**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

//TIP To <b>Run</b> code, press <shortcut actionId="Run"/> or
// click the <icon src="AllIcons.Actions.Execute"/> icon in the gutter.
public class MaximumMultipleSum {
    public static void main(String[] args) {
        List<Integer> list = readInput();
        for(Integer n: list){
            int max = 0, maxX = 0, maxK = 0;

            for(int x=2;x<=n;x++){
                //System.out.println("For x: " + x);
                int low =1; int high =100;

                while(low<=high){
                    // here mid is k
                    int mid = low + (high-low)/2;
                    int val = x*mid*(mid+1)/2;

                    if(mid*x<=n && val>=max){
                        max = val;
                        maxX = x;
                        maxK = mid;
                        //System.out.println("Trying K : " + mid + " max Sum: " + val + " x: " + x);
                    }

                    if(val<=n){
                        low = mid+1;
                    }
                    else{
                        high = mid-1;
                    }
                }

            }
            System.out.println(maxX);
        }
    }

    public static List<Integer> readInput(){
        Scanner scanner = new Scanner(System.in);
        int numberOfLines = scanner.nextInt();
        scanner.nextLine(); // Consume the newline after the number

        List<Integer> data = new ArrayList<>();

        for (int i = 0; i < numberOfLines; i++) {
            data.add(scanner.nextInt());
        }

        return data;
    }

}

**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 to find the correct maximum sum of multiples. The code uses a binary search approach to find the maximum sum, but it has some potential issues in its logic.

Here are the steps to think about the test input:

1. **Edge Cases for Small Values of \( n \)**: Test with \( n \) values that are small but still within the valid range (2 to 100). This can help identify issues in the initial conditions or boundary handling.
2. **Edge Cases for Large Values of \( n \)**: Test with \( n \) values that are closer to the upper limit of 100. This can help identify issues in the binary search logic.
3. **Special Values**: Test with values of \( n \) that have specific properties, such as being a multiple of small integers, to ensure the code handles these cases correctly.
4. **Multiple Test Cases**: Include multiple test cases to ensure the code can handle different scenarios.

Given these considerations, a challenging test input that exposes potential bugs in the code could be:

```plaintext
5
2
10
15
99
100
```

This input includes:
- The smallest valid \( n \) (2).
- A small \( n \) with a clear optimal \( x \) (10).
- A medium \( n \) that has been provided in the example (15).
- A large \( n \) close to the upper limit (99).
- The largest valid \( n \) (100).

These test cases will help in verifying if the code correctly handles different ranges and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
import java.util.Scanner;

public class Main
{
	public static void main(String[] args) {
	    Scanner s=new Scanner(System.in);
	    int t=s.nextInt();
	    for(int i=0;i<t;i++){
	       int n=s.nextInt();
	     if(n<=5)
	       System.out.println(n);
	       else
	       System.out.println(2);
	    }

	}
}

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

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

Think step by step.

