**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

Buggy Code:
def balanced_shuffle(s):
    n = len(s)

    # Prepare list to hold the prefix balance, position, and character
    entries = []

    # Calculate prefix balances
    balance = 0
    for i, char in enumerate(s):
        if char == '(':
            balance += 1
        else:  # char == ')'
            balance -= 1
        # Append the current balance AFTER this character, position, and the character itself
        entries.append((balance, i, char))

    # Adjust balance values to reflect the state before the character
    for i in range(len(entries)):
        if entries[i][2] == ')':
            entries[i] = (entries[i][0] + 1, entries[i][1], entries[i][2])
        else:
            entries[i] = (entries[i][0] - 1, entries[i][1], entries[i][2])

    # Sort entries by balance (increasing) and by position (decreasing)
    entries.sort(key=lambda x: (x[0], -x[1]))

    # Build the result using the character from sorted entries
    result = ''.join(entry[2] for entry in entries)

    return result

def main():
    # Read input from standard input
    input_sequence = input().strip()

    # Get the balanced shuffle output
    output_sequence = balanced_shuffle(input_sequence)

    # Output result to standard output
    print(output_sequence)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the problem and the buggy code in detail.

### Analysis of the Buggy Code
1. **Prefix Balance Calculation**: The code correctly calculates the prefix balance for each character in the input string.
2. **Adjust Balance Values**: The code attempts to adjust the balance values to reflect the state before the character. However, this adjustment is incorrect:
   - For `)`, it adds 1 to the balance.
   - For `(`, it subtracts 1 from the balance.
   This is the wrong way to adjust the balance. The correct way should be:
   - For `)`, it should subtract 1 from the balance.
   - For `(`, it should add 1 to the balance.
3. **Sorting**: The code sorts the entries by balance (increasing) and by position (decreasing), which is correct.

### Challenging Test Input
To expose this bug, we need an input where the incorrect balance adjustment leads to a wrong result. A good candidate is an input where the prefix balance values are close to each other, and the positions play a crucial role in the sorting.

### Test Input
Consider the following balanced parentheses sequence:
```plaintext
(()(()))
```

### Expected Output
The expected output for the input `(()(()))` is:
```plaintext
(()(()))
```

### Explanation
1. **Correct Balance Adjustment**:
   - For `(`, the balance before the character is the current balance.
   - For `)`, the balance before the character is the current balance + 1.

   For the input `(()(()))`:
   - Position 1: `(`, balance = 0
   - Position 2: `)`, balance = 1 (0 + 1)
   - Position 3: `(`, balance = 0
   - Position 4: `(`, balance = 1
   - Position 5: `)`, balance = 2 (1 + 1)
   - Position 6: `)`, balance = 2 (1 + 1)
   - Position 7: `)`, balance = 3 (2 + 1)
   - Position 8: `)`, balance = 3 (2 + 1)

   Correct entries:
   - (0, 0, '(')
   - (1, 1, ')')
   - (0, 2, '(')
   - (1, 3, '(')
   - (2, 4, ')')
   - (2, 5, ')')
   - (3, 6, ')')
   - (3, 7, ')')

2. **Sorting**:
   - Sort by balance (increasing) and by position (decreasing):
   - (0, 2, '(')
   - (0, 0, '(')
   - (1, 3, '(')
   - (1, 1, ')')
   - (2, 5, ')')
   - (2, 4, ')')
   - (3, 7, ')')
   - (3, 6, ')')

3. **Result**:
   - `(()(()))`

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

This input will expose the bug in the balance adjustment part of the code, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

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

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        String s=sc.nextLine();
        int []a=new int[s.length()];
        Map<Integer,List<Integer>>mp=new HashMap<>(); //to store balance factor: corresponding indices of array
        a[0]=0;  //1st element
        int i=1;
        char[]ch=s.toCharArray(); //for calculating balance factor
        List<Integer>first=new ArrayList<>(); //to add 1st ele into map
        first.add(0);
        mp.put(0,first);
        for(int j=1;j<s.length();j++)
            {
                if(ch[j-1]=='(') //calculating balance factor each index
                {
                    a[i]=a[i-1]+1;
                }
                else
                {
                    a[i]=a[i-1]-1;

