+1
        print(s)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the logic of the code fails to correctly compute the minimum number of operations required. The main issue with the buggy code is the loop and the way it calculates the value of `k` and `s`.

### Analysis of the Buggy Code
- The code initializes `k` to 5 and `s` to 2.
- It then iterates through the range `n`, doubling `k` and adding 1 to `k` in each iteration.
- The loop breaks when `k` is greater than or equal to `n`.
- The final value of `s` is printed, which is supposed to be the minimum number of operations.

### Identifying the Bug
- The initial value of `k` is 5, which is arbitrary and not based on any logical reasoning.
- The loop condition and the way `k` is updated are not aligned with the problem's requirements.
- The code does not correctly handle the transition from the first type of operation to the second type of operation.

### Edge Cases to Consider
1. **Small Values of `n`**: The code might fail for small values of `n` where the initial value of `k` is not appropriate.
2. **Large Values of `n`**: The code might fail to terminate or produce incorrect results for large values of `n` due to the arbitrary initial value of `k`.

### Test Input
To expose the bug, we can use a small value of `n` where the initial value of `k` is not appropriate. For example, `n = 3`.

### Output
```plaintext
1
3
```

### Explanation
- The test input contains one test case with `n = 3`.
- The correct minimum number of operations for `n = 3` is 2.
- The buggy code will likely produce an incorrect result because the initial value of `k` is 5, which is not suitable for `n = 3`.

This test input should help expose the bug 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: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import sys
input = sys.stdin.readline

MAX_M = 1000
coverage = [0]*(MAX_M+1)
coverage[1] = 1
for m in range(2, MAX_M+1):
    coverage[m] = coverage[m-1] + (m*(m+1))//2

def min_seeds(n):
    for m in range(1, MAX_M+1):
        if coverage[m] >= n:
            return m
    return MAX_M

t = int(input().strip())
for _ in range(t):
    n = int(input().strip())
    print(min_seeds(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 buggy code, we need to focus on edge cases and scenarios where the faulty logic in the code is most likely to fail. The buggy code incorrectly calculates the coverage for the array, leading to incorrect results for certain values of \( n \).

### Step-by-Step Analysis:

1. **Understanding the Bug**:
   - The buggy code calculates the coverage using the formula \( \text{coverage}[m] = \text{coverage}[m-1] + \frac{m \times (m+1)}{2} \). This formula is incorrect because it does not correctly represent the minimum number of operations needed to cover the array.
   - The correct formula for the minimum number of operations needed to cover an array of length \( n \) is \( \left\lceil \frac{n + 1}{2} \right\rceil \).

2. **Edge Cases**:
   - Small values of \( n \) (e.g., 1, 2, 3, 4) are already tested in the example.
   - Larger values of \( n \) where the incorrect formula leads to a significant discrepancy.

3. **Challenging Input**:
   - We need to find a value of \( n \) where the incorrect coverage calculation significantly deviates from the correct result.
   - For \( n = 10 \), the correct answer should be 5 (since \( \left\lceil \frac{10 + 1}{2} \right\rceil = 5 \)), but the buggy code might give a different result.

### Challenging Test Input:
```plaintext
1
10
```

This test input will expose the bug in the buggy code because the incorrect coverage calculation will not match the correct minimum number of operations needed to cover the array of length 10.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0){
            int n = sc.nextInt();
            if(n == 1){
                System.out.println(1);
            }
            else{
                if(n <= 4){
                    System.out.println(2);
                }
                else{
                    int ans = 1;
                    int mul = 2;
                    while(mul <= n){
                        mul *= 2;
                        ans++;
                    }
                    System.out.println(ans - 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
5
1
2
4
5
6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import java.util.Scanner;

public class MinimumOperationsAlternative {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for (int i = 0;i<t; i++) {
            int n = in.nextInt();
            if (n <=2) {
                System.out.println(n);
            } else if(n>2 && n<5){
                System.out.println((n + 2) / 3);
            }else{
                System.out.println((n + 4) / 5);
            }
        }
        in.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. The code provided has different conditions for different ranges of `n`:

1. For `n <= 2`, it prints `n`.
2. For `2 < n < 5`, it prints `(n + 2) / 3`.
3. For `n >= 5`, it prints `(n + 4) / 5`.

We need to test the boundaries and values within these ranges to ensure the code behaves correctly.

### Step-by-Step Analysis:

1. **For `n <= 2`**:
   - `n = 1` and `n = 2` are straightforward and should work correctly.

2. **For `2 < n < 5`**:
   - `n = 3` and `n = 4` should be tested.
   - The formula `(n + 2) / 3` should be checked for correctness.

3. **For `n >= 5`**:
   - `n = 5` and `n = 6` should be tested.
   - The formula `(n + 4) / 5` should be checked for correctness.

### Test Cases:
- **`n = 1`**: Expected output: `1`
- **`n = 2`**: Expected output: `2`
- **`n = 3`**: Expected output: `2`
- **`n = 4`**: Expected output: `2`
- **`n = 5`**: Expected output: `2`
- **`n = 6`**: Expected output: `2`

### Additional Edge Cases:
- **`n = 7`**: Expected output: `2` (This should be `3` according to the correct formula, but the code might fail here)
- **`n = 9`**: Expected output: `3` (This should be `3` according to the correct formula, but the code might fail here)

### Chosen Test Input:
To maximize the failure probability, we will include the boundary values and a few values just beyond the boundaries to ensure the formula is applied correctly.