Response:
```plaintext
3
6
10
14
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
#include<iostream>
#include<string.h>
using namespace std;
class ss
{
public:
    void result()
    {
        char s[5500];
        cin>> s;
        int v=0;
        int c=0;
        int y=0;
        int sy=0;
        int ng=0;
        int sl;
        int len = strlen(s);
        for(int i=0; i<len; i++)
        {
            if(s[i]=='A' || s[i]=='E' || s[i]=='I' || s[i]=='O' || s[i]=='U')
            {
                v+=1;
            }
            else if(s[i]=='Y')
            {
                y+=1;
            }
            else if(s[i]=='N' && s[i+1]=='G')
            {
                ng+=1;
            }
            else
            {
                c+=1;
            }
            c=c+ng;
        }
        while (y> 0 && (v == 0 || c < 2))
        {
            if (c < 2)
            {
                ++c;
            }
            else if (v == 0)
            {
                ++v;
            }
            --y;
        }
        sy=min(c/2,v);
        sl=sy*3;

        cout << sl<<endl;

    }
};
int main()
{
    ss obj;
    obj.result();
}

**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
NGYAEIOU
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

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

using namespace std;

int main() {
    string s;
    cin >> s;

    int vowels = 0, consonants = 0, ngCount = 0;

    int n = s.size();

    // Count vowels and consonants, and handle "NG" as a special consonant
    for (int i = 0; i < n; i++) {
        char ch = s[i];
        if (ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') {
            vowels++;
        } else if (ch == 'Y') {
            vowels++; // Y can be treated as a vowel
            consonants++; // Y can also be treated as a consonant
        } else if (ch == 'N' && i + 1 < n && s[i + 1] == 'G') {
            ngCount++;  // NG is treated as a single consonant
            i++; // Skip the next character 'G'
        } else {
            consonants++;
        }
    }

    // Total consonants include regular consonants and "NG"
    consonants += ngCount;

    // The maximum number of syllables is constrained by the number of vowels and consonants
    int syllables = min(vowels, consonants / 2);

    // The length of the word is 3 * the number of syllables
    cout << syllables * 3 << 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
NGNGNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

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

int main() {
    string S;
    cin >> S;

    // Preprocess the input string
    unordered_map<char, int> freq;
    int n = S.size();
    int ng_count = 0;

    for (int i = 0; i < n; ++i) {
        if (i + 1 < n && S[i] == 'N' && S[i + 1] == 'G') {
            // Count "NG" as a single consonant
            ng_count++;
            i++; // Skip the next character since it's part of "NG"
        } else {
            freq[S[i]]++;
        }
    }

    // Count vowels, consonants, and Ys
    int vowels = freq['A'] + freq['E'] + freq['I'] + freq['O'] + freq['U'];
    int consonants = ng_count; // Start with "NG" count
    int ys = freq['Y'];

    for (char c = 'A'; c <= 'Z'; ++c) {
        if (c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U' && c != 'Y') {
            consonants += freq[c];
        }
    }

    // Maximize syllables
    int syllables = 0;
    while (vowels > 0 && consonants >= 2) {
        syllables++;
        vowels--;          // Use one vowel
        consonants -= 2;   // Use two consonants
    }

    // Use Ys if necessary
    while (ys > 0 && syllables > 0) {
        if (vowels == 0) {
            vowels++;
            ys--;
        } else if (consonants < 2) {
            consonants += 2;
            ys--;
        }
    }

    // Recalculate syllables if Ys were used
    while (vowels > 0 && consonants >= 2) {
        syllables++;
        vowels--;
        consonants -= 2;
    }

    // Output the result
    cout << syllables * 3 << 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
NGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
def longest_word_length(s):
    vowels = set("AEIOU")
    consonants = set("BCDFGHJKLMNPQRSTVWXZ")
    y_char = "Y"

    vowel_count = 0
    consonant_count = 0
    y_count = 0

    for char in s:
        if char in vowels:
            vowel_count += 1
        elif char in consonants:
            consonant_count += 1
        elif char == y_char:
            y_count += 1

    ng_count = s.count("NG")
    consonant_count += ng_count

    max_length = 0
    for y_as_vowel in range(y_count + 1):
        current_vowel_count = vowel_count + y_as_vowel
        current_consonant_count = consonant_count + (y_count - y_as_vowel)
        syllables = min(current_consonant_count // 2, current_vowel_count)
        max_length = max(max_length, syllables * 3)

    return max_length

s = input().strip()
result = longest_word_length(s)
print(result)

**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
NGYNGYNG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
def longest_word_length(S):
    # Define vowels and classify characters
    vowels = {'A', 'E', 'I', 'O', 'U'}
    consonants = set("BCDFGHJKLMNPQRSTVWXZ")  # All consonants except Y
    vowel_count = 0
    consonant_count = 0
    ng_count = 0
    y_count = 0

    # Count vowels, consonants, and "NG" pairs
    i = 0
    while i < len(S):
        if S[i] in vowels:
            vowel_count += 1
        elif S[i] in consonants:
            consonant_count += 1
        elif S[i] == 'Y':
            y_count += 1
        if i < len(S) - 1 and S[i] == 'N' and S[i + 1] == 'G':
            ng_count += 1
            i += 1  # Skip the next character ('G') since "NG" is treated as one consonant
        i += 1

    # "NG" counts as a consonant
    consonant_count += ng_count

    # Maximize the number of syllables by using Y optimally
    max_syllables = 0
    for y_as_vowels in range(y_count + 1):  # Y can act as vowels from 0 to y_count
        available_vowels = vowel_count + y_as_vowels
        available_consonants = consonant_count + (y_count - y_as_vowels)
        syllables = min(available_consonants // 2, available_vowels)
        max_syllables = max(max_syllables, syllables)
    return max_syllables * 3

S = input().strip()
result = longest_word_length(S)
print(result)

**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
NGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
def main():
    S = input().strip()
    # First, count the total number of each letter
    count = {}
    for c in S:
        count[c] = count.get(c, 0) + 1
    vowels = set('AEIOU')
    # Total vowels are A, E, I, O, U and Y can be a vowel
    total_vowels = 0
    for v in vowels:
        total_vowels += count.get(v, 0)
    y_consonant = count.get('Y', 0)  # Y can be either vowel or consonant
    # Now count total consonants: all letters except A, E, I, O, U plus Y as consonant
    total_consonants = 0
    for c in count:
        if c not in vowels:
            if c == 'Y':
                total_consonants += y_consonant  # Y as consonant
            else:
                total_consonants += count[c]
    # Now handle NG, which can be treated as a single consonant
    # Search for 'NG' in the string
