**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
# import random
# import itertools
# from sys import stdin, stdout
# import operator
# import collections
import bisect
import collections
import heapq
import math
import sys
from sys import maxsize
# from decimal import Decimal

#sys.setrecursionlimit(10**6)

p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()

# mod = 998244353
# fval = [1]
# for i in range(1,1002):
#     x = (fval[-1]*i)%mod
#     fval.append(x)
# #
# ifac = [1]*(1002)
# ifac[1001] = pow(fval[1001], -1, mod)
# for i in range(1001-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%mod
# print(fval)
# print(ifac)
# _________________ For taking Input from the Text Files __________________
# import os.path
#
# if (os.path.exists('input.txt')):
#     sys.stdin = open("input.txt", "r")
#     sys.stdout = open("output.txt", "w")
#     sys.stderr = open("error.txt", "w")

def drank(d, processing, da, rank):
    tmp = 10 ** 9
    if len(d[da]) == 1:
        return 1
    for di in d[da]:
        if processing[di - 1] == 0:
            processing[di - 1] = 1
            tmp = min(tmp, drank(d, processing, di, rank))
            processing[di - 1] = 0
    rank[da - 1] = tmp + 1
    return tmp + 1

# def primeFactors(n, d):
#     if (n % 2 == 0):
#         d[2] = 0
#     while n % 2 == 0:
#         d[2] += 1
#         n = n // 2
#     for i in range(3, int(math.sqrt(n)) + 1, 2):
#         if (n % i == 0):
#             d[i] = 0
#         while n % i == 0:
#             d[i] += 1
#             n = n // i
#     if n > 2:
#         d[n] = 1
# def powerofadivisor(n,div):
#     x = 0
#     while n%div == 0:
#         n//=div
#         x+=1
#     return x

#import heapq as hq
#import math

def gcd(a,b):
    if b==0:
        x = 1
        y = 0
        return x, y, a
    x, y, g = gcd(b, a%b)
    return y, x- (a//b)*y, g

# mod = 998244353
#
# is_prime = [True]*(10**7+5)
# primes = []
# def primeseive(n):
#     is_prime[0] = is_prime[1] = False
#     for i in range(4,n+1,2): is_prime[i] = False
#     i = 3
#     while i*i<=n:
#         if is_prime[i] == True:
#             for j in range(i*i,n+1,i):
#                 is_prime[j] = False
#         i+=2
    # for i in range(n):
    #     if is_prime[i]: primes.append(i)
import heapq

def solve():
    m, x = li()
    c = li()
    for i in range(1,m):
        c[i]-=x
    v = []
    heapq.heapify(v)
    cval = 0
    ans = 0
    for i in range(1,m):
        cval+= c[i]
        if cval>0:
            if v == []:
                cval-=(x+c[i])
            else:
                if -v[0]+x>=c[i]:
                    cval += (v[0] - x)
                    heapq.heappop(v)
                    heapq.heappush(v,-c[i])

        else:
            if c[i]>0: heapq.heappush(v, -c[i])
            ans+=1
    return ans

def main():
    # primeseive(10**7+2)
    # print(is_prime)
    for _ in range(ii()):
        sys.stdout.write(str(solve()) + "\n")
        #solve()
        # si()
        #print("? {} {}".format(low, mid), flush=True)
        # print(solve(), flush=True)
    # print(solve())

    #     z += str(ans) + '\n'
    # print(len(ans), ' '.join(map(str, ans)), sep='\n')
    # stdout.write(z)

# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided buggy code, let's analyze the problem and the code:

### Problem Analysis:
1. **Monthly Earnings and Happiness Costs:**
   - Charlie earns `x` pounds per month.
   - Each month, he can buy one unit of happiness for `c_i` pounds.
   - Money earned in month `i` can only be spent in a later month `j > i`.

2. **Objective:**
   - Maximize the number of units of happiness Charlie can buy over `m` months.

### Buggy Code Analysis:
- The code attempts to adjust the costs by subtracting `x` from each `c[i]` for `i > 1`.
- It uses a heap to manage the costs and tries to optimize the purchase of happiness units.

