**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
t=int(input())
for t1 in range(t):
    a,b,c=map(int,input().split())
    h=0
    w = 1
    if a!=0:
        while a>w:
            a-=w
            w*=2
            h+=1
    if a!=0:
        h+=1
        wt=w-a
        w+=a
    else:
        wt=0
    if b>=wt:
        b-=wt
        wt=0
    if b!=0:
        h+=int(b/w + 0.9999)
    c-=w
    h+=1
    if c!=0:
        h=0
    print(h-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:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The code has several issues, including incorrect handling of the number of children and the calculation of the tree height. Let's consider the following points:

1. **Handling of `a` (vertices with 2 children):** The code uses a while loop to reduce `a` and increase `w` and `h`. This loop can fail if `a` is not handled correctly.
2. **Handling of `b` (vertices with 1 child):** The code subtracts `wt` from `b` and then calculates the height based on `b` and `w`. This can fail if `b` is not handled correctly.
3. **Handling of `c` (vertices with 0 children):** The code subtracts `w` from `c` and then adds 1 to `h`. This can fail if `c` is not handled correctly.

Let's generate a test input that maximizes the failure probability in the faulty solution:

- **Case 1:** `a = 1`, `b = 0`, `c = 0`
  - This case should return `-1` because a tree with one vertex with 2 children and no other vertices is impossible.
- **Case 2:** `a = 0`, `b = 1`, `c = 0`
  - This case should return `1` because a tree with one vertex with 1 child and no other vertices is possible.
- **Case 3:** `a = 0`, `b = 0`, `c = 1`
  - This case should return `0` because a tree with one vertex with 0 children is possible.
- **Case 4:** `a = 1`, `b = 1`, `c = 1`
  - This case should return `2` because a tree with one vertex with 2 children, one vertex with 1 child, and one vertex with 0 children is possible.
- **Case 5:** `a = 2`, `b = 1`, `c = 3`
  - This case should return `2` as per the example in the problem statement.
- **Case 6:** `a = 0`, `b = 3`, `c = 1`
  - This case should return `3` because a tree with three vertices with 1 child and one vertex with 0 children is possible.

Here is the test input:

```plaintext
6
1 0 0
0 1 0
0 0 1
1 1 1
2 1 3
0 3 1
```

This test input includes a variety of edge cases and should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
# Template Codeforces
import sys
import math
input = sys.stdin.readline

    ############ ---- Input Functions ---- ############
def inp(): #integer
    return(int(input()))
def inlt(): # list
    return(list(map(int,input().split())))
def insr(): # string
    s = input()
    return(list(s[:len(s) - 1]))
def invr(): # separated int
    return(map(int,input().split()))

    # sys.stdout.write(str('yes') + "\n")

############ ---- Problem Solution ---- ############
def main():
    cases = inp()
    for _ in range(cases):
        a, b, c = inlt()

        if a+1 <= b+c and c - (a+1 - b) == b:
            if a == 0 and b == 0:
                tp = 0
            else:
                if a != 0:
                    n_a = int(math.log2(a)) + 1
                    if b != 0:
                        if (2*(a - 2**(n_a-1)-1) + (2**n_a - a - 1)) == 0:
                            n_b = math.ceil(b - (2**n_a - a - 1))
                        else:
                            n_b = math.ceil((b - (2**n_a - a - 1))/(2*(a - 2**(n_a-1)-1) + (2**n_a - a - 1)))
                    else:
                        n_b = 0
                else:
                    n_a = 0
                    n_b = b

                tp = n_a + n_b
        else:
            tp = -1

        sys.stdout.write(str(tp) + "\n")

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. Here are the steps to identify such a test case:

1. **Understand the Bug**: The code seems to have issues with the logic for calculating the height of the tree, especially when dealing with the constraints \(a + 1 \leq b + c\) and the conditions for forming a valid tree.

2. **Edge Cases**:
   - **Case 1**: When \(a = 0\) and \(b = 0\), the tree should be a single vertex with height 0.
   - **Case 2**: When \(a = 0\) and \(b \neq 0\), the tree should have a height of 1 if \(c \geq b\).
   - **Case 3**: When \(a \neq 0\) and \(b = 0\), the tree should have a height of at least 2.
   - **Case 4**: When \(a + 1 > b + c\), the tree is invalid and should return -1.
   - **Case 5**: When \(a + 1 = b + c\), the tree should have a height of 1.

3. **Complex Case**: A more complex case where \(a\), \(b\), and \(c\) are all non-zero and the height calculation is non-trivial.

4. **Test Case Selection**: Choose a test case that is small but complex enough to challenge the logic in the code.

Given these considerations, let's create a test input that includes a mix of edge cases and a complex case:

```plaintext
6
0 0 1
0 1 1
1 0 2
2 1 3
3 1 4
24 36 48
```

This test input includes:
- A single vertex tree (0 0 1)
- A tree with one edge (0 1 1)
- A tree with one vertex with 2 children (1 0 2)
- A tree with a valid configuration (2 1 3)
- A more complex valid tree (3 1 4)
- A large invalid tree (24 36 48)

This should help expose any issues in the logic for calculating the height of the tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import math

t = int(input())

def solve(a, b, c):
	if c != a + 1:
		return -1
	if a == 0:
		return b
	layers2 = int(math.log2(a))
	#print('l2', layers2)
	r = 2**(layers2+1) - 1 - a
	originalr = r
	#print('r', r)
	d = min(b, r)
	b -= d
	r -= d
	d = min(c, r)
	c -= d
	r -= d
	#print('b', b)
	#print('c', c)
	layers1 = 0
	if b > 0:
		layers1 = int(math.ceil(b / (2**(layers2+1) - originalr)))
		#print('l1', layers1)
		c -= min(c, (2**(layers2+1) - originalr) - b % (2**(layers2+1) - originalr))
		#print('c', c)
	if c > 0:
		layers1 += 1
	#print('l1', layers1)
	return layers2 + layers1

for i in range(t):
	a, b, c = map(int, input().split())
	print(solve(a, b, c))

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

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

Think step by step.

Response:
```plaintext
3
0 2 0
1 0 1
2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class F {
    static FastInputReader in = new FastInputReader();
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public static void main(String[] args) throws IOException {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.flush();
    }

    public static void solve() throws IOException {
        int a = in.nextInt();
        int b = in.nextInt();
        int c = in.nextInt();

        if (a == c - 1) {
            int cap = 1;
            int level = 1;
            while (cap <= a) {
                a -= cap;
                level++;
                cap *= 2;
            }

            while (cap - a < b) {
                b -= cap - a;
                level++;
                cap = cap + a;
                a = 0;
            }

            if (b > 0) {
                level++;
            }
            out.println(--level);
        } else {
            out.println(-1);
        }
    }

    static class FastInputReader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

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

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

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

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

        BigInteger nextBigInteger() throws IOException {
            return new BigInteger(next());
        }

        BigDecimal nextBigDecimal() throws IOException {
            return new BigDecimal(next());
        }

        char nextChar() throws IOException {
            String s = next();
            if (s.length() != 1) {
                throw new IOException("nextChar expecting token of length 1. Instead found: " + s);
            }
            return s.charAt(0);
        }

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

        String nextLine() throws IOException {
            String str;
            if (tokenizer != null && tokenizer.hasMoreTokens()) {
                str = tokenizer.nextToken("\n");
            } else {
                str = reader.readLine();
            }
            return str;
        }

        int[] nextArrayInt(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        long[] nextArrayLong(int n) throws IOException {
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        double[] nextArrayDouble(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }

        char[] nextArrayChar(int n) throws IOException {
            return next().toCharArray();
        }

        BigInteger[] nextArrayBigInteger(int n) throws IOException {
            BigInteger[] arr = new BigInteger[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextBigInteger();
            }
            return arr;
        }

        BigDecimal[] nextArrayBigDecimal(int n) throws IOException {
            BigDecimal[] arr = new BigDecimal[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextBigDecimal();
            }
            return arr;
        }

        String[] nextArrayString(int n) throws IOException {
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = next();
            }
            return arr;
        }
    }

}

**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
2 0 1
```

### Explanation:
This test case is designed to expose a specific bug in the given buggy code. Here's the breakdown:

1. **Input Values**:
   - `a = 2` (vertices with 2 children)
   - `b = 0` (vertices with 1 child)
   - `c = 1` (vertices with 0 children)

2. **Expected Behavior**:
   - The tree should have 3 vertices in total.
   - One vertex has 2 children, and those children are leaves (no children).

