**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yet Another Real Number Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: Three r there are's in strawberry.

You are given an array $b$ of length $m$. You can perform the following operation any number of times (possibly zero):

  * Choose two distinct indices $i$ and $j$ where $\bf{1\le i < j\le m}$ and $b_i$ is even, divide $b_i$ by $2$ and multiply $b_j$ by $2$.

Your task is to maximize the sum of the array after performing any number of such operations. Since it could be large, output this sum modulo $10^9+7$.

Since this problem is too easy, you are given an array $a$ of length $n$ and need to solve the problem for each prefix of $a$.

In other words, denoting the maximum sum of $b$ after performing any number of such operations as $f(b)$, you need to output $f([a_1])$, $f([a_1,a_2])$, $\ldots$, $f([a_1,a_2,\ldots,a_n])$ modulo $10^9+7$ respectively.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of $a$.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the starting values of array $a$.

It is guaranteed that the sum of $n$ over all test cases will not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers representing the answer for each prefix of $a$ modulo $10^9+7$.

Examples:
input:
3
10
1 2 3 4 5 6 7 8 9 10
11
1 6 9 4 7 4 4 10 3 2 3
4
527792568 502211460 850237282 374773208
output:
1 3 8 13 46 59 126 149 1174 1311
1 7 22 26 70 74 150 1303 1306 1308 1568
527792568 83665723 399119771 773892979

Note: For each prefix in the first example, a possible array after operations is:

  * $[1]$ and the sum is $1$;    * $[1, 2]$ and the sum is $3$;    * $[1, 1, 6]$ and the sum is $8$;    * $[1, 1, 3, 8]$ and the sum is $13$;    * $[1, 1, 3, 1, 40]$ and the sum is $46$;    * $[1, 1, 3, 1, 5, 48]$ and the sum is $59$;    * $[1, 1, 3, 1, 5, 3, 112]$ and the sum is $126$;    * $[1, 1, 3, 1, 5, 3, 7, 128]$ and the sum is $149$;    * $[1, 1, 3, 1, 5, 3, 7, 1, 1152]$ and the sum is $1174$;    * $[1, 1, 3, 1, 5, 3, 7, 1, 9, 1280]$ and the sum is $1311$.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    static class DisjointSet {
        List<Integer> rank = new ArrayList<>();
        List<Integer> parent = new ArrayList<>();
        List<Integer> size = new ArrayList<>();
        public DisjointSet(int n) {
            for (int i = 0; i <= n; i++) {
                rank.add(0);
                parent.add(i);
                size.add(1);
            }
        }

        public int findUPar(int node) {
            if (node == parent.get(node)) {
                return node;
            }
            int ulp = findUPar(parent.get(node));
            parent.set(node, ulp);
            return parent.get(node);
        }

        public void unionBySize(int u, int v) {
            int ulp_u = findUPar(u);
            int ulp_v = findUPar(v);
            if (ulp_u == ulp_v) return;
            if (size.get(ulp_u) < size.get(ulp_v)) {
                parent.set(ulp_u, ulp_v);
                size.set(ulp_v, size.get(ulp_v) + size.get(ulp_u));
            } else {
                parent.set(ulp_v, ulp_u);
                size.set(ulp_u, size.get(ulp_u) + size.get(ulp_v));
            }
        }
    }
    static long power(long a, long b, long m) {
        a %= m;
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = (res * a) % m;
            a = (a * a) % m;
            b >>= 1;
        }
        return res;
    }
    static long gcd(long a, long b)  {
        return b == 0? a: gcd(b, a % b);
    }
    static int MOD = 1000000007;
    static long modinv(long a) {
        return a <= 1 ? a : MOD - (MOD / a) * modinv(MOD % a) % MOD;
    }

    static class pair {
        long i;
        long x;
        long y;
        pair(long a, long b, long c) {
            this.i = a;
            this.x = b;
            this.y = c;
        }
    }
    public static void main(String args[]) throws IOException {
        FastReader sc = new FastReader();
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            long a[] = new long[n];
            for(int i=0; i<n; i++) {
                a[i] = sc.nextLong();
            }
            long factor[][] = new long[n][2];

            for(int i=0; i<n; i++) {
                long x = a[i];
                long y = 0;
                while(x%2 == 0) {
                    x = x/2;
                    y++;
                }
                factor[i][0] = y;
                factor[i][1] = x;
            }

            Stack<pair> s = new Stack<>();
            long ans = a[0];
            //if(factor[0][0] > 1) {
                s.push(new pair(0, factor[0][0], factor[0][1]));
            //}

