**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
/*Problem Link -> */
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define pi pair<int, int>
#define pll pair<ll, ll>
#define Yes cout << "YES" << '\n'
#define No cout << "NO" << '\n'
#define nl '\n'
#define loop(i, s, n) for (ll i = s; i <= n; i++)
using namespace std;
using namespace __gnu_pbds;
using namespace chrono;
template <typename T>
using pbds = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using pbdms = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
/*Debug*/
#ifdef Rakib_18
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void init_code() {
#ifdef Rakib_18
	freopen("Error.txt", "w", stderr);
#endif
}
const ll mod = 1e9 + 7;
ll inv(ll i) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;}
bool isPrime(ll n) {if (n <= 1)return false; if (n <= 3)return true; if (n % 2 == 0 || n % 3 == 0)return false; for (ll i = 5; i * i <= n; i += 6) {if (n % i == 0 || n % (i + 2) == 0)return false;} return true;}
ll lcm(ll a, ll b) {return (a / __gcd(a, b)) * b;}
int nXOR(int n) {if (n % 4 == 0)return n; if (n % 4 == 1)return 1; if (n % 4 == 2)return n + 1; return 0;}
/*--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
/*--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
ll go(int n) {
	if (n == 1 || n == 3) {
		return -1;
	}
	else if (n == 2) {
		return 66;
	}
	else if (n % 2 == 0) {
		string s = "3366";
		for (int i = 0; i < n - 4; i++) {
			s = '3' + s;
		}
		return stoi(s);
	}
	else {
		string s = "36366";
		for (int i = 0; i < n - 5; i++) {
			s = '3' + s;
		}
		return stoi(s);
	}
}
void solve()
{
	int n;
	cin >> n;
	cout << go(n) << nl;
}
int main()
{
	init_code();
	fastio();
	int t = 1;
	cin >> t;
	auto start1 = high_resolution_clock::now();
	while (t--)
	{
		solve();
	}
	auto stop1 = high_resolution_clock::now();
	auto duration = duration_cast<microseconds>(stop1 - start1);
#ifdef Rakib_18
	cerr << "Time: " << duration . count() / 1000 << " ms" << endl;
#endif
	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 expose the bug in the provided code, we need to consider edge cases and scenarios where the code might fail. The code provided has a specific logic for generating numbers based on the length `n`. Let's analyze the code and identify potential issues:

1. **n = 1 and n = 3**: The code correctly returns `-1` for these cases.
2. **n = 2**: The code correctly returns `66`.
3. **n % 2 == 0**: The code generates a string starting with "3366" and prepends '3's to make the length `n`.
4. **n % 2 != 0**: The code generates a string starting with "36366" and prepends '3's to make the length `n`.

The potential issue lies in the assumption that the generated string will always be divisible by both 33 and 66. However, this assumption may not hold for all values of `n`.

### Test Case to Expose the Bug

We need a test case where the generated string is not divisible by both 33 and 66. Let's consider `n = 6`:

- The code will generate the string "333366" for `n = 6`.
- We need to check if "333366" is divisible by both 33 and 66.

Let's verify:

- **Divisibility by 33**: A number is divisible by 33 if it is divisible by both 3 and 11.
  - **Divisibility by 3**: Sum of digits of "333366" = 3 + 3 + 3 + 3 + 6 + 6 = 24. Since 24 is divisible by 3, "333366" is divisible by 3.
  - **Divisibility by 11**: Alternating sum of digits = (3 + 3 + 6) - (3 + 3 + 6) = 12 - 12 = 0. Since 0 is divisible by 11, "333366" is divisible by 11.
- **Divisibility by 66**: A number is divisible by 66 if it is divisible by both 2 and 33.
  - **Divisibility by 2**: The last digit is 6, so "333366" is divisible by 2.
  - Since "333366" is divisible by both 2 and 33, it is divisible by 66.

Since "333366" is divisible by both 33 and 66, this test case does not expose the bug. Let's try a different length.

### Test Case for n = 7

- The code will generate the string "3336366" for `n = 7`.
- We need to check if "3336366" is divisible by both 33 and 66.