3. **Correct Output**:
   - The minimum height of the tree should be 2. The root has 2 children, and those children are leaves.

4. **Buggy Code Behavior**:
   - The buggy code has a condition `if (a == c - 1)`, which is incorrect for this case.
   - For `a = 2` and `c = 1`, `a != c - 1` (2 != 0), so the code will output `-1` instead of the correct height `2`.

This test case will help identify the bug in the condition and ensure that the code correctly handles cases where the number of vertices with 2 children is not equal to the number of leaves minus one.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import java.text.CollationElementIterator;
import java.util.*;

public class ProblemF {

//   static class FenwickTree{
//        int maxSize;
//        int tree[];
//        int numberOfElements;
//        FenwickTree(int n){
//            maxSize= n;
//            tree = new int[n+1];
//            numberOfElements = 0;
//        }
//
//        void add(int n,int delta){
//            while (n<=maxSize){
//                tree[n]+=delta;
//                n+=(n&(-n));
//            }
//            numberOfElements++;
//        }
//        long getCeilCount(int index){
//            int floorCount = 0;
//            while(index>0){
//                floorCount+=tree[index];
//                index -= (index&(-index));
//            }
//            return numberOfElements-floorCount;
//        }
//
//    }

    static long mod = (long)(1e9+7);
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();

            int nodeSpace = a+1;

            if(b>=1 && c==0){
                System.out.println(-1);
                continue;
            }

            if(c>nodeSpace || c<nodeSpace-2*b) {
                System.out.println(-1);
                continue;
            }

            if(c!=a+1){
                System.out.println(-1);
                continue;
            }

            if(a==0){
                System.out.println(b);
                continue;
            }

            int fullNodeHeight = (int)Math.ceil((Math.log(a)/(double)Math.log(2)));
            int l = (1<<fullNodeHeight)-nodeSpace;
            int halfNodeHeight = (b>l)?(b-l)/(nodeSpace)+1:0;
            int res = fullNodeHeight+halfNodeHeight;
            System.out.println(res);
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the code and identify potential edge cases and scenarios where the code might fail.

### Analysis of the Buggy Code

1. **Condition 1: `if(b>=1 && c==0)`**
   - This condition checks if there is at least one vertex with 1 child and no vertices with 0 children. If this is true, the code outputs `-1`.
   - **Potential Issue:** This condition might be too strict. For example, a tree with one vertex having 1 child and no leaf nodes is not impossible.

2. **Condition 2: `if(c>nodeSpace || c<nodeSpace-2*b)`**
   - This condition checks if the number of leaf nodes is within a valid range.
   - **Potential Issue:** The range calculation might be incorrect, leading to false negatives.

3. **Condition 3: `if(c!=a+1)`**
   - This condition checks if the number of leaf nodes is exactly `a + 1`.
   - **Potential Issue:** This condition might be too strict, especially for cases where `b` and `c` can vary.

4. **Condition 4: `if(a==0)`**
   - This condition handles the case where there are no vertices with 2 children.
   - **Potential Issue:** The output for this condition is `b`, which might not always be correct.

### Challenging Test Input

To expose the bugs, we can focus on the following edge cases:
- **Case 1:** A tree with one vertex having 1 child and no leaf nodes.
- **Case 2:** A tree where the number of leaf nodes is just outside the range calculated by the code.
- **Case 3:** A tree with no vertices having 2 children but a valid number of leaf nodes.

### Test Input

```plaintext
3
1 1 0
0 2 2
0 3 1
```

### Explanation

1. **Test Case 1: `1 1 0`**
   - This test case has one vertex with 1 child and no leaf nodes.
   - The correct output should be `1` (a single edge tree).
   - The buggy code outputs `-1` due to the condition `if(b>=1 && c==0)`.

2. **Test Case 2: `0 2 2`**
   - This test case has no vertices with 2 children, two vertices with 1 child, and two leaf nodes.
   - The correct output should be `2` (a linear chain of 4 vertices).
   - The buggy code might output an incorrect value due to the condition `if(a==0)`.

3. **Test Case 3: `0 3 1`**
   - This test case has no vertices with 2 children, three vertices with 1 child, and one leaf node.
   - The correct output should be `3` (a linear chain of 4 vertices with one additional vertex).
   - The buggy code outputs `3` but might fail in other scenarios due to the condition `if(c!=a+1)`.

This test input should effectively expose the bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
#define ll long long
#define f(i, a, b) for (ll i = a; i < b; i++)
#define fr(i, a, b) for (ll i = b-1; i >= a; i--)
#define fa(ele, x) for (auto ele : x)
#define sorta(v,n) sort(v,v+n)
#define sortv(v) sort(v.begin(),v.end())
#define F first
#define S second
#define MP make_pair
#define PB push_back
#define vi vector<int>
#define vl vector<ll>
#define vs vector<string>
#define vpii vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define sei set<int>
#define sel set<ll>
#define mii map<int, int>
#define mll map<ll, ll>
#define mipii map<int, pair<int, int>>
#define mic map<int, char>
#define qi queue<int>
#define ql queue<ll>
#define qpii queue<pair<int, int>>
const ll mod = 1e9 + 7;
const ll INF = LONG_LONG_MAX;
const ll N = 1e5 + 5;
using namespace __gnu_pbds;
template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ordered_set <ll> or_s;

int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
ll t;
cin>>t;
while(t--){
    ll a,b,c;
    cin>>a>>b>>c;
    if(c!=a+1){
        cout<<-1<<endl;
    }
    else if(c==1) cout<<b<<endl;
    else {
        ll h = log2(a)+1;
        // cout<<h<<endl;
        ll k = (1<<h)-1;
        // cout<<k<<endl;
        c-=(a+1);
        b-=(k-a);

        ll div = (1<<(h-1));
        // cout<<div<<endl;
        h+=(b/(div));
        if(b%div) h++;
        cout<<h<<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 and scenarios that are likely to cause the code to fail. Let's analyze the conditions and the logic of the code to identify potential issues:

1. **Invalid Tree Conditions:**
   - The code checks if `c != a + 1` and outputs `-1` if this condition is not met. This is incorrect because `c` should be equal to `a + 1` only when there are no vertices with 1 child (`b == 0`).
   - The correct condition should be `c >= a + 1` for the tree to be valid.

2. **Special Cases:**
   - The code handles the case `c == 1` by outputting `b`. This is correct but can be tested for edge cases.
   - The code calculates the height using `log2(a) + 1` and then adjusts it based on `b` and `c`. This logic can be flawed if `a` is zero or very small.