            System.out.print(ans+" ");
            for(int i=1; i<n; i++) {
                long x2 = factor[i][0];
                long y2 = factor[i][1];

                while(!s.isEmpty() && (s.peek().y) <= (power(2, x2, MOD)*y2)%MOD) {
                    long x1 = s.peek().x;
                    long y1 = s.peek().y;
                    s.pop();
                    ans = (ans - ((power(2, x1, MOD) - 1) * y1) % MOD + MOD) % MOD;
                    x2 = (x2 + x1)%MOD;
                }
                //System.out.println(x2*y2);
                ans = ans + ((power(2, x2, MOD) * y2) % MOD) % MOD;
                //ans %= MOD;
                System.out.print(ans+" ");
                //if(x2 > 1) {
                    s.push(new pair(i, x2, y2));
                //}
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

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

int main(){
    char s[500001];
    int pb[500000];
    cin>>s;
    pb[0]=0;
    pb[1]=1;
    int max=0;
    for(int i=2; s[i]; i++){
        if(s[i-1]=='(') pb[i]=pb[i-1]+1;
        else pb[i]=pb[i-1]-1;
        if(pb[i]>max) max=pb[i];
    }
    for(int j=0; j<=max; j++){
        for(int i=strlen(s)-1; i>=0; i--){
            if(pb[i]==j) cout<<s[i];
        }

    }
    cout<<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
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
	string s;
	cin>>s;
	int cnt=0;
	int close=0;
	int n=s.size();
		for(int i=0;i<n;i++){
		if(s[i]=='('){
			cnt++;
		}
		else{
			close++;
		}
	}
	string str="";
	string a="()";
	if(cnt>=1 && close>=1 ){
		str+=a;
		cnt--;
		close--;
		int i=0;
		while(cnt--){
			str.push_back('(');
			if(i>0)str.push_back(')');
			i++;
		}
		str.push_back(')');
	}
	cout<<str<<endl;

	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code has a few issues:

1. **Incorrect Handling of Balances**: The code does not correctly compute the balance of the prefix for each character and sort the columns accordingly.
2. **Incorrect Construction of the Result**: The code attempts to construct the result string by appending characters in a fixed pattern, which does not follow the balanced shuffle operation as described.

Given these issues, a good test input should be a balanced parentheses sequence that, when processed correctly, should produce a different result than what the buggy code would generate. A small-scale test input that can expose these issues is:

```plaintext
((()))
```

Let's analyze why this input is effective:

1. **Correct Balance**: The input `((()))` is a balanced parentheses sequence with a balance of 0.
2. **Prefix Balances**: The prefix balances for `((()))` are:
   - Position 1: Balance 1
   - Position 2: Balance 2
   - Position 3: Balance 3
   - Position 4: Balance 2
   - Position 5: Balance 1
   - Position 6: Balance 0

   When sorted in increasing order of prefix balance, breaking ties in decreasing order of position, the table becomes:
   - Balance 0: Position 6
   - Balance 1: Position 5, Position 1
   - Balance 2: Position 4, Position 2
   - Balance 3: Position 3

   The sorted columns would give us the sequence `(()())`.

3. **Buggy Code Output**: The buggy code will output `()()()` instead of `(()())`, which is incorrect.

Therefore, the test input `((()))` should effectively expose the bug in the given buggy code.

```plaintext
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    string s; cin>>s;
    int n = s.size();
    vector<pair<int,int>> v(n); vector<char> res(n);
    v[0]={0,1};
    for(int i=1; i<s.size() ;i++){
        int count = 0;
         for(int j=0 ; j<i ; j++){
            if(s[j]=='(') count+=1;
            else count -= 1;
         }
         v[i]={count,i};
         res[i] = s[i];
    }
    sort(v.begin(), v.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
        return (a.first < b.first) || (a.first == b.first && a.second > b.second);
    });
    for(int i=0 ; i<v.size();i++){
        cout<<res[v[i].second];
    }
    cout<<endl;
}

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

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

Think step by step.

Response:
```plaintext
(((())))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
from sys import stdin, stdout

s = stdin.readline().rstrip()
n = len(s)

out = ['']*(n>>1)

b = 0
for c in s[::-1]:
    if c == '(':
        b -= 1
    else:
        b += 1
    out[b] += c

stdout.write(''.join(out))

**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
(((())))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
from sys import stdin, stdout

s = stdin.readline().rstrip()

out = ['']*(len(s)>>1)
b = 0

for c in s:
    out[b] = c + out[b]
    if c == ')':
        b -= 1
    else:
        b += 1

stdout.write(''.join(out))

**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
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader h = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		StringTokenizer st = new StringTokenizer(h.readLine());
    String a=st.nextToken();
    int[] count=new int[a.length()];
    count[0]=0;int max=-1;boolean flag=false;
    for (int i = 1; i < count.length; i++) {
      char c=a.charAt(i-1);
      if (c=='('){count[i]=count[i-1]+1;}
      if (c==')'){count[i]=count[i-1]-1;flag=true;}
      max=Math.max(max,count[i]);
    }
    StringBuilder ans=new StringBuilder();
    if (!flag) {
      for (int i = 0; i < count.length; i++) {
        pw.print('(');
      }
    } else {
      for (int i = 0; i <= max; i++) {
      for (int j = a.length()-1; j>=0; j--) {
        if (count[j]==i) {
          ans.append(a.charAt(j));
          //pw.println(i+" "+j+" "+a.charAt(j));
        }
      }
    }
    }

    pw.print(ans);
		h.close();
		pw.close();
	}
}

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

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

Think step by step.