                }
              mp.putIfAbsent(a[i], new ArrayList<>()); // map: ( bf : index list )
            mp.get(a[i]).add(i); //for eg: 1 2 3 index has bf as 5 map looks like (5:1,2,3)
                i++;
            }
        Arrays.sort(a); //sorting bf as asc order
        Set<Integer>st=new HashSet<>(); // to avoid repetition of bf as we are adding every char of balance factor at once in line  41-44
       for(int l:a)
           {
               if(!st.contains(l)) // if bf occurs 1st time we print all indices of it so need to iterate on same bf again
               {

                   List<Integer>ele=mp.get(l); //extracting list of indices
                   Collections.reverse(ele); // as given in the que, indices should be in desc order
                   st.add(l); //adding into set to avoid iterating on same bf again
                   for(int ele2:ele)
                       {
                           System.out.print(s.charAt(ele2)); // printing char at indices
                       }
               }
           }

    }
}

**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 Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

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

public class BalancedUnshuffle {

	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s = br.readLine();
		int[][] a = new int[s.length()][3];
		for(int j=0; j<s.length(); j++) {
			if(s.charAt(j)=='(') a[j][0] = 1;
			else a[j][0] = -1;
			a[j][2] = j;
			if(j!=0) a[j][1] = a[j-1][1]+a[j-1][0];
		}
		Arrays.sort(a, (p,q) -> p[1]!=q[1] ? p[1]-q[1] : q[2]-p[2]);

		for(int j=0; j<s.length(); j++) {
			if(a[j][0]==1) System.out.print("(");
			else System.out.print(")");
		}
	}

}

**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 that can cause the sorting logic to fail. The main issue in the code is in the calculation and usage of the prefix balance. Specifically, the prefix balance is not being calculated correctly, which can lead to incorrect sorting.

Let's consider a small-scale test input that can highlight this issue:

1. **Input with a simple balanced sequence**: This will help verify the basic functionality.
2. **Input with a sequence that has multiple positions with the same prefix balance**: This will test the tie-breaking logic.
3. **Input with a sequence that has a complex structure**: This will test the overall correctness of the algorithm.

A good test input that can expose the bug is a sequence with multiple positions having the same prefix balance and a non-trivial structure. For example, the sequence `(()())` has a prefix balance that is the same for multiple positions, and it is a balanced sequence.

Here is the test input:

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

This input will help verify if the sorting logic correctly handles multiple positions with the same prefix balance and ensures that the output is the correct preimage of the balanced shuffle operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
struct Shuffle
{
	int balance;
	int p;
	char c;
};
bool cmp(Shuffle a,Shuffle b)
{
	if(a.balance==b.balance)
		return a.p>b.p;
	else
	return a.balance<b.balance;
}
int main()
{
	vector<Shuffle> v;
	string s;
	cin>>s;
	int n=0;
	for(int i=0;i<s.size();i++)
	{
		v.push_back({n,i,s[i]});
		if(s[i]=='(')
		n++;
		else
		n--;
	}
	sort(v.begin(),v.end(),cmp);
	for(int i=0;i<v.size();i++)
	{
		cout<<v[i].c;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can highlight issues in the sorting and balance calculation logic. The buggy code sorts the columns based on the prefix balance and position, but it might not handle certain edge cases correctly.

Here are some key points to consider:
1. **Balanced Sequences with Repeated Patterns**: Sequences that have repeated patterns can challenge the sorting logic.
2. **Sequences with Long Runs of Opening or Closing Parentheses**: These can test the balance calculation and sorting logic.
3. **Sequences with Mixed Balances**: Sequences where the balance changes frequently can expose issues in the sorting logic.

Given these points, a small-scale test input that can challenge the code is a balanced parentheses sequence with a mixed balance and repeated patterns. Here is a suitable test input:

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

This input has a mixed balance and repeated patterns, which can help expose any issues in the sorting logic of the buggy code. Let's break down why this input is challenging:

1. **Mixed Balance**: The sequence has a mix of opening and closing parentheses, which can test the balance calculation.
2. **Repeated Patterns**: The sequence "(()(()))" has a repeated pattern "(()" which can challenge the sorting logic.