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

This test input will help expose the bug in the code by testing the boundaries and just beyond the boundaries of the different conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
#include <bits/stdc++.h>
#include <vector>
#include <math.h>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <cmath>

using namespace std;

#define vi vector<int>
#define pb push_back
#define M 1000000007
#define endl "\n"
#define int long long
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep1(i, k, n) for(int i = k; i <= n; i++)
#define print(a) rep(i,a.size()) cout << a[i] << ' ';cout << endl;
#define read(a,n) vi a(n);rep(i,n) cin >> a[i];
#define inp(a) int a; cin >> a;
#define all(a) a.begin(), a.end()
#define pi pair<int,int>
#define mi map<int,int>
#define f first
#define s second
const int N = 2e5 + 5;
const int max_logn = 20;
const int inf = 1e9+19;
const int MOD = 1e18;

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

bool isPrime(int n) {
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    return true;
}

bool isPowerOfTwo(int n) {
    if (n == 0)
        return false;
    return ((int)ceil(log2(n)) == (int)floor(log2(n)));
}

int powm(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1)
            res = (res * a) % MOD ;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res % MOD;
}

int divide(int a, int b) {
    return (a % M) * powm(b, M - 2) % M;
}

int mul(int a, int b) {
    return ((a % M) * (b % M)) % M;
}

// int add(int a, int b) {
//     return (a % M + b%M) %M;
// }

void check(bool b){
    cout << (b ? "YES":"NO") << endl;
}

// vi dis(N,0);

// vi  vis(N);
// void bfs(int node){

//     queue<int> q;
//     q.push(node);
//     dis[node]=0;
//     vis[node]=1;
//     while(!q.empty()){
//         int d=q.front();
//         q.pop();
//         for(auto child:g[d]){
//             if(!vis[child]){
//                 vis[child]=1;
//                 dis[child]=dis[d]+1;
//                 q.push(child);
//             }
//         }
//     }
// }

//LCA
 // void dfs(int node, int p=-1){
 //    par[node]=p;
 //    for(auto child:adj[node]){
 //        par[child]=node;
 //        dfs(child, node);
 //    }
 // }
 // vector<int> (int v){
 //    vi ans;
 //    while(v!=-1){
 //        ans.pb(v);
 //        v=par[v];
 //    }
 //    reverse(ans.begin(),ans.end());
 //    return ans;
 // }

//  TYPICAL STRUCTURE OF FLOYYD WARSHAL SHORTEST DISTANCE BETWEEN EACH PAIR OF NODES
// for(int k=0;k<n;k++){
//         int k_add=del[k];
//         for(int i=1;i<=n;i++){
//             for(int j=1;j<=n;j++){
//                 a[i][j]=min(a[i][j],a[i][k_add]+a[k_add][j]);
//             }
//         }
//  }

// DIJKSTRA ( SHORTEST DISTANCE FROM A NODE TO OTHER)
// vector<pair<int,int>> g[N];
// vector<int> dist(N,INF);
// void dijkstra(int src){
//     vector<int> vis(N);

//     set<pair<int ,int>> s;  //first element for distance ,second node number
//     dist[src]=0;
//     s.insert({0,src});
//     while(s.size() > 0){ // minimum distance wala hi operate huga
//         auto node=*s.begin();
//         int v=node.first;
//         int dis=node.second;
//         s.erase(s.begin());
//         if(vis[v]) continue;
//         vis[v]=1;
//         for(auto child:g[v]){
//             int child_wt= child.second;
//             int child_v=child.first;
//             if(dist[v]+ child_wt< dist[child_v]){
//                 dist[child_v]=dist[v]+child_wt;
//                 s.insert({dist[child_v],child_v});
//             }
//         }
//     }
// }

//                    DSU
// int parent[N];
// int siz[N];
// multiset<int> sizes;
// void make(int node){
//     parent[node]=node;
//     siz[node]=1;
//     sizes.insert(1);
// }
// int find(int node){
//     if(parent[node]==node) return node;
//     else{
//         return parent[node]=find(parent[node]);
//     }
// }
// void merge(int a, int b){
//     sizes.erase(sizes.find(siz[a]));
//     sizes.erase(sizes.find(siz[b]));
//     sizes.insert(siz[a]+siz[b]);
// }
// void Union(int a, int b){
//     a=find(a);
//     b=find(b);
//     if(a!=b){
//         if(siz[a]<siz[b]) swap(a,b);
//         parent[b]=a;
//         merge(a,b);
//         siz[a]+=siz[b];
//     }
// }
// __builtin_clz(x): the number of zeros at the beginning of the number
// __builtin_ctz(x): the number of zeros at the end of the number
// __builtin_popcountll(x): the number of ones in the number
// __builtin_parity(x): the parity (even or odd) of the number of ones
// __builtin_ffs(int) finds the index of the first (most right) set bit
// *find_by_order(k): It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time.To find the first element k must be zero.
// order_of_key(k) : It returns to the number of items that are strictly smaller than our item k in O(logn) time.

int loga(int base,int num){
    return (int)ceil(log2(num)/log2(base));
}
int ans = 0;
int ask(int l, int r) {
    cout << "? " << l << ' ' << r << endl;
    cout.flush();
    inp(x);
    return x;
}
// int n, m, k;
// vector<int> adj[N];
// vector<int> vis(N);
// vector<pi> temp;
// //vector<int> cnt;
// void dfs(int node, vector<int> &c) {
//     vis[node] = 1;

//     temp.pb({node, c[node]});
//     for (auto child : adj[node]) {
//         if (!vis[child]) {
//             dfs(child, c);
//         }
//     }
// }
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int ans =0;
        if (n==2){
            cout << 2 << endl;
            continue;
        }
        int c=1;
        ans =1;
        while(c<(n+1)/2){
            ans++;
            c=2*c+1;
        }

        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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or specific scenarios that might cause the code to fail. Let's analyze the code and the problem constraints:

1. **Basic Cases**:
   - The code handles small cases like `n = 1`, `n = 2`, and `n = 4` correctly.
   - For `n = 2`, it directly outputs `2`.

2. **General Case**:
   - The code uses a loop to calculate the minimum number of operations needed.
   - The loop increments `ans` and updates `c` until `c` is greater than or equal to `(n + 1) / 2`.

3. **Potential Bugs**:
   - The loop might not handle large values of `n` correctly.
   - The condition `c < (n + 1) / 2` might not always be accurate for certain values of `n`.

Let's generate a test case that could expose this issue. A good candidate is a value of `n` that is just above a power of 2, as this might cause the loop to behave unexpectedly.

### Chosen Test Case
We will use `n = 7` as a test case. This value is just above 4, which is a power of 2, and should require a specific number of operations to be painted completely.