3. **Height Calculation:**
   - The height calculation logic is complex and can fail if the values of `a`, `b`, and `c` are not well-balanced. For example, if `b` is much larger than `a` and `c`, the height calculation might not be accurate.

Given these points, let's generate a test input that challenges these aspects:

1. **Invalid Tree Condition:**
   - Test with `c < a + 1` to ensure the code outputs `-1`.

2. **Special Case:**
   - Test with `c == 1` and different values of `b` to ensure the code handles this correctly.

3. **Height Calculation:**
   - Test with small values of `a` and large values of `b` to ensure the height calculation is accurate.

Here is the test input:

```plaintext
3
2 1 2
0 3 1
1 5 2
```

Explanation:
1. **Test Case 1:**
   - `a = 2`, `b = 1`, `c = 2`
   - This is an invalid tree because `c < a + 1`. The correct output should be `-1`.

2. **Test Case 2:**
   - `a = 0`, `b = 3`, `c = 1`
   - This is a special case where `c == 1`. The correct output should be `3`.

3. **Test Case 3:**
   - `a = 1`, `b = 5`, `c = 2`
   - This tests the height calculation with a small `a` and a large `b`. The correct output should be `3`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
#include <iomanip>
#include <numeric>
 #include <algorithm>
#include<random>

#define rep(i, n) for (ll i = 0; i < n; i++)
#define sor(a) sort(a.begin(), a.end())
#define rev(a) reverse(a.begin(), a.end())
#define prique priority_queue<int, vector<int>, greater<int>>
#define all(G) G.begin(), G.end()
using ll = long long;
using ull = unsigned long long;

const ll inf = 1e18;

using namespace std;

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

    using F=ll;
    using S=ll;
    S op(S a,S b){
        return min(a,b);
    }
    S e(){
        return 0;
    }
    S mapping(F f,S x){
        return f+x;
    }
    F composition(F f,F g){
        return f+g;
    }
    F id(){
        return 0;
    }

    using pl=pair<ll,ll>;

class SegmentTree {
private:
    vector<long long> tree;
    vector<long long> arr;
    int n;

    void buildTree(int node, int start, int end) {
        if (start == end) {
            tree[node] = arr[start];
        } else {
            int mid = (start + end) / 2;
            buildTree(2 * node, start, mid);
            buildTree(2 * node + 1, mid + 1, end);
            tree[node] = min(tree[2 * node], tree[2 * node + 1]);
        }
    }

    long long query(int node, int start, int end, int l, int r) {
        if (r < start || end < l) return LLONG_MAX; // outside the range
        if (l <= start && end <= r) return tree[node]; // inside the range
        int mid = (start + end) / 2;
        return min(query(2 * node, start, mid, l, r), query(2 * node + 1, mid + 1, end, l, r));
    }

    void update(int node, int start, int end, int idx, long long val) {
        if (start == end) {
            tree[node] = val;
            arr[idx] = val;
        } else {
            int mid = (start + end) / 2;
            if (idx <= mid)
                update(2 * node, start, mid, idx, val);
            else
                update(2 * node + 1, mid + 1, end, idx, val);
            tree[node] = min(tree[2 * node], tree[2 * node + 1]);
        }
    }

    long long getValue(int node, int start, int end, int idx) {
        if (start == end) return tree[node];
        int mid = (start + end) / 2;
        if (idx <= mid)
            return getValue(2 * node, start, mid, idx);
        else
            return getValue(2 * node + 1, mid + 1, end, idx);
    }

public:
    SegmentTree(const vector<long long>& input) {
        arr = input;
        n = input.size();
        tree.assign(4 * n, 0); // Size of tree will be at most 4 * input size
        buildTree(1, 0, n - 1);
    }

    long long query(int l, int r) {
        return query(1, 0, n - 1, l, r);
    }

    void update(int idx, long long val) {
        update(1, 0, n - 1, idx, val);
    }
    long long getValue(int idx) {
        return getValue(1, 0, n - 1, idx);
    }
};

 bool inside(ll x, ll y, ll h, ll w) {
     bool a = x < h && y < w;
     bool b = x >= 0 && y >= 0;
     return a && b;
 }

 string to_bit(ll x){
    string s="";
    while(x){
        s+=to_string(x%2);
        x/=2;
    }
    ll h=31-s.size();
    for(ll i=0;i<h;i++)s+="0";
    reverse(all(s));
    return s;

 }

void solve(){
  ll a,b,c;
    cin>>a>>b>>c;
    if(a+1!=c){
        cout<<-1<<endl;
        return;
    }
    ll ans=0;
    ll leaf=1;
    ll left=0;

    while(a){
        ll j=min(leaf,a);
        left=leaf-j;
        leaf+=j;
        a-=j;
        ans++;
    }
    b-=left;
    ll temp=b/leaf;
    if(b%leaf!=0)temp++;
    ans+=temp;

    cout<<ans<<endl;

}

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

 }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the conditions under which the code might fail. The key issues to consider are:

1. **Invalid Tree Conditions**: The code incorrectly checks if `a + 1 != c` to determine if the tree is invalid. This condition is not correct. The correct condition should be based on the total number of vertices and the structure of the tree.

2. **Tree Structure**: The code should ensure that the number of vertices with 2 children, 1 child, and 0 children can form a valid tree. The correct condition is:
   - \(a + b + c \geq 1\)
   - The total number of edges should be \(a + b\), and the total number of vertices should be \(a + b + c\). This means \(a + b = a + b + c - 1\), which simplifies to \(c = 1\).

3. **Minimum Height Calculation**: The code should correctly calculate the minimum height of the tree. The current logic for calculating the height might not be optimal for all cases.

Given these considerations, let's create a test input that exposes these issues:

### Test Input
```plaintext
3
2 1 3
1 0 2
1 1 3
```

### Explanation
1. **Test Case 1: (2, 1, 3)**
   - This is a valid input where \(a = 2\), \(b = 1\), and \(c = 3\). The tree can be formed with a height of 2, as shown in the example.
   - The code should output 2.

2. **Test Case 2: (1, 0, 2)**
   - This is a valid input where \(a = 1\), \(b = 0\), and \(c = 2\). The tree can be formed with a height of 1.
   - The code should output 1.

3. **Test Case 3: (1, 1, 3)**
   - This is an invalid input because \(a + 1 = 2\) and \(c = 3\), which does not satisfy the correct condition for a valid tree.
   - The code should output -1.

### Expected Output
```plaintext
2
1
-1
```

This test input will help identify the issues in the buggy code by covering valid and invalid cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
a = int(input())
for _ in range(0, a):
    a,b,c = map(int, input().split(' '))
    p = 1
    for i in range(1000000):
        if a>0:
            if a>=p:
                a-=p
                p=p*2
            elif a+b>=p:
                np = p
                p = p+a
                b-=np-a
                a = 0
            elif a+b+c>=p:
                p=p+a-(p-a-b)
                c -= p-a-b
                a = 0
                b = 0
            else:
                print(-1)
                break
        elif b>0:
            if b>=p:
                b-=p
            elif b+c>=p:
                np = p
                p-=(c-b)
                c -= np-b
                b = 0
            else:
                print(-1)
                break
        elif c>0:
            if c >p:
                print(-1)
                break
            elif c == p:
                print(i)
                break
            else:
                print(-1)
                break
        else:
            print(-1)
            break

**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 scenarios where the code might fail. Here are some considerations:

1. **Invalid Input**: Check if the code handles cases where no valid tree can be formed.
2. **Zero Vertices with Children**: Check if the code handles cases where all vertices have no children.
3. **Balanced Trees**: Check if the code can handle balanced trees with a specific structure.
4. **Unbalanced Trees**: Check if the code can handle unbalanced trees with a specific structure.
5. **Large Input**: Check if the code can handle large values of `a`, `b`, and `c`.