By using this input, we can ensure that the buggy code's sorting and balance calculation logic is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

Buggy Code:
//#pragma GCC optimize("Ofast,O3,unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define bp __builtin_popcountll
#define mp make_pair
#define pb push_back
#define vec vector
#define ll long long
#define F first
#define S second
#define ld long double
#define pii pair <int, int>
#define elif else if
#define sz(a) int32_t(a.size())
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define fl fflush(stdout)
#define ex exit(0)
#define int ll
#define FOR(i, l, r) for(int i = l; i <= r; ++i)
#define ROF(i, l, r) for(int i = r; i >= l; --i)
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(228355);
mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 10;
ld f[N], x[N]; int n;

bool cmp(pii a, pii b){
    if(a.F < b.F) return 1;
    if(a.F == b.F && a.S > b.S) return 1;
    return 0;
}

string get(string s){
    vector <pii> v;
    v.pb(mp(0, 0));

    int x = 0;
    FOR(i, 0, sz(s)){
        if(i != 0) v.pb(mp(x, i));
        if(s[i] == '(') x++;
        else x--;
    }
    v.pop_back();
    sort(all(v), cmp);

    map <int, int> m;
    for(auto to: v) m[to.F]++;

    string res;
    for(auto to: v){
        res += s[to.S];
    }
    return res;
}

void solve(){
    string s; cin >> s;
    string b = s;
    string last = s;
    FOR(i, 1, 10000){
        s = get(s);
        if(s == b) break;
        last = s;
    }
    cout << last;
}

main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int t = 1;
    //cin >> t;

    while(t--) solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

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

def check(a, t):
    while 1:
        a += t
        if x[a]:
            break
    return a

def calc(s):
    c = 0
    for i in s:
        if i == '(':
            c += 1
        else:
            c -= 1
        if c < 0:
            return 0
    return 1

s = input()[:-1]
n = len(s)
i = i1 = 0
j = j1 = n-1
d = ['']*n
x = [1]*n
while i < j:
    if calc(s[i+1:j]):
        d[i1] = '('
        d[i1+1] = ')'
        i1 += 2
        i = check(i, 1)
        j = check(j, -1)
    elif s[i+1] == ')':
        d[i1] = '('
        d[j1] = ')'
        i1 += 1
        j1 -= 1
        for _ in range(2):
            i = check(i, 1)
    else:
        ii = i
        while 1:
            ii += 1
            if ii == ')':
                x[ii] = 0
                break
        i = check(i, 1)
        d[j1] = ')'
        d[j1-1] = '('
        j1 -= 2
print(''.join(d))

**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 Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

Buggy Code:
import java.util.*;

public class Main {
    static class Pair implements Comparable<Pair> {
        int first, second;

        Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }

        public int compareTo(Pair other) {
            if (this.first != other.first) {
                return Integer.compare(this.first, other.first);
            } else {
                // Sort in decreasing order of position for ties in prefix balance
                return Integer.compare(other.second, this.second);
            }
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.next();
        int n = s.length();
        int[] a = new int[n];
        int k = 0;
        for (int i = 0; i < n; i++) {
            a[i] = k;
            if (s.charAt(i) == '(') {
                k++;
            } else {
                k--;
            }
        }
        List<Pair> v = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            v.add(new Pair(a[i], i));
        }
        Collections.sort(v);
        for (Pair u : v) {
            System.out.print(s.charAt(u.second));
        }
        scanner.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, we need to focus on edge cases where the sorting logic might fail. The key issue in the buggy code is the sorting of pairs based on prefix balance and position. The code sorts in increasing order of prefix balance and in decreasing order of position for ties in prefix balance.

### Edge Case Analysis:
1. **Equal Prefix Balances with Different Positions**: The sorting logic should handle cases where multiple positions have the same prefix balance. The positions should be sorted in decreasing order for these cases.
2. **Balanced Parentheses Sequences with Complex Structure**: Sequences that have nested and interleaved balanced parentheses can be challenging to handle correctly.

### Challenging Test Input:
Let's consider a balanced parentheses sequence that has multiple positions with the same prefix balance and a complex structure. A good candidate is a sequence with nested and interleaved balanced parentheses.

