**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

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

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
# import sys,threading,math,functools
# sys.setrecursionlimit(600000)
# threading.stack_size(10**8)
def primes(n):
    ans=[]
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            ans.append(i)
            while(n%i==0):
                n//=i
        if n==1:
            break
    if n!=1:
        ans.append(n)
    return ans

def countSetBits(n):
    ans=0
    while(n):
        ans+=n%2
        n//=2
    return ans

def factors(n):
    ans=[]
    rev=[]
    for i in range(1,int(n**0.5)+1):
        if n%i==0:
            if i*i==n:
                ans.append(i)
                continue
            ans.append(i)
            rev.append(n//i)

    return ans+rev[::-1]
def factorial(x,mod):
    if x==1 or x==0 or x>=mod:
        return 1
    ans=1
    for i in range(2,x+1):
        ans*=i
        if ans%mod==0:
            return 0
        ans%=mod
    return ans
    return ((x%mod)*factorial(x-1,mod))%mod

def primeFactorsRepeated(number):
    prime_factors=[]
    i=2
    while(i*i<=number and number!=1):
        while(number%i==0):
            # print(number,i)
            prime_factors.append(i)
            number//=i
        i+=1
    if number!=1:
        prime_factors.append(number)
    return prime_factors

def primeFactors(number):
    prime_factors=set()
    i=2
    while(i*i<=number and number!=1):
        while(number%i==0):
            # print(number,i)
            prime_factors.add(i)
            number//=i
        i+=1
    if number!=1:
        prime_factors.add(number)
    return list(prime_factors)

def primes(num):
    prime = [True for i in range(num+1)]
    p = 2
    while (p * p <= num):
        if (prime[p] == True):
            for i in range(p * p, num+1, p):
                prime[i] = False
        p += 1
    return prime
def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den, p - 2, p)) % p
def nonCoPrimes(n):
    res = n
    i = 2
    while i*i<=n:
        if n%i==0:
            res/=i
            res*=(i-1)

            while n%i==0:
                n/=i
        i+=1
    if n>1:
        res/=n
        res*=(n-1)
    return int(res)

def next_permutation(s):
        arr=[i for i in s]
        # print(arr)
        N=len(arr)
        ind = 0
        l = [i for i in arr]
        # step 1 : find the index from right from where changes will happen
        # it is where first the incresing order is breaking from right to left in last
        index=N-1
        while(index>0 and l[index]<=l[index-1]):
            index-=1
        # print(index)
        # now either index is greater than its before element or it is 0
        # if index==0: just reverse the string so wait for step-3
        if index:
            index-=1
            # step-2: find next greatest number from right and swap
            for temp in range(N-1,index,-1):
                if l[temp]>l[index]:
                    break
            # print(temp,index)
            l[temp],l[index]=l[index],l[temp]
            left=index+1
        else:
            left=0
        # print("l: ",l)

        # step-3: reverse the decreasing order to get the smallest string after index
        right=N-1
        while(left<right):
            l[left],l[right]=l[right],l[left]
            left+=1
            right-=1
        # print("l2 : ",l)

        return l

class fenvik_tree:

    def __init__(self,n):
        self.arr=[0]*(n+1)
        self.len=n
    def add(self,ind,v):
        ind+=1
        while(ind<=self.len):
            self.arr[ind]+=v
            ind+= ind & (-ind)
    def query(self,ind):
        ind+=1
        ans=0
        while(ind):
            ans+=self.arr[ind]
            ind-= ind & (-ind)
        return ans

#  python a.py <inputf.in>outputf.in
#  list(map(int,input().split()))
# from heapq import heappush, heappop

import math
from collections import defaultdict
from bisect import bisect_left,bisect_right
mod=10**9+7
for _ in range(int(input())):
    n,c = map(int,input().split())
    arr = list(map(int,input().split()))
    arr[0]+=c
    dp=[arr[i] for i in range(n)]
    maxi,ind=arr[0],0
    for i in range(1,n):
        dp[i]+=dp[i-1]
        if arr[i]>maxi:
            maxi,ind=arr[i],i
    for i in range(n):
        if ind==i:
            print(0)
        elif ind<i:
            print(i)
        else:
            ans=i+1
            if dp[i]>maxi:
                ans=i
            print(ans)

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

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

Think step by step.

Response:
```plaintext
1
3 0
5 5 5
```

### Explanation:
This test case is designed to expose potential issues in the code where all candidates have the same number of fans, and there are no undecided voters. Here’s the reasoning:

1. **Equal Votes for All Candidates:**
   - Each candidate has 5 fans, and there are no undecided voters. This means that initially, each candidate will have 5 votes.
   - The candidate with the lowest number (Candidate 1) will win by default because they all have the same number of votes.

2. **Excluding Candidates:**
   - **For Candidate 1 to win:** No candidates need to be excluded because Candidate 1 already wins.
   - **For Candidate 2 to win:** Candidate 1 must be excluded, so that Candidate 2 can win with 5 votes. This requires excluding 1 candidate.
   - **For Candidate 3 to win:** Both Candidate 1 and Candidate 2 must be excluded, so that Candidate 3 can win with 5 votes. This requires excluding 2 candidates.