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

### Code Analysis
1. **Prefix Balance Calculation**:
   - The code calculates the prefix balance for each character in the input string.
   - The balance is correctly updated for each character.

2. **Sorting and Building the Result**:
   - The code attempts to sort the characters based on the prefix balance and position.
   - The sorting logic is flawed. It iterates over the prefix balances and appends characters to the result string.

### Potential Bugs
- **Incorrect Sorting Logic**: The code sorts the characters by prefix balance and then appends them in reverse order of positions. This does not correctly implement the balanced shuffle operation.
- **Edge Cases**:
  - Empty string (though the problem guarantees a non-empty string).
  - Strings with a single type of parentheses (e.g., "()" or "(()())").
  - Strings with a balanced structure but complex nested patterns (e.g., "(((())))").

### Challenging Test Input
To expose the bug, we need an input that has a complex nested structure and a balance that changes frequently. A good candidate is a string with multiple levels of nested parentheses.

### Test Input
```plaintext
(((())))
```

### Explanation
- **Input**: "(((())))"
- **Expected Output**: "(()())()"
- **Actual Output from Buggy Code**: "(((())))"

The expected output is "(()())()", but the buggy code will output "(((())))" because it does not correctly handle the sorting and appending of characters based on the prefix balance and position.

This test input should effectively expose the bug in the sorting logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.util.Scanner;
public class CodeForces {
    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        String s=sc.next();
        int n=s.length();
        int balance=0;
        int saldo[]=new int [n];
        int tope=0;
        String sub[]=new String[n/2];
        for(int i=0;i<n;i++){
            saldo[i]=balance;
            if(balance>tope){
                tope=balance;
            }
            if(s.charAt(i)=='('){
                balance++;
            }else{
                balance--;
            }

            if(sub[saldo[i]]!=null){
                sub[saldo[i]]=Character.toString(s.charAt(i))+sub[saldo[i]];
            }else{
                sub[saldo[i]]=Character.toString(s.charAt(i));
            }

        }
        String t="";
        for(int i=0;i<tope+1;i++){
            t=t+sub[i];
        }
        System.out.println(t);
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's consider the following steps:

1. **Understand the Bug**: The bug in the code is in how it handles the `sub` array. Specifically, it incorrectly updates and concatenates the characters based on the balance values, leading to incorrect shuffling.

2. **Edge Cases**:
   - **Short Balanced Sequences**: Simple and short sequences can help in quickly identifying issues.
   - **Sequences with Many Balanced Subsequences**: These can test how the code handles multiple balanced parts.
   - **Sequences with Nested Parentheses**: These can test the balance calculation and shuffling logic.

3. **Generate a Test Input**:
   - We need a sequence that is balanced and has a clear structure to make it easy to verify the expected output.
   - A sequence with nested parentheses and multiple balanced parts will be particularly challenging.

Let's consider the sequence `(((())))`. This sequence has a clear nested structure and should be shuffled to `()()()()`.

### Test Input
```plaintext
(((())))
```

### Explanation
- The sequence `(((())))` has a balance of 0 and is balanced.
- The prefix balances are:
  - Position 1: Balance 1
  - Position 2: Balance 2
  - Position 3: Balance 3
  - Position 4: Balance 4
  - Position 5: Balance 3
  - Position 6: Balance 2
  - Position 7: Balance 1
  - Position 8: Balance 0
- Sorting the columns by balance and breaking ties by position in decreasing order gives:
  - Balance 0: Position 8
  - Balance 1: Position 1, 7
  - Balance 2: Position 2, 6
  - Balance 3: Position 3, 5
  - Balance 4: Position 4
- The shuffled sequence should be `()()()()`.

This test input should help expose the bug in the code by demonstrating the incorrect handling of nested parentheses and balance values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
s = input()
temp = []
index = 0
chi_so = 1
for i in range (0,len(s)):
    if s[i] == "(":
        temp.append([index,chi_so,"("])
        index +=1
        chi_so+=1
    else:
        temp.append([index,chi_so,")"])
        index -=1
        chi_so+=1
temp.sort()
bien = 0
for i in range (len(temp)-1):
    if temp[i][0] < temp[i+1][0]:
        for k in range (i,bien-1,-1):
            print(temp[k][2],end='')
        bien = i+1
print(temp[-1][2])

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, let's consider the following aspects:

1. **Prefix Balance Calculation**: The code correctly calculates the prefix balance and the index for each character in the input string.
2. **Sorting**: The code sorts the columns based on the prefix balance and position, which should be correct.
3. **Output Generation**: The issue lies in the output generation part, specifically in the nested loops that print the characters. The nested loops are not correctly handling the order of characters after sorting.