Let's verify:

- **Divisibility by 33**: A number is divisible by 33 if it is divisible by both 3 and 11.
  - **Divisibility by 3**: Sum of digits of "3336366" = 3 + 3 + 3 + 6 + 3 + 6 + 6 = 30. Since 30 is divisible by 3, "3336366" is divisible by 3.
  - **Divisibility by 11**: Alternating sum of digits = (3 + 3 + 3 + 6) - (3 + 6 + 6) = 15 - 15 = 0. Since 0 is divisible by 11, "3336366" is divisible by 11.
- **Divisibility by 66**: A number is divisible by 66 if it is divisible by both 2 and 33.
  - **Divisibility by 2**: The last digit is 6, so "3336366" is divisible by 2.
  - Since "3336366" is divisible by both 2 and 33, it is divisible by 66.

Since "3336366" is divisible by both 33 and 66, this test case does not expose the bug either. Let's try a different length.

### Test Case for n = 8

- The code will generate the string "33333366" for `n = 8`.
- We need to check if "33333366" is divisible by both 33 and 66.

Let's verify:

- **Divisibility by 33**: A number is divisible by 33 if it is divisible by both 3 and 11.
  - **Divisibility by 3**: Sum of digits of "33333366" = 3 + 3 + 3 + 3 + 3 + 3 + 6 + 6 = 30. Since 30 is divisible by 3, "33333366" is divisible by 3.
  - **Divisibility by 11**: Alternating sum of digits = (3 + 3 + 3 + 3) - (3 + 3 + 6 + 6) = 12 - 18 = -6. Since -6 is not divisible by 11, "33333366" is not divisible by 11.
- **Divisibility by 66**: A number is divisible by 66 if it is divisible by both 2 and 33.
  - **Divisibility by 2**: The last digit is 6, so "33333366" is divisible by 2.
  - Since "33333366" is not divisible by 11, it is not divisible by 33, and therefore not divisible by 66.

This test case should expose the bug in the code.

### Test Input

```plaintext
1
8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

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

void solve() {
    ll n;
    cin>>n;
    if(n<2 || n==3){
        cout<<-1<<endl;
        return;
    }
    if(n==2){
        cout<<66<<endl;
        return;
    }
    string base="366";
    if(n%2==0){
        for(ll i=3;i<n;i++){
            base="3"+base;
        }
    }
    else{
        base="36"+base;
        for(ll i=5;i<n;i++){
            base="3"+base;
        }
    }
    ll ans=stoll(base);
    cout<<ans<<endl;

}

int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

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

// Function to generate all numbers of 'digits' length made up of only 3s and 6s
void generateNumbers(int digits, int currentNum, int currentLength, vector<int>& numbers) {
    if (currentLength == digits) {
        numbers.push_back(currentNum);
        return;
    }
    generateNumbers(digits, currentNum * 10 + 3, currentLength + 1, numbers);
    generateNumbers(digits, currentNum * 10 + 6, currentLength + 1, numbers);
}

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

    for (int i = 0; i < tc; i++) {
        int digits;
        cin >> digits;

        vector<int> numbers;
        generateNumbers(digits, 0, 0, numbers); // Generate numbers with only 3 and 6

        int result = -1;
        for (int num : numbers) {
            if (num % 66 == 0) { // Check if the number is divisible by 66
                result = num;
                break;
            }
        }
        cout << result << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
10
1
2
3
4
5
6
7
8
9
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
t = eval(input())

for i in range(t):
    n = int(input())
    if n == 1 or n ==3:
        print(-1)
        continue
    if n == 2:
        print(66)
        continue
    if n == 4:
        print(3366)
        continue
    aaa = "3"*(n-4)
    ans = aaa + "6366"
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
t = int(input())

for _ in range(t):
  z = int(input())
  forr = int('1'+('0'*(z-1))+'0')
  if z == 2:
    print(66)
  else:
    for i in range(0,forr,66):
      if ''.join(sorted(set(str(i)))) == '36' and len(str(i))==z:
        print(i)
        break
    else:
      print(-1)

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

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

Think step by step.

Response:
```plaintext
6
1
2
3
10
11
12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""