    ng_count = 0
    i = 0
    while i < len(S) - 1:
        if S[i] == 'N' and S[i + 1] == 'G':
            ng_count += 1
            i += 2
        else:
            i += 1
    total_consonants -= ng_count  # Remove individual N and G counts
    total_consonants += ng_count  # Add back as NG
    # Now, each syllable requires 1 vowel and 2 consonants
    # Y can be used as vowel or consonant, but for simplicity, we'll assume Y as consonant first
    # So total syllables is min(total_consonants // 2, total_vowels + y_consonant)
    total_syllables = min(total_consonants // 2, total_vowels + y_consonant)
    # The maximum length is total_syllables * 3
    if total_syllables <= 0:
        print(0)
    else:
        print(total_syllables * 3)

if __name__ == "__main__":
    main()

# y9pK3Mw5jeLHOah70it

**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
NGYNGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

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

public class Main {
    static Set<Character> vowels = new HashSet<>(Arrays.asList('A', 'E', 'I', 'O', 'U'));

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String S = br.readLine();
        System.out.println(solve(S));
    }

    static int solve(String s) {
        int[] freq = new int[26];
        for (char c : s.toCharArray()) {
            freq[c - 'A']++;
        }

        // NG 쌍의 개수를 먼저 계산
        int ngPairs = Math.min(freq['N' - 'A'], freq['G' - 'A']);
        freq['N' - 'A'] -= ngPairs;
        freq['G' - 'A'] -= ngPairs;

        // 모음 개수 계산
        int vowelCount = 0;
        for (char v : vowels) {
            vowelCount += freq[v - 'A'];
        }

        // 남은 자음 개수 계산
        int consonantCount = 0;
        for (int i = 0; i < 26; i++) {
            char c = (char)('A' + i);
            if (!vowels.contains(c) && c != 'Y') {
                consonantCount += freq[i];
            }
        }

        int yCount = freq['Y' - 'A'];
        int maxSyllables = 0;

        // Y를 모음과 자음으로 분배하여 최적의 경우 찾기
        for (int yAsVowel = 0; yAsVowel <= yCount; yAsVowel++) {
            int yAsConsonant = yCount - yAsVowel;
            int totalVowels = vowelCount + yAsVowel;
            int totalConsonants = consonantCount + yAsConsonant;

            int syllables = Math.min(totalVowels, totalConsonants / 2);
            maxSyllables = Math.max(maxSyllables, syllables);
        }

        if (maxSyllables == 0) return 0;

        // 특별한 경우: NGENG와 같이 NG를 포함하는 한 음절
        if (ngPairs > 0 && maxSyllables == 1) {
            return 5;
        }

        // 일반적인 경우: 각 음절은 3글자
        return maxSyllables * 3;
    }
}

**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
NGNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
import java.util.*;

public class ScrambledScrabble {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String S = sc.nextLine().strip();

        Set<Character> vowels = new HashSet<>(Arrays.asList('A', 'E', 'I', 'O', 'U'));
        Set<Character> consonants = new HashSet<>(Arrays.asList('B', 'C', 'D', 'F', 'G', 'H', 'J', 'K', 'L', 'M', 'N', 'P', 'Q', 'R', 'S', 'T', 'V', 'W', 'X', 'Z'));

        int vowelCount = 0;
        int consonantCount = 0;
        int ngCount = 0;
        int yAsVowel = 0;
        int yAsConsonant = 0;

        int i = 0;
        while (i < S.length()) {
            if (i + 1 < S.length() && S.charAt(i) == 'N' && S.charAt(i + 1) == 'G') {
                ngCount++;
                consonantCount++;
                i += 2;
            } else if (vowels.contains(S.charAt(i))) {
                vowelCount++;
                i++;
            } else if (S.charAt(i) == 'Y') {
                yAsVowel++;
                yAsConsonant++;
                i++;
            } else {
                consonantCount++;
                i++;
            }
        }

        vowelCount += yAsVowel;
        consonantCount += yAsConsonant;

        if (vowelCount == 0 || consonantCount < 2) {
            System.out.println(0);
        } else {
            int maxSyllables = Math.min(vowelCount, consonantCount / 2);
            System.out.println(maxSyllables * 3);
        }

        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
NGYNGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
import re

x = input()
lst = re.findall('[AEIOUY]', x)
lst1 = re.findall('[^AEIOU]', x)

rem = 1
lst2 = list()

if 'NG' in x :
    lst2 = re.findall('NG+' , x)
    for i in lst1:
        if rem == len(lst1):
            continue
        if i =='N' and lst1[rem]=='G':
            lst1.pop(rem)
            lst1.pop(rem -1)
        rem = rem + 1
    for i in lst2 :
        lst1.append(i[0])
        lst1.append(i[1])

count = 0

while len(lst1) > 0 and len(lst) > 0:
    lst.pop(-1)
    lst1.pop(-1)
    try:
        lst1.pop(-1)
    except:
        break
    count = count + 3

if len(lst1) > 0 and len(lst) == 0 and len(lst2) > 0:
    count = count + len(lst2)
    print(count)
    exit()
if len(lst1) > 0 and len(lst) == 0:
    count = count + 1
print(count)

**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
NGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
int ans,f,y,Y,n,g,ng;
int main(){
    string s;
    cin >> s;
    for(int i = 0;i < s.length();i++)
    {
        if(s[i]=='Y') y++;
        else if(s[i]=='N') n++;
        else if(s[i]=='G') g++;
        else if(s[i]=='A'||s[i]=='E'||s[i]=='I'||s[i]=='O'||s[i]=='U')
            Y++;
        else f++;
    }
    ng = min(n,g);
    f += max(n,g) - ng;
    //cout << "f:" << f << " Y:" << Y << " y:" << y << " ng:" << ng << endl;
    if(Y*2>= ng)//1
    {
        Y -= ng/2;
        ans += (ng/2) * 5;
        ng %= 2;
        if(ng&&(Y-f/2)==1)
        {
        	f+=2;
        	ng--;
        }
        if(ng&&f&&Y)
        {
            ng--;
            f--;
            Y--;
            ans+=4;
        }
        if(ng&&y&&f)
        {
        	y--;
        	f--;
        	ng--;
        	ans+=4;
        }

        if (Y * 2 >= f)//3
        {
           Y -= f/2;
           ans += (f/2)*3;
           f%=2;
           if(Y * 2 >= y)//5
           {
              ans+= y/2*3;
              Y -= y/2;
              y%=2;
              if(Y&&y&&f) ans+=3;
           }
           else//6
           {

                ans += Y * 3;
                y -= Y * 2;
                Y = 0;
                ans += y/3*3;//7

           }

        }
        else//4
        {
            f -= Y * 2;
            ans += Y * 3;
            Y = 0;
            if(f >= y * 2)//8
            {
                ans += y * 3;
            }
            else // 9
            {
                y -= f / 2;
                ans += f/2*3;
                f %= 2;
                if(f&&y>=2)
                {
                    ans ++;
                    f --;
                    y -= 2;
                }
                ans += y/3*3;//10
            }
        }
    }
    else//2
    {
        ng -= Y*2;
        ans+=Y*5;
        Y = 0;
        if(y*2 >= ng)//12
        {
            ans += ng/2*5;
            y-=ng/2;
            ng %= 2;
            if(ng&&(Y-f/2)==1)
            {
        	    f+=2;
        	    ng--;
            }
	        if(ng&&f&&Y)
	        {
	            ng--;
	            f--;
	            Y--;
	            ans+=4;
	        }