### Edge Case Considerations:
- **Single Character Strings**: These are trivial and unlikely to expose bugs.
- **Balanced Parentheses with Equal Prefix Balances**: This can help check if the sorting and output generation handle ties correctly.
- **Complex Nested Parentheses**: These can help check if the code correctly handles nested structures.

### Test Input:
A balanced parentheses sequence with a complex nested structure that can expose issues in the output generation logic. For example, a sequence with multiple levels of nesting and equal prefix balances.

```plaintext
(((())))
```

### Explanation:
- **Input**: `(((())))`
- **Expected Output**: `(()(()))`

This input has a complex nested structure, and the balanced shuffle should rearrange it to `(()(()))`. The provided buggy code might fail to correctly handle the nested structure and output the correct sequence.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
# import math
# for _ in range(int(input())):
s=input()

# d=dict()
# temp=0
# j=1
# for i in s:
#     d[j]=temp
#     j+=1
#     if i=='(':
#         temp+=1
#     else:
#         temp-=1

# a=sorted(d.values())
# mini=0
# for i in a :
#     for j in

################

d=dict()
temp=0
j=1
for i in s:

    if i=='(':
        d[(temp,j)]='('
        temp+=1
    else:
        d[(temp,j)]=')'
        temp-=1
    j+=1

j,c=-1,0
l=len(d)
s=''
while(True):
    if c>=l:
        break
    j+=1
    for i in range(l,0,-1):
        m=d.get((j,i),0)
        if m!=0:
            s+=m
            c+=1

print(s)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test inputs that are at the boundaries of the problem constraints.
2. **Complexity**: Test inputs that require multiple levels of nested parentheses.
3. **Symmetry**: Test inputs that have symmetric properties that might be handled incorrectly.
4. **Special Patterns**: Test inputs that have specific patterns that might cause the sorting or shuffling logic to fail.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug might be a sequence with a balance of 0 but with a complex nested structure. Let's consider the following test input:

```plaintext
(((())))
```

### Explanation:
- **Length**: The input is small (8 characters) but complex enough to test the shuffling logic.
- **Balance**: The input is a balanced parentheses sequence with a balance of 0.
- **Nested Structure**: The input has a deeply nested structure, which can help in testing the sorting and shuffling logic.

### Expected Output:
The balanced shuffle of `(((())))` should be `()()()()`.

### Why This Input?
- **Complexity**: The deeply nested structure ensures that the prefix balances are non-trivial and require correct handling.
- **Symmetry**: The input is symmetric, which can help in testing the correctness of the sorting and shuffling logic.

Let's see if this input can expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
def balanced(s):
    prefix =[0]; index =0
    for i in s:
        if i == "(":
            prefix.append(prefix[index]+1); index+=1
        else:
            prefix.append(prefix[index] - 1);index +=1
    prefix.pop(); mx = max(prefix); pt = 0; s1 = ""

    while pt <=mx:
        for i in range(len(prefix)-1,-1,-1):
            if prefix[i] == pt:
                s1 = s1 + s[i]
        pt+=1
    return s1

def main():
    s = input()
    print(balanced(s))
main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test with the smallest possible balanced parentheses sequences.
2. **Complex Sequences**: Test with sequences that have a variety of nested and non-nested structures.
3. **Longer Sequences**: Test with sequences that are close to the maximum length to ensure the code handles large inputs correctly.

Given the constraints, we will focus on a small-scale test input that is complete and valid. Let's consider a sequence that has a mix of nested and non-nested structures to ensure the balanced shuffle operation is correctly applied.

A good candidate is a sequence that has a balance of 0 and contains both nested and non-nested parentheses. For example, a sequence like "(()())" has a balance of 0 and contains both nested and non-nested structures.

Here is the test input:

```plaintext
(()())
```

This input should be challenging enough to expose any issues in the buggy code. Let's break down why:

1. **Balance of 0**: The sequence "(()())" has a balance of 0, which is a requirement for a balanced parentheses sequence.
2. **Nested and Non-nested**: The sequence contains both nested and non-nested structures, which will test the prefix balance calculation and the sorting logic.
3. **Small Scale**: The sequence is small enough to be manageable and to ensure that the output is not too long.