"""

t = int(input())
for _ in range(t):
    n = int(input())
    i = 1
    while True:
        x = i * 66
        if len(str(x)) > n:
            print(-1)
            break
        elif len(str(x)) == n and (
            (
                set(str(x))
                == set(
                    [
                        "3",
                    ]
                )
            )
            or (set(str(x)) == set(["6"]))
            or (set(str(x)) == set(["3", "6"]))
        ):
            print(x)
            break
        else:
            i += 1

"""
/***
6
1
2
3
4
5
7
---
-1
66
-1
3366
36366
3336366
***/
"""

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

1. **Small Values of \( n \)**: The code should handle small values of \( n \) correctly, especially \( n = 1 \) and \( n = 2 \).
2. **Odd Values of \( n \)**: The code should handle odd values of \( n \) correctly, as some odd values might not have valid solutions.
3. **Large Values of \( n \)**: The code should handle large values of \( n \) efficiently, ensuring that it does not run into performance issues.
4. **Specific Patterns**: The code should correctly identify and handle specific patterns of digits (3s and 6s) that are divisible by both 33 and 66.

Given these considerations, a small-scale test input that can expose potential bugs is:

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

This test input includes:
- \( n = 1 \): No valid number exists.
- \( n = 2 \): The smallest valid number is 66.
- \( n = 3 \): No valid number exists.
- \( n = 4 \): The smallest valid number is 3366.

This input is small enough to be manageable and complete, and it covers a range of scenarios that can help identify 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: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
import java.util.Scanner;
public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();
        int[] output = new int[cases];
        for (int i = 0; i < cases; i++) {
            int x = sc.nextInt();
            int x1 = (int) Math.pow(10, x - 1);
            int x2 = (int) Math.pow(10, x) - 1;
            boolean found = false;
            for (int b = x1; b <= x2; b++) {
                if (b % 66 == 0) {
                    int b1 = b;
                    int check = 0;
                    for (int j = 0; j < x; j++) {
                        int temp = b1 % 10;
                        if (temp == 3 || temp == 6) {
                            check++;
                        }
                        b1 /= 10;
                    }
                    if (check == x) {
                        output[i] = b;
                        found = true;
                        break;
                    }
                }
            }
            if (!found) {
                output[i] = -1;
            }
        }
        for (int i = 0; i < cases; i++) {
            System.out.print(output[i] + "\n");
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1
2
3
10
15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
import java.util.*;

public class codeforces{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        int[] arr = new int[t];
        for (int i=0 ; i<t ; i++){
            arr[i]=scan.nextInt();
        }
        for (int i=0 ; i<t ; i++){
            System.out.println(recurse(arr[i],0));
        }
    }

    public static int recurse(int len , int num){
        if (len==0){
            if (num%66==0){
                return num;
            }
            return -1;
        }
        int val = recurse(len-1,num*10+3);
        if (val!=-1){
            return val;
        }
        val = recurse(len-1,num*10+6);
        return val;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

void solve()
{
    ll n, res = 0;
    cin>>n;
    if(n == 1 || n == 3)
    {
        cout<<-1<<endl;
    }
    else
    {
        string s;
        if(n%2 == 0)
        {
            for(int i = 0; i < n-2; i++)
            {
                s += '3';
            }
        }
        else
        {
            for(int i = 0; i < n - 4; i++)
            {
                s += '3';
            }
            s += '6';
            s += '3';
        }
        s += '6';
        s += '6';
        res = stoll(s);
        cout<<res<<endl;
    }
}

int main()
{
    int trial;
    cin>>trial;
    for(int i = 0; i < trial; i++)
    {
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

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

#define ll long long
#define vll vector<ll>
#define f(i,a,b) for(ll i = a;i<b;i++)
#define all(v) v.begin(), v.end()
#define print(x) cout<<x<<endl
#define test(t) ll t; cin>>t; while(t--)

using namespace std;

int main()
{
  test(t)
  {
    int n;
    cin>>n;
    int res = 0;
    if(n==1||n==3)
    {
      cout<<-1<<endl;
    }
    else if(n==2) print(66);
    else if(n==5) print(36366);
    else if(n%2==0)
    {
    for(int i = 1;i<=n-2;i++)
    {
      res=res*10+3;
    }
    res=res*100+66;
    print(res);
    }
    else{
      for(int i = 1;i<=n-4;i++)
    {
      res=res*10+3;
    }
    res=res*10000+6366;
    print(res);
    }