	        if(ng&&y&&f)
	        {
	        	y--;
	        	f--;
	        	ng--;
	        	ans+=4;
	        }

            if(f >= y * 2)//13
            {
                ans += y * 3;

            }
            else//14
            {
                y -= f/2;
                ans+=f/2*3;
                f%=2;
                if(f&& y>=2)
                {
                    f--;
                    y-=2;
                    ans+=3;
                }
                ans+=y/3*3;//15;
            }

        }
        else//11
        {
            ans+=y*5;
        }

    }
    cout << 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
NGYNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

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

void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
}

const int mod= 1e9 + 7;
#define ll long long
#define pb push_back

bool isVowel(char ch){
    return ch=='A'||ch=='E'||ch=='I'||ch=='O'||ch=='U';
}

int solve(string &S){

    int c=0,v=0,y=0,n=0,g=0;

    for(char ch:S){
        if(ch=='Y') y+=1;
        else if(isVowel(ch)) v+=1;
        else{
            c+=1;
            if(ch=='N') n++;
            if(ch=='G') g++;
        }
    }

    int ng=min(n,g);
    c=c-2*ng;
    //c,ng,v,y

    int count=0;

    while(ng>1){

        if(v>0){
            v-=1;ng-=2;count+=5;
        }
        else if(y>0){
            y-=1;ng-=2;count+=5;
        }
        else{
            return count;
        }

    }

    //remaining ng
    int temp=0;
    if(ng==1){
        if(v>1 && c>1) {
            v-=2;c-=2;ng-=1;temp+=6;
        }
        else if(y>1 && c>1) {
            y-=2;c-=2;ng-=1;temp+=6;
        }
        else if(v==1){
            if(c>0){
                v-=1;ng-=1;c-=1;temp+=4;
            }
            else {
                v-=1;ng-=1;temp+=3;
            }
        }
        else if(y==1){
                if(c>0){
                y-=1;ng-=1;c-=1;temp+=4;
            }
            else {
                y-=1;ng-=1;temp+=3;
            }
        }
    }
    count+=temp;
    int ans=min((int)c/2,v);
    for(int i=0;i<=y;i++){
        ans=max(ans,min(v+i,(int)(c+(y-i))/2));
    }

    count+=(3*ans);

    return count;

}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    fastio();

    string S;
    cin>>S;

    int result=solve(S);
    cout<<result;

    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
NGNGAEIOU
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
import re

# Read input string
string = input()

# Find all 'NG' substrings
ng = re.findall('NG', string)

# Preprocess the string to remove 'NG'
string = string.replace('NG', '')

# Initialize variables
v = []
c = []
the_Y = 0
the_output = []

# Process the string to separate vowels, consonants, and handle 'Y'
for i in string:
    if i in ('A', 'E', 'I', 'O', 'U'):
        v.append(i)
        the_Y = 0
    elif i == 'Y':
        the_Y += 1
        if the_Y == 1:
            v.append(i)
        elif the_Y == 2:
            c.append(i)
        elif the_Y == 3:
            c.append(i)
            the_Y = 0
    else:
        c.append(i)
        the_Y = 0

# Add 'NG' elements back to the consonants
c.extend(ng)

# If there are no vowels or not enough consonants, print 0
if not v or len(c) < 2:
    print(0)
else:
    # Merge the consonants and vowels into the output
    while v and c:
        the_output.append(c.pop() + v.pop())

    # If consonants are left, add them to the output
    if c :
        the_output.append(c.pop())
    else:
        the_output.pop()

    # Join the list into a single string and print its length
    print(len(''.join(the_output)))

**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
YYNGYY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;

public class Main1 {

	public static void main(String[] args) {

		Scanner sc=new Scanner(System.in);
		//System.out.println("Enter String");
		String input=sc.nextLine();

		HashSet<Character> h=new HashSet<>();
		h.add('A');h.add('E');h.add('I');h.add('O');h.add('U');

		int count_y=0;

		//int count_NG=0;

		HashMap<Character, Integer> hashmap=new HashMap<>();
		int vowelCount=0;
		int ConsonetCount=0;
		for(char c: input.toCharArray()) {
			if(h.contains(c)) {
				vowelCount+=1;
			}
			else {
				if(c=='Y') {
					count_y+=1;
				}
				else {
					ConsonetCount+=1;
				}

			}
			hashmap.put(c, hashmap.getOrDefault(c, 0)+1);
		}

//		for(Map.Entry<Character, Integer> m: hashmap.entrySet()) {
//			System.out.printf("%c %d\n",m.getKey(),m.getValue());
//		}

		//System.out.printf("Total Consonents %d\n",ConsonetCount);
		//System.out.printf("Total Vowels %d\n",vowelCount);

		int ans=0;

		int minNg=Math.min(hashmap.getOrDefault('N', 0),
				hashmap.getOrDefault('G', 0));

		int TotalCC=ConsonetCount;

		ConsonetCount-=minNg-minNg;

		if(vowelCount==0) {
			ans=hashmap.getOrDefault('Y', 0)<3?0:(hashmap.getOrDefault('Y', 0)/3)*3;
		}

		if(TotalCC==vowelCount*2) {
			ans=TotalCC+vowelCount;
			TotalCC=0;
			vowelCount=0;
		}

		if((vowelCount>0) && (vowelCount*2<=minNg) && minNg>0) {
			ans+=(vowelCount*4+vowelCount);
			minNg-=vowelCount*2;
			vowelCount-=vowelCount;

		}

		else if(TotalCC>vowelCount*2) {

//			System.out.printf("First minNg minNg %d ans %d\n",minNg,ans);
//			System.out.printf("Total Consonents %d\n",ConsonetCount);
//			System.out.printf("Total Vowels %d\n",vowelCount);

			int FirstMinVal=Math.min(Math.min(minNg, vowelCount),ConsonetCount);
			ans+=FirstMinVal*4;
			minNg-=FirstMinVal;
			ConsonetCount-=FirstMinVal;
			vowelCount-=FirstMinVal;
			TotalCC-=FirstMinVal-FirstMinVal;

//
//			System.out.printf("After First minNg minNg %d ans %d\n",minNg,ans);
//			System.out.printf("Total Consonents %d\n",ConsonetCount);
//			System.out.printf("Total Vowels %d\n",vowelCount);
//
			while( (vowelCount>0) && !(vowelCount/2<=minNg) && minNg>0 ) {

				ans+=3;
				minNg--;
				vowelCount--;

			}

			if((vowelCount>0) && (vowelCount*2<=minNg) && minNg>0) {
				ans+=(vowelCount*4+vowelCount);
				minNg-=vowelCount*2;
				vowelCount-=vowelCount;

			}
//
//			System.out.printf("After Second minNg minNg %d ans %d\n",minNg,ans);
//			System.out.printf("Total Consonents %d\n",ConsonetCount);
//			System.out.printf("Total Vowels %d\n",vowelCount);
//
//
		}

		ConsonetCount+=minNg*2;
// ----------------------------------------------------------------------------------------------

		if(ConsonetCount>0 && vowelCount>0 && (ConsonetCount+vowelCount)>=3){
			int minCons=ConsonetCount/2;
			int minVal=Math.min(vowelCount, minCons);
			ans+=minVal+minVal+minVal;
			ConsonetCount=ConsonetCount-minVal-minVal;
			vowelCount-=minVal;
			if(ConsonetCount>0) {
				int minY=Math.min(ConsonetCount/2, hashmap.getOrDefault('Y',0));
				//System.out.printf("We are getting %d\n",minY);
				ans+=(minY*3);
			}
		}

		if(count_y>0 && (count_y+vowelCount)>=3){
			int minCons=count_y/2;
			int minVal=Math.min(vowelCount, minCons);
			ans+=minVal+minVal+minVal;
			count_y=count_y-minVal-minVal;
			vowelCount-=minVal;
			if(count_y>0) {
				ans+=((count_y/3)*3);
			}
		}

		System.out.println(ans);

	}

}

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

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

Think step by step.

Response:
```plaintext
NGNGNG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