By using this test input, we can verify if the buggy code correctly applies the balanced shuffle operation and produces the expected output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int balance;
    int position;
    char character;
} Entry;

int main() {
    char s[500001];
    scanf("%s", s);

    int len = strlen(s);
    Entry entries[len];

    int balance = 0;

    // Calculate prefix balances and fill entries array
    for (int i = 0; i < len; i++) {
        entries[i].balance = balance;
        entries[i].position = i;
        entries[i].character = s[i];
        if (s[i] == '(') {
            balance++;
        } else {
            balance--;
        }
    }

    // Sort based on prefix balance and position
    for (int i = 0; i < len - 1; i++) {
        for (int j = 0; j < len - i - 1; j++) {
            if (entries[j].balance > entries[j + 1].balance ||
                (entries[j].balance == entries[j + 1].balance && entries[j].position < entries[j + 1].position)) {
                // Swap entries
                Entry temp = entries[j];
                entries[j] = entries[j + 1];
                entries[j + 1] = temp;
            }
        }
    }

    // Print the sorted characters
    for (int i = 0; i < len; i++) {
        printf("%c", entries[i].character);
    }
    printf("\n");

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a sorting issue where it sorts based on prefix balance and position. The key points to consider are:

1. **Prefix Balance Calculation**: Ensure that the prefix balance is calculated correctly.
2. **Sorting Criteria**: The sorting criteria should correctly handle ties based on position in descending order.

A good test case to expose the bug would be a balanced parentheses sequence where the prefix balances and positions create a scenario where the sorting logic fails. Specifically, we need a sequence where the positions of the parentheses create a tie in the prefix balance that the sorting logic incorrectly handles.

Let's consider the following test case:

```plaintext
(()(()))
```

This input has a balanced sequence with the following prefix balances and positions:

| Position | Character | Prefix Balance |
|----------|-----------|----------------|
| 1        | (         | 0              |
| 2        | (         | 1              |
| 3        | (         | 2              |
| 4        | )         | 1              |
| 5        | (         | 2              |
| 6        | )         | 1              |
| 7        | (         | 2              |
| 8        | )         | 1              |

When sorted based on prefix balance and position, the entries should be:

| Prefix Balance | Position | Character |
|----------------|----------|-----------|
| 0              | 1        | (         |
| 1              | 4        | )         |
| 1              | 6        | )         |
| 1              | 8        | )         |
| 2              | 2        | (         |
| 2              | 5        | (         |
| 2              | 7        | (         |
| 3              | 3        | (         |

The expected output should be:

```plaintext
()(()())
```

However, the provided code sorts the positions incorrectly when there is a tie in the prefix balance. The correct output should be:

```plaintext
()(()())
```

Let's test this input:

```plaintext
(()(()))
```

This input should expose the bug in the sorting logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
#include<iostream>

using namespace std;
int balpref(char s[],int i, int &o, int &c){
    if (s[i-1]=='('){
        o++;
    }
    if (s[i-1]==')'){
        c++;
    }
    return o - c;
}

int main(){
    string k;
    cin >>k;
    int n = k.length();
    char s[n];
    for (int i=0;i<n;i++){
        s[i] = k[i];
    }

    int a[n];
    int o=0,c=0;
    for (int i = 0; i <n; i++){
        a[i] = balpref(s,i,o,c);
    }
    char t[n];
    int index_of_t = 0;
    int i = 0;
    int index = n-1;
    while (true){
        if (a[index]==i){
            t[index_of_t] = s[index];
            index_of_t++;
        }
        index--;
        if (index == -1){
            index = n-1;
            i++;
        }
        if(i==((n/2)+1)){break;}
    }
    for (int i=0;i<n;i++){
        cout << t[i];
    }
}

**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
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class BalancedShuffleEasy_1970A1 {
    public static void main(String args[]) throws IOException {
        int t, n, arr[], res, count, m;
        char[] in;
        boolean check;
        Temp result[];
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        in = br.readLine().toCharArray();
        count = 0;
        result = new Temp[in.length];
        check = true;

        // creates object temp and initialize elements in the array.
        for (int i = 0; i < in.length; i++) {
            result[i] = new Temp(in[i], count);
            if(in[i] == '(')
                count++;
            else count--;
        }

        // reverse the array so it's already in decreasing position order
        for(int i = 0; i < result.length / 2; i++){
            Temp x = result[i];
            result[i] = result[in.length - 1 - i];
            result[in.length - 1 - i] = x;
        }

        // shift elements by count
        while(check){
            check = false;
            for(int i = 0; i < result.length - 1; i++){
                if(result[i].getCount() > result[i + 1].getCount()){
                    check = true;
                    Temp x = result[i];
                    result[i] = result[i + 1];
                    result[i + 1] = x;
                }

            }
        }

        for(Temp x : result){
            System.out.print(x.getPar());
        }
        System.out.println();

    }
}

class Temp {
    private char parenthesis;
    private int count;

    public Temp (char p, int c){
        parenthesis = p;
        count = c;
    }

    public char getPar(){
        return parenthesis;
    }

    public int getCount(){
        return 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:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can highlight the issues in the sorting and balancing logic. The key points to consider are:

1. **Prefix Balance Handling**: The code should correctly handle the prefix balance for each character in the input string.
2. **Sorting Logic**: The sorting should be done in increasing order of prefix balance, with ties broken in decreasing order of position.