### Expected Output
For `n = 7`, the correct output should be `3`. Here's why:
- Initially, the array is `[0, 0, 0, 0, 0, 0, 0]`.
- We can perform the following operations:
  1. Set `a[1] = 1` (array becomes `[1, 0, 0, 0, 0, 0, 0]`).
  2. Set `a[7] = 1` (array becomes `[1, 0, 0, 0, 0, 0, 1]`).
  3. Perform the second type of operation on the segment `[1, 7]` (array becomes `[1, 1, 1, 1, 1, 1, 1]`).

### Test Input
```plaintext
1
7
```

This test case should expose any issues in the loop logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

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

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n, c=1;
        cin>>n;
        if(n==1)
        cout<<1<<'\n';
        else if(n==2)
        cout<<2<<'\n';
        else
        {
        while(n>2)
        {
            c++;
            n = n/2;
        }
        cout<<c<<'\n';
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has a flaw in its logic for determining the minimum number of operations needed to make all elements of the array equal to one.

### Analysis of the Buggy Code:
1. **For `n == 1` and `n == 2`**: The code correctly outputs 1 and 2 respectively.
2. **For `n > 2`**: The code uses a while loop to repeatedly divide `n` by 2 and increment a counter `c`. This logic is incorrect because it does not account for the actual number of operations needed to paint the array.

### Key Points to Consider:
- **Odd and Even Lengths**: The code might fail for specific odd and even lengths of the array.
- **Powers of 2**: The code might fail for lengths that are powers of 2.
- **Small Values**: Small values of `n` can help in quickly identifying the issue.

### Challenging Test Input:
Let's create a test input that includes:
1. A small value of `n` that is not a power of 2.
2. A value of `n` that is a power of 2.
3. A value of `n` that is slightly larger than a power of 2.
4. A value of `n` that is slightly smaller than a power of 2.

```plaintext
4
3
4
5
7
```

### Explanation:
- **n = 3**: The correct output should be 2. The array can be painted as follows: `[1, 0, 0]` -> `[1, 1, 0]` -> `[1, 1, 1]`.
- **n = 4**: The correct output should be 2. The array can be painted as follows: `[1, 0, 0, 0]` -> `[1, 0, 0, 1]` -> `[1, 1, 1, 1]`.
- **n = 5**: The correct output should be 3. The array can be painted as follows: `[1, 0, 0, 0, 0]` -> `[1, 0, 0, 0, 1]` -> `[1, 1, 0, 1, 1]` -> `[1, 1, 1, 1, 1]`.
- **n = 7**: The correct output should be 4. The array can be painted as follows: `[1, 0, 0, 0, 0, 0, 0]` -> `[1, 0, 0, 0, 0, 0, 1]` -> `[1, 1, 0, 0, 0, 1, 1]` -> `[1, 1, 1, 0, 1, 1, 1]` -> `[1, 1, 1, 1, 1, 1, 1]`.

This test input will help in exposing the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    lst=[0]*n
    lst[0]=1
    l=1
    count=1
    step1=1
    while True:
        r=(2*(count)) + (l+1)
        if r>=n :
            if lst[-1]==0:
                step1+=1
            break
        lst[r-1]=1
        count+=1
        step1+=1
        count+=(r-l)-1
        l=r
    print (step1)

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

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

Think step by step.

Response:
To expose the bug in the provided code, let's analyze the logic and identify potential edge cases or scenarios that could cause the code to fail.

### Analysis of the Buggy Code

1. **Initialization**:
   - The code initializes the list `lst` with zeros and sets the first element to 1.
   - It starts with `l = 1` and `count = 1`.

2. **Loop**:
   - The loop calculates `r` as `r = (2 * count) + (l + 1)`.
   - It checks if `r >= n` to determine if the loop should break.
   - If `r` is within bounds, it sets `lst[r-1]` to 1 and updates `count` and `l`.

3. **Edge Cases**:
   - **Small Arrays**: Arrays of length 1, 2, or 4, which are already handled in the examples.
   - **Large Arrays**: Arrays of length close to the upper limit (100,000) to see if the logic breaks.
   - **Odd and Even Lengths**: Arrays of both odd and even lengths to check if the logic handles them correctly.
   - **Specific Patterns**: Arrays where the initial placement of 1s might cause issues.

### Challenging Test Input

To challenge the code, let's create a test case with an array length that is not trivial and might expose the logic flaw. A good candidate is an array length that is just above the initial placement logic, such as 5 or 7.

### Test Input
```plaintext
1
5
```

### Explanation
- **Test Case 1**: `n = 5`
  - The array starts as `[0, 0, 0, 0, 0]`.
  - After the first operation, it becomes `[1, 0, 0, 0, 0]`.
  - The next operation should ideally place a 1 in a position that allows the second type of operation to be used effectively.

### Expected Output
- The expected output for `n = 5` should be `3`:
  - First operation: `[1, 0, 0, 0, 0]`
  - Second operation: `[1, 0, 0, 1, 0]`
  - Third operation: `[1, 1, 1, 1, 0]` (or similar valid sequence)

This test input should help expose the bug in the provided code by challenging the logic for placing the 1s and performing the second type of operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import sys

INF = float("inf")

def solution(n) -> int:

    if n == 1:
        return 1
    if n == 2:
        return 2

    ops = 0
    val = 0

    while val < n:
        val = (val + 1) * 2
        ops += 1

    return ops

def problem():
    n_test_cases = int(input().strip())

    for _ in range(n_test_cases):

        n = int(input().strip())
        print(solution(n))

problem()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider the logic and the conditions under which the code might fail. The code uses a loop to double the value of `val` and increments the number of operations until `val` is at least `n`. This approach is flawed because it does not correctly handle the condition for the second type of operation, which requires that the sum of the segment is at least half the length of the segment.

### Steps to Identify the Bug:
1. **Initial Conditions**: The code correctly handles the base cases for `n = 1` and `n = 2`.
2. **Loop Logic**: The loop in the `solution` function increments `val` by doubling it and increments the operation count. This approach is incorrect because it does not consider the correct way to apply the second type of operation.

### Test Case to Expose the Bug:
We need a test case where the correct number of operations is not the result of the doubling logic. Specifically, we should test a value of `n` where the correct number of operations is not a simple power of two or a multiple of a power of two.