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

public class Main {
    static long MOD = (long) (1e9+7);
    static Scanner in = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out, false);
    public static void main(String[] args) throws Exception {

        Main main = new Main();
        int T = 1;//in.nextInt();
        while (T-- > 0){
            main.solve();
        }
        out.flush();
    }
    void solve() {
        String s = in.next();
        int a = 0;
        int b = 0;
        int y = 0;
        int n = 0;
        int g = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'){
                a++;
            }else if(c == 'Y'){
                y++;
            }else if(c == 'N'){
                n++;
            }else if(c == 'G'){
                g++;
            }else {
                b++;
            }
        }
        //
        int ans = 0;
        for (int i = 0; i <= y ; i++) {//y作为辅音的个数
            for (int j = 0; j <= Math.min(n, g) ; j++) {//n+g作为整体的个数
                int b0 = b + (y - i) + j + Math.abs(n - g);//辅音
                int a0 = a + i;
                if(a0 == 0 || b0 < 2) continue;
                int cur;
                if(a0 >= b0){
                    cur = b0 * 2 - 1 + j;
                }else {
                    cur = a0 * 2 + 1 + j;
                }
                ans = Math.max(ans, cur);
            }
        }
        out.println(ans);

    }

}

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

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

Think step by step.

Response:
```plaintext
NGYNGYNGYNG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace __gnu_pbds;
using ordered_multiset=tree<int,null_type,less_equal<int>,
rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
string a;
void solve(){
    cin>>a;
    unordered_map<char,int> Map;
    for(int i=0;i<a.length();i++){
        Map[a[i]]++;
    }
    int cnt_yuan=Map['A']+Map['E']+Map['I']+Map['O']+Map['U'];
    int cnt_y=Map['Y'];
    int cnt_fu=a.length()-cnt_yuan-cnt_y;
    int cnt_min_change=min(Map['N'],Map['G']);
    if(cnt_yuan*2>=cnt_fu+cnt_y){
        int change_able=0;
        if((cnt_fu+cnt_y)%2==1){
            cnt_fu--;
            change_able=1;
            change_able=min(change_able,cnt_min_change);
        }
        cout<<((cnt_fu+cnt_y)/2)*3+change_able;
        return;
    }
    if((cnt_yuan+cnt_y)*2<=cnt_fu){
        int change_able=cnt_fu-(cnt_yuan+cnt_y)*2;
        change_able=min(change_able,cnt_min_change);
        cout<<(cnt_yuan+cnt_y)*3+change_able;
        return;
    }
    int s=(cnt_fu+cnt_y+cnt_yuan)/3;
    int change_able=(cnt_fu+cnt_y+cnt_yuan)%3;
    change_able=min(change_able,cnt_min_change);
    cout<<(s)*3+change_able;
}
int main(){
    int T=1;
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases and scenarios that might cause it to fail.

### Analysis of the Buggy Code
1. **Character Counting**: The code counts the number of vowels, consonants, and 'Y' characters.
2. **Handling 'NG'**: The code attempts to count the 'NG' pairs as a single consonant.
3. **Syllable Formation**: The code tries to form syllables by balancing the counts of vowels, consonants, and 'Y' characters.

### Potential Issues
1. **Handling 'NG' Pairs**: The code might not correctly count 'NG' pairs, especially if there are multiple 'N' and 'G' characters in the string.
2. **Balancing Vowels and Consonants**: The code might not correctly balance the number of vowels and consonants, especially when there are more consonants than vowels.
3. **Edge Cases**: The code might fail for edge cases where the string is very short or contains only a few specific characters.

### Challenging Test Input
To expose these issues, we can create a test input that:
1. Contains multiple 'N' and 'G' characters to test the 'NG' handling.
2. Has a mix of vowels, consonants, and 'Y' characters to test the syllable formation.
3. Is short enough to be a valid input but complex enough to challenge the code.

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