A good test input should challenge the sorting and prefix balance calculation. One such input is a string that has a complex structure with multiple nested parentheses, which can help us verify if the sorting and balance calculation are correct.

Let's consider the input `(()(()))`. This input is a balanced parentheses sequence, and its balanced shuffle should be `()(()())`.

However, to make it more challenging and to expose potential bugs, we can use a more complex input like `(((())))`. The expected output for this input should be `(()(()))`.

Here is the test input:
```plaintext
(((())))
```

This input has a nested structure that should challenge the sorting and balance calculation logic in the code. If the code correctly handles this input, it will produce the expected output `(()(()))`. If it fails, it will help us identify the bug in the sorting or balance calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;

public class BalancedShuffleEasy_1970A1 {
    public static void main(String args[]) throws IOException {
        int arr[], res, count, maxC;
        char[] in;
        StringBuilder result = new StringBuilder();
        boolean check;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // i replaced arrays with lists
        ArrayList<Temp> startL = new ArrayList<>();
        ArrayList<Temp> resL = new ArrayList<>();

        in = br.readLine().toCharArray();
        count = 0;
        check = true;
        maxC = 0;

        // creates object temp and initialize elements in the array.
        for(int i = 0; i < in.length; i++) {
            startL.add(new Temp(in[i], count));

            if(in[i] == '(')
                count++;
            else count--;

            if(maxC < count)
                maxC = count;
        }

        // i need to reverse the list in an efficient way
        Collections.reverse(startL);

//        int nDeleted = 0;
        // iterate for every count
        for(int i = 0; i <= maxC; i++){
            for(int k = 0; k < startL.size(); k++){
                Temp toScrutinize = startL.get(k);
                if(toScrutinize.getCount() == i){
//                    resL.add(toScrutinize);
                    result.append(toScrutinize.getPar());
                    startL.remove(toScrutinize);
                    i--;
                }
            }
        }

        System.out.println(result);

    }
}

class Temp {
    private final char parenthesis;
    private final int count;

    public Temp (char p, int c){
        parenthesis = p;
        count = c;
    }

    public char getPar(){
        return parenthesis;
    }

    public int getCount(){
        return 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
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;

void swap(int *a, int *b)
{
  ll temp = *a;
  *a = *b;
  *b = temp;
}

void merge(int arr[], int low, int mid, int high) {
    vector<int> temp; // temporary array
    int left = low;      // starting index of left half of arr
    int right = mid + 1;   // starting index of right half of arr

    //storing elements in the temporary array in a sorted manner//

    while (left <= mid && right <= high) {
        if (arr[left] <= arr[right]) {
            temp.push_back(arr[left]);
            left++;
        }
        else {
            temp.push_back(arr[right]);
            right++;
        }
    }

    // if elements on the left half are still left //

    while (left <= mid) {
        temp.push_back(arr[left]);
        left++;
    }

    //  if elements on the right half are still left //
    while (right <= high) {
        temp.push_back(arr[right]);
        right++;
    }

    // transfering all elements from temporary to arr //
    for (int i = low; i <= high; i++) {
        arr[i] = temp[i - low];
    }
}

void mergeSort(int arr[], int low, int high) {
    if (low >= high) return;
    int mid = (low + high) / 2 ;
    mergeSort(arr, low, mid);  // left half
    mergeSort(arr, mid + 1, high); // right half
    merge(arr, low, mid, high);  // merging sorted halves
}

int main(){
   string s;
   cin >> s;
   int b[s.size()];
   int sum=0;
   for(int i=0;i<s.size();i++)
   {
      if(s[i]=='(')
      {
         b[i] = sum;
         sum++;
      }
      else
      {
         b[i] = sum;
         sum--;
      }
   }
   int c[s.size()];
   for(int i=0;i<s.size();i++)
   {
      c[i] = b[i];
   }
   mergeSort(c,0,s.size()-1);
   for(int i=s.size()-1;i>=0;i--)
   {
      if(b[i]==c[0]){
        cout << s[i];
      }
   }
   for(int i=1;i<s.size();i++)
   {
      if(c[i]!=c[i-1]){
        //cout << "here " << c[i] << " " << c[i-1] << endl;
        for(int j=s.size()-1;j>=0;j--)
        {
          //cout << b[j] << " " << c[i] << endl;
          if(b[j]==c[i]){
            cout << s[j];
          }
        }
      }
   }

return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the following aspects:

1. **Balance of the Sequence**: The input must be a balanced parentheses sequence.
2. **Edge Cases**: Consider sequences with repeated patterns or specific structures that might cause the sorting logic to fail.
3. **Small Scale**: Ensure the input is small enough to fit within the output limit but large enough to be meaningful.