### Example Test Case:
Let's consider `n = 3`. The correct number of operations is 2, but the code will incorrectly return 3.

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

### Explanation:
- For `n = 3`, the correct sequence of operations is:
  1. Perform an operation of the first type with `i = 1`. The array becomes `[1, 0, 0]`.
  2. Perform an operation of the first type with `i = 3`. The array becomes `[1, 0, 1]`.
  3. Perform an operation of the second type with `l = 1` and `r = 3`. The array becomes `[1, 1, 1]`.

The correct number of operations is 2, but the code will return 3 because it incorrectly doubles the value of `val` and increments the operation count.

### Expected Output:
```plaintext
2
```

This test case should 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: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import java.util.Scanner;

public class paintastripin
{
    static int check(int n){
        int count=1;
        //int n=sc.nextInt();
        if(n<4){
            return n;
        }
        else{
            int last=1;
            while(last<n){
                int cal=(last+1)*2;
                count++;
                last=cal;
            }
        }
        return count;
    }
    public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int test=sc.nextInt();
    while(test-->0){
        int n=sc.nextInt();
        System.out.println(check(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 provided buggy code, let's analyze the code and identify potential issues:

1. **Understanding the Buggy Code:**
   - The function `check` is supposed to calculate the minimum number of operations of the first type needed to make all elements of the array equal to one.
   - The code uses a loop to double the length of the segment that can be filled with ones, starting from 1.
   - The logic in the loop is to double the `last` value and increment the `count` each time.

2. **Identifying the Bug:**
   - The bug in the code is that it incorrectly calculates the number of operations needed. Specifically, it does not correctly handle the case where `n` is not a power of 2 plus 1.
   - The code increments `count` for each doubling, but it does not correctly account for the remaining elements that need to be filled.

3. **Edge Cases and Challenging Inputs:**
   - **Power of 2 minus 1:** These cases should require a specific number of operations, and the code should handle them correctly.
   - **Odd and even numbers:** The code should handle both odd and even lengths correctly.
   - **Small values:** Small values of `n` should be handled as per the examples provided.
   - **Large values:** The code should not fail for large values of `n` within the constraints.

4. **Generating the Test Input:**
   - We will include a mix of small and large values, including edge cases like powers of 2 minus 1 and odd/even numbers.

```plaintext
4
1
3
7
15
```

This test input includes:
- `1`: A small value that should require 1 operation.
- `3`: A value that should require 2 operations.
- `7`: A value that should require 4 operations.
- `15`: A value that should require 8 operations.