  }
  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 expose the bug in the given code, we need to focus on edge cases and scenarios where the code might fail. The code has specific conditions for certain values of `n`, and it handles even and odd lengths differently. We should test the following scenarios:

1. Small values of `n` that are explicitly handled in the code (e.g., `n = 1`, `n = 2`, `n = 3`, `n = 5`).
2. Larger even and odd values of `n` to check the general logic.
3. Values of `n` that are not explicitly handled but should still produce a valid result.

Let's create a test input that includes these scenarios:

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

This test input covers:
- `n = 1` (should return `-1` because no such number exists)
- `n = 2` (should return `66` because `66` is the smallest number that meets the criteria)
- `n = 3` (should return `-1` because no such number exists)
- `n = 4` (should return `3366` because it is the smallest number that meets the criteria)
- `n = 5` (should return `36366` as specified in the code)
- `n = 7` (should return a valid number, but the code might fail to generate it correctly)

This test input should effectively expose any bugs in the code, especially in the handling of odd and even lengths.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
from sys import stdin,stdout
from math import *
from collections import *
from pprint import pprint
import random
def read_list_line(elem_func=int):
    return [elem_func(x) for x in stdin.readline().strip().split()]

def read_line(elem_func=str):
    return elem_func(stdin.readline().strip())

def print_list_line(the_list,elem_func=str,sep=' ',end=""):
    st=(sep.join([elem_func(x) for x in the_list]))+end
    stdout.write(st)
    return st

def print_line(st):
    stdout.write(str(st)+"\n")
    return st

ntest = read_line(int)

def main():
    n = read_line(int)
    f=[]
    for _ in range(n+2):
        f.append([[None]*66 for __ in range(33)])
    f[0][0][0]=''
    for i in range(n):
        for rem33 in range(33):
            for rem66 in range(66):
                if  (f[i][rem33][rem66] != None):
                    s=f[i][rem33][rem66]
                    nr33=(rem33*10+6)%33
                    nr66=(rem66*10+6)%66
                    s6=s+'6'
                    if (f[i+1][nr33][nr66]==None):
                        f[i+1][nr33][nr66]=s6
                    else:
                        f[i+1][nr33][nr66]=min(f[i+1][nr33][nr66],s6)
                    nr33 = (rem33*10+3)%33
                    nr66 = (rem66*10+3)%66
                    s3=s+'3'
                    if (f[i+1][nr33][nr66]==None):
                        f[i+1][nr33][nr66]=s3
                    else:
                        f[i+1][nr33][nr66]=min(f[i+1][nr33][nr66],s3)

    if (f[n][0][0]==None):
        print_line('-1')
    else:
        print_line(f[n][0][0])
    return 0

for test_case in range(1,1+ntest):
    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
500
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
t=int(input())
while t>0:
    t-=1
    n=int(input())
    a=int('3'* n)
    b=int('6'* n)
    d=0
    for i in range(a,b+1):
        if set(str(i)).issubset({'3','6'}) and i%33==0 and i%66==0:
            print(i)
            d=1
            break
    if d==0:
        print(-1)

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

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

Think step by step.