Let's create a test input that includes a mix of these cases:

```plaintext
5
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
```

### Explanation:

1. **Test Case 1: `2 1 3`**
   - This is a balanced tree with 2 vertices having 2 children, 1 vertex having 1 child, and 3 vertices having 0 children.
   - Expected output: 2

2. **Test Case 2: `0 0 1`**
   - This is a single vertex with no children.
   - Expected output: 0

3. **Test Case 3: `0 1 1`**
   - This is a tree with 1 vertex having 1 child and 1 vertex having 0 children.
   - Expected output: 1

4. **Test Case 4: `1 0 2`**
   - This is a tree with 1 vertex having 2 children and 2 vertices having 0 children.
   - Expected output: 1

5. **Test Case 5: `1 1 3`**
   - This is a tree with 1 vertex having 2 children, 1 vertex having 1 child, and 3 vertices having 0 children.
   - Expected output: -1 (no valid tree can be formed)

This test input should help expose any issues in the buggy code, particularly in handling invalid cases and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
temp = 1
running_list = [0]
while temp < 500000:
    running_list.append(temp + 0)
    temp *= 2

def quality_assurace(temp_a,temp_b,temp_c):
    a = temp_a + 0
    b = temp_b + 0
    c = temp_c + 0
    left_nodes = 0
    for i in range(len(running_list)):
        if a >= running_list[i]:
            a-=running_list[i]
            left_nodes = running_list[i + 1]
        else:
            left_nodes += a
            break
    if c != left_nodes:
        return -1
    height = i
    if a != 0:
        if left_nodes - a >= b:
            b -= left_nodes - a
        height += 1
    height += b // left_nodes
    if b % left_nodes != 0:
        height += 1
    height -= 1
    return height

t = int(input())
while t:
    t-=1
    a,b,c = map(int,input().split())
    height = quality_assurace(a,b,c)
    print(height)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are some considerations:

1. **Zero Vertices with 2 Children (a = 0)**: This can cause issues if the algorithm does not handle the case where there are no vertices with 2 children correctly.
2. **Zero Vertices with 1 Child (b = 0)**: This can cause issues if the algorithm does not handle the case where there are no vertices with 1 child correctly.
3. **Zero Vertices with 0 Children (c = 0)**: This can cause issues if the algorithm does not handle the case where there are no leaves correctly.
4. **All Vertices with 2 Children (a > 0, b = 0, c = 0)**: This can cause issues if the algorithm does not handle the case where all vertices have 2 children correctly.
5. **All Vertices with 1 Child (a = 0, b > 0, c = 0)**: This can cause issues if the algorithm does not handle the case where all vertices have 1 child correctly.
6. **All Vertices with 0 Children (a = 0, b = 0, c > 0)**: This can cause issues if the algorithm does not handle the case where all vertices are leaves correctly.
7. **Balanced Cases**: Cases where the number of vertices with 2, 1, and 0 children are balanced can also be challenging.

Given these considerations, a small-scale test input that includes a mix of these edge cases is:

```plaintext
4
0 0 1
0 1 1
1 0 2
2 1 3
```

This test input includes:
- A single vertex with 0 children (a = 0, b = 0, c = 1)
- A single vertex with 1 child and 1 leaf (a = 0, b = 1, c = 1)
- A single vertex with 2 children and 2 leaves (a = 1, b = 0, c = 2)
- A balanced case with 2 vertices with 2 children, 1 vertex with 1 child, and 3 leaves (a = 2, b = 1, c = 3)

These test cases should help expose any bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

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

public class Main

{
	static FastReader sc = new FastReader();
	static long dp[][];
	// static boolean v[][][];
	// static int mod=998244353;;
	static int mod = 1000000007;
	static long max;
	static int bit[];
	static int bit1[];
	static int cnt;
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		// StringBuffer sb=new StringBuffer("");
		int ttt = 1;
		ttt = i();
		outer: while (ttt-- > 0) {

			int a=i(),b=i(),c=i();
			Queue<Integer> q=new LinkedList<>();
			q.add(0);
			ArrayList<Integer> A[]=new ArrayList[a+b+c+10];
			for(int i=0;i<a+b+c+10;i++){
				A[i]=new ArrayList<>();
			}
			int curr=1;
			while(a-->0){
				int y=q.remove();
				A[y].add(curr);
				A[y].add(curr+1);
				q.add(curr++);
				q.add(curr++);
			}
			while(b-->0){
				int y=q.remove();
				A[y].add(curr);
				q.add(curr++);
			}
			cnt=0;
			int height=go(A, 0, -1);
			if(cnt==c){
				System.out.println(height);
			}
			else{
				System.out.println(-1);
			}

		}
		// System.out.println(sb.toString());
		out.close();

		// CHECK FOR N=1 //CHECK FOR M=0
		// CHECK FOR N=1 //CHECK FOR M=0
		// CHECK FOR N=1

	}
	static int go(ArrayList<Integer> A[],int i,int par){
		int max=0;
		if(A[i].size()==0){
			cnt++;
			return 0;
		}
		for(int child : A[i]){
			if(child!=par){
				int op1=go(A, child, i);
				max=max(op1,max);
			}
		}
		return 1+max;
	}

	static class Pair implements Comparable<Pair> {
		int x;
		int y;
		int z;

		Pair(int x, int y) {
			this.x = x;
			this.y = y;
			// this.z=z;

		}

		@Override
		public int compareTo(Pair o) {
			if (this.x > o.x)
				return 1;
			else if (this.x < o.x)
				return -1;
			else {
				if (this.y > o.y)
					return -1;
				else if (this.y < o.y)
					return 1;
				else
					return 0;
			}
		}

		public int hashCode() {
			long ans = 1;
			ans = x * 31 + y * 13;
			return (int) ans;
		}

		// @Override
		public boolean equals(Object o) {
			if (this == o) {
				return true;
			}
			if (o == null) {
				return false;
			}
			if (this.getClass() != o.getClass()) {
				return false;
			}
			Pair other = (Pair) o;
			if (this.x != other.x || this.y != other.y) {
				return false;
			}
			return true;
		}
		//
		/* FOR TREE MAP PAIR USE */

		// public int compareTo(Pair o) {
		// if (x > o.x) {
		// return 1;
		// }
		// if (x < o.x) {
		// return -1;
		// }
		// if (y > o.y) {
		// return 1;
		// }
		// if (y < o.y) {
		// return -1;
		// }
		// return 0;
		// }

	}

	// static int find(int A[],int a) {
	// if(A[a]==a)
	// return a;
	// return A[a]=find(A, A[a]);
	// }
	// FENWICK TREE
	static void update(int i, int x) {
		for (; i < bit.length; i += (i & -i))
			bit[i] += x;
	}

	static int sum(int i) {
		int ans = 0;
		for (; i > 0; i -= (i & -i))
			ans += bit[i];
		return ans;
	}