These values will help expose the bug in the code by testing its handling of different lengths and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import java.util.*;
import java.io.*;
/*
⠀⠀⠀⠀⠀⠀  ⠀⠀⠀⠀⠀⠀⣾⡳⣼⣆⠀⠀⢹⡄⠹⣷⣄⢠⠇⠻⣷⣶⢀⣸⣿⡾⡏⠀⠰⣿⣰⠏⠀⣀⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⡀⣀⣀⣀⡹⣟⡪⢟⣷⠦⠬⣿⣦⣌⡙⠿⡆⠻⡌⠿⣦⣿⣿⣿⣿⣦⣿⡿⠟⠚⠉⠀⠉⠳⣄⡀
⠀⠀⠀⠀⠀⠀⠀⡀⢀⣼⣟⠛⠛⠙⠛⠉⠻⢶⣮⢿⣯⡙⢶⡌⠲⢤⡑⠀⠈⠛⠟⢿⣿⠛⣿⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⣆
⠀⠀⠀⠀⠀⡸⠯⣙⠛⢉⣉⣙⣿⣿⡳⢶⣦⣝⢿⣆⠉⠻⣄⠈⢆⢵⡈⠀⠀⢰⡆⠀⣼⠓⠀⠀⠀   Nah    ⠈⣷
⠀⠀⠀⠖⠉⠻⣟⡿⣿⣭⢽⣽⣶⣈⢛⣾⣿⣧⠀⠙⠓⠀⠑⢦⡀⠹⣧⢂⠀⣿⡇⢀⣿⠺⠇⠀    I'd⠀     ⣿
⠀⠀⠀⠀⠐⠈⠉⢛⣿⣿⣶⣤⣈⠉⣰⣗⡈⢛⣇⠀⣵⡀⠀⠘⣿⡄⢻⣤⠀⢻⡇⣼⣧⣿⡄⠀⠀   Win⠀    ⠀⡿
⠀⠀⠀⠀⠀⣠⣾⣿⢍⡉⠛⠻⣷⡆⠨⣿⣭⣤⣍⠀⢹⣷⡀⠀⠹⣿⡄⠈⠀⢿⠁⣿⣿⠏            ⠀⠀⠀⣇
⠀⣿⣇⣠⣾⣿⣛⣲⣿⠛⠀⠀⢀⣸⣿⣿⣟⣮⡻⣷⣤⡙⢟⡀⠀⠙⢧⠀⠀⠎⠀⠉⠁⠰⣿⠀⠀         ⠀⢀⡿
⠀⠈⢻⣿⣿⣽⣿⣿⣿⣴⡏⠚⢛⣈⣍⠛⠛⠿⢦⣌⢙⠻⡆⠁⠀⠀⠀⣴⣦⠀⠀⠀⠐⢳⢻⣦⣀⠀⠀⠀⠀⠀⠀⠀⠀⢀⠮
⠀⠀⠈⠙⣿⣧⣶⣿⠿⣧⣴⣿⢻⡉⠀⢀⣠⣴⣾⡟⠿⠃⠁⣠⣤⡶⣾⡟⠅⠀⣀⡄⠀⣾⢸⣿⣏⢻⢶⣦⣤⣤⣄⢶⣾⣿⣡⣤⡄
⠀⠀⣠⣞⣋⣿⣿⣾⣿⡿⡛⣹⡟⣤⢰⡿⠟⠉⣀⣀⣤⣤⡠⠙⢁⣾⡿⠂⠀⣿⠟⣁⠀⣹⠀⣹⣿⡟⣼⣿⣿⣌⣿⣞⣿⣿⠁
⠀⢠⡿⢛⢟⣿⣿⣿⣿⣿⣿⡟⣼⣿⣟⢓⠛⣿⣏⣿⣵⣗⣵⣴⣿⢟⡵⣣⣼⣿⢟⣵⣶⢻⣶⣿⠀⠀⣈⢻⣿⣿⣿⢿⣾⢿⣧
⠀⠘⠃⢸⣿⡾⣿⣿⣿⣿⣯⣿⣿⣿⣶⣿⣿⣟⣾⡿⣫⣿⣿⣿⣽⣿⣿⣿⣿⢫⣾⣿⣿⣿⣿⣿⣴⡆⣻⣿⡏⣿⢻⣧⣿⡿⣿⡆
⠀⠀⠀⠜⣿⣾⢿⣿⣿⣿⣾⣿⣿⣿⣿⣿⣿⣭⣿⣖⣿⢿⣿⡿⣿⣿⣿⡿⢡⢯⣿⣿⣿⣿⣿⣿⣿⣧⡿⣾⣷⣿⣿⢿⣿⡇⠉⠁
⠀⠀⠀⠀⣿⣥⣾⣿⣿⣿⣿⣿⣿⣿⡇⣭⣿⣿⣿⣿⠃⠞⠟⣸⣿⠏⣸⣧⣀⠿⢿⣿⣿⣟⣿⣿⣿⣿⣽⣿⢿⣿⣿⣿⣿⠁
⠀⠀⠀⠈⠛⣹⣿⣿⣿⣿⢿⣿⣿⣿⣿⣿⣟⣿⣿⡿⢶⣦⣄⣿⠏⠀⣿⣟⣿⣶⠾⣿⣟⣋⣛⣿⣿⣿⣿⡇⣻⣿⣿⣿⡏
⠀⠀⠀⠀⠟⠛⠫⣿⣿⣿⣿⣿⡿⣧⠛⣿⠛⣿⣿⣿⣷⡌⠹⡟⠀⠀⠉⡟⠋⢠⣾⣿⣿⣿⡟⣿⣿⣿⣿⢀⣿⣿⣿⣿⣧
⠀⠀⠀⠀⠀⠀⠘⠋⣾⣷⣿⣿⣧⠙⠀⠙⢣⠝⠛⠋⣽⣷⢦⠇⠀⠀⠘⠁⣤⣾⣿⠝⠛⠉⠘⢻⣿⣿⢿⣼⣷⡟⢻⣷⠉
⠀⠀⠀⠀⠀⠀⠀⠐⠟⢻⣿⣿⣿⡀⠀⠀⠀⠀⠀⠀⠀⠉⠀⠀⠀⠀⠀⠀⠈⠛⠀⠀⠀⠀⠀⣾⠟⠀⢸⣷⣿⡇⠀⠛
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠛⠁⠀⢹⣇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣿⣿⣿⡧
⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠈⣿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣾⢻⡿⠈⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣇⠀⠀⠀⠀⠀⠀⠀⠀⠲⣄⠀⡄⠆⠀⠀⠀⠀⠀⠀⠀⠀⣼⡏
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣿⣷⡀⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⣀⠀⠀⣠⣾⣿⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⢻⣆⠀⠛⠁⠶⣶⣶⣶⣶⣶⣶⡶⠆⠘⠋⣠⡾⢫⣾⡟
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⠛⠀⠙⣷⡀⠀⠀⠙⠛⠛⠛⠛⠋⠁⠀⢀⣴⠋⠀⣾⣿⡇
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣤⣿⣰⣦⡀⠸⣿⣦⡀⠀⠀⠀⠀⠀⠀⢀⣴⡟⠁⠀⠐⢻⣿
⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⣴⣾⣿⣿⣿⡄⢺⣿⡄⠹⣿⠻⢦⣤⣤⣤⣤⣶⣿⡟⢀⣀⠀⠀⢸⣿⣦⣄⡀
⠀⠀⠀⠀⢀⣠⣴⣾⣿⣿⣿⣿⣿⣿⣿⣿⣮⣿⣿⡀⠹⡷⣦⣀⡀⡀⢸⣿⠏⢠⣾⣿⠀⠀⣾⣿⣿⣿⣿⣶⣄⣀
⣀⣤⣴⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠘⣷⣻⡟⠀⡼⠁⣴⣿⣿⣯⣥⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣶⣤⣀
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣶⣯⣿⣤⣤⣤⣬⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣶⣤⣄
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
*/

/*Write you code here -------- */
public class Main {
    public static void main(String[] args) throws IOException {
        MyTemplate.FastIO jio = new MyTemplate().new FastIO();
        int t=jio.nextInt();
        while(t-->0){
            int n=jio.nextInt();
            int cnt=1;
            if(n==1) {jio.println(1);continue;}
            if(n>1 && n<=4) {jio.println(2);continue;}
            while((1<<cnt)-1<=n){
                cnt++;
            }
            jio.println(cnt-1);
        }
        jio.close();
    }
}

/*end here-------------------- */
class MyTemplate {
    class Calc{
        public static ArrayList<Integer> generatePrimes(int n) {
            // Create an array to track prime status
            boolean[] isPrime = new boolean[n + 1];
            ArrayList<Integer> primes = new ArrayList<>();

            // Initialize all numbers as prime
            for (int i = 2; i <= n; i++) {
                isPrime[i] = true;
            }

            // Sieve of Eratosthenes algorithm
            for (int p = 2; p * p <= n; p++) {
                if (isPrime[p]) {
                    // Mark all multiples of p as non-prime
                    for (int multiple = p * p; multiple <= n; multiple += p) {
                        isPrime[multiple] = false;
                    }
                }
            }

            // Collect all prime numbers
            for (int i = 2; i <= n; i++) {
                if (isPrime[i]) {
                    primes.add(i);
                }
            }

            return primes;
        }

        public static long GCD(long a,long b){
            if(a<b){
                long temp=a;
                a=b;
                b=temp;
            }
            if(b==0){
                return a;
            }
            return GCD(b,a%b);
        }
        public static int GCD(int a,int b){
            if(a<b){
                int temp=a;
                a=b;
                b=temp;
            }
            if(b==0){
                return a;
            }
            return GCD(b,a%b);
        }
        public static void swap(int a,int b){
            int temp=a;
            a=b;
            b=temp;
        }
        public static void swap(long a,long b){
            long temp=a;
            a=b;
            b=temp;
        }
    }

    class FastIO {
        private BufferedReader br;
        private BufferedWriter bw;
        private StringTokenizer st;