Response:
```plaintext
1
11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
import java.util.*;

public class MyClass{

public static void main(String[]args){

	Scanner sc=new Scanner(System.in);

	int t=sc.nextInt();

	while(t>0){
		int n=sc.nextInt();

		if(n==1 || n==3) System.out.println(-1);
		else if(n==2) System.out.println(66);
		else if(n==4) System.out.println(3366);
		else if(n==5) System.out.println(36366);

		else {
			String str="36366";

			int diff=n-str.length();

			String ans="";

			for(int i=0;i<diff;i++) ans+="3";

			ans+=str;

			if(Integer.parseInt(ans)%33==0 && Integer.parseInt(ans)%66==0) System.out.println(ans);
			else System.out.println(-1);
		}

		t--;
	}

}

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
import java.util.*;

public class Test {
    public static void generateNumbers(int n, String current, List<String> results) {
        if (current.length() == n) {
            results.add(current);
            return;
        }
        generateNumbers(n, current + "3", results);
        generateNumbers(n, current + "6", results);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            List<String> results = new ArrayList<>();
            int ans =-1;
            generateNumbers(n, "", results);
            for(int i=0;i<results.size();i++){
                int x = Integer.parseInt(results.get(i));
                if(x%66==0){
                    ans = x;
                    break;
                }
            }
            System.out.println(ans);

        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
#include <iostream>

using namespace std;

int main(){
	int t;
	cin >> t;
	while(t -- ){
		int n;
		cin >> n;
		if(n == 1 || n == 3) cout << -1 << endl;
		else if(n == 2) cout << 66 << endl;
		else{
			long long ans = 0;
			if(n % 2 == 0){
				for(int i = 1; i <= n - 2; i ++ ){
					ans = ans * 10 + 3;
				}
				ans = ans * 100 + 66;
			}
			else{
				for(int i = 1; i <= n - 4; i ++ ){
					ans = ans * 10 + 3;
				}
				ans = ans * 10000 + 6366;
			}
			cout << ans << endl;
		}
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define NO cout << "NO\n"
#define YES cout << "YES\n"
void solve() {
    int n; cin >> n;
    if(n == 2) {
        cout << "66\n";
        return;
    }
    if(n == 4) {
        cout << "3366\n";
        return;
    }
    if(n == 3 || n == 1) {
        cout << -1 << "\n";
        return;
    }
    cout << string(max(n-4,n-5), '3') + "6366" << endl;
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tst; cin >> tst;
    while(tst--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
from itertools import product

t=int(input())
for _ in range(t):
	n = int(input())
	t=0
	def generate_permutations(a, b, n):
		a, b = str(a), str(b)
		permutations = [''.join(p) for p in product((a, b), repeat=n)]
		return permutations
	result=generate_permutations(3, 6, n)
	result=sorted(result)
	for i in result:
		i=int(i)
		if i%11==0 and i%2==0 and i%3==0:
			print(i)
			t+=1
			break
	if t==0:
		print(-1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
t=int(input())
while t>0:
    t-=1
    n=int(input())
    a=int('3'* n)
    b=int('6'* n)
    d=0
    for i in range(a,b+1):
        if set(str(i)).issubset({'3','6'}) and i%33==0 and i%66==0:
            print(i)
            d+=1
            break

    if d==0:
        print(-1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

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

public class Solution {

    public static void solve(int n) {
        if (n < 2 || n == 3) {
            System.out.println(-1);
            return;
        }
        else if(n==2){
            System.out.println("66");
        }
        else if(n%2==1){
            int rep = n-4;
            String s = "3".repeat(rep);
            s+="6366";
            System.out.println(s);
            return;
        } else if(((n & (n - 1)) == 0) || n%3==0){
            String s = "3".repeat(n/2);
            s+="6".repeat(n/2);
            System.out.println(s);
            return;
        }
        else System.out.println(-1);

    }

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

        while (t-- > 0) {
            int n = sc.nextInt();
            solve(n);

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everyone Loves Tres
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are 3 heroes and 3 villains, so 6 people in total.

Given a positive integer $n$. Find the smallest integer whose decimal representation has length $n$ and consists only of $3$s and $6$s such that it is divisible by both $33$ and $66$. If no such integer exists, print $-1$.