### Potential Bugs:
1. **Cost Adjustment:**
   - The code subtracts `x` from `c[i]` for `i > 1`, but it should subtract `x` from `c[i]` for `i >= 1` to account for the first month as well.

2. **Heap Management:**
   - The logic for managing the heap and adjusting the cumulative cost (`cval`) might be flawed, especially when handling the condition `cval > 0`.

### Test Case to Expose the Bug:
To expose the bug, we need a test case where the cost adjustment and heap management logic fails. A small-scale test case that highlights the issue is:

- **Test Case:**
  - `m = 4`
  - `x = 2`
  - `c = [1, 2, 3, 4]`

### Explanation:
- **Month 1:**
  - Cost: 1
  - Earnings: 2
  - Charlie can buy happiness this month.
- **Month 2:**
  - Cost: 2
  - Earnings: 2
  - Charlie can buy happiness this month.
- **Month 3:**
  - Cost: 3
  - Earnings: 2
  - Charlie cannot buy happiness this month.
- **Month 4:**
  - Cost: 4
  - Earnings: 2
  - Charlie cannot buy happiness this month.

### Expected Output:
- The maximum number of units of happiness Charlie can buy is 2.

### Test Input:
```plaintext
1
4 2
1 2 3 4
```

This test case should expose the bug in the cost adjustment and heap management logic. The buggy code might incorrectly handle the costs or the heap, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
def main():
    import sys
    data = sys.stdin.read().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    outputs = []
    for _ in range(t):
        m = int(data[idx])
        x = int(data[idx + 1])
        idx += 2
        c = []
        for _ in range(m):
            c.append(int(data[idx]))
            idx += 1
        # Create list of (cost, month index)
        months = [(c[i], i+1) for i in range(m)]
        # Sort by cost ascending
        months.sort(key=lambda x: x[0])
        balance = 0
        pointer = 1
        happiness = 0
        for cost, i in months:
            if i < pointer:
                continue  # Cannot move back
            # Accumulate earnings from pointer to i
            earnings_to_add = x * (i - pointer)
            balance += earnings_to_add
            pointer = i
            if balance >= cost:
                balance -= cost
                happiness += 1
        outputs.append(str(happiness))
    print('\n'.join(outputs))

if __name__ == '__main__':
    main()

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

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

Think step by step.

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

1. **Edge Cases:**
   - Very low monthly salary.
   - Costs that are very high compared to the monthly salary.
   - Costs that are very close to each other, causing sorting issues.
   - Scenarios where the costs are in a decreasing order but the earnings are insufficient to buy happiness in earlier months.

2. **Scenarios to Test:**
   - A scenario where the monthly salary is 1 and the costs are all 1, but the number of months is large.
   - A scenario where the costs are in a non-increasing order and the monthly salary is just enough to buy happiness in the last few months.
   - A scenario where the costs are in a non-decreasing order and the monthly salary is just enough to buy happiness in the first few months.

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

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

### Explanation:
1. **First Test Case:**
   - **m = 10, x = 1**
   - **c = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]**
   - Charlie earns 1 pound per month and the cost to buy happiness is 1 pound every month. The buggy code should correctly handle this case, but it's a good starting point.

2. **Second Test Case:**
   - **m = 10, x = 1**
   - **c = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]**
   - The costs are in decreasing order. Charlie can only buy happiness in the last few months when he has accumulated enough earnings. The buggy code might fail to correctly handle the pointer and balance management in this scenario.

3. **Third Test Case:**
   - **m = 10, x = 1**
   - **c = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]**
   - The costs are in increasing order. Charlie can only buy happiness in the first few months. The buggy code might fail to correctly handle the pointer and balance management in this scenario.