        // Constructor to initialize input and output
        public FastIO() {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        // Method to read next line
        public String nextLine() throws IOException {
            return br.readLine();
        }

        // Method to read next token
        public String next() throws IOException {
            while (st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

        // Method to read an integer
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        // Method to read a long
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        // Method to read a double
        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        // Method to read an integer array
        public int[] readIntArray(int size) throws IOException {
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        // Method to read a long array
        public long[] readLongArray(int size) throws IOException {
            long[] arr = new long[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        // Method to print a string without a newline
        public void print(String str) throws IOException {
            bw.write(str);
            bw.flush();
        }

        // Method to print a string followed by a newline
        public void println(String str) throws IOException {
            bw.write(str + "\n");
            bw.flush();
        }

        // Method to print an integer followed by a newline
        public void println(int num) throws IOException {
            bw.write(num + "\n");
            bw.flush();
        }

        // Method to print a long followed by a newline
        public void println(long num) throws IOException {
            bw.write(num + "\n");
            bw.flush();
        }

        // Method to print an integer array
        public void printArray(int[] arr) throws IOException {
            for (int i = 0; i < arr.length; i++) {
                bw.write(arr[i] + (i == arr.length - 1 ? "" : " "));
            }
            bw.write("\n");
            bw.flush();
        }

        // Method to print a long array
        public void printArray(long[] arr) throws IOException {
            for (int i = 0; i < arr.length; i++) {
                bw.write(arr[i] + (i == arr.length - 1 ? "" : " "));
            }
            bw.write("\n");
            bw.flush();
        }

        // Flush the output stream
        public void flush() throws IOException {
            bw.flush();
        }

        // Close input and output streams
        public void close() throws IOException {
            br.close();
            bw.close();
        }
    }
}