Input: The first line contains a single integer $t$ ($1\le t\le 500$) — the number of test cases.

The only line of each test case contains a single integer $n$ ($1\le n\le 500$) — the length of the decimal representation.

Output: For each test case, output the smallest required integer if such an integer exists and $-1$ otherwise.

Examples:
input:
6
1
2
3
4
5
7
output:
-1
66
-1
3366
36366
3336366

Note: For $n=1$, no such integer exists as neither $3$ nor $6$ is divisible by $33$.

For $n=2$, $66$ consists only of $6$s and it is divisible by both $33$ and $66$.

For $n=3$, no such integer exists. Only $363$ is divisible by $33$, but it is not divisible by $66$.

For $n=4$, $3366$ and $6666$ are divisible by both $33$ and $66$, and $3366$ is the smallest.

Buggy Code:
import java.util.*;

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

        for (int k = 0; k < t; k++) {
            int n = s.nextInt();
            long lb = (long) Math.pow(10, n - 1);
            long ub = (long) Math.pow(10, n) - 1;

            long minNum = Long.MAX_VALUE;
            boolean found = false;

            for (int i = 0; i < (1 << n); i++) {
                StringBuilder num = new StringBuilder();
                for (int j = 0; j < n; j++) {
                    num.append((i & (1 << j)) == 0 ? '3' : '6');
                }

                long val = Long.parseLong(num.toString());

                if (val >= lb && val <= ub && val % 66 == 0) {
                    found = true;
                    minNum = Math.min(minNum, val);
                }
            }

            System.out.println(found ? minNum : "-1");
        }

        s.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
6
1
2
3
4
5
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
#define int long long
typedef pair<int,int> pii;
const int mod=998244353;
const int inf=1e9+7;
const int N=1e6+10;
int depth[N];
int du[N];
int n;
int q[N],e[N*2],ne[N*2],idx;
int bs[25];
void init(){
    bs[0]=1;bs[1]=2;
    for(int i=2;i<=22;i++){
        bs[i]=bs[i-1]*2;
    }
}
void add(int a,int b){
    e[idx]=b;ne[idx]=q[a];q[a]=idx++;
}
void dfs(int u){
    vector<int> re;
    for(int i=q[u];i!=-1;i=ne[i]){
        int j=e[i];
        dfs(j);
        re.push_back(depth[j]);
    }
    if(du[u]==0) {
        depth[u]=0;
        return;
    }
    int k=0;
    for(int i=1;i<=20;i++){
        if(bs[i]>=du[u]){
            k=i;
            break;
        }
    }
    int dy=bs[k]-du[u];
    sort(re.begin(),re.end());
    int o=re.size()-1;
    if(k==1) {
        depth[u]=re[o]+1;
        return;
    }
    for(int i=0;i<dy;i++){
        depth[u]=max(depth[u],re[o--]+k-1);
    }
    for(int i=o;i>=0;i--){
        depth[u]=max(depth[u],re[i]+k);
    }
}
void solve(){
    cin>>n;
    for(int i=0;i<=n+5;i++) {
        q[i]=-1;
        depth[i]=0;
        du[i]=0;
    }
    idx=0;
    for(int i=2;i<=n;i++) {
        int x;cin>>x;
        du[x]++;
        add(x,i);
    }
    dfs(1);
    cout<<depth[1]<<'\n';
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    init();
    int t;cin>>t;
    for(int i=1;i<=t;i++) {
        if(i==4416){
            int n;cin>>n;
            string s;
            char a;
            for(int i=1;i<=n;i++){
                cin>>a;
                s+=a;
            }
            cout<<n<<a<<endl;
        }
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
using namespace std;

const int N=1000000+10;
int p[N];
vector <int> g[N];

int DFS(int v) {

    vector<int> z;
    for (int j=0;j<g[v].size();j++)
        z.push_back(DFS(g[v][j]));

    if (z.size()==0) return 0;

    sort(z.begin(),z.end());

    int next_cnt=0;
    int cur_cnt=0;
    int cur=0;

    for (int j=0;j<z.size();j++) {
        if (z[j]!=cur) {
            cur_cnt+=next_cnt;
            next_cnt=0;
            cur=z[j];
        }

        cur_cnt++;
        next_cnt+=(cur_cnt)/2;
        cur_cnt%=2;
    }

    //cout<<v<<" --> "<<cur<<" : "<<cur_cnt<<" "<<next_cnt<<endl;

    next_cnt+=cur_cnt;
    cur++;
    while (next_cnt>1) {
        cur++;
        next_cnt/=2;
    }
    return cur;
}

main(){
    int t;
    scanf("%d",&t);
    while (t--) {
        int n;
        scanf("%d",&n);
        for (int i=1;i<=n;i++) g[i].clear();

        for (int i=2;i<=n;i++) {
            scanf("%d",&p[i]);
            g[p[i]].push_back(i);
        }

        printf("%d\n",DFS(1));
    }
}

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

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

Think step by step.

Response:
```plaintext
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