3. **Output:**
   - The expected output for this test case should be:
     ```
     0 1 2
     ```

This test case is small and complete, and it should help identify issues in the code where it might not handle equal votes correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
import java.util.*;

public class ProblemD {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int  n = sc.nextInt();
            long c = sc.nextLong();
            long arr[] = new long[n];
            for(int i=0;i<n;i++){
                arr[i] = sc.nextLong();
            }

            arr[0]+=c;

            long maxi = 0;
            long maxIndex = 0;
            for(int i=0;i<n;i++) {
                if(arr[i]>maxi){
                    maxi = arr[i];
                    maxIndex = i;
                }
            }

            long prefixSum = 0;
            for(int i=0;i<n;i++){
                prefixSum+=arr[i];

                if(i>maxIndex)
                    System.out.print(i+" ");
                else if(i==maxIndex)
                    System.out.print(0+" ");
                else{
                    if(prefixSum>maxi)
                        System.out.print(i +" ");
                    else
                        System.out.print((i+1) +" ");
                }
            }

            System.out.println();
        }
    }

}

/*
   5,1,10,7
   if index<maxIndex
    if(prefixSum>max)
      ans = index
    else
       ans = index+1
   else if index>maxIndex
      ans = index+1
   else
      ans = 0

    8,10,7,1
    1,0,2,3

    2,2,2,3,3,3
    1,1
*/

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

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

Think step by step.

Response:
```plaintext
3
3 0
1 1 1
4 0
1 2 2 1
5 0
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

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

public class Vikash_45{
	static FastScanner sc = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int t = sc.nextInt();
	    while(t-->0) {
	    	vk45();
	    	out.flush();
	    }
	    out.close();
	}

/*-----------------------------------------------------------------------------------------------------------------------------*/

	static void vk45() {
		int n = sc.nextInt();
		int c = sc.nextInt();

		long arr[] = sc.readArrayLong(n);
		arr[0]+=c;

		long left[] = new long[n];
		left[n-1] = arr[n-1];
		for(int i=n-2; i>=0; i--) {
			left[i] = Math.max(arr[i],left[i+1]);
		}

//		out.println(Arrays.toString(left));

		long right[] = new long[n];
		right[0] = arr[0];
		for(int i=1; i<n; i++) {
			right[i] = Math.max(arr[i],right[i-1]);
		}

//		out.println(Arrays.toString(right));
		long sum = 0;
		for(int i=0; i<n; i++) {
			sum+=arr[i];
			long l = 0;
			long r = 0;
			if(i-1>=0) {
				if(right[i-1]>=arr[i]) {
					r = 1;
				}
			}
			if(i+1<n) {
				if(left[i+1]>arr[i]) {
					l = 1;
				}
			}

			if(r==0) {
				if(l==0) {
					out.print("0 ");
				}else {
					out.print("1 ");
				}
			}else {
				if(i+1<n && sum<left[i+1]) {
					out.print((i+1)+" ");
				}else {
					out.print(i+" ");
				}

			}
		}

		out.println();

	}

	static class pair implements Comparable<pair>{
		int a;
		int b;
		int c;
		public pair(int a,int b,int c) {
			this.a = a;
			this.b = b;
			this.c = c;
		}
		@Override
		public int compareTo(pair p) {
			if(this.a==p.a) {
				return this.b-p.b;
			}
			return this.a-p.a;
		}
	}

/*-----------------------------------------------------------------------------------------------------------------------------*/

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		long[] readArrayLong(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's analyze the code and identify potential issues:

1. **Initial Addition of Undecided Votes**: The code adds all undecided votes to the first candidate's votes (`arr[0] += c`). This might not always be correct, especially when the first candidate is not the one who will win.

2. **Left and Right Arrays**: The code calculates the maximum votes to the left and right of each candidate. However, it does not correctly handle the scenario where the candidate being considered is the one with the highest votes.

3. **Sum Calculation**: The code sums up all the votes and uses this sum in a condition that might not always be valid.

4. **Output Calculation**: The code's logic for determining the number of candidates to exclude is flawed, especially when the candidate being considered is not the one with the highest votes.

Given these potential issues, we can create a test case that exposes these bugs. Let's consider a scenario where the first candidate has the lowest votes, and the last candidate has the highest votes. This will challenge the initial addition of undecided votes and the logic for determining the number of candidates to exclude.

Here's a small-scale test input that should expose the bugs:

```plaintext
1
3 10
1 2 3
```

### Explanation:
- **Test Case**: There is one test case.
- **Candidates and Undecided Votes**: There are 3 candidates, and there are 10 undecided votes.
- **Votes for Each Candidate**: The first candidate has 1 vote, the second candidate has 2 votes, and the third candidate has 3 votes.

### Expected Output:
- **Candidate 1**: To win, we need to exclude the second and third candidates, so the answer is `2`.
- **Candidate 2**: To win, we need to exclude the third candidate, so the answer is `1`.
- **Candidate 3**: The third candidate already has the most votes, so the answer is `0`.

The expected output for this test case is:
```plaintext
2 1 0
```

This test case should expose the bug in the initial addition of undecided votes and the logic for determining the number of candidates to exclude.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define FOR(a, c) for (int(a) = 0; (a) < (c); (a)++)
#define FORL(a, b, c) for (int(a) = (b); (a) <= (c); (a)++)
#define FORR(a, b, c) for (int(a) = (b); (a) >= (c); (a)--)
typedef long long int ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef pair<int, int> pi;
#define read(v) FOR(i,v.size()){cin>>v[i];}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        int n;
        ll c;
        cin>>n>>c;
        vl a(n);
        read(a);
        ll mx=*max_element(a.begin()+1,a.end());
        ll mxc=max(a[0]+c,mx);
        ll win,sum=c;
        if(mxc==a[0]+c){
            win=0;
        }
        else{
            win=find(a.begin()+1,a.end(),mx)-a.begin();
        }
        FOR(i,n){
            int ans;
            if(i==win){
                ans=0;
            }
            else if(sum+a[i]>=mx){
                ans=i;
            }
            else{
                ans=i+1;
            }
            cout<<ans<<" ";
            sum+=a[i];
        }
        cout<<endl;

    }
    return 0;
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose issues in the buggy code by creating a situation where all candidates have the same number of fans and there are no undecided voters. The expected output for this test case is `0 1 2`, but the buggy code may not handle this scenario correctly due to its logic for determining the winner and the number of candidates to exclude.

1. **Test Case Description:**
   - There is 1 test case.
   - There are 3 candidates, and 0 undecided voters.
   - Each candidate has 1 fan.

2. **Expected Output:**
   - For candidate 1 to win: No candidates need to be excluded (`0`).
   - For candidate 2 to win: Candidate 1 needs to be excluded (`1`).
   - For candidate 3 to win: Candidates 1 and 2 need to be excluded (`2`).

3. **Buggy Code Analysis:**
   - The buggy code may not correctly handle the case where all candidates have the same number of fans.
   - It might incorrectly determine the winner or the number of candidates to exclude.

By using this test case, you can verify if the buggy code correctly handles scenarios where all candidates have the same number of fans and there are no undecided voters.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define int ll
using namespace std;
constexpr int N=2e5+7;
constexpr int M=2e3+7;
int n,c;
int a[N],f[N][20];;
int ans[N];
int presum[N];
int sufmax[N];
int query(int l,int r)
{
	int len=r-l+1;
	int k=log(len)/log(2);
	int res=max(f[l][k],f[r-(1<<k)+1][k]);
	return res;
}
bool check(int mid,int idx)
{
	if(mid==idx) return 0;
	if(query(mid,idx-1)>=a[idx]) return 1;
	//if(presum[mid-1]+c+a[mid]>=a[idx]) return 1;
	return 0;
}
bool check2(int mid,int idx)
{
	if(presum[mid]>a[idx]) return 0;
	if(presum[mid]==a[idx]) return 1;
	return 1;
}
void solve()
{
	cin>>n>>c;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		presum[i]=presum[i-1]+a[i];
	}
	for(int i=1;i<=n;i++)
	{
		presum[i]+=c;
	}
	sufmax[n+1]=0;
	for(int i=n;i>=1;i--)
	{
		sufmax[i]=max(sufmax[i+1],a[i]);
	}
	for(int j=0;j<20;j++)
	{
		for(int i=1;i+(1<<j)-1<=n;i++)
		{
			if(j==0) f[i][j]=a[i];
			else f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
		}
	}

	for(int i=1;i<=n;i++)
	{
		int l=1,r=i;
		// 找小于a[i]的位置
		while(l<r)
		{
			int mid=(l+r)>>1;
			if(check(mid,i)) l=mid+1;
			else r=mid;
		}
		//cout<<"l: "<<l<<endl;

		if(l==i)
		{
			//cout<<l<<' '<<i<<endl;
			ans[i]=i-1;
			//if(i==1) cout<<ans[i]<<"******"<<endl;
		}
		else
		{
			// 找比a[i]小的位置中 [l,i-1] 中小于a[i]的presum的值
			int j=1;
			for(j=l;j<=i-1;j++)
			{
				if(presum[j]<a[i]) break;
				//if(i==3) cout<<"j: "<<j<<endl;
			}
			ans[i]=j-1;
			//if(i==3) cout<<"j: "<<j<<"l: "<<l<<endl;
		}

		int sum=0;
		if(ans[i]==i-1)
		{
			sum=(a[i]+presum[i-1]);
			if(i==1) sum+=c;
		}
		else sum=a[i];
		if(sufmax[i+1]<=sum) continue;
		ans[i]++;
	}

	for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
	cout<<endl;
}
signed main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int T=1; cin>>T;
	while(T--){solve();}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input for the given buggy code, we need to focus on edge cases and scenarios that can expose potential issues in the algorithm. Here are the steps to consider:

1. **Edge Cases for Input Constraints:**
   - Test with the minimum and maximum values for `n` and `c`.
   - Test with zero fans for some candidates.

2. **Complex Scenarios:**
   - Test with candidates having equal fans.
   - Test with a large number of undecided people.
   - Test with a scenario where excluding one candidate changes the winner significantly.

3. **Small-Scale Test Input:**
   - Ensure the test input is small enough to be manageable and complete.

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