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

### Explanation:
- **Prefix Balances**:
  - Position 1: 0
  - Position 2: 1
  - Position 3: 2
  - Position 4: 1
  - Position 5: 2
  - Position 6: 3
  - Position 7: 2
  - Position 8: 1
  - Position 9: 0

- **Sorted Pairs**:
  - (0, 1)
  - (1, 8)
  - (1, 4)
  - (2, 3)
  - (2, 7)
  - (2, 5)
  - (3, 6)
  - (0, 9)

- **Expected Output**:
  - The sorted pairs should form the sequence: `()(()())`

This test input should expose any issues in the sorting logic, particularly the handling of positions with the same prefix balance. If the code correctly sorts the pairs and reconstructs the sequence, it will produce the correct output. If not, it will likely produce an incorrect sequence, indicating a bug in the sorting logic.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Balanced Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    static class Pair implements Comparable{
        Integer x ;
        Integer y ;
        Integer z ;

        public 	Pair(Integer i, Integer j,Integer k) {
            this.x = i;
            this.y = j;
            this.z = k;
        }
        public int compareTo(Object o) {
            Pair p = (Pair) o ;
            if (this.x.compareTo(p.x) != 0)
                return this.x.compareTo(p.x);
            else
                return -this.y.compareTo(p.y);
        }

        public String toString(){
            return "("+this.x + "," + this.y + ")";
        }
    }

    public static void solve() throws IOException {
        char a [] = sc.next().toCharArray();
        int n = a.length;
        int cnt = 0;
        Pair ans [] = new Pair[n];
        for(int i = 0; i < n ; i++){
            ans[i] = new Pair(cnt,i + 1, a[i] == '(' ? 1 : -1);
            if(a[i] == '(')
                cnt++;
            else
                cnt--;
        }
        Arrays.sort(ans);
        for(Pair p : ans)
            pw.print(p.z == 1 ? "(" : ")");
    }

    public static void main(String[] args) throws IOException {
        int t = 1;
        while(t -- > 0){
            solve();
        }
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

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

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

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

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}

**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 that can highlight issues in the sorting and balancing logic. The code uses a custom `Pair` class to store the balance, position, and character type. It then sorts these pairs based on the balance and, in case of a tie, the position in descending order.

### Key Considerations:
1. **Balance Calculation**: Ensure the balance is calculated correctly.
2. **Sorting Logic**: Ensure the sorting logic correctly handles ties in balance and positions.
3. **Character Type**: Ensure the character type is correctly stored and printed.

### Edge Case Analysis:
1. **Empty Sequence**: Not applicable since the input is guaranteed to be non-empty.
2. **Single Character**: Not applicable since the input is guaranteed to be balanced.
3. **Balanced Sequence with Ties in Balance**:
   - A sequence where multiple characters have the same balance but different positions.
4. **Complex Balanced Sequence**:
   - A sequence that has a complex structure, such as nested and interleaved balanced parentheses.

### Test Input:
A balanced sequence that has multiple characters with the same balance but different positions will help expose the sorting logic. A good candidate is a sequence with nested and interleaved balanced parentheses.

### Chosen Test Input:
```plaintext
(()(()))
```

### Explanation:
- The input `(()(()))` is a balanced sequence.
- The balance and position table for this input is:
  - Prefix balance: 0, 1, 2, 1, 2, 3, 2, 1
  - Positions: 1, 2, 3, 4, 5, 6, 7, 8
  - Characters: (, (, (, ), (, ), ), )

- The sorted table based on the balance and position in descending order will be:
  - Prefix balance: 0, 1, 1, 1, 2, 2, 2, 3
  - Positions: 1, 8, 4, 2, 7, 5, 3, 6
  - Characters: (, ), ), (, ), (, (, ), )

- The resulting sequence should be `(()(()))`.

This input will test the sorting logic and the handling of ties in balance, which is a key part of the algorithm. If the code correctly handles this input, it will produce the correct output. If not, it will expose the bug in 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 Unshuffle (Medium)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The differences with the easy version of this problem are highlighted in bold.

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.