Buggy Code:
#include <iostream>

using namespace std;
//ifstream cin("input.in");
//ofstream cout("output.out");
struct nod
{
    unsigned int x;
    nod * p;    ;
};

void adauga(nod * & r, unsigned int d)
{
    nod * q;
    q = new nod;
    q->x=d;
    q->p=r;
    r=q;
}
void dealocare(nod * & r)
{
    if(r!=nullptr)
    {
        dealocare(r->p);
        delete r;
        r=nullptr;
    }
}
unsigned short log_sarac(unsigned int nr)
{
    unsigned int masca=1,aux;
    unsigned short rasp=0,act=0;
    if(nr <= 2)
        return 1;
    while(nr >=masca)
    {
        aux=nr&masca;
        if(aux != 0)
            rasp=act;
        masca=masca<<1;
        act++;
    }
    return rasp+1;
}
nod * vect[1000000];

unsigned short adanc(unsigned int indice)
{
    if(vect[indice] == nullptr)
        return 0;
    nod * help=vect[indice];
    unsigned int nrcop=0;
    unsigned short depth=0;
    while (help!=nullptr)
    {
        nrcop++;
        depth=max(depth,adanc(help->x));
        help=help->p;
    }
    depth++;
    depth=max(depth,log_sarac(nrcop));
    return depth;
}
int main()
{
    unsigned int t,n,val,raspuns;
    cin>>t;
    while(t)
    {
        t--;
        cin>>n;
        raspuns=0;
        for(unsigned int i=1;i<n;i++)
        {
            cin>>val;
            adauga(vect[val-1],i);
        }
        raspuns=adanc(0);
        cout<<raspuns<<'\n';
        for(int i=0;i<n;i++)
            dealocare(vect[i]);
    }
    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 consider the following:

1. **Edge Cases**: Test cases that push the limits of the algorithm, such as the smallest and largest possible input sizes.
2. **Complex Tree Structures**: Test cases with complex tree structures that require multiple operations to become isomorphic to a perfect binary tree.
3. **Leaf Nodes and Branching**: Test cases with varying numbers of leaf nodes and branching factors.

Given the constraints and the nature of the problem, a good test case would be a tree that has a structure which is not immediately a perfect binary tree and requires multiple operations to transform into one.

### Chosen Test Input
Let's consider a tree with 7 vertices. This tree is not a perfect binary tree, and it requires some operations to transform it into a perfect binary tree of depth 3.

### Test Input
```plaintext
1
7
1 1 2 2 3 3
```

### Explanation
- **Input**: `1` indicates one test case.
- **Number of Vertices**: `7` indicates the tree has 7 vertices.
- **Parent Array**: `1 1 2 2 3 3` indicates the parent-child relationships:
  - Vertex 2 and 3 are children of vertex 1.
  - Vertex 4 and 5 are children of vertex 2.
  - Vertex 6 and 7 are children of vertex 3.

### Tree Structure
```
    1
   / \
  2   3
 / \ / \
4  5 6  7
```

### Expected Output
The expected output for this test case is `3`, as the tree is already isomorphic to a perfect binary tree of depth 3.

### Why This Test Input is Challenging
- **Non-Perfect Binary Tree**: The input tree is not a perfect binary tree, which requires the algorithm to correctly determine the minimum depth.
- **Balanced Structure**: The tree is balanced, but not a perfect binary tree, which tests the algorithm's ability to handle balanced but not perfectly structured trees.
- **Small Size**: The tree is small enough to be manageable and easy to verify, but complex enough to challenge the algorithm.

This test input should help expose any bugs in the provided buggy code, particularly in the logic that determines the minimum depth of the perfect binary tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Chloe's Trees
time_limit_ms: 3500
memory_limit_mb: 512
Description: With just a few hours left until Penchick and Chloe leave for Singapore, they could hardly wait to see the towering trees at the Singapore Botanic Gardens! Attempting to contain their excitement, Penchick crafted a rooted tree to keep Chloe and himself busy.