```plaintext
1
5 0
1 1 1 1 1
```

### Explanation:
- **Number of Test Cases:** 1
- **Number of Candidates:** 5
- **Number of Undecided People:** 0
- **Fans for Each Candidate:** 1, 1, 1, 1, 1

### Why This Test Input is Challenging:
- **Equal Fans:** All candidates have the same number of fans. This scenario tests how the algorithm handles ties and the rule that the candidate with the lowest number wins.
- **No Undecided People:** This eliminates the effect of undecided voters, focusing on the core logic of candidate exclusion and vote counting.
- **Small Scale:** The input is small enough to be easily manageable and complete, but it still covers a non-trivial case where multiple candidates have the same number of fans.

This test input should help expose any issues in the buggy code, particularly in how it handles ties and the logic for determining the minimum number of candidates to exclude for each candidate to win.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
import bisect
from collections import Counter

for _ in range(int(input())):
    n, c = map(int, input().split())
    a = list(map(int, input().split()))
    ans = []
    a_sort = a.copy()
    a_sort.sort()
    a_sort_pre = [0]
    for i in range(n):
        a_sort_pre.append(a_sort_pre[i - 1] + a[i])
    cc = Counter()
    for i in range(n):
        if i == 0:
            if a[i] + c < a_sort[-1]:
                j = n - 1
                s = a[i] + c
                res = i
                while s < a_sort[j]:
                    s += a_sort[j]
                    j -= 1
                    res += 1
                ans.append(res)
            else:
                ans.append(0)
        elif a[i] == a_sort[-1] and cc[a[i]] == 0 and a[0] + c < a_sort[-1]:
            ans.append(0)
        else:
            if a_sort_pre[i] + c + a[i] < a_sort[-1]:
                j = n - 1
                s = a_sort_pre[i] + c + a[i]
                res = i
                while s < a_sort[j] and j > i:
                    s += a_sort[j]
                    j -= 1
                    res += 1
                ans.append(res)
            else:
                ans.append(i)
        cc[a[i]] += 1
    print(*ans)

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

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

Think step by step.

Response:
```plaintext
3
3 0
1 1 1
4 0
1 2 3 4
5 10
10 20 30 40 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
import sys
from itertools import accumulate

# region fastio
input = lambda: sys.stdin.readline().rstrip()
sint = lambda: int(input())
mint = lambda: map(int, input().split())
ints = lambda: list(map(int, input().split()))
# endregion fastio

# MOD = 998_244_353
# MOD = 10 ** 9 + 7
# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL
# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))

def solve() -> None:
    n, c = mint()
    nums = ints()
    pres = list(accumulate(nums, initial=0))
    ans = [0] * n
    suff = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        suff[i] = max(nums[i], suff[i + 1])
    mx = -1
    for i, x in enumerate(nums):
        if i == 0:
            x += c
        y = x
        if y <= mx:
            ans[i] += i
            y += pres[i] + c
        if y < suff[i + 1]:
            ans[i] += 1
        mx = max(mx, x)
    print(*ans)

for _ in range(int(input())):
    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
1
4 5
5 5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
import java.util.*;

public class Elections {
    public static void main(String[] args) {
        int i = 0;
        List<String> ll = new ArrayList<>();
        Scanner sc = new Scanner(System.in);
        while (true) {
            if (i == 0) {
                i = Integer.valueOf(sc.nextLine()) * 2;
            } else {
                ll.add(sc.nextLine());
                if (--i == 0) {
                    break;
                }
            }
        }
        int p = ll.size();
        for (int k = 0; k < p; k += 2) {
            int c = Integer.valueOf(ll.get(k).split(" ")[1]);
            String[] ss = ll.get(k + 1).split(" ");
            int[] arr = new int[ss.length];
            for (int q = 0; q < ss.length; q++) {
                arr[q] = Integer.valueOf(ss[q]);
            }
            elections(arr, c);
        }
    }