/*

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣬⢧⣬⡒⠀⠀⠀⢀⣠⠝⠉⠉⠉⠉⠁⠀⠸⠟⠙⠹⣷⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣈⡹⠒⠚⠁⣠⠞⣉⠀⢀⡴⠚⠉⠁⡀⠀⠀⠀⠀⠀⠀⠉⠙⠲⠶⣶⣤⣀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣀⣀⡤⠴⠞⠁⠀⠀⢀⢞⡡⠊⠁⡰⢋⣠⣶⠖⠋⠁⠀⠠⡄⠀⠀⠀⠀⠀⠀⢠⡀⣀⣉⠭⠿⠟⠋⠉
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠻⢯⣍⡻⠭⠭⠤⠄⣀⠴⠂⠀⣰⠟⠉⠀⣠⡾⠛⢩⣷⠃⠀⠀⠀⠀⠀⢱⠀⡆⡀⣠⠀⠀⠘⢿⠮⣷⣤⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠉⣛⡿⣣⢯⠄⡴⢫⡅⠀⢀⡴⠃⠀⠀⣾⠇⠀⠀⢸⠀⠀⠀⠸⡄⡇⡇⣿⡇⠀⠀⠈⢷⡄⠙⢿⣦⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⡾⠋⣰⠟⣡⢞⡵⠋⠀⢠⠎⠀⡤⠀⢠⡟⠀⠀⠀⡇⢠⠀⡀⠀⣷⣷⣧⢿⡗⢲⡀⠀⢸⡿⡄⠀⠙⢾⡆
⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⢴⣚⣡⠞⠞⠁⣾⠵⠋⢀⢆⣴⠏⣠⠊⢀⣠⡼⠀⠀⠀⠀⡷⠘⣷⡇⢰⣿⣿⣳⡼⣧⠸⣷⠀⠀⠇⠙⢦⡀⠈⢿⡄
⠀⠀⠀⠀⠀⠀⠀⠈⠉⠉⠉⠁⣼⠃⣠⡾⠉⠀⠀⢀⡿⠋⣸⠞⢁⡠⠋⠁⠀⠀⠀⠀⣴⠃⣸⣿⡃⢸⣱⡸⢻⣿⣿⣿⣿⢸⠀⢤⠀⢀⠙⢦⡀⠹⣦⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣾⡷⣾⣿⠃⠀⡄⢀⡜⠁⠐⡧⠴⠊⠀⠀⠀⢀⠀⢠⠖⠛⢠⣿⡏⡇⣸⣿⣿⣾⣿⣿⣿⣿⣸⡄⠈⣦⠈⣧⠠⣷⡦⣝⡟⠦⣄
⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⡿⠟⢰⣿⠃⠀⣼⢠⡟⡀⠀⠀⡇⢀⡇⢰⠀⢀⡎⢠⡏⠀⡤⢸⣿⡇⣧⠿⣯⣿⣿⣿⣿⣿⣿⣷⣷⢀⡈⢧⣹⢦⡙⢷⡀⠉⠙⠛
⠀⠀⠀⠀⠀⠀⠀⠤⠒⠃⠀⣰⢟⣥⠃⢠⣧⡟⣰⡇⠀⠀⢀⡞⣰⣧⠀⡜⣰⢻⠃⢸⣸⠈⢻⣳⣇⡿⣿⠘⢿⣿⣼⣿⣏⣿⣿⡇⠟⣦⡉⢳⡘⠓⢿⣦⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣞⣵⣟⠁⢀⣾⢿⡹⣹⠀⢠⢀⠞⣰⣻⠃⣰⡷⠃⣸⡆⡏⡿⠀⢸⠉⢹⡇⣿⣿⣼⣿⡿⣿⣿⣼⣹⡀⢠⣿⣿⠸⡗⢶⣶⣯⡿⠶
⠀⠀⠀⠀⠀⠀⠀⢀⣴⠿⠋⣞⡼⢀⡞⢹⣿⡇⣿⠀⡟⣏⡼⣱⣯⡶⠿⠁⡄⢹⣿⢧⠃⠀⢸⠀⣼⡏⢙⣿⠻⣿⣷⣽⣿⣿⡻⡇⢸⢯⡇⣷⡿⣄
⠀⠀⠀⠀⠀⠀⠀⠉⠁⠀⣰⡿⣣⣿⠀⣼⣿⣿⡟⢸⢀⣿⠱⢋⡟⠀⢀⠆⡇⣼⡏⠸⢠⠃⢻⣴⣿⠁⢸⣿⠀⣿⢻⣿⢿⡍⡛⣇⣾⣦⣷⡈⢿⠯⢷⠄
⠀⠀⠀⠀⠀⠀⠀⠀⠀⣰⣯⡟⢩⡟⣰⠟⢦⣯⣇⣼⣿⡇⠀⣿⡇⠀⣼⣿⢳⠀⡇⡀⣾⢰⣼⣿⣿⣀⣿⣿⡄⣿⣿⣿⠀⡇⣅⣿⣵⠇⣿⣿⣮⣧⡀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣀
⠀⠀⠀⠀⠀⠀⠀⠀⠘⠛⠛⠀⢸⣵⢻⣄⣆⣿⣿⣿⣿⡇⢀⣿⡇⢠⣻⢹⣾⣰⢷⢷⣿⢸⣿⣿⣻⣿⡿⠿⣿⣿⠾⡏⢰⠃⣿⡿⡟⢿⣏⣻⡈⠙⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣿⠇⢸⣿⣾⣷⣻⣏⣿⣇⣼⡿⡇⣸⣿⣾⣷⣿⣩⣿⣯⡿⢡⣿⣿⣿⣿⣿⣿⣯⣄⣹⡟⣤⣿⣾⢀⣾⠏⢿⣧
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠛⠁⠀⢸⣿⠻⣏⡟⣿⣿⣿⣿⣡⣿⣿⣿⣿⣿⣿⣿⣿⣿⡓⠃⢘⣿⠟⣿⣭⢳⠉⠹⠿⢠⣿⡏⣹⣿⣿⠀⠀⠈⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠾⠃⠀⠙⣦⣻⣿⠿⠙⡿⢿⣿⣽⠋⢯⣿⡏⠀⠻⠃⣤⣻⢏⣀⠛⠟⢋⣤⠗⣠⡄⣿⡴⠋⠛⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠙⣿⠲⠀⠀⠘⠋⠛⠀⠈⠿⠁⠀⠀⠀⠐⠌⠓⠛⣿⠿⡿⠓⠀⠉⢠⣟⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣴⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠘⢻⣟⡭⠁⠀⢀⣾⣿⣧⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⣾⣿⣿⣿⣿⡌⢷⡀⢤⡄⠀⠀⠀⠀⠳⣄⢀⡤⠀⠀⠀⠀⠀⣿⣷⠀⢀⣼⣿⣿⣿⣿⣦⡀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣴⣾⣿⣿⣿⣿⣿⣿⣿⡇⠀⢿⣄⠋⠲⢶⣤⣀⣀⣤⣤⣶⣶⣶⣶⣶⠖⠋⠀⣠⠞⢹⣿⣿⣿⣿⣿⡇
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢈⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠛⢦⡀⡓⠿⣿⣟⣻⡿⢖⣚⣿⠟⠁⢀⣴⠞⠁⠀⣾⣿⣿⣿⣿⣿⠇
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣼⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠄⠀⠘⡎⣿⣷⣤⡈⠛⠛⠻⠛⠛⠉⢁⣴⣿⠋⠐⣠⣾⣿⣿⣿⣿⣿⣿⣧⡀
⠀⠀⠀⠀⠀⠀⠀⠀⣀⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣄⠀⠀⢁⠘⣿⢷⣤⣀⣀⠀⣀⣠⣿⠿⡻⣡⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣶⣄⡀
⠀⠀⠀⠀⠀⣠⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡖⠘⠦⢼⣟⠸⠛⢻⣻⣽⣟⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⣻⣿⣿⣿⣿⣿⣿⣿⣶⣄
⠀⠀⢀⣴⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡀⠀⣬⠽⣷⣶⠿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⣄
⣴⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣶⣋⣀⣿⠏⠰⢺⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⣀
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣿⣿⡿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡬⣿⣿⢯⣷⣶⠝⠘⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡘⠛⢤⣀
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣌⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯⣷⣶⢿⣷⣶⢤⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡄⠀⠉
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*/

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases where the code might fail. The code seems to be using a logarithmic approach to determine the minimum number of operations, but it might have issues with specific edge cases.

Let's analyze the code:

1. **For \( n = 1 \)**: The code correctly prints 1.
2. **For \( 2 \le n \le 4 \)**: The code correctly prints 2.
3. **For \( n > 4 \)**: The code uses a loop to find the smallest \( cnt \) such that \( (1 << cnt) - 1 \le n \), and then prints \( cnt - 1 \).

The potential issue lies in the loop condition and the final calculation. The loop condition \( (1 << cnt) - 1 \le n \) might not always accurately determine the minimum number of operations.

### Edge Case Analysis

1. **Power of 2 minus 1**: Values like 7, 15, 31, etc., which are one less than a power of 2, might cause issues because the loop might not correctly handle these cases.
2. **Just below a power of 2**: Values like 6, 14, 30, etc., which are just below a power of 2, might also cause issues.

### Challenging Test Input

Let's test the code with \( n = 7 \):

- For \( n = 7 \), the correct output should be 3 because:
  - First, place 1 at index 1: [1, 0, 0, 0, 0, 0, 0]
  - Second, place 1 at index 7: [1, 0, 0, 0, 0, 0, 1]
  - Third, use the second operation on the segment [1, 7] to fill the rest: [1, 1, 1, 1, 1, 1, 1]

### Test Input

```plaintext
1
7
```

This test case should expose the bug if the code incorrectly calculates the minimum number of operations for \( n = 7 \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

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

/* Author : Sanaul Islam Siyam */

#define ll long long int
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long>
#define mii map<int, int>
#define si set<int>
#define sc set<char>
#define endl "\n"
#define optimise() ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);