	// END
	static void add(HashMap<Long, Integer> map, long v) {
		if (!map.containsKey(v)) {
			map.put(v, 1);
		} else {
			map.put(v, map.get(v) + 1);
		}
	}

	static void remove(HashMap<Long, Integer> map, long v) {
		if (map.containsKey(v)) {
			map.put(v, map.get(v) - 1);
			if (map.get(v) == 0)
				map.remove(v);
		}
	}

	static void add(HashMap<Integer, Integer> map, int v) {
		if (!map.containsKey(v)) {
			map.put(v, 1);
		} else {
			map.put(v, map.get(v) + 1);
		}
	}

	static void remove(HashMap<Integer, Integer> map, int v) {
		if (map.containsKey(v)) {
			map.put(v, map.get(v) - 1);
			if (map.get(v) == 0)
				map.remove(v);
		}
	}

	static void add(TreeMap<Integer, Integer> map, int v) {
		if (!map.containsKey(v)) {
			map.put(v, 1);
		} else {
			map.put(v, map.get(v) + 1);
		}
	}

	static void add(TreeMap<Long, Integer> map, long v) {
		if (!map.containsKey(v)) {
			map.put(v, 1);
		} else {
			map.put(v, map.get(v) + 1);
		}
	}

	static void remove(TreeMap<Integer, Integer> map, int v) {
		if (map.containsKey(v)) {
			map.put(v, map.get(v) - 1);
			if (map.get(v) == 0)
				map.remove(v);
		}
	}

	static void remove(TreeMap<Long, Integer> map, long v) {
		if (map.containsKey(v)) {
			map.put(v, map.get(v) - 1);
			if (map.get(v) == 0)
				map.remove(v);
		}
	}

	static long modInverse(long n, int p) {
		return power(n, p - 2, p);
	}

	static int[] copy(int A[]) {
		int B[] = new int[A.length];
		for (int i = 0; i < A.length; i++) {
			B[i] = A[i];
		}
		return B;
	}

	static long[] copy(long A[]) {
		long B[] = new long[A.length];
		for (int i = 0; i < A.length; i++) {
			B[i] = A[i];
		}
		return B;
	}

	static int[] input(int n) {
		int A[] = new int[n];
		for (int i = 0; i < n; i++) {
			A[i] = sc.nextInt();
		}
		return A;
	}

	static long[] inputL(int n) {
		long A[] = new long[n];
		for (int i = 0; i < n; i++) {
			A[i] = sc.nextLong();
		}
		return A;
	}

	static String[] inputS(int n) {
		String A[] = new String[n];
		for (int i = 0; i < n; i++) {
			A[i] = sc.next();
		}
		return A;
	}

	static long sum(int A[]) {
		long sum = 0;
		for (int i : A) {
			sum += i;
		}
		return sum;
	}

	static long sum(long A[]) {
		long sum = 0;
		for (long i : A) {
			sum += i;
		}
		return sum;
	}

	static void reverse(long A[]) {
		int n = A.length;
		long B[] = new long[n];
		for (int i = 0; i < n; i++) {
			B[i] = A[n - i - 1];
		}
		for (int i = 0; i < n; i++)
			A[i] = B[i];

	}

	static void reverse(int A[]) {
		int n = A.length;
		int B[] = new int[n];
		for (int i = 0; i < n; i++) {
			B[i] = A[n - i - 1];
		}
		for (int i = 0; i < n; i++)
			A[i] = B[i];

	}

	static void input(int A[], int B[]) {
		for (int i = 0; i < A.length; i++) {
			A[i] = sc.nextInt();
			B[i] = sc.nextInt();
		}
	}

	static int[][] input(int n, int m) {
		int A[][] = new int[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				A[i][j] = i();
			}
		}
		return A;
	}

	static char[][] charinput(int n, int m) {
		char A[][] = new char[n][m];
		for (int i = 0; i < n; i++) {
			String s = s();
			for (int j = 0; j < m; j++) {
				A[i][j] = s.charAt(j);
			}
		}
		return A;
	}

	static int nextPowerOf2(int n) {
		if (n == 0)
			return 1;
		n--;
		n |= n >> 1;
		n |= n >> 2;
		n |= n >> 4;
		n |= n >> 8;
		n |= n >> 16;
		n++;

		return n;
	}

	static int highestPowerof2(int x) {

		x |= x >> 1;
		x |= x >> 2;
		x |= x >> 4;
		x |= x >> 8;
		x |= x >> 16;

		return x ^ (x >> 1);

	}

	static long highestPowerof2(long x) {

		x |= x >> 1;
		x |= x >> 2;
		x |= x >> 4;
		x |= x >> 8;
		x |= x >> 16;
		return x ^ (x >> 1);

	}

	static int max(int A[]) {
		int max = Integer.MIN_VALUE;
		for (int i = 0; i < A.length; i++) {
			max = Math.max(max, A[i]);
		}
		return max;
	}

	static int min(int A[]) {
		int min = Integer.MAX_VALUE;
		for (int i = 0; i < A.length; i++) {
			min = Math.min(min, A[i]);
		}
		return min;
	}

	static long max(long A[]) {
		long max = Long.MIN_VALUE;
		for (int i = 0; i < A.length; i++) {
			max = Math.max(max, A[i]);
		}
		return max;
	}

	static long min(long A[]) {
		long min = Long.MAX_VALUE;
		for (int i = 0; i < A.length; i++) {
			min = Math.min(min, A[i]);
		}
		return min;
	}

	static long[] prefix(long A[]) {
		long p[] = new long[A.length];
		p[0] = A[0];
		for (int i = 1; i < A.length; i++)
			p[i] = p[i - 1] + A[i];
		return p;
	}

	static long[] prefix(int A[]) {
		long p[] = new long[A.length];
		p[0] = A[0];
		for (int i = 1; i < A.length; i++)
			p[i] = p[i - 1] + A[i];
		return p;
	}

	static long[] suffix(long A[]) {
		long p[] = new long[A.length];
		p[A.length - 1] = A[A.length - 1];
		for (int i = A.length - 2; i >= 0; i--)
			p[i] = p[i + 1] + A[i];
		return p;
	}

	static long[] suffix(int A[]) {
		long p[] = new long[A.length];
		p[A.length - 1] = A[A.length - 1];
		for (int i = A.length - 2; i >= 0; i--)
			p[i] = p[i + 1] + A[i];
		return p;
	}

	static void fill(int dp[]) {
		Arrays.fill(dp, -1);
	}

	static void fill(int dp[][]) {
		for (int i = 0; i < dp.length; i++)
			Arrays.fill(dp[i], -1);
	}

	static void fill(int dp[][][]) {
		for (int i = 0; i < dp.length; i++) {
			for (int j = 0; j < dp[0].length; j++) {
				Arrays.fill(dp[i][j], -1);
			}
		}
	}

	static void fill(int dp[][][][]) {
		for (int i = 0; i < dp.length; i++) {
			for (int j = 0; j < dp[0].length; j++) {
				for (int k = 0; k < dp[0][0].length; k++) {
					Arrays.fill(dp[i][j][k], -1);
				}
			}
		}
	}

	static void fill(long dp[]) {
		Arrays.fill(dp, -1);
	}

	static void fill(long dp[][]) {
		for (int i = 0; i < dp.length; i++)
			Arrays.fill(dp[i], -1);
	}

	static void fill(long dp[][][]) {
		for (int i = 0; i < dp.length; i++) {
			for (int j = 0; j < dp[0].length; j++) {
				Arrays.fill(dp[i][j], -1);
			}
		}
	}

	static void fill(long dp[][][][]) {
		for (int i = 0; i < dp.length; i++) {
			for (int j = 0; j < dp[0].length; j++) {
				for (int k = 0; k < dp[0][0].length; k++) {
					Arrays.fill(dp[i][j][k], -1);
				}
			}
		}
	}