These test cases should help expose any issues with the pointer and balance management in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
# import random
# import itertools
# from sys import stdin, stdout
# import operator
# import collections
import bisect
import collections
import heapq
import math
import sys
from sys import maxsize
# from decimal import Decimal

#sys.setrecursionlimit(10**6)

p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()

# mod = 998244353
# fval = [1]
# for i in range(1,1002):
#     x = (fval[-1]*i)%mod
#     fval.append(x)
# #
# ifac = [1]*(1002)
# ifac[1001] = pow(fval[1001], -1, mod)
# for i in range(1001-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%mod
# print(fval)
# print(ifac)
# _________________ For taking Input from the Text Files __________________
# import os.path
#
# if (os.path.exists('input.txt')):
#     sys.stdin = open("input.txt", "r")
#     sys.stdout = open("output.txt", "w")
#     sys.stderr = open("error.txt", "w")

def drank(d, processing, da, rank):
    tmp = 10 ** 9
    if len(d[da]) == 1:
        return 1
    for di in d[da]:
        if processing[di - 1] == 0:
            processing[di - 1] = 1
            tmp = min(tmp, drank(d, processing, di, rank))
            processing[di - 1] = 0
    rank[da - 1] = tmp + 1
    return tmp + 1

# def primeFactors(n, d):
#     if (n % 2 == 0):
#         d[2] = 0
#     while n % 2 == 0:
#         d[2] += 1
#         n = n // 2
#     for i in range(3, int(math.sqrt(n)) + 1, 2):
#         if (n % i == 0):
#             d[i] = 0
#         while n % i == 0:
#             d[i] += 1
#             n = n // i
#     if n > 2:
#         d[n] = 1
# def powerofadivisor(n,div):
#     x = 0
#     while n%div == 0:
#         n//=div
#         x+=1
#     return x

#import heapq as hq
#import math

def gcd(a,b):
    if b==0:
        x = 1
        y = 0
        return x, y, a
    x, y, g = gcd(b, a%b)
    return y, x- (a//b)*y, g

# mod = 998244353
#
# is_prime = [True]*(10**7+5)
# primes = []
# def primeseive(n):
#     is_prime[0] = is_prime[1] = False
#     for i in range(4,n+1,2): is_prime[i] = False
#     i = 3
#     while i*i<=n:
#         if is_prime[i] == True:
#             for j in range(i*i,n+1,i):
#                 is_prime[j] = False
#         i+=2
    # for i in range(n):
    #     if is_prime[i]: primes.append(i)
import heapq

def solve():
    m, x = li()
    c = li()
    for i in range(1,m):
        c[i]-=x
    v = []
    heapq.heapify(v)
    cval = 0
    ans = 0
    for i in range(1,m):
        cval+= c[i]
        if cval>0:
            if v == []:
                cval-=(x+c[i])
            else:
                if -v[0]>c[i]:
                    cval += (v[0] - x)
                    heapq.heappop(v)
                    heapq.heappush(v,-c[i])

        else:
            if c[i]>0: heapq.heappush(v, -c[i])
            ans+=1
    return ans

def main():
    # primeseive(10**7+2)
    # print(is_prime)
    for _ in range(ii()):
        sys.stdout.write(str(solve()) + "\n")
        #solve()
        # si()
        #print("? {} {}".format(low, mid), flush=True)
        # print(solve(), flush=True)
    # print(solve())

    #     z += str(ans) + '\n'
    # print(len(ans), ' '.join(map(str, ans)), sep='\n')
    # stdout.write(z)

# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios that are likely to cause issues. The key aspects to focus on are:

1. **Initial Conditions**: Ensure the initial conditions (number of months and monthly salary) are set to values that can trigger edge cases.
2. **Cost Distribution**: The cost of happiness in each month should be distributed in a way that can expose logical errors in the algorithm, such as incorrect handling of costs or incorrect accumulation of savings.

### Step-by-Step Reasoning:

1. **Number of Test Cases**: Start with a small number of test cases to keep the input manageable.
2. **Number of Months (m)**: Choose a value that is large enough to test the algorithm's efficiency and correctness but small enough to fit the output limit.
3. **Monthly Salary (x)**: Set the monthly salary to a value that is not trivially large or small, to ensure the algorithm has to handle savings and costs more carefully.
4. **Cost of Happiness (c)**: Create a sequence of costs that can challenge the algorithm. For example, a sequence where the costs are initially high and then drop, or a sequence with alternating high and low costs.

### Example Test Input:

Let's create a test input with 3 test cases:

1. **Test Case 1**:
   - **m = 5, x = 3**
   - **c = [5, 4, 3, 2, 1]**
   - This case tests a scenario where costs decrease over time, which should allow Charlie to buy more happiness as he saves money.