#define f(i,s,e) for(long long int i=s;i<e;i++)
#define cf(i,s,e) for(long long int i=s;i<=e;i++)
#define rf(i,e,s) for(long long int i=e-1;i>=s;i--)
#define all(v) v.begin(), v.end()
#define pb push_back
#define eb emplace_back

template <class T>
void print_v(vector<T> &v) { cout << "{"; for (auto x : v) cout << x << ","; cout << "\b}"; }

#define MOD 1000000007
#define PI 3.1415926535897932384626433832795
#define read(type) readInt<type>()
ll min(ll a,int b) { if (a<b) return a; return b; }
ll min(int a,ll b) { if (a<b) return a; return b; }
ll max(ll a,int b) { if (a>b) return a; return b; }
ll max(int a,ll b) { if (a>b) return a; return b; }
ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
string to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }
string to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }
bool prime(ll a) { if (a==1) return 0; for (int i=2;i<=round(sqrt(a));++i) if (a%i==0) return 0; return 1; }
void yes() { cout<<"YES\n"; }
void no() { cout<<"NO\n"; }

/*  All Required define Pre-Processors and typedef Constants */
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;

void hulk ()
{
    int n;
    cin>>n;
    if(n==1)
    {
        cout<<1<<endl;
        return;
    }
    if(n<5)
    {
        cout<<2<<endl;
        return;
    }

    cout << (int)(log2(n) + 1) -1<< endl;

}

int main()
{
    optimise();
    int tc;
    cin>>tc;

    while(tc--){
        hulk();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1
2
4
20
21
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll mod = 1e9 + 7;
ll MOD=998244353;

void soln() {
    ll n;
    cin>>n;
    if(n==1){
        cout<<1<<endl;
        return;
    }else if(n<=5){
        cout<<2<<endl;
        return;
    }else{
        ll ct=2;
        ll sum=5,temp=5;
        while(temp<n){
            temp=(2*sum+1);
            ct++;
            sum=temp;
        }

        cout<<ct<<endl;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll t;
    cin >> t;
    while (t--) {
        soln();
    }
    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
4
1
2
4
13
```

**Explanation:**

1. **Test Case 1: n = 1**
   - This is a simple edge case where the array has only one element. The expected output is 1, as you need to perform one operation of the first type to make the single element 1.

2. **Test Case 2: n = 2**
   - This is another simple edge case where the array has two elements. The expected output is 2, as you need to perform two operations of the first type to make both elements 1.

3. **Test Case 3: n = 4**
   - This is a slightly more complex case where the array has four elements. The expected output is 2, as you can perform two operations of the first type and then use the second type to fill the entire array.

4. **Test Case 4: n = 13**
   - This is a more challenging case that might expose the bug in the code. The expected output is 4, but the buggy code might not handle this correctly due to the logic in the while loop.

These test cases cover a range of scenarios, including the smallest possible array, a small array, a medium-sized array, and a larger array that could potentially reveal issues in the logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import math
for _ in range(int(input())):
    a = int(input())
    if a == 1:
        print(1)
    elif a == 2:
        print(2)
    else:
        print(max(1, (int(a ** 0.5)) // 2 + 1) * 2 - 2 + a % 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 test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. The buggy code has a specific formula that might not handle all cases correctly, especially when the length of the array is a specific value that doesn't fit the pattern expected by the formula.

Let's analyze the code:
- For \( n = 1 \) and \( n = 2 \), the code handles these cases correctly.
- For \( n > 2 \), the code uses the formula `max(1, (int(a ** 0.5)) // 2 + 1) * 2 - 2 + a % 2`.

We need to find a value of \( n \) that might cause this formula to produce an incorrect result. One way to do this is to test values where the square root and integer division might lead to incorrect results.

### Step-by-Step Analysis:
1. **Small Values:**
   - For \( n = 3 \):
     - The formula becomes `max(1, (int(3 ** 0.5)) // 2 + 1) * 2 - 2 + 3 % 2`
     - `int(3 ** 0.5) = 1`
     - `1 // 2 + 1 = 1`
     - `max(1, 1) * 2 - 2 + 1 = 1 * 2 - 2 + 1 = 1`
     - The correct answer is 2 (one operation to set \( a_1 \) and one to set \( a_2 \) and \( a_3 \) using the second operation).

2. **Larger Values:**
   - For \( n = 5 \):
     - The formula becomes `max(1, (int(5 ** 0.5)) // 2 + 1) * 2 - 2 + 5 % 2`
     - `int(5 ** 0.5) = 2`
     - `2 // 2 + 1 = 2`
     - `max(1, 2) * 2 - 2 + 1 = 2 * 2 - 2 + 1 = 3`
     - The correct answer is 3 (one operation to set \( a_1 \), one to set \( a_3 \), and one to set \( a_2 \) and \( a_4 \) and \( a_5 \) using the second operation).

3. **Edge Case:**
   - For \( n = 6 \):
     - The formula becomes `max(1, (int(6 ** 0.5)) // 2 + 1) * 2 - 2 + 6 % 2`
     - `int(6 ** 0.5) = 2`
     - `2 // 2 + 1 = 2`
     - `max(1, 2) * 2 - 2 + 0 = 2 * 2 - 2 + 0 = 2`
     - The correct answer is 3 (one operation to set \( a_1 \), one to set \( a_4 \), and one to set \( a_2 \) and \( a_3 \) and \( a_5 \) and \( a_6 \) using the second operation).

### Test Input:
```plaintext
3
3
5
6
```

This test input includes values where the buggy code's formula might produce incorrect results, particularly for \( n = 3 \) and \( n = 6 \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
import math

t = int(input())
for i in range(t):
    n = int(input())

    if (n == 1):
        print(1)
    elif (n > 1 and n <=4):
        print(2)
    else:
        print(math.ceil(n/5))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

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 one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

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

public class Contest {

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

        int testcases = sc.nextInt();

        for (int z = 0; z < testcases; z++) {

            int n = sc.nextInt();
            if (n > 2) {
                n = n / 2 + n % 2;
            }

            int steps = 0;

            while (n > 0) {
                if (steps < 2) {
                    n--;
                    steps++;
                } else {
                    n -= 4;
                    steps++;
                }
            }

            System.out.println(steps);

        }

        sc.close();
    }
}

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

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

Think step by step.

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