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

	static int min(int a, int b, int c) {
		return Math.min(a, Math.min(b, c));
	}

	static int min(int a, int b, int c, int d) {
		return Math.min(a, Math.min(b, Math.min(c, d)));
	}

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

	static int max(int a, int b, int c) {
		return Math.max(a, Math.max(b, c));
	}

	static int max(int a, int b, int c, int d) {
		return Math.max(a, Math.max(b, Math.max(c, d)));
	}

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

	static long min(long a, long b, long c) {
		return Math.min(a, Math.min(b, c));
	}

	static long min(long a, long b, long c, long d) {
		return Math.min(a, Math.min(b, Math.min(c, d)));
	}

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

	static long max(long a, long b, long c) {
		return Math.max(a, Math.max(b, c));
	}

	static long max(long a, long b, long c, long d) {
		return Math.max(a, Math.max(b, Math.max(c, d)));
	}

	static long power(long x, long y, long p) {

		if (y == 0)
			return 1;
		if (x == 0)
			return 0;
		long res = 1;
		x = x % p;

		while (y > 0) {

			if (y % 2 == 1)
				res = (res * x) % p;

			y = y >> 1;
			x = (x * x) % p;
		}

		return res;
	}

	static long power(long x, long y) {

		if (y == 0)
			return 1;
		if (x == 0)
			return 0;
		long res = 1;

		while (y > 0) {

			if (y % 2 == 1)
				res = (res * x);

			y = y >> 1;
			x = (x * x);
		}

		return res;
	}

	static void print(int A[]) {
		for (int i : A) {
			out.print(i + " ");
		}
		out.println();
	}

	static void print(long A[]) {
		for (long i : A) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	static long mod(long x) {
		return ((x % mod + mod) % mod);
	}

	static String reverse(String s) {
		StringBuffer p = new StringBuffer(s);
		p.reverse();
		return p.toString();
	}

	static int i() {
		return sc.nextInt();
	}

	static String s() {
		return sc.next();
	}

	static long l() {
		return sc.nextLong();
	}

	static void sort(int[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			int tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}

	static void sort(long[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			long tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}

	static String sort(String s) {
		Character ch[] = new Character[s.length()];
		for (int i = 0; i < s.length(); i++) {
			ch[i] = s.charAt(i);
		}
		Arrays.sort(ch);
		StringBuffer st = new StringBuffer("");
		for (int i = 0; i < s.length(); i++) {
			st.append(ch[i]);
		}
		return st.toString();
	}

	static HashMap<Integer, Integer> hash(int A[]) {
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		for (int i : A) {
			if (map.containsKey(i)) {
				map.put(i, map.get(i) + 1);
			} else {
				map.put(i, 1);
			}
		}
		return map;
	}

	static HashMap<Long, Integer> hash(long A[]) {
		HashMap<Long, Integer> map = new HashMap<Long, Integer>();
		for (long i : A) {
			if (map.containsKey(i)) {
				map.put(i, map.get(i) + 1);
			} else {
				map.put(i, 1);
			}
		}
		return map;
	}

	static TreeMap<Integer, Integer> tree(int A[]) {
		TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
		for (int i : A) {
			if (map.containsKey(i)) {
				map.put(i, map.get(i) + 1);
			} else {
				map.put(i, 1);
			}
		}
		return map;
	}

	static TreeMap<Long, Integer> tree(long A[]) {
		TreeMap<Long, Integer> map = new TreeMap<Long, Integer>();
		for (long i : A) {
			if (map.containsKey(i)) {
				map.put(i, map.get(i) + 1);
			} else {
				map.put(i, 1);
			}
		}
		return map;
	}

	static boolean prime(int n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;
		double sq = Math.sqrt(n);

		for (int i = 5; i <= sq; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;
		return true;
	}

	static boolean prime(long n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;
		double sq = Math.sqrt(n);

		for (int i = 5; i <= sq; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;
		return true;
	}

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

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}

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

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

Think step by step.

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

1. **Edge Cases for Input Constraints:**
   - Test with the minimum values of `a`, `b`, and `c` to ensure the code handles small inputs correctly.
   - Test with the maximum values of `a`, `b`, and `c` to ensure the code handles large inputs without memory or time issues.
   - Test with `a + b + c` equal to 1 to see how the code handles a single vertex.

2. **Special Cases:**
   - Test with `a = 0`, `b = 0`, and `c = 1` to see if the code correctly handles a single vertex with no children.
   - Test with `a = 1`, `b = 0`, and `c = 2` to see if the code correctly handles a tree with a single parent and two leaf nodes.
   - Test with `a = 0`, `b = 1`, and `c = 1` to see if the code correctly handles a tree with a single parent and one leaf node.

3. **Invalid Cases:**
   - Test with `a + b + c = 0` to ensure the code correctly outputs `-1`.
   - Test with `a = 1`, `b = 0`, and `c = 0` to see if the code correctly handles a single vertex with one child, which is impossible.

4. **Complex Cases:**
   - Test with `a = 2`, `b = 1`, and `c = 3` to see if the code correctly handles a tree with a specific structure as described in the problem statement.
   - Test with `a = 3`, `b = 1`, and `c = 4` to see if the code correctly handles a more complex tree structure.

Based on these considerations, a challenging test input that exposes the bug in the buggy code is:

```plaintext
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
```

This test input includes a variety of edge cases and special cases that should help identify any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import java.util.*;

public class c5 {

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

        int t = Sc.nextInt();
        while (t-- > 0){
            int  a = Sc.nextInt();
            int  b = Sc.nextInt();
            int  c = Sc.nextInt();
            if (a +1 != c){
                System.out.println(-1);
            } else if (a == 0) {
                System.out.println(b);
            } else {
                double double_h = (Math.log10(c) / Math.log10(2));
                int height = (int) Math.ceil(double_h);
                int minA  = (int) double_h;
                b -= (c - Math.pow(2,minA));
                if (b > 0){
                    b = (int) Math.ceil(b/ (double)c);
                    height += b;
                }
                System.out.println(height);
            }
        }
    }
}

**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 and scenarios that are likely to cause the code to fail. Here are the key points to consider:

1. **Invalid Tree Conditions**: The code incorrectly assumes that `a + 1 != c` means no valid tree can be formed. This is not always true. We need to provide a test case where `a + 1 != c` but a valid tree can still be formed.

2. **Special Cases**: The code has special handling for `a == 0`. We need to test cases where `a == 0` and `b` and `c` are non-zero.

3. **Height Calculation**: The height calculation in the code is flawed. We need to provide a test case where the height calculation is incorrect.

Let's construct a test input that covers these points:

1. **Invalid Tree Condition**: A test case where `a + 1 != c` but a valid tree can be formed.
2. **Special Case**: A test case where `a == 0` and `b` and `c` are non-zero.
3. **Height Calculation**: A test case where the height calculation in the code is incorrect.

Here is the test input:

```plaintext
3
1 2 2
0 2 2
2 2 2
```

Explanation:

1. **Test Case 1: `1 2 2`**
   - `a + 1 = 2` and `c = 2`, so `a + 1 == c`.
   - A valid tree can be formed with 1 vertex with 2 children, 2 vertices with 1 child, and 2 leaves.
   - Expected height: 2.

2. **Test Case 2: `0 2 2`**
   - `a == 0`.
   - A valid tree can be formed with 2 vertices with 1 child and 2 leaves.
   - Expected height: 2.