2. **Test Case 2**:
   - **m = 5, x = 3**
   - **c = [1, 2, 3, 4, 5]**
   - This case tests a scenario where costs increase over time, which should limit the number of happiness units Charlie can buy.

3. **Test Case 3**:
   - **m = 6, x = 2**
   - **c = [2, 1, 3, 4, 2, 1]**
   - This case tests a scenario with alternating high and low costs, which should challenge the algorithm to make optimal decisions.

### Final Test Input:

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

This test input is designed to expose potential issues in the buggy code by testing different cost distributions and ensuring that the algorithm handles savings and costs correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//System.out.println(1<<25);
		//System.out.println(1000000);

		int t = sc.nextint();
		while (t-- > 0) {
			p=false;
			if (t==2711-1703) {
				p=true;
			}
			sovle1();
		}
		out.close();
	}
	static boolean p;
	static Map<Integer, List<Integer>>map;
	static void sovle() {
		int n=sc.nextint();
		String []arr=new String[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i]=sc.next();
		}
		Map<String,Long>map=new HashMap<String, Long>();
		for (int i = 2; i < arr.length; i++) {
			String a = arr[i];
			String b=arr[i-1];
			String c=arr[i-2];
			map.put(a+"+"+b+"+ ", 0l);
			map.put(a+"+ +"+c, 0l);
			map.put(" +"+b+"+"+c, 0l);
			map.put(a+"+"+b+"+"+c, 0l);
			//System.out.println(a+b+c);
		}
		long ans=0;
		for (int i = 2; i < arr.length; i++) {
			String a = arr[i];
			String b=arr[i-1];
			String c=arr[i-2];
			ans+=map.get(a+"+"+b+"+ ")-map.get(a+"+"+b+"+"+c);
			ans+=map.get(" +"+b+"+"+c)-map.get(a+"+"+b+"+"+c);
			ans+=map.get(a+"+ +"+c)-map.get(a+"+"+b+"+"+c);
			map.put(a+"+"+b+"+ ", map.get(a+"+"+b+"+ ")+1);
			map.put(a+"+ +"+c, map.get(a+"+ +"+c)+1);
			map.put(" +"+b+"+"+c, map.get(" +"+b+"+"+c)+1);
			map.put(a+"+"+b+"+"+c, map.get(a+"+"+b+"+"+c)+1);
			//System.out.println(map);
			//System.out.println(ans);
		}
		out.println(ans);
	}
	static String bu(String s) {
		String jiaString="";
		for (int i = 0; i < 7-s.length(); i++) {
			jiaString+="0";
		}
		return jiaString+s;
	}
	static int []dx={0,0,1,-1};
	static int []dy={1,-1,0,0};

	static int max;

	static void sovle1() {
		int n=sc.nextint();
		int x=sc.nextint();
		int []arr=new int[n+1];
		int sum=0;
		Queue<int[]>queue=new PriorityQueue<int[]>((a,b)->a[0]-b[0]);
		for (int i = 1; i < arr.length; i++) {
			arr[i]=sc.nextint();
			int []add=new int[2];
			add[0]=arr[i];add[1]=i;
			queue.add(add);
		}
		c=new long[n+1];
		int ans=0;
		long sheng=1l*(n-1)*x;
		int max=0;
		while (!queue.isEmpty()) {
			int []v=queue.poll();
			if (1l*(v[1]-1)*x-sum(v[1])>=v[0]&&v[1]>=max) {
				//System.out.println(v[1]);
				ans++;add(v[1], v[0]);sheng-=v[0];
				if (1l*(v[1]-1)*x-sum(v[1])-v[0]<v[0]) {
					max=v[1];
				}
			}
		}
		out.println(ans);
	}
	static 	int gcd(int a,int b) {
		return b>0 ? gcd(b,a%b):a;
	}

	static long pow(long a,long b) {
		if (b==0) {
			return 1;
		}else {
			long c=pow(a, b/2);
			if (b%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*a%mod;
			}
		}
	}
	static long mod = 998244353;
	static long[] c;