    static void elections(int[] nums, long c) {
        int n = nums.length;
        if (n == 1) {
            System.out.println(0);
            return;
        }
        TreeSet<int[]> set = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; i++) {
            arr[i][0] = nums[i];
            arr[i][1] = i;
            set.add(arr[i]);
        }
        long sum = 0;
        TreeSet<int[]> set1 = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        for (int i = 0; i < n; i++) {
            set.remove(arr[i]);
            if (set.size() == 0) {
                if (arr[i][0] > sum + arr[0][0] + c) {
                    System.out.print(0 + " ");
                } else {
                    System.out.print(i + " ");
                }
                break;
            }
            int[] fir = set.first();
            if (i == 0) {
                if (arr[i][0] + c >= fir[0]) {
                    System.out.print(0 + " ");
                } else {
                    System.out.print(1 + " ");
                }
            } else if (i == 1) {
                if (arr[i][0] >= fir[0]) {
                    if (arr[i][0] <= arr[0][0] + c) {
                        System.out.print(1 + " ");
                    } else {
                        System.out.print(0 + " ");
                    }
                } else {
                    if (arr[i][0] + arr[0][0] + c >= fir[0]) {
                        System.out.print(1 + " ");
                    } else {
                        System.out.print(2 + " ");
                    }
                }
                set1.add(arr[i]);
                sum += arr[i][0];
            } else {
                int[] fir1 = set1.first();
                if (arr[i][0] >= fir[0]) {
                    if (arr[i][0] <= fir1[0] || arr[i][0] <= arr[0][0] + c) {
                        System.out.print(i + " ");
                    } else {
                        System.out.print(0 + " ");
                    }
                } else {
                    if (arr[i][0] + sum + arr[0][0] + c >= fir[0]) {
                        System.out.print(i + " ");
                    } else {
                        System.out.print(i + 1 + " ");
                    }
                }
                set1.add(arr[i]);
                sum += arr[i][0];
            }
        }
        System.out.println();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4 10
10 10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
//  package problems;

import java.util.*;

public class CommonTemplate {
    static class Pair {

        // Pair attributes
        public long index;
        public long value;

        // Constructor to initialize pair
        public Pair(long index, long value)
        {
            // This keyword refers to current instance
            this.index = index;
            this.value = value;
        }
    }

    static int lcmOfTwoNumber(int a,int b){
        // Checking for the largest
        // Number between them
        int ans = Math.max(a, b);

        // Checking for  smallest number that
        // can de divided by both numbers
        while (ans % a != 0 || ans % b != 0) {
            ans++;
        }
        return ans;
    }
    static ArrayList<Integer> primeFactors(int n)
    {
        // Print the number of 2s that divide n
        ArrayList<Integer> l1 = new ArrayList<>();
        while (n % 2 == 0) {
//            System.out.print(2 + " ");
            l1.add(2);
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            // While i divides n, print i and divide n
            while (n % i == 0) {
//                System.out.print(i + " ");
                l1.add(i);
                n /= i;
            }
        }

        // This condition is to handle the case when
        // n is a prime number greater than 2
        if (n > 2)l1.add(n);
//            System.out.print(n);
        return l1;
    }

    static Integer[] sortedIndices(long[] array){
        // Create an array of indices
        Integer[] indices = new Integer[array.length];
        for (int i = 0; i < array.length; i++) {
            indices[i] = i;
        }

        // Sort indices based on the values in the original array
        Arrays.sort(indices, Comparator.comparingLong(a -> array[a]));

        return indices;
    }
    static long gcd(long a, long b)
    {
        return b == 0? a:gcd(b, a % b);
    }

    static long LcmOfArray(int[] arr, int idx)
    {
        if (idx == arr.length - 1){
            return arr[idx];
        }
        long a = arr[idx];
        long b = LcmOfArray(arr, idx+1);
        return (a*b/gcd(a,b)); //
    }
    static int mod(int idx,int n){
        if (idx>=n){
            idx-=n;
        }
        if (idx<0){
            idx+=n;
        }
        return idx;
    }

    private static int countDigits(long l) {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }
    static int[] numToNumArray(long x){
        StringBuilder str = new StringBuilder(String.valueOf(x));
        int[] arr = new int[str.length()];
        for (int i = 0; i < str.length(); i++) {
            int ch = str.charAt(i)-'0';
            arr[i] = ch;
        }
        return arr;
    }
    static int trailingZero(long n){
        int ct = 0;
        while (n%10==0){
            ct++;
            n/=10;

        }
        return ct;
    }
    static Scanner sc = new Scanner(System.in);
    public static void main(String[] args) {
        int q = sc.nextInt();
        loop1:
        while (q-->0){
                        solve();
        }
    }
    public static void solve(){
        int n = sc.nextInt();
        boolean flag1 = true;
        long c = sc.nextLong(), max = Integer.MIN_VALUE;
        long[] arr = new long[n];
        HashMap<Integer,Long> h1=  new HashMap<>();
        for(int i=0; i<n;i++) {
        	arr[i] =  sc.nextLong();
            if (i==0){
                h1.put(i,0L);
            }else {
                h1.put(i,max);
            }
        	max = Math.max(max,arr[i]);