### Analysis of the Buggy Code
- The code computes the balance of each prefix and stores it in an array `b`.
- It then sorts this array `c` based on the balance values.
- The final output is generated by iterating through the sorted array and printing the corresponding characters from the original string.

### Potential Bugs
- The sorting logic might not correctly handle ties in balance values.
- The final output might not correctly reconstruct the balanced shuffle if the positions are not properly managed.

### Challenging Test Input
A test input that can expose the bug is a balanced parentheses sequence with a specific pattern that might cause the sorting to fail. For example, a sequence with a lot of repeated sub-patterns or a sequence where the balance values are the same for multiple positions.

### Example Test Input
```plaintext
(()(()))
```

This input is small and balanced, but it has a pattern that can cause the sorting logic to fail if the positions are not correctly handled. The expected output is `()(()())`, but the buggy code might not produce this correctly due to the way it handles ties in balance values.

### Verification
- The balance values for the input `(()(()))` are: `0 1 2 1 2 3 2 1`.
- The sorted balance values should be: `0 1 1 1 2 2 2 3`.
- The positions corresponding to these balance values should be correctly mapped to the original string to produce the correct output.

By using this test input, we can verify if the buggy code correctly handles the sorting and reconstruction of the balanced shuffle.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import os
import sys
from io import BytesIO, IOBase

BUFSIZE: int = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

a,l=input(),[]
b=0
for i in range(len(a)):
    l.append(b)
    if i!=0:
        if a[i]=='(':
            b+=1
        else:
            b-=1
    else:
        if a[i]=='(':
            b+=1
        else:
            b-=1
i=0
while i<=max(l):
    for j in range(len(l)-1,-1,-1):
        if l[j]==i:
            print(a[j],end='')
    i+=1

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the code's logic. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code computes the balance of the prefix for each character and then sorts the characters based on the balance and position. The main issue seems to be in the way the balance is computed and how the characters are printed in the sorted order.

2. **Edge Cases**:
   - **Empty String**: Although the problem guarantees a non-empty input, it's good to consider how the code handles minimal cases.
   - **Single Character**: A string with a single character "(" or ")".
   - **Balanced Sequences with Minimal Complexity**: Sequences like "()" or "(())".
   - **Complex Balanced Sequences**: Sequences with multiple levels of nesting and interleaved parentheses, such as "((()))" or "(()())".
   - **Large Number of Nested Parentheses**: Sequences with a large number of nested parentheses, but still within the limit of 500,000 characters.

3. **Challenging Input**: We need a small-scale input that can still expose the bug. A good candidate is a sequence with multiple levels of nested parentheses, as it will test the balance computation and the sorting logic.

4. **Generate the Input**:
   - A sequence like "((()))" has a clear balance and should be straightforward to handle.
   - A sequence like "(()())" has interleaved parentheses, which can be more challenging to handle correctly.

Given these considerations, a small-scale but challenging test input is:

```plaintext
(()())
```

This input has interleaved parentheses and should expose any issues in the balance computation and the sorting logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import os
import sys
from io import BytesIO, IOBase

BUFSIZE: int = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

a,b,l=input(),0,[]
for i in range(len(a)):
    l.append(b)
    if a[i]=='(':
        b+=1
    else:
        b-=1
i=0
while i<=max(l):
    for j in range(len(l)-1,-1,-1):
        if l[j]==i:
            print(a[j],end='')
    i+=1

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

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

Think step by step.

Response:
To create a challenging test input for the given buggy code, we need to focus on edge cases that can expose potential issues in the algorithm. Here are some considerations:

1. **Empty Sequence**: Although the problem guarantees a non-empty input, it's good to consider edge cases.
2. **Single Character**: Input with a single character, either '(' or ')', to see how the code handles it.
3. **Balanced Sequence with Minimum Length**: A sequence with the minimum possible length that is balanced, e.g., "()".
4. **Balanced Sequence with Maximum Depth**: A sequence with the maximum possible depth, e.g., "((()))".
5. **Balanced Sequence with Alternating Characters**: A sequence that alternates between '(' and ')', e.g., "(()())".
6. **Balanced Sequence with Multiple Components**: A sequence that has multiple balanced components, e.g., "()()()".

Given the constraints and the need for a small-scale test input, let's focus on a balanced sequence with multiple components and a moderate depth. This will help us check if the code correctly handles the sorting and rearranging of the characters.