```plaintext
NGNGNGAEIOUY
```

### Explanation
- **'NG' Handling**: The string contains multiple 'N' and 'G' characters, which should be counted as single consonants.
- **Vowel and Consonant Balance**: The string has a mix of vowels ('A', 'E', 'I', 'O', 'U') and consonants, including 'Y'.
- **Short and Complex**: The string is short (12 characters) but complex enough to test the code's ability to form syllables correctly.

This input should help expose any issues in the code's handling of 'NG' pairs and the balance of vowels and consonants.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS 1
#pragma GCC optimize("Ofast")

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>

#define int long long
#define lowbit(x) (x & -x)
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
const int N = 2e6 + 10;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f3f3f3f3fLL;
typedef __int128 ll;

template<typename T>
inline T Min(T x, T y) {
	if (x <= y) return x;
	return y;
}
template<typename T>
inline T Max(T x, T y) {
	if (x >= y) return x;
	return y;
}
template<typename T>
inline T gcd(T x, T y) {
	if (y == 0) return x;
	return gcd(y, x % y);
}
template<typename T>
inline T lcm(T x, T y) {
	return x * y / gcd(x, y);
}

namespace IO {
	char buf[1 << 20], *p1, *p2;
	#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin), p1 == p2) ? 0 : *p1++)
	inline void read128(__int128& x) {
		bool f = 1; x = 0; char ch = getchar();
		while (ch < '0' || ch>'9') { if (ch == '-') f = !f; ch = getchar(); }
		while (ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar(); }
		x = (f ? x : -x); return;
	}
	inline void write128(__int128 x) {
		if (x < 0) putchar('-'), x = -x;
		if (x > 9) write128(x / 10);
		putchar(x % 10 + '0'); return;
	}

	inline void readD(double& x) { scanf("%lf", &x); }

	inline void writeD(double x) { printf("%lf", x); }

	inline void writeD(double x, int len) { printf("%.*lf", len, x); }

	template<typename T>inline void readT(T& x) {
		bool f = 1; x = 0; char ch = getchar();
		while (ch < '0' || ch>'9') { if (ch == '-') f = !f; ch = getchar(); }
		while (ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar(); }
		x = (f ? x : -x); return;
	}
	template<typename T>
	inline void writeT(T x) {
		if (x < 0) putchar('-'), x = -x;
		if (x > 9) writeT(x / 10);
		putchar(x % 10 + '0'); return;
	}

	inline void readS(std::string& s) {
		char ch = getchar();
		while (ch == ' ' || ch == '\n') ch = getchar();
		while (ch != ' ' && ch != '\n') s += ch, ch = getchar();
	}

	inline void readSL(std::string& s) {
		char ch = getchar();
		while (ch == '\n') ch = getchar();
		while (ch != '\n') s += ch, ch = getchar();
	}

	inline void writeS(std::string s) {
		int len = s.length();
		for (int i = 0; i < len; i++)
			putchar(s[i]);
	}

	inline int readC(char s[]) {
		int tot = 0; char ch = getchar();
		while (ch == ' ' || ch == '\n') ch = getchar();
		while (ch != ' ' && ch != '\n') s[tot++] = ch, ch = getchar();
		return tot;
	}
	inline void writeC(char s[], int len) {
		for (int i = 0; i < len; i++)
			putchar(s[i]);
	}

	inline void readc(char& c) {
		c = getchar();
		while (c == ' ' || c == '\n')
			c = getchar();
	}inline void writec(char c) { putchar(c); }

	const std::string c = "c";
	const std::string i = "i";
	const std::string j = "j";
	const std::string X = "x";
	const std::string y = "y";
	const std::string b = "b";
	const std::string s = "s";
	const std::string Ss = "Ss";

	template<class T>
	inline void read(T& x) {
		if (typeid(x).name() == i) readT(x);
		else if (typeid(x).name() == j) readT(x);
		else if (typeid(x).name() == X) readT(x);
		else if (typeid(x).name() == y) readT(x);
		if (typeid(x).name() == b) {
			int k; readT(k);
			x = (k > 0 ? true : false);
		}
		else if (typeid(x).name() == s) readT(x);
		else if (typeid(x).name() == c) { char ch; readc(ch); x = ch; }
	}
	template<class T, class... Ts>
	inline void read(T& x, Ts&... xx) {
		if (typeid(x).name() == i) readT(x);
		else if (typeid(x).name() == j) readT(x);
		else if (typeid(x).name() == X) readT(x);
		else if (typeid(x).name() == y) readT(x);
		if (typeid(x).name() == b) {
			int k; readT(k);
			x = (k > 0 ? true : false);
		}
		else if (typeid(x).name() == s) readT(x);
		else if (typeid(x).name() == c) { char ch; readc(ch); x = ch; }
		read(xx...);
	}
	inline void read(std::string& x) { readS(x); }

	inline void read(char x[]) { readC(x); }

	inline void read(double& x) { readD(x); }

	inline void write(double x) { writeD(x); }

	inline void write(double x, int len) { writeD(x, len); }

	inline void write(std::string x) { writeS(x); }

	inline void write(char x[]) { writeC(x, strlen(x)); }

	inline void write(double x, char a) { writeD(x); putchar(a); }

	inline void write(std::string x, char a) { writeS(x); putchar(a); }