        }
        int idx = 0, jdx = 0;
        for(int i=0; i<n;i++) {
           if (arr[i]==max){
               idx = i;
           }
           if (arr[i]==max&&flag1){
               jdx = i;
               flag1 = false;
           }
        }
        long[] ps = new long[n];
        ps[0] = arr[0];
        for (int i = 1; i <n; i++) {
            ps[i] = ps[i-1]+arr[i];
        }
        long[] brr = new long[n];
        for (int i = 0; i < n; i++) {
            long ct = 0;
            if (i==0||i==jdx){
                if (i==jdx){
                    if (arr[i]>h1.get(i)+c){
                        ct = 0;
                    }else {
                        ct = i;
                    }
                }else {
                    if (ps[i]+c>=max){
                        ct+= i;
                    }else {
                        ct+= i+1;
                    }
                }
            }else {
                if (i<jdx){
                    if (ps[i]+c>=max){
                        ct+= i;
                    }else {
                        ct+= i+1;
                    }
                }else {
                    ct = i;
                }
            }
            brr[i] = ct;
        }
//        System.out.println(h1);
//        System.out.println(idx+" "+jdx);
        for (long e:brr){
            System.out.print(e+" ");
        }
        System.out.println();
    }
    public static int binarySearch(long[] arr, long x) {
        // System.out.println(x);
        int left = 0;
        int right = arr.length - 1;
        int result = -1; // Innitialize result to -1

        while (left <= right) {
            int mid = left + (right - left) / 2;

            // If arr[mid] is less than or equal to x, update result and search in right half
            if (arr[mid] <= x) {
                result = mid;
                left = mid + 1;
            } else {
                // If arr[mid] is greater than x, search in left half
                right = mid - 1;
            }
        }

        return result;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider a few key aspects of the problem:

1. **Handling of Undecided Votes**: The code should correctly account for the undecided votes and how they influence the final vote count.
2. **Candidates with Equal Votes**: The code should handle scenarios where multiple candidates have the same number of votes and the candidate with the lowest index wins.
3. **Excluding Candidates**: The code should correctly determine the minimum number of candidates to exclude for each candidate to win.

Given these points, let's create a test input that includes:
- A candidate with a large number of fans.
- A candidate with a small number of fans.
- A candidate with a moderate number of fans.
- Undecided votes that can tip the balance.

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

```plaintext
1
3 10
5 10 15
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Candidates (n)**: 3
- **Number of Undecided People (c)**: 10
- **Votes for Each Candidate**: [5, 10, 15]

### Expected Output:
- **Candidate 1 (index 0)**: To win, we need to exclude candidate 2 (index 2), who has 15 fans. This will make candidate 1 the highest with 15 votes (5 original + 10 undecided). So, the answer is 1.
- **Candidate 2 (index 1)**: To win, we need to exclude candidate 2 (index 2), who has 15 fans. This will make candidate 2 the highest with 15 votes (10 original + 5 undecided from candidate 1). So, the answer is 1.
- **Candidate 3 (index 2)**: To win, we don't need to exclude any candidate because candidate 3 already has the highest number of votes. So, the answer is 0.

The expected output for this test case is:
```plaintext
1 1 0
```

This test input should help in identifying any issues with the handling of undecided votes and the logic for excluding candidates to ensure the correct candidate wins.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
t = int(input())

while(t):
    n, c = map(int, input().split())
    a = list(map(int, input().split()[:n]))

    if(n == 1):
        print(0)
    elif(n == 2):
        if(a[1] > a[0] + c):
            print(*[1, 0])
        else:
            print(*[0, 1])
    else:
        res = []
        max_val = max(a)

        if(a[0] + c >= max_val):
            res.append(0)
        else:
            res.append(1)

        if(a[1] == max_val and a[0] < max_val):
            res.append(0)
        else:
            if(a[1] + a[0] + c >= max_val):
                res.append(1)
            else:
                res.append(2)

        cur_sum = a[0] + a[1]
        cur_max = max(a[0], a[1])
        for i in range(2, n):
            if(a[i] == max_val and a[i] > cur_max + c):
                res.append(0)
            else:
                to_add = i
                if(cur_sum + c + a[i] >= max_val):
                    res.append(to_add)
                else:
                    res.append(to_add + 1)
            cur_sum += a[i]
            cur_max = max(cur_max, a[i])

        print(*res)

    t -= 1

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

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

Think step by step.

Response:
```plaintext
3
3 0
1 2 2
4 1
1 2 2 2
5 0
1 2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

Buggy Code:
import sys
from itertools import accumulate

# region fastio
input = lambda: sys.stdin.readline().rstrip()
sint = lambda: int(input())
mint = lambda: map(int, input().split())
ints = lambda: list(map(int, input().split()))
# endregion fastio

# MOD = 998_244_353
# MOD = 10 ** 9 + 7
# DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1)) #URDL
# DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))

def solve() -> None:
    n, c = mint()
    nums = ints()
    pres = list(accumulate(nums, initial=0))
    pres[1] += c
    ans = [0] * n
    suff = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        suff[i] = max(nums[i], suff[i + 1])
    pref = [-1] * (n + 1)
    for i, x in enumerate(nums):
        if i == 0:
            x += c
        pref[i + 1] = max(pref[i], x)
        if x <= pref[i] or x < suff[i + 1]:
            ans[i] += i
            x += pres[i]
        if x < suff[i + 1]:
            ans[i] += 1
    s = 0
    stk = []
    for i in range(n - 1, 0, -1):
        while stk and stk[-1] <= nums[i]:
            s -= stk.pop()
        if s + nums[0] + c < nums[i] and pref[i] < nums[i] and len(stk) < ans[i]:
            ans[i] = len(stk)
        stk.append(nums[i])
        s += nums[i]
    print(*ans)

for _ in range(int(input())):
    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
1
3 0
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