Penchick has a rooted tree$$$^{\text{∗}}$$$ consisting of $$$n$$$ vertices, numbered from $$$1$$$ to $$$n$$$, with vertex $$$1$$$ as the root, and Chloe can select a non-negative integer $$$d$$$ to create a perfect binary tree$$$^{\text{†}}$$$ of depth $$$d$$$.

Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic$$$^{\text{‡}}$$$ to Penchick's tree. To meet this condition, Chloe can perform the following operation on her own tree any number of times:

- Select an edge $$$(u,v)$$$, where $$$u$$$ is the parent of $$$v$$$.
- Remove vertex $$$v$$$ and all the edges connected to $$$v$$$, then connect all of $$$v$$$'s previous children directly to $$$u$$$.

In particular, doing an operation on an edge $$$(u, v)$$$ where $$$v$$$ is a leaf will delete vertex $$$v$$$ without adding any new edges.

Since constructing a perfect binary tree can be time-consuming, Chloe wants to choose the minimum $$$d$$$ such that a perfect binary tree of depth $$$d$$$ can be made isomorphic to Penchick's tree using the above operation. Note that she can't change the roots of the trees.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the number of vertices in Penchick's tree.

The second line of each test case contains $$$n-1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i \leq i-1$$$) — the parent of vertex $$$i$$$.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output a single integer on each line: the minimum depth of Chloe's perfect binary tree.

Examples:
input:
5
6
1 2 2 1 1
15
1 1 2 2 3 3 4 4 5 5 6 6 7 7
5
1 2 2 2
7
1 1 2 1 1 2
10
1 1 1 2 2 2 4 3 3
output:
2
3
3
3
3

Note: For the first test case, create a perfect binary tree with depth $$$2$$$.

Consider carrying out the operation on edge $$$AC$$$. Then the edges $$$AC$$$, $$$CF$$$, and $$$CG$$$ are removed, and edges $$$AF$$$ and $$$AG$$$ are added.

The resulting tree is isomorphic to the tree given in the input. It can be proven that no sequence of operations carried out on a binary tree of depth less than $$$2$$$ can lead to a tree isomorphic to the tree given in the input.

In the second test case, the tree is already isomorphic to a perfect binary tree of depth $$$3$$$.

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

def dfs(node):
    d = 0
    fh = []
    count_on = 0

    for v in tree[node]:
        x = dfs(v)
        fh.append(x - 1)
        d = max(d, x)

    for k in fh:
        while k < len(xx) and xx[k] == node:
            count_on -= 1
            k += 1

        if k == len(xx):
            xx.append(0)
        count_on += (xx[k] != node)
        xx[k] = node
        d = max(d, k)

    if d < len(xx) and xx[d] == node and count_on > 1:
        d += 1

    return d + 1

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))

    tree = [[] for _ in range(n + 1)]
    for i in range(n - 1):
        tree[a[i]].append(i + 2)

    xx = []
    print(dfs(1) - 1)

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

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

Think step by step.

Response:
```plaintext
3
4
1 1 1
6
1 2 2 3 3
8
1 1 2 2 3 3 4
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** 4
   - **Parent Array:** 1 1 1
   - **Tree Structure:**
     ```
     1
    /|\
   2 3 4
     ```
   - **Expected Output:** 2
   - **Explanation:** This tree is already a perfect binary tree of depth 2.

2. **Test Case 2:**
   - **Input:** 6
   - **Parent Array:** 1 2 2 3 3
   - **Tree Structure:**
     ```
     1
     |
     2
    / \
   3   4
    \   \
     5   6
     ```
   - **Expected Output:** 3
   - **Explanation:** This tree can be transformed into a perfect binary tree of depth 3 by removing the edges (3, 5) and (4, 6).

3. **Test Case 3:**
   - **Input:** 8
   - **Parent Array:** 1 1 2 2 3 3 4
   - **Tree Structure:**
     ```
     1
    / \
   2   3
  / \ / \
 4  5 6  7
     |
     8
     ```
   - **Expected Output:** 3
   - **Explanation:** This tree can be transformed into a perfect binary tree of depth 3 by removing the edge (7, 8).

These test cases are designed to cover different tree structures and ensure that the buggy code is thoroughly tested for edge cases and scenarios where it might fail.