Surprisingly, it turns out that the balanced shuffle of any balanced parentheses sequence is always another balanced parentheses sequence (we will omit the proof for brevity). Even more surprisingly, the balanced shuffles of two different balanced parentheses sequences are always different, therefore the balanced shuffle operation is a bijection on the set of balanced parentheses sequences of any given length (we will omit this proof, too).

You are given a balanced parentheses sequence. Find its preimage: the balanced parentheses sequence the balanced shuffle of which is equal to the given sequence.

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 $$$1\,000$$$.

Output: Print the balanced parentheses sequence $$$t$$$ such that the balanced shuffle of $$$t$$$ is equal to $$$s$$$. It is guaranteed that the answer always exists and is unique.

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

Note:

Buggy Code:
def shuffle(x):
	balance = 0
	items = []

	index = 0
	for c in x:
		items.append( (balance, -index, c) )
		index += 1
		balance += 1 if c == '(' else -1

	items.sort()
	return "".join( [i[2] for i in items] )

class Tree:
	def __init__(self):
		self.children = []
root = Tree()
layer = [root]

pos = [0]
for x in input():
	if x == '(':
		pos[-1] += 1
	else:
		pos.append(0)
j = 0
while layer:
	new_layer = []
	for i in range(len(layer)-1,-1,-1):
		n = layer[i]
		for _ in range(pos[j]):
			t = Tree()
			n.children.append( t )
			new_layer.append(t)
		j += 1
	layer = new_layer
cs = []
def do(t):
	cs.append('(')
	for y in t.children:
		do(y)
	cs.append(')')
do(root)
print( "".join(cs[1:-1] ) )

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases that might cause the code to fail. Here are some key points to consider:

1. **Balanced Parentheses Sequences**: The input must be a balanced parentheses sequence.
2. **Edge Cases**: Short sequences, sequences with a high number of nested parentheses, and sequences with alternating patterns.
3. **Unique Preimage**: The preimage should be unique and valid.

Let's consider a few scenarios:

- **Short Sequences**: These can help identify if the code handles small inputs correctly.
- **Highly Nested Sequences**: These can test if the code correctly handles deep nesting.
- **Alternating Patterns**: These can test if the code correctly handles sequences that alternate between opening and closing parentheses.

Given these points, a challenging test input could be a sequence that has a mix of nesting and alternating patterns. Here is a small-scale test input:

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

This input is balanced and has a mix of nested and alternating patterns, 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: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;

#define ll long long

int main() {
    int t = 0;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        int count = 0;

        vector<int> changed;

        int prev_val = -1;
        for (int i = 0; i < n; i++) {
            int val;
            cin >> val;
            if (val < prev_val) {
                changed.push_back(prev_val - val);
            }
            else {
                prev_val = val;
            }
        }

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

        int prev = 0;

        for (int i = 0; i < changed.size(); i++) {
            count += (changed[i] - prev)*((changed.size() - i) + 1);
            prev = changed[i];
        }

        cout << count << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

// #define ll long long int
#define inf INT_MAX
#define ninf INT_MIN
#define lmax LONG_MAX
#define lmin LONG_MIN
#define vi vector<int>
#define vll vector<long long int>
const int mod=1e9+7;

#define next "\n"
#define srt(x) sort((x).begin(), (x).end())
#define dsrt(x) sort((x).begin(), (x).end(), greater<int>())
#define all(x) (x).begin(), (x).end()

// ============================================= debug code  ============================================= //

int task=0;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define nline "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define sz(x) ((int)(x).size())
// #define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

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

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

// =========================================== debug code ends =========================================== //

void solve(){
    int n;
    cin>>n;

    vi arr(n);
    for(int& i: arr){
        cin>>i;
    }
    int prev_max= arr[0];
    ll coins=0;
    for(int i=1; i<n; i++){
        if(arr[i]>=prev_max){
            prev_max= arr[i];
            continue;
        }

        if(arr[i]+coins>=prev_max){
            continue;
        }

        coins+= 1LL*(prev_max-arr[i]);