	inline void write(char x[], char a) { writeC(x, strlen(x)); putchar(a); }

	inline void write(double x, int len, char a) { writeD(x, len); putchar(a); }

	template<class T>inline void write(T x, char a) {
		if (typeid(x).name() == i) writeT(x);
		else if (typeid(x).name() == j) writeT(x);
		else if (typeid(x).name() == X) writeT(x);
		else if (typeid(x).name() == y) writeT(x);
		if (typeid(x).name() == b) {
			if (x) writeT(1);
			else writeT(0);
		}
		else if (typeid(x).name() == s) writeT(x);
		else if (typeid(x).name() == c) { writec(x); }
		putchar(a);
	}
	inline void write(double x, std::string a) { writeD(x); writeS(a); }

	inline void write(std::string x, std::string a) { writeS(x); writeS(a); }

	inline void write(char x[], std::string a) { writeC(x, strlen(x)); writeS(a); }

	inline void write(double x, int len, std::string a) { writeD(x, len); writeS(a); }

	inline void write(char a, std::string s) { writec(a); writeS(s); }

	template<typename T>inline void write(T x, std::string a) {
		if (typeid(x).name() == i) writeT(x);
		else if (typeid(x).name() == j) writeT(x);
		else if (typeid(x).name() == X) writeT(x);
		else if (typeid(x).name() == y) writeT(x);
		if (typeid(x).name() == b) {
			if (x) writeT(1);
			else writeT(0);
		}
		else if (typeid(x).name() == s) writeT(x);
		else if (typeid(x).name() == c) { writec(x); }
		writeS(a);
	}
	template<typename T>inline void write(T x) {
		if (typeid(x).name() == i) writeT(x);
		else if (typeid(x).name() == j) writeT(x);
		else if (typeid(x).name() == X) writeT(x);
		else if (typeid(x).name() == y) writeT(x);
		if (typeid(x).name() == b) {
			if (x) writeT(1);
			else writeT(0);
		}
		else if (typeid(x).name() == s) writeT(x);
		else if (typeid(x).name() == c) writec(x);
	}
	template<typename T, typename... Ts>
	inline void write(T x, Ts... xx) {
		if (typeid(x).name() == i) writeT(x);
		else if (typeid(x).name() == j) writeT(x);
		else if (typeid(x).name() == X) writeT(x);
		else if (typeid(x).name() == y) writeT(x);
		if (typeid(x).name() == b) {
			if (x) writeT(1);
			else writeT(0);
		}
		else if (typeid(x).name() == s) writeT(x);
		else if (typeid(x).name() == c) { writec(x); }
		write(xx...);
	}
	inline void put(std::string s) {
		int len = s.length();
		for (int i = 0; i < len; i++) {
			putchar(s[i]);
		}
	}
}

std::string s;

void solve() {
	IO::readS(s);
	int cnt = 0, p = 0, q = 0, ans = 0, a = 0, b = 0;
	for (register int i = 0; i < s.size(); ++i) {
		char chr = s[i];
		if (chr == 'N')			++a;
		else if (chr == 'G')	++b;
		switch(chr) {
			case 'A': {
				++p;
				break;
			}
			case 'E': {
				++p;
				break;
			}
			case 'I': {
				++p;
				break;
			}
			case 'O': {
				++p;
				break;
			}
			case 'U': {
				++p;
				break;
			}
			case 'Y': {
				++cnt;
				break;
			}
			default: {
				++q;
				break;
			}
		}
	}
	if (q > 2 * p) {
		int tmp = 3 * p;
		ans += tmp;
		q -= 2 * p;
		p = 0;
		while (cnt && q / 2) {
			--cnt, q -= 2;
			ans += 3;
		}
		if (cnt) {
			cnt += q;
			int tmp = 3 * (cnt / 3);
			ans += tmp;
			cnt -= tmp;
			int l = cnt, r = Min(a, b);
			while (l && r) {
				--l, --r;
				++ans;
			}
		}
		else {
			int l = q, r = Min(a, b);
			while (l && r) {
				--l, --r;
				++ans;
			}
		}
		IO::write(ans, '\n');
		return;
	}
	else if (q == 2 * p) {
		ans += 3 * p;
		int tmp = 3 * (cnt / 3);
		ans += tmp;
		cnt -= tmp;
		int l = cnt, r = Min(a, b);
		while (l && r) {
			--l, --r;
			++ans;
		}
		IO::write(ans, '\n');
		return;
	}
	else {
		int tmp = (q / 2);
		ans += 3 * tmp;
		q -= 2 * tmp;
		p -= tmp;
		if (q && Min(a, b))	++ans, --q;
		while (cnt / 2 && p) {
			--p, cnt -= 2;
			ans += 3;
		}
		ans += 3 * (cnt / 3);
		IO::write(ans, '\n');
	}
	return;
}

signed main() {

	int t = 1;
//	IO::read(t);
	while (t--) {
		solve();
	}
	return 0;
}
/*
1
15
1 -3
9 -2
-4 -8
-9 5
2 6
-10 -5
9 -2
6 0
4 -7
-2 -5
7 7
7 7
-2 -9
9 7
2 10
*/

**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 consider the following aspects:

1. **Edge Cases for Vowels and Consonants**:
   - Strings with a high number of vowels and consonants.
   - Strings with only vowels or only consonants.
   - Strings with a mix of vowels, consonants, and the special 'NG' consonant.

2. **Special Cases for 'Y'**:
   - Strings with multiple 'Y' characters.
   - Strings where 'Y' is the only character.

3. **Special Cases for 'NG'**:
   - Strings with multiple 'NG' sequences.
   - Strings where 'NG' is the only sequence.

4. **Combination of All**:
   - Strings that combine all the above elements in a way that might expose the bug.

Given these considerations, a string that includes a mix of vowels, consonants, 'Y', and 'NG' sequences, and is designed to challenge the algorithm's ability to form valid syllables, would be:

```plaintext
NGYYYNGAEIOUYNG
```

This string includes:
- Multiple 'NG' sequences.
- Multiple 'Y' characters.
- A mix of vowels and consonants.

This input will test the code's ability to handle the 'NG' as a single consonant, the versatility of 'Y', and the formation of syllables with the given constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
from math import ceil
letters = {i:0 for i in "abcdefghijklmnopqrstuvwxyz"}
s = input()
for i in s:
    letters[i.lower()] += 1
maxx = 0
letters['ng'] = 0
for i in range(min(letters["n"],letters["g"])+1):
    if i != 0:
        letters["ng"] += 1
        letters["n"] -= 1
        letters["g"] -= 1
    vowels = sum([j for i,j in letters.items() if i in "aeiou"])
    consonants = sum([j for i,j in letters.items() if i not in "aeiouy"])
    y = letters['y']
    diff =  2*vowels - consonants
    if diff < 0:
        ys = min(y,ceil(-diff/2))
        vowels += ys
        y -= ys
    diff = 2*vowels - consonants
    if diff > 0:
        ys = min(y,diff)
        consonants += ys
        y -= ys
    vowels += y//3
    consonants += 2*(y//3)
    maxx = max(maxx,3*(min(vowels,consonants//2))+letters['ng'])
print(maxx)

**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
NGN
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
from math import ceil
letters = {i:0 for i in "abcdefghijklmnopqrstuvwxyz"}
s = input()
for i in s:
    letters[i.lower()] += 1
maxx = 0
letters['ng'] = 0
for i in range(min(letters["n"],letters["g"])+1):
    if i != 0:
        letters["ng"] += 1
        letters["n"] -= 1
        letters["g"] -= 1
    vowels = sum([j for i,j in letters.items() if i in "aeiou"])
    consonants = sum([j for i,j in letters.items() if i not in "aeiouy"])
    y = letters['y']
    diff =  2*vowels - consonants
    if diff < 0:
        ys = min(y,ceil(-diff/2))
        vowels += ys
        y -= ys
    diff = 2*vowels - consonants
    if diff > 0:
        ys = min(y,diff)
        consonants += ys
        y -= ys
    vowels += y//3
    consonants += 2*y//3
    maxx = max(maxx,3*(min(vowels,consonants//2))+letters['ng'])
print(maxx)

**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
NGNGNGY
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
//package Practice;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class scrabble {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine().toLowerCase();
        int cCount = 0;
        int vCount = 0;
        int nCount = 0;
        int gCount = 0;
        int yCount = 0;
        for (int i = 0; i < s.length(); i++) {
            char c= s.charAt(i);
            if (c == 'a' || c == 'i' || c == 'e' || c == 'o' || c == 'u') {
                vCount++;
            }
            else if (c == 'y') {
                yCount++;
            }
            else {
                cCount++;
            }
            if (c == 'n') {
                nCount++;
            }
            if (c == 'g') {
                gCount++;
            }
        }
        int cStart = cCount;
        int ans = 0;
        while (true) {
            if (cCount >= 2 && vCount >= 1) {
                cCount -= 2;
                vCount--;
                ans += 3;
            }
            else if (yCount > 0) {
                if (vCount == 0) {
                    vCount++;
                    yCount--;
                }
                else {
                    cCount++;
                    yCount--;
                }
            }
            else {
                break;
            }
        }
        if (cCount > 0) {
            if (nCount > 0 && gCount > 0)  {
                // add 1 for each, min of cCount and nCount and gCount
                int cUsed = ans / 3 * 2;
                int nonNG = cStart - gCount - nCount;
                int nUsed = 0;
                int gUsed = 0;
                while (nonNG < cUsed) {
                    if (nCount > gCount) {
                        nCount--;
                        nUsed++;
                    }
                    else {
                        gCount--;
                        gUsed++;
                    }
                    nonNG++;
                }
                //System.out.println("gUsed:" + gUsed);
                //System.out.println("nUsed:" + nUsed);
                //System.out.println("nCount:" + nCount);
                //System.out.println("gCount: " + gCount);
                //System.out.println("ans initial:" + ans);
                //ans += Math.min(gUsed, nCount);
                //System.out.println("ans initial 2:" + ans);
                //ans += Math.min(nUsed, gCount);
                //System.out.println("ans final:" + ans);
                if (gUsed > 0 || nUsed > 0) {
                    ans += Math.min(gUsed, nCount);
                    //System.out.println("ans initial 2:" + ans);
                    ans += Math.min(nUsed, gCount);
                }
                else {
                    ans += Math.min(gCount, nCount);
                }
            }
        }
        System.out.println(ans);
        br.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
NGYNGYNGY
```