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);

    public static void solve() throws IOException {
        int n = sc.nextInt(), c = sc.nextInt();
        if(n == 200000){
            while (true){

            }
//            pw.println();
//                return;
        }
        Integer a [] = sc.nextIntegerArray(n);
        Integer b [] = a.clone();
        int ans [] = new int [n];
        Arrays.sort(b);
        long prefix [] = new long [n + 1];
        for(int i = 1; i <= n ; i++)
            prefix[i] = prefix[i - 1] + b[i - 1];
        long Max = b[n - 1];
        long cnt = 0;
        for(int i = 0; i < n ; i++){
            int low = 1, high = n;
            if(0l + a[i] + cnt + c >= Max){
                ans[i] = i;
            }
            else {
                while (low <= high) {
                    int mid = (low + high) / 2;
                    if (0l + a[i] + prefix[n] - prefix[mid - 1] + cnt + c >= Max)
                        low = mid + 1;
                    else
                        high = mid - 1;
                }
                ans[i] = i + n - high + 1;
            }
            cnt += a[i];
        }
        for(int i = 0 ; i < n ; i++){
            if(a[i] + c >= Max){
                ans[i] = 0;
                break;
            }
        }
        for(int i : ans)
            pw.print(i+" ");
        pw.println();
    }

    public static void main(String[] args) throws IOException{
        int t = sc.nextInt();
        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 generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases that can cause the algorithm to fail. Here are the steps to think through the problem:

1. **Understand the Bug**: The code has a potential infinite loop when `n == 200000`. This suggests that large inputs can cause issues. However, since we need a small-scale test input, we should focus on other potential bugs.

2. **Identify Edge Cases**:
   - **Small Number of Candidates**: Test with a small number of candidates to ensure the logic handles small inputs correctly.
   - **Equal Number of Fans**: Test with candidates having the same number of fans to check how the code handles ties.
   - **Undecided People**: Test with a significant number of undecided people to see how the code handles the distribution of votes.
   - **Single Candidate**: Test with only one candidate to ensure the code handles this edge case correctly.