	static int lowbit(int x) {
		return x & -x;
	}

	static void add(int x, long v) {
		while (x < c.length) {
			// System.out.println(x);
			c[x] += v;
			x += lowbit(x);
		}
	}

	static long sum(int x) {
		long ans = 0;
		while (x > 0) {
			ans += c[x];
			x -= lowbit(x);
		}
		return ans;
	}
}

class kattio extends PrintWriter {
	private BufferedReader r;
	private StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String o) throws IOException {
		super(o);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exceptfion
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}
	public 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 generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Zero Cost for Happiness:** If the cost of happiness is zero for some months, the algorithm should be able to handle it correctly.
2. **High Initial Costs:** If the initial months have very high costs, the algorithm should be able to handle the delay in buying happiness.
3. **Very Low Monthly Salary:** If the monthly salary is very low, the algorithm should be able to handle the constraints properly.
4. **Long Periods with No Affordability:** If there are long periods where Charlie cannot afford any happiness, the algorithm should handle this correctly.
5. **Sudden Drops in Costs:** If the costs suddenly drop after a long period of high costs, the algorithm should be able to take advantage of this.

Let's create a small-scale test input that combines these edge cases:

```plaintext
3
3 1
10 10 1
5 2
10 10 10 10 1
6 3
1 10 10 10 10 1
```

### Explanation:
1. **Test Case 1:**
   - **m = 3, x = 1**
   - **Costs: [10, 10, 1]**
   - Charlie earns 1 pound per month. In the first two months, the cost is 10 pounds, which he cannot afford. In the third month, the cost is 1 pound, which he can afford. The expected output is 1.

2. **Test Case 2:**
   - **m = 5, x = 2**
   - **Costs: [10, 10, 10, 10, 1]**
   - Charlie earns 2 pounds per month. In the first four months, the cost is 10 pounds, which he cannot afford. In the fifth month, the cost is 1 pound, which he can afford. The expected output is 1.

3. **Test Case 3:**
   - **m = 6, x = 3**
   - **Costs: [1, 10, 10, 10, 10, 1]**
   - Charlie earns 3 pounds per month. In the first month, the cost is 1 pound, which he can afford. In the next four months, the cost is 10 pounds, which he cannot afford. In the sixth month, the cost is 1 pound, which he can afford. The expected output is 2.

This test input should expose any issues in the buggy code by testing various edge cases and constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class G {
  static int HAS_TESTCASES = 1;
  static String INPUT = "1";
  class Pair implements Comparable<Pair> {
    int mon;
    int c;
    Pair(int mon, int c) {
      this.mon=mon;
      this.c=c;
    }
    public String toString() {
      return String.format("(%d,%d)", mon, c);
    }
    public int compareTo(Pair p2) {
      if (p2.c!=this.c) {
        return this.c-p2.c;
      }
      return this.mon-p2.mon;
    }
  }
  void solve() {
    int m = ni();
    int x = ni();
    int[] c = new int[m];
    for (int i = 0; i<m; i++) {
      c[i]=ni();
    }
    List<Pair> pairs = new ArrayList<>();
    for (int i = 0; i<m; i++) {
      pairs.add(new Pair(i*x, c[i]));
    }
    Collections.sort(pairs);
    int solution = 0;
    int spent = 0;
    for (int i = 0; i<m; i++) {
      Pair p = pairs.get(i);
      if (p.c+spent<=p.mon) {
        solution++;
        spent+=p.c;
      }
    }
    out.println(solution);
  }
  static int ni() {
    return Integer.parseInt(ns());
  }
  static String ns() {
    t();
    return st.nextToken();
  }
  static void t() {
    while (!st.hasMoreTokens()) {
      try {
        st=new StringTokenizer(br.readLine());
      } catch (Exception e) {
        throw new RuntimeException("out of tokens!");
      }
    }
  }
  static BufferedReader br;
  static StringTokenizer st = new StringTokenizer("");
  static PrintWriter out = new PrintWriter(System.out);
  public static void main(String[] a) {
    if (System.getProperty("ONLINE_JUDGE")==null) {
      System.setIn(new ByteArrayInputStream(INPUT.getBytes()));
    }
    br=new BufferedReader(new InputStreamReader(System.in));
    int t = HAS_TESTCASES>0 ? ni() : 1;
    for (int i = 1; i<=t; i++) {
      new G().solve();
    }
    out.flush();
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