Here is a small-scale test input that fits the criteria:

```plaintext
(()(()()))
```

This input has a balanced sequence with multiple components and a moderate depth, which should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.util.Scanner;
public class CodeForces {
    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        String s=sc.next();
        int n=s.length();
        int balance=0;
        int saldo[]=new int [n];
        int tope=0;
        for(int i=0;i<n;i++){
            saldo[i]=balance;
            if(balance>tope){
                tope=balance;
            }
            if(s.charAt(i)=='('){
                balance++;
            }else{
                balance--;
            }
        }
        String t="";
        for(int i=0;i<=tope;i++){
            for(int j=n-1;j>=0;j--){
                if(saldo[j]==i){
                    t=t+s.charAt(j);
                }
            }
        }
        System.out.println(t);
    }
}

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

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

Think step by step.

Response:
```plaintext
((()))
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Shuffle (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: A parentheses sequence is a string consisting of characters "(" and ")", for example "(()((".

A balanced parentheses sequence is a parentheses sequence which can become a valid mathematical expression after inserting numbers and operations into it, for example "(()(()))".

The balance of a parentheses sequence is defined as the number of opening parentheses "(" minus the number of closing parentheses ")". For example, the balance of the sequence "(()((" is 3.

A balanced parentheses sequence can also be defined as a parentheses sequence with balance 0 such that each of its prefixes has a non-negative balance.

We define the balanced shuffle operation that takes a parentheses sequence and returns a parentheses sequence as follows: first, for every character of the input sequence, we compute the balance of the prefix of the sequence before that character and write those down in a table together with the positions of the characters in the input sequence, for example:

Prefix balance01212321Position12345678Character(()(()))

Then, we sort the columns of this table in increasing order of prefix balance, breaking ties in decreasing order of position. In the above example, we get:

Prefix balance01112223Position18427536Character()(()())

The last row of this table forms another parentheses sequence, in this case "()(()())". This sequence is called the result of applying the balanced shuffle operation to the input sequence, or in short just the balanced shuffle of the input sequence.

You are given a balanced parentheses sequence. Print its balanced shuffle.

Input: The only line of input contains a string $$$s$$$ consisting only of characters "(" and ")". This string is guaranteed to be a non-empty balanced parentheses sequence with its length not exceeding $$$500\,000$$$.

Output: Print the parentheses sequence $$$t$$$ — the balanced shuffle of $$$s$$$.

Examples:
input:
(()(()))
output:
()(()())

Note:

Buggy Code:
import java.util.Scanner;

public class Cp1 {

  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int times = 1;
    while (times-- > 0) {
      String s=scan.next();int n=s.length();
      int []v=new int[n];int max=-1;
      for(int i=1;i<n;i++)
      {
        if(s.charAt(i-1)=='(')
        {
          v[i]=v[i-1]+1;
          if(v[i]>max)max=v[i];
          continue;
        }
        v[i]=v[i-1]-1;
      }
      int []x=new int[n];int ind=0;
      for(int i=0;i<=max;i++)
      {
        for(int j=n-1;j>=0;j--)
        {
           if(v[j]==i)
            x[ind++]=j;

        }
      }
      for(int i=0;i<x.length;i++)
      {
        System.out.print(s.charAt(x[i]));
      }
      System.out.println();

    }

  }
}

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

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

Think step by step.

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

### Analysis of the Buggy Code

1. **Balance Calculation**:
   - The balance is calculated correctly using the array `v`.
   - However, the initial value of `v[0]` is not set, which could lead to incorrect balance values if the first character is `(`.

2. **Sorting and Indexing**:
   - The code sorts the indices based on the balance and position.
   - The sorting logic is implemented using a nested loop, which is correct in logic but could be optimized.

3. **Edge Cases**:
   - **Single Character**: The code should handle a single character input correctly.
   - **Empty Sequence**: The input is guaranteed to be non-empty, so this is not a concern.
   - **Balanced Sequence with Long Runs of Open or Close Parentheses**: This could expose issues in the balance calculation and sorting logic.

### Challenging Test Input

To create a challenging test input, we need a sequence that maximizes the chance of exposing the bug. A sequence with a long run of opening parentheses followed by a long run of closing parentheses can be particularly challenging.

#### Test Input
```plaintext
(((()))))
```

### Explanation

- **Balance Calculation**:
  - The balance array `v` should be `[0, 1, 2, 3, 4, 3, 2, 1, 0]`.
  - The initial value of `v[0]` is 0, which is correct.

- **Sorting and Indexing**:
  - The sorted indices based on balance and position should be `[0, 8, 7, 6, 5, 4, 3, 2, 1]`.
  - This should result in the output `()()()()`.

### Expected Output
```plaintext
()()()()
```

This test input will help verify if the balance calculation and sorting logic are implemented correctly, especially in handling sequences with long runs of parentheses.