3. **Construct the Test Input**:
   - **Test Case 1**: Small number of candidates with equal number of fans and some undecided people.
   - **Test Case 2**: Small number of candidates with varying number of fans and no undecided people.
   - **Test Case 3**: Single candidate with some undecided people.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
3 2
1 1 1
4 0
3 2 1 0
1 5
0
```

Explanation:
- **Test Case 1**: 3 candidates with 1 fan each and 2 undecided people. This tests how the code handles equal number of fans and undecided people.
- **Test Case 2**: 4 candidates with varying number of fans and no undecided people. This tests how the code handles different numbers of fans.
- **Test Case 3**: 1 candidate with 5 undecided people. This tests the edge case of a single candidate.

This input should help expose any issues in the code's logic for handling these specific scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Elections
time_limit_ms: 2000
memory_limit_mb: 256
Description: Elections are taking place in Berland. There are $n$ candidates participating in the elections, numbered from $1$ to $n$. The $i$-th candidate has $a_i$ fans who will vote for him. Additionally, there are $c$ people who are undecided about their favorite candidate, let's call them undecided. Undecided people will vote for the candidate with the lowest number.

The candidate who receives the maximum number of votes wins the elections, and if multiple candidates receive the same maximum number of votes, the candidate with the lowest number among them wins.

You found these elections too boring and predictable, so you decided to exclude some candidates from them. If you do not allow candidate number $i$ to participate in the elections, all $a_i$ of his fans will become undecided, and will vote for the candidate with the lowest number.

You are curious to find, for each $i$ from $1$ to $n$, the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win the elections.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $c$ ($1 \le n \le 2 \cdot 10^5$, $0 \le c \le 10^9$) — the number of candidates in the elections and the number of undecided people.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the number of fans for each candidate.

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

Output: For each test case, output $n$ integers, the $i$-th of which should be equal to the minimum number of candidates that need to be excluded from the elections for candidate number $i$ to win.

Examples:
input:
5
3 1
2 0 3
2 3
0 10
5 3
5 4 3 2 1
4 5
3 10 7 1
6 0
2 2 2 3 3 3
output:
0 1 2
1 0
0 1 2 3 4
1 0 2 3
1 1 2 0 4 5

Note: In the first test case:

  * If all candidates are allowed, candidate number $1$ will receive $3$ votes ($1$ undecided person will vote for him), candidate number $2$ will receive $0$ votes, and candidate number $3$ will receive $3$ votes. Therefore, candidate number $1$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $0$.   * If candidate number $1$ is not allowed, his $2$ fans will become undecided. Then candidate number $2$ will receive $3$ votes ($3$ undecided people will vote for him) and candidate number $3$ will receive $3$ votes. Therefore, candidate number $2$ wins (he received the same number of votes as candidate $3$, but his number is lower), so the answer for him is $1$.   * If candidates with numbers $1$ and $2$ are not allowed, candidate number $3$ wins, so the answer for him is $2$.

In the second test case, candidate number $1$ will win if candidate number $2$ is not allowed to participate.

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

public class b2
{
    static long MOD = 998244353;
    public static void main (String[] args) throws IOException {
        FastReader sc = new FastReader();
        int t = sc.nextInt();

        loop: for (int z = 0; z < t; z++) {
            int n = sc.nextInt();
            long c = sc.nextInt();
            long[] a = new long[n];
            long[] sort = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextLong();
                sort[i] = a[i];
            }

            TreeMap<Long, Integer> map = new TreeMap<>();
            Arrays.sort(sort);
            ArrayList<Integer> count = new ArrayList<>();
            count.add(1);
            map.put(sort[0], 0);
            for (int i = 1; i < n; i++) {
                if(sort[i] == sort[i-1]){
                    count.set(count.size()-1, count.get(count.size()-1) + 1);
                }else{
                    map.put(sort[i], count.size());
                    count.add(1);
                }
            }
            map.put(Long.MAX_VALUE, Integer.MAX_VALUE);

            long[] counts = new long[count.size()];
            for (int i = 0; i < count.size(); i++) {
                counts[i] = count.get(i);
            }

//            map.forEach((i,j) -> System.out.println(i + " ; " + j));

            FenwickTree ft = new FenwickTree(counts);

            long sumFromPrev = c;
            long mostPrev = a[0] + c;
            StringBuilder ans = new StringBuilder();
            for (int i = 0; i < n; i++) {
                long kos2 = Integer.MAX_VALUE, kos1 = Integer.MAX_VALUE;
                if(a[i] > mostPrev){
                    int ind = map.higherEntry(a[i]).getValue();
                    if(ind == Integer.MAX_VALUE){
                        kos1 = 0;
                    }else {
                        kos1 = ft.query(ind + 1, n);
                    }
                }

                kos2 = i;
                long votes = a[i] + sumFromPrev;
//                System.out.println("votes = " + votes);
                int ind = map.higherEntry(votes).getValue();
                if(ind != Integer.MAX_VALUE && ft.query(ind + 1, counts.length) > 0){
                    kos2 ++;
                }

                ft.update(map.get(a[i]) + 1, -1);
                sumFromPrev += a[i];
                mostPrev = Math.max(mostPrev, a[i]);
                ans.append(Math.min(kos1, kos2) + " ");
            }

            System.out.println(ans);
        }
    }
    //One Indexed, Also the update just adds and subtracts
    static class FenwickTree {
        int size;
        long ft[];
        public FenwickTree(int n) {
            this.ft = new long[n+1];
            this.size = n;
        }
        public FenwickTree(long[] arr) {
            this(arr.length);
            for (int i = 0; i < arr.length; i++) {
                update(i+1, arr[i]);
            }
        }

        void update(int p, long val) {
            for (; p <= size; p += p&(-p))
                ft[p] += val;
        }

        long query(int p) {
            long sum = 0;
            for (; p > 0; p -= p&(-p)) sum += ft[p];
            return sum;
        }

        long query(int a, int b) {
            return query(b) - query(a-1);
        }
    }
    static final long mod_inverse(long num) {
        return binExp(num, MOD - 2);
    }

    static final long binExp(long base, long exp) {
        base %= MOD;
        long result = 1;
        while (exp > 0) {
            if ((exp & 1) > 0) {
                result *= base;
                result %= MOD;
            }
            base *= base;
            base %= MOD;
            exp >>= 1;
        }
        return result;
    }

    static class FastReader{
        BufferedReader br;
        StringTokenizer st;

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

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

        int nextInt() throws IOException {return Integer.parseInt(next());}
        long nextLong() throws IOException {return Long.parseLong(next());}
        double nextDouble() throws IOException {return Double.parseDouble(next());}

        String nextLine() throws IOException {
            String str = "";
            if(st.hasMoreTokens()){
                str = st.nextToken("\n");
            }
            else{
                str = br.readLine();
            }
            return str;
        }

        int[] intArray(int n) throws IOException {
            int[] arr = new int[n];

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

            return arr;
        }
    }
}

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

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

Think step by step.

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

**Explanation:**

This test case is designed to expose potential issues in the buggy code:

1. **All candidates have 0 fans**: This scenario checks how the code handles cases where no candidate has any initial fans. The undecided voters (which are 0 in this case) will not change the outcome.

2. **No undecided voters**: This ensures that the code must rely solely on the fans of the candidates to determine the winner.

3. **Small number of candidates**: With only 3 candidates, the test case is small enough to be manageable and easy to verify manually.

**Expected Output:**
```plaintext
0 0 0
```

**Reasoning:**
- Since all candidates have 0 fans and there are no undecided voters, no candidates need to be excluded for any candidate to win. Each candidate will have 0 votes, and the candidate with the lowest number will win by default. Therefore, the answer for each candidate is 0.