struct segtree {
    struct node {
        ll mn = 0, idx = 0;
        node() {}
        node(int _mn, int _idx) : mn(_mn), idx(_idx) {}
    };
    ll n, N;
    vector<node> st;
    vector<ll> lazy;
    ll l(ll p) {
        return p << 1;
    }
    ll r(ll p) {
        return p << 1 | 1;
    }

    ll unit = 1e9;
    segtree() {}
    segtree(ll sz) : n(sz), N(1 << (__lg(sz) + 1)), st(2 * N), lazy(2 * N, 0) {}

    node conq(node a, node b) {
        if (a.mn < b.mn) return a;
        if (b.mn < a.mn) return b;
        return node(a.mn, min(a.idx, b.idx));
    }

    void build(int p, int L, int R, const vector<ll>& a) {
        if (L == R) {
            st[p] = node(a[L], L);
        } else {
            int m = (L + R) / 2;
            build(l(p), L, m, a);
            build(r(p), m + 1, R, a);
            st[p] = conq(st[l(p)], st[r(p)]);
        }
    }

    void propagate(int p, int L, int R) {
        if (lazy[p] != 0) {
            st[p].mn += lazy[p];
            if (L != R) {
                lazy[l(p)] += lazy[p];
                lazy[r(p)] += lazy[p];
            }
            lazy[p] = 0;
        }
    }

    node query(int p, int L, int R, int i, int j) {
        propagate(p, L, R);
        if (j < L || i > R) return node(unit, 0);
        if (L >= i && R <= j) return st[p];
        int m = (L + R) / 2;
        return conq(query(l(p), L, m, i, j), query(r(p), m + 1, R, i, j));
    }

    void update(int p, int L, int R, int i, int j, ll val) {
        propagate(p, L, R);
        if (j < L || i > R) return;
        if (L >= i && R <= j) {
            lazy[p] += val;
            propagate(p, L, R);
        } else {
            ll m = (L + R) / 2;
            update(l(p), L, m, i, j, val);
            update(r(p), m + 1, R, i, j, val);
            st[p] = conq(st[l(p)], st[r(p)]);
        }
    }
    void update(int i, int j, ll val) {
        update(1, 0, n - 1, i, j, val);
    }
    node query(int i, int j) {
        return query(1, 0, n - 1, i, j);
    }
};

void solve() {
    ll m, x;
    cin >> m >> x;
    vector<ll> c(m);
    for (auto& i : c) cin >> i;
    segtree st(m);
    st.build(1, 0, m - 1, c);
    ll cur = 0, ans = 0;
    for (int i = 0; i < m - 1; i++) {
        cur += x;
        while (true) {
            const auto& q = st.query(i + 1, m - 1);
            if (q.mn <= cur) {
                ans++, cur -= q.mn;
                st.update(q.idx, q.idx, 1e18);
            } else break;
        }
    }
    cout << ans << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to challenge the buggy code by focusing on a small-scale scenario where the monthly salary is 1 pound and the cost of happiness units increases each month. This setup is particularly challenging because:

1. **Monthly Salary is 1 Pound:** Charlie can only save 1 pound per month, which limits the amount of happiness he can buy each month.
2. **Increasing Cost of Happiness:** The cost of happiness increases linearly from 1 to 5 pounds. This means that Charlie will be able to buy happiness only in the first few months when the cost is low.
3. **No Borrowing:** Charlie cannot borrow, so he must save money over multiple months to afford the higher costs.

### Expected Behavior:
- In the first month, Charlie earns 1 pound and can buy 1 unit of happiness (cost = 1).
- In the second month, Charlie earns another 1 pound, but he cannot buy happiness because the cost is 2 pounds.
- In the third month, Charlie has 2 pounds saved and can buy 1 unit of happiness (cost = 3).
- In the fourth month, Charlie earns another 1 pound, but he cannot buy happiness because the cost is 4 pounds.
- In the fifth month, Charlie has 3 pounds saved and can buy 1 unit of happiness (cost = 5).