3. **Test Case 3: `2 2 2`**
   - `a + 1 = 3` and `c = 2`, so `a + 1 != c`.
   - A valid tree can be formed with 2 vertices with 2 children, 2 vertices with 1 child, and 2 leaves.
   - Expected height: 3.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
/*
    Author : Sunny Kumar
    Descr  : Always Ignore
*/

#include <bits/stdc++.h>
#pragma GCC optimize("02")

using namespace std;

#define fastio()                           \
    std::ios_base::sync_with_stdio(false); \
    std::cin.tie(NULL);                    \
    std::cout.tie(NULL)

#define nline \
    std::cout << "\n"
#define YES \
    std::cout << "YES"
#define NO \
    std::cout << "NO"
#define two_input(a, b) \
    std::cin >> a >> b;
#define thr_input(a, b, c) \
    std::cin >> a >> b >> c;
#define four_input(a, b, c, d) \
    std::cin >> a >> b >> c >> d;

#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18

#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define sz(x) ((int)(x).size())
#define max(a, b) (a < b ? b : a)
#define min(a, b) ((a > b) ? b : a)
#define FOR(a, c) for (int a = 0; a < c; a++)
#define FORL(a, b, c) for (int a = b; a <= c; a++)
#define FORR(a, b, c) for (int a = b; a >= c; a--)
#define allass(x) sort(x.begin(), x.end());
#define alldsc(x) sort(x.rbegin(), x.rend());
#define revass(x) reverse(x.begin(), x.end());
#define revdsc(x) reverse(x.rbegin(), x.rend());
#define sum_int(x) accumulate(x.begin(), x.end(), 0)
#define sum_long_long(x) accumulate(x.begin(), x.end(), 0ll)
#define lower__bound(x, val) lower_bound(x.begin(), x.end(), val) - x.begin()
#define upper__bound(x, val) upper_bound(x.begin(), x.end(), val) - x.begin()
#define sqr_veci(x) for_each(x.begin(), x.end(), [](int &n)({ n *= n; }))
#define sqr_vecl(x) for_each(x.begin(), x.end(), [](long long &n)({ n *= n; }))

#define take_input \
    std::cin >>
#define read_output \
    std::cout <<

typedef unsigned long long ull;
typedef long long ll;
typedef float flt;
typedef long long int lli;
typedef double dd;
typedef long double lld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;

int start_time; // Time Variable

void Start_Time()
{
    start_time = clock();
}

#ifndef ONLINE_JUDGE
#define debug(sunny)                        \
    cerr << "[" << #sunny << "] = " << " "; \
    _display_output(sunny);                 \
    cerr << endl;
#else
#define debug(sunny)
#endif

void _display_output(int t) { cerr << t; }
void _display_output(long t) { cerr << t; }
void _display_output(ll t) { cerr << t; }
void _display_output(ull t) { cerr << t; }
void _display_output(lld t) { cerr << t; }
void _display_output(bool t) { cerr << (t ? "true" : "false"); }
void _display_output(flt t) { cerr << t; }
void _display_output(double t) { cerr << t; }
void _display_output(char t) { cerr << t; }
void _display_output(string t) { cerr << t; }

void End_Time()
{
    int Time_Taken = double(clock() - start_time) / CLOCKS_PER_SEC;
    // cerr << " : [";
    debug(Time_Taken);
    // cerr << "s]";
}

#define power_value(a, b, val) \
    ll res = 1;                \
    while (b--)                \
    {                          \
        res *= a;              \
    }                          \
    val = res;

#define square_root(a, val)    \
    ll root = 1;               \
    while ((root * root) <= a) \
    {                          \
        root++;                \
    }                          \
    val = root - 1;

#define log_base(bs, exp, val) \
    int cnt = 0;               \
    while (exp)                \
    {                          \
        exp /= bs;             \
        cnt++;                 \
    }                          \
    val = cnt - 1;

template <class T, class V>
void _display_output(pair<T, V> p)
{
    cerr << "{";
    _display_output(p.ff);
    cerr << ",";
    _display_output(p.ss);
    cerr << "}";
}
template <class T>
void _display_output(vector<T> vec)
{
    cerr << "[ ";
    for (T i : vec)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _display_output(vector<pair<T, V>> vec_pair)
{
    cerr << "[ ";
    for (auto i : vec_pair)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _display_output(set<T> ordered_set)
{
    cerr << "[ ";
    for (T i : ordered_set)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _display_output(unordered_set<T> hash)
{
    cerr << "[ ";
    for (T i : hash)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _display_output(multiset<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _display_output(map<T, V> v)
{
    cerr << "[ ";
    for (auto i : v)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _display_output(multimap<T, V> v)
{
    cerr << "[ ";
    for (auto i : v)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _display_output(unordered_map<T, V> hash)
{
    cerr << "[ ";
    for (auto i : hash)
    {
        _display_output(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _display_output(priority_queue<T> heap)
{
    vector<T> heap_vector;
    while (!heap.empty())
    {
        heap_vector.pb(heap.top());
        heap.pop();
    }
    debug(heap_vector);
}
template <class T>
void _display_output(priority_queue<T, vector<T>, greater<T>> heap)
{
    vector<T> heap_vector;
    while (!heap.empty())
    {
        heap_vector.pb(heap.top());
        heap.pop();
    }
    debug(heap_vector);
}
template <class T, class V>
void _display_output(priority_queue<pair<T, V>> heap)
{
    vector<pair<T, V>> heap_vector;
    while (!heap.empty())
    {
        heap_vector.pb({heap.top().ff, heap.top().ss});
        heap.pop();
    }
    debug(heap_vector);
}
template <class T, class V>
void _display_output(priority_queue<pair<T, V>, vector<pair<T, V>>, greater<pair<T, V>>> heap)
{
    vector<pair<T, V>> heap_vector;
    while (!heap.empty())
    {
        heap_vector.pb({heap.top().ff, heap.top().ss});
        heap.pop();
    }
    debug(heap_vector);
}
template <class T>
void _display_output(queue<T> que)
{
    cerr << "[ ";
    while (!que.empty())
    {
        _display_output(que.front());
        que.pop();
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _display_output(stack<T> stck)
{
    cerr << "[ ";
    while (!stck.empty())
    {
        _display_output(stck.top());
        stck.pop();
        cerr << " ";
    }
    cerr << "]";
}

void Sanhayu546__()
{
    int testcase_always_ignore;

    take_input testcase_always_ignore;
    // testcase_always_ignore = 1;

    while (testcase_always_ignore--)
    {
        int a, b, c;
        thr_input(a, b, c);
        int initially_depth = 0;
        int number_node = 0;
        int a_node = 0, c_node = 0;
        int remaining_node = 0;
        if (a == 1)
        {
            a_node = 1;
            c_node = 2;
            initially_depth = 1;
        }
        while (a > a_node)
        {
            if ((a_node + (number_node * 2)) < a)
            {
                if (number_node == 0)
                {
                    number_node++;
                    a_node++;
                }
                else
                {
                    number_node *= 2;
                    a_node += number_node;
                }
            }
            else
            {
                remaining_node = a - a_node;
                a_node += remaining_node;
                c_node = 2 * remaining_node + (2 * number_node - remaining_node);
                initially_depth++;
                break;
            }
            initially_depth++;
        }
        debug(initially_depth);
        debug(a_node);
        debug(c_node);
        debug(number_node);
        if (a == 0)
        {
            if (b == 0)
            {
                if (c == 1)
                    read_output 0;
                else
                    read_output - 1;
            }
            else
            {
                if (c == 1)
                    read_output b;
                else
                    read_output - 1;
            }
        }
        else
        {
            if (b == 0)
            {
                if (c == c_node)
                    read_output initially_depth;
                else
                    read_output - 1;
            }
            else
            {
                if (c > c_node)
                    read_output - 1;
                else
                {
                    int cd = c_node - (1 << (initially_depth - 1));
                    int xd = (1 << (initially_depth - 1)) - cd;
                    debug(xd);
                    xd = b - xd;
                    int val = ceil(float(xd) / float(c));
                    read_output val + initially_depth;
                }
            }
        }
        nline;
    }
    return;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("Output.txt", "w", stderr);
#endif
    fastio();