    }

    prev_max= arr[0];
    for(int i=1; i<n; i++){
        if(arr[i]>=prev_max){
            prev_max= arr[i];
            continue;
        }
        coins+= 1LL*(prev_max-arr[i]);
    }

    cout<<coins;
}

int main() {

#ifndef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    //for getting input from input.txt
    freopen("input.txt","r", stdin);
    //for writing output to output.txt
    freopen("output.txt","w",stdout);

    //debugger
    freopen("debug.txt", "w", stderr);
#endif

    //solve();return 0;

    int t;
    cin >> t;
    while (t--) {
        task++;
        solve();
        cout<<nline;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

#define ll long long

using namespace std;

void setIO(string name = ""){
	cin.tie(0)->sync_with_stdio(0);
	if ((ll)(name.size())){
		freopen((name + ".in").c_str(), "r", stdin);
		freopen((name + ".out").c_str(), "w", stdout);
	}
}

template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v) { for (auto &it : v) cin >> it; return istream; }
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second << '\n'); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

ll modPow(ll a, ll b, ll m) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % m; a = (a * a) % m; b = b >> 1;} return res;}
bool isPrime(ll n) { if(n < 2) return false; for(ll k = 2; k * k <= n; k++) if(n % k == 0) return false; return true; }

const ll MOD = 1e9+7;

void solve()
{
	int N;
	cin >> N;
	vector<int> A(N);
	cin >> A;
	vector<int> ans2;
	int anc = A[0];
	int ans = 0;
	for (int i=0; i<N; i++){
		if (A[i] < anc){
			// ans += (anc - A[i]);
			// cout << "anc" << ' ' << anc-A[i] << endl;
			ans2.push_back(anc - A[i]);
		}else{
			anc = A[i];
			// cout << "Skip" << endl;
		}
	}
	sort(ans2.begin(), ans2.end());
	int prev = 0;
	for (int i=0; i<ans2.size(); i++){
		ans += ((ans2.size() - i) + 1) *(ans2[i] - prev);
		prev = ans2[i];
	}
	cout << ans << '\n';
}

int main()
{
	setIO("");
	ll T;
	cin >> T;
	// T = 1;
	while (T--)
	{
		solve();
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios where the code might fail. The buggy code has a few potential issues:

1. **Handling of consecutive elements that need to be incremented**: The code might not correctly handle cases where multiple consecutive elements need to be incremented to make the array non-decreasing.
2. **Optimal choice of `k`**: The code might not choose the optimal `k` values to minimize the total cost.
3. **Edge cases with small arrays**: The code might fail for small arrays where the logic might not be thoroughly tested.

Let's construct a test case that maximizes the failure probability:

1. **Consecutive elements that need to be incremented**: We can create an array where several consecutive elements need to be incremented to make the array non-decreasing.
2. **Optimal choice of `k`**: We can create a scenario where the optimal choice of `k` is not straightforward.

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

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

### Explanation:
1. **Test Case 1:**
   - Array: `[4, 3, 2, 1]`
   - This array is in reverse order, and all elements need to be incremented to make it non-decreasing. The optimal solution involves multiple operations, and the code should handle this correctly.

2. **Test Case 2:**
   - Array: `[1, 2, 3, 4, 5]`
   - This array is already non-decreasing, so the expected output is `0`. This tests the basic case where no operations are needed.

3. **Test Case 3:**
   - Array: `[1, 5, 2, 6, 3, 7]`
   - This array has multiple elements that need to be incremented to make it non-decreasing. The optimal solution involves a mix of different `k` values, and the code should handle this correctly.

These test cases should help identify any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
def asd(arr, n):
    if arr == [344, 12, 37, 60, 311, 613, 365, 328, 675]:
        return 1490 + 331
    else:
        mx = arr[0]
        s = 0
        for i in arr:
            if i < mx:
                s += mx - i
            else:
                mx = i
        return s + 1
for _ in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))
    if arr == sorted(arr):
        print(0)
    else:
        print(asd(arr, n))

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

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

Think step by step.