    Start_Time();

    Sanhayu546__();

    nline;
    End_Time();
}

**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
2 0 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

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

int racist(int n) {
  int america = 0;
  while (pow(2, america) - 1 < n) {
    america++;
  }
  return america;
}

int main() {
  int test_cases = 0;
  int t;
  cin >> t;
  while (t--) {
    test_cases++;
    int a, b, c;
    cin >> a >> b >> c;
    if (a != c - 1) {
      cout << -1 << endl;
      continue;
    }
    if (test_cases == 969) {
      cout << a << ' ' << b << ' ' << c << endl;
    }
    if (test_cases > 10) {
      continue;
    }
    int left = pow(2, racist(a)) - 1 - a;
    if (left >= b) {
      cout << ceil(log2(c)) << endl;
      continue;
    }
    else {
      int skibidi = 2*pow(2, racist(a)-1) - left;
      int height = ceil(log2(c));
      while (b > 0) {
        height++;
        b -= skibidi;
      }
      cout << height << endl;
    }
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Condition Check for Valid Tree:**
   - The code checks if `a != c - 1`. This is incorrect because the relationship between `a` and `c` should be more complex. The correct condition should be that the total number of children should match the total number of vertices minus one (i.e., `2a + b = a + b + c - 1`).

2. **Height Calculation:**
   - The code uses a function `racist` to calculate the minimum height, but the logic is flawed. It uses `pow(2, america) - 1` to determine the height, which is not the correct way to calculate the minimum height of a tree.

3. **Edge Cases:**
   - The code has a special case for `test_cases == 969`, which is likely a debug statement and should not affect the logic.
   - The code has a condition `if (test_cases > 10) continue;` which skips processing for test cases beyond the 10th one. This is a serious issue as it will fail to process valid test cases.

Given these issues, a challenging test input should:

- Use values of `a`, `b`, and `c` that do not satisfy the incorrect condition `a != c - 1`.
- Use values that challenge the height calculation logic.
- Ensure the sum of `a + b + c` is valid and within the constraints.

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

```plaintext
3
2 1 3
1 0 0
0 3 1
```

Explanation:
- **Test Case 1: `2 1 3`**
  - This is a valid tree with 2 vertices having 2 children, 1 vertex having 1 child, and 3 vertices having 0 children.
  - The correct height is 2, but the buggy code might fail due to the incorrect condition check.

- **Test Case 2: `1 0 0`**
  - This is an invalid tree because a tree with 1 vertex having 2 children and no vertices with 0 children cannot exist.
  - The correct output should be -1, but the buggy code might fail to handle this case correctly.

- **Test Case 3: `0 3 1`**
  - This is a valid tree with 3 vertices having 1 child and 1 vertex having 0 children.
  - The correct height is 3, but the buggy code might fail due to the incorrect height calculation logic.

This test input should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 0, 1, 2, Tree!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Find the minimum height of a rooted tree$$$^{\dagger}$$$ with $$$a+b+c$$$ vertices that satisfies the following conditions:

- $$$a$$$ vertices have exactly $$$2$$$ children,
- $$$b$$$ vertices have exactly $$$1$$$ child, and
- $$$c$$$ vertices have exactly $$$0$$$ children.

The tree above is rooted at the top vertex, and each vertex is labeled with the number of children it has. Here $$$a=2$$$, $$$b=1$$$, $$$c=3$$$, and the height is $$$2$$$.

$$$^{\dagger}$$$ A rooted tree is a connected graph without cycles, with a special vertex called the root. In a rooted tree, among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child.

The distance between two vertices in a tree is the number of edges in the shortest path between them. The height of a rooted tree is the maximum distance from a vertex to the root.

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

The only line of each test case contains three integers $$$a$$$, $$$b$$$, and $$$c$$$ ($$$0 \leq a, b, c \leq 10^5$$$; $$$1 \leq a + b + c$$$).

The sum of $$$a + b + c$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, if no such tree exists, output $$$-1$$$. Otherwise, output one integer — the minimum height of a tree satisfying the conditions in the statement.

Examples:
input:
10
2 1 3
0 0 1
0 1 1
1 0 2
1 1 3
3 1 4
8 17 9
24 36 48
1 0 0
0 3 1
output:
2
0
1
1
-1
3
6
-1
-1
3

Note: The first test case is pictured in the statement. It can be proven that you can't get a height smaller than $$$2$$$.

In the second test case, you can form a tree with a single vertex and no edges. It has height $$$0$$$, which is clearly optimal.

In the third test case, you can form a tree with two vertices joined by a single edge. It has height $$$1$$$, which is clearly optimal.

Buggy Code:
import sys, threading
import math
from os import path
from collections import deque, defaultdict, Counter
from bisect import *
from string import ascii_lowercase
from functools import cmp_to_key
from random import randint
from heapq import *
from array import array
from types import GeneratorType

def readInts():
    x = list(map(int, (sys.stdin.readline().rstrip().split())))
    return x[0] if len(x) == 1 else x

def readList(type=int):
    x = sys.stdin.readline()
    x = list(map(type, x.rstrip('\n\r').split()))
    return x

def readStr():
    x = sys.stdin.readline().rstrip('\r\n')
    return x

write = sys.stdout.write
read = sys.stdin.readline

MAXN = 1123456

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}

    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]

    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]

    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item

    def getkeys(self):
        return [self.random ^ i for i in self.dict]

    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

def lcm(a, b):
    return (a*b)//(math.gcd(a,b))

def btcnt(n):
    cnt = 0
    while n > 0:
        if n&1:
            cnt += 1

        n >>= 1

    return cnt

def mod(n):
    return n%(1000000007)

def solve(t):
    # print(f'Case #{t}: ', end = '')
    a, b, c = readInts()
    num = c
    lvl = 1
    left = 0
    done = False
    while num > 1:
        nxt = btcnt(num)
        nlvl = int(math.log(num, 2))
        if not done:
            tmp = num
            p = 0
            while tmp > 0:
                if tmp&1:
                    left += nlvl-p
                p += 1
                tmp >>= 1

            done = True

        lvl += nlvl
        b -= int(num&1)
        a -= (1 << nlvl) - 1
        if a < 0 or b < 0:
            print(-1)
            return

        num = nxt

    ans = lvl-1
    if a != 0:
        print(-1)
        return

    if b > 0 and c == 0:
        print(-1)
        return

    b -= left
    if b <= 0:
        print(ans)
    else:
        ans += (b+c-1)//c
        print(ans)

def main():
    t = 1
    if path.exists("/Users/arijitbhaumik/Library/Application Support/Sublime Text/Packages/User/input.txt"):
        sys.stdin = open("/Users/arijitbhaumik/Library/Application Support/Sublime Text/Packages/User/input.txt", 'r')
        sys.stdout = open("/Users/arijitbhaumik/Library/Application Support/Sublime Text/Packages/User/output.txt", 'w')
    # sys.setrecursionlimit(100000)
    t = readInts()
    for i in range(t):
        solve(i+1)

if __name__ == '__main__':
    main()

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

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

Think step by step.

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