Response:
```plaintext
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
loop = int(input())

for i in range(loop):
    n = int(input())
    a = list(map(int, input().split()))
    store=0
    to_print=0
    first_instant=0

    for j in range(n - 1 ):
        dy=j+1

        if a[j]>a[dy]:
            diff = a[j] - a[dy]
            a[dy] += diff
            store += diff
            to_print+=1
            if to_print==1:
                first_instant+=diff
    if to_print>=1:
        print(store+first_instant)
    else:
        print(0)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
a = []
for i in range(int(input())):
    n = int(input())
    l = list(map(int, input().split()))
    mx = 0
    for i in range(1,n):
        mx = max(mx,l[i-1]-l[i])
    ans = 0
    for i in range(1,n):
        if l[i] < l[i-1]:
            ans+=(l[i-1]-l[i])
            l [i] = l[i-1]
    a.append(ans+mx)
for i in a : print(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
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();

        while(t--!=0){

            int n = sc.nextInt();
            int arr[] = new int[n];
            for(int i=0; i<n; i++){
                arr[i] = sc.nextInt();
            }

         //   boolean b = false;
            int coin = 0;
            int s = -1;
            long sValue = 99999999999999L;
            for(int i=1; i<n; i++){

                if(arr[i] < arr[i-1]){
               ///     b = true;

               if(arr[i] < sValue){

                sValue = arr[i];
                s = i;

            }

               coin += (Math.abs(arr[i]-arr[i-1]));

                    arr[i] = arr[i-1];

                } //else b = false;

            }
            if(coin != 0)
            coin += Math.abs(arr[s]-sValue);
            System.out.println(coin);

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    while(t-->0){
      int k = 0;
      int maxInc = 0;

      int n = sc.nextInt();
      int[] arr = new int[n];
      for(int i = 0; i < n; i++){
        arr[i] = sc.nextInt();
        if(i!=0){
          int inc = arr[i-1] - arr[i];
          if(inc< 0) continue;

          maxInc = Math.max(maxInc, inc);

          k += inc;
          arr[i] = arr[i-1];
        }
      }
      System.out.println(k ==0 ? 0 : k + maxInc);
    }
    sc.close();
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        long long arr[n];
        for(int i=0; i<n; i++){
            cin>>arr[i];
        }
        long long num=arr[0];
        vector<long long> store;

        for(int i=1; i<n; i++){
            if(num>arr[i]){
                store.push_back(num-arr[i]);
            }else num=arr[i];
        }
        sort(store.begin(), store.end());
        int size=store.size();
        int coins=size+1;
        long long ans=0, already=0;
        for(int i=0; i<size; i++){
            int sup=store[i]-already;
            ans+=coins*(sup);
            already+=sup;
            // for(int j=i+1; j<size; j++){
            //     store[j]-=store[i];
            // }
            coins--;
        }
        cout<<ans<<endl;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might fail to correctly calculate the minimum number of coins needed to make the array non-decreasing. The code seems to have issues with handling large differences between elements and the way it calculates the number of coins required.

Let's consider the following test input:

```plaintext
1
5
1 1000000000 1 1000000000 1
```

### Explanation:
1. **Initial Array**: `[1, 1000000000, 1, 1000000000, 1]`
2. **Expected Output**: The minimum number of coins needed to make the array non-decreasing.

### Step-by-Step Analysis:
1. **First Element**: `1` (no action needed)
2. **Second Element**: `1000000000` (no action needed)
3. **Third Element**: `1` (needs to be increased by `999999999` to match the second element)
4. **Fourth Element**: `1000000000` (no action needed)
5. **Fifth Element**: `1` (needs to be increased by `999999999` to match the fourth element)

### Calculation:
- For the third element: `1` needs to be increased by `999999999` (costs `2` coins).
- For the fifth element: `1` needs to be increased by `999999999` (costs `2` coins).

### Total Cost:
- The total number of coins needed is `2 + 2 = 4`.

### Expected Output:
```plaintext
4
```

This test input should expose the bug in the code by requiring a large number of operations to make the array non-decreasing, which the code might not handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

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

  while(t--)
  {
    int n;
    cin >> n;

    int a[n];
    for(int i = 0; i < n; i++)
    {
      cin >> a[i];
    }

    int k = 0 ;

    int big = a[0];

    int ideal[n];

    ideal[0] = a[0];

    for(int i = 1; i < n; i++)
    {
      if(a[i] >= big)
      {
        big = a[i];
      }

      ideal[i] = big;
    }

    for(int i = 0; i < n; i++)
    {
      if(ideal[i] != a[i])
      {
        k = k + 1;
      }
    }

    int diff[k];

    int x = 0;

    for(int i = 0; i < n; i++)
    {
      if(ideal[i] != a[i])
      {
        diff[x] = ideal[i] - a[i];
        x++;
      }
    }

    sort(diff, diff + k);

    long long coins = 0;

    for(int i = 0; i < k; i++)
    {

      int temp = diff[i];
      coins = coins + diff[i]*(k - i + 1);

      for(int j = i; j < k; j++)
      {
        diff[j] = diff[j] - temp;
      }
    }

    cout << coins << 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
1
5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

Buggy Code:
test_cast = int(input())
answerpool = []
for i in range(test_cast):
    lengthofeach = int(input()) -1
    test = list(map(int,input().split()))
    cost = 0
    while True:
        indice = set()
        for i in range(lengthofeach):
            n =i+1
            while test[i] > test[n]:
                indice.add(n)
                n +=1
                if n == lengthofeach +1:
                    break
        if indice == set():
            break
        for i in indice:
            test[i] +=1
        cost += 1+len(indice)
    answerpool.append(cost)
for i in answerpool:
    print(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
3
5
1 2 3 4 5
5
5 4 3 2 1
5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: K-Sort
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array of integers $a$ of length $n$.

You can apply the following operation any number of times (maybe, zero):

  * First, choose an integer $k$ such that $1 \le k \le n$ and pay $k + 1$ coins.    * Then, choose exactly $k$ indices such that $1 \le i_1 < i_2 < \ldots < i_k \le n$.    * Then, for each $x$ from $1$ to $k$, increase $a_{i_x}$ by $1$.

Find the minimum number of coins needed to make $a$ non-decreasing. That is, $a_1 \le a_2 \le \ldots \le a_n$.

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

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

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum number of coins needed to make $a$ non-decreasing.

Examples:
input:
5
3
1 7 9
5
2 1 4 7 6
4
1 3 2 4
1
179
9
344 12 37 60 311 613 365 328 675
output:
0
3
2
0
1821

Note: In the first test case, $a$ is already sorted, so you don't have to spend any coins.

In the second test case, the optimal sequence of operations is:

  * Choose $k = 2$ and the indices $2$ and $5$: $[ 2, \color{red}{1}, 4, 7, \color{red}{6} ] \rightarrow [2, 2, 4, 7, 7]$. This costs $3$ coins.

It can be proven that it is not possible to make $a$ non-decreasing by spending less than $3$ coins.

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

for _ in range(int(input())):
    n=int(input())
    arr=list(map(int,input().split()))
    if n==1:
        print(0)
    else:
        x=0
        maxi=arr[0]
        one,two=0,0
        work=True
        for i in range(1,len(arr)):
            if arr[i]<maxi:
                x+=maxi-arr[i]
                if work:
                    one=maxi
                    two=arr[i]
                    work=False
            else:
                maxi=arr[i]
        print(x+one-two)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios where the logic of the code might fail. The code seems to be trying to calculate the minimum number of coins needed to make the array non-decreasing, but it has some flaws in its logic. Specifically, it might not correctly handle cases where the array has multiple dips and the cost calculation is not accurate.

Here are some steps to think through the test case:

1. **Single Element Array**: This is already handled correctly by the code.
2. **Already Sorted Array**: This is also handled correctly by the code.
3. **Array with a Single Dip**: This should be handled correctly, but we need to ensure the cost calculation is accurate.
4. **Array with Multiple Dips**: This is a more complex case and might expose the bug.
5. **Array with Large Values**: This can help test the edge cases with large values.

Let's create a test case with multiple dips to expose the bug:

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

Explanation:
- The array `5 3 4 6 2 7` has multiple dips: `5 -> 3`, `4 -> 2`.
- The correct solution should be to:
  - Increase `3` to `5` (cost 3 coins).
  - Increase `2` to `6` (cost 4 coins).
  - Total cost: 3 + 4 = 7 coins.

The buggy code might not correctly handle the multiple dips and might miscalculate the total cost. This test case should expose the bug.