### Expected Output:
```plaintext
3
```

This test case should expose the bug in the code if it fails to correctly handle the incremental savings and the increasing cost of happiness units.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define oo 100000000000000000
const int mx = 2e5+5;
int c[mx];
void solve() {
	int m,x; cin >> m >> x;
	vector<pair<int,int>> v;
	for(int i = 1; i <= m; ++i) {
		cin >> c[i];
		v.push_back({c[i],i});
	}
	sort(v.begin(),v.end());
	int cost = 0, res = 0;
	for(int i = 0; i < m; ++i) {
		int cur = (v[i].second-1)*x - cost;
		if(cur >= v[i].first) {
			++res;
			cost += v[i].first;
		}
	}
	cout << res;
}

signed main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t; cin >> t;
	while(t--) {
		solve();
		cout << '\n';
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
from collections import defaultdict, Counter
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
import random
import sys
from io import BytesIO, IOBase
import os
RANDOM = random.randrange(2**62)

def Wrapper(x):
  return x ^ RANDOM
"""
TO DO

lexicographically largest (good problem. overcomplicated it for the longest time.
Reminds me of that balloon problem in cp1 im bad at hashmap problems but good to learn from it)
do D1 next
naq problem b (easy to do min length counting distinct strings was hard)
1900 div 1 c dp (couldnt do will go back to it later)
G from last div 3

"""

from collections import defaultdict
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
import random

RANDOM = random.randrange(2**62)

def Wrapper(x):
  return x ^ RANDOM

INF = 10**18
MOD = 9302023

def main():
    # basic idea is to iterate through happiness levels find out which
    # the minimum cost to attain those happiness levels
    # and update states accordingly: dp[i + hp] = (x * i) - dp[j] - cost[i]
    # dp[hp] = min cost to gain hp
    for i in range(int(input())):
        m, x = map(int, input().split())
        cost = [int(x) for x in input().split()]
        hp = []
        heapify(hp)
        money = 0
        for j in range(m):
            if money >= cost[j]:
                money -= cost[j]
                heappush(hp, -cost[j])
            else:
                if hp:
                  c = -heappop(hp)
                  if c >= cost[j]:
                      money += c
                  else:
                      heappush(hp, -c)
                      continue
                  if money >= cost[j]:
                      money -= cost[j]
                      heappush(hp, -cost[j])
                  else:
                      heappush(hp, -c)
            money += x
        print(len(hp))

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = 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")

# endregion

if __name__ == "__main__":

    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think about:

1. **Understand the Bug**: The buggy code uses a heap to manage the costs of happiness units. It attempts to maximize the number of happiness units by always trying to buy the cheapest available unit that can be afforded. However, it might not correctly handle cases where the money available is not sufficient to buy the cheapest unit, leading to incorrect results.

2. **Identify Edge Cases**:
   - **Zero or One Month**: Test cases with very few months to ensure the algorithm handles small inputs correctly.
   - **High Costs**: Test cases where the costs are very high relative to the monthly salary.
   - **Low Salary**: Test cases where the monthly salary is very low.
   - **Uniform Costs**: Test cases where the costs are uniform to see if the algorithm can handle such cases.
   - **Random Costs**: Test cases with random costs to ensure the algorithm can handle varied inputs.

3. **Generate Test Input**:
   - We will create a test input with a small number of months and a low salary to ensure the algorithm is tested under conditions where it might fail due to insufficient funds.

Here is a small-scale test input that is complete and valid:

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

This test input includes:
- A test case with 3 months and a salary of 1 pound, where the costs are all 2 pounds. This should result in 0 units of happiness.
- A test case with 4 months and a salary of 2 pounds, where the costs are 3, 4, 2, and 1 pounds. This should result in 2 units of happiness.
- A test case with 2 months and a salary of 3 pounds, where the costs are 1 and 5 pounds. This should result in 1 unit of happiness.

These cases will help expose any issues in the algorithm's logic for handling insufficient funds and varying costs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
def main():
    import sys
    data = sys.stdin.read().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    outputs = []
    for _ in range(t):
        m = int(data[idx])
        x = int(data[idx + 1])
        idx += 2
        c = []
        for _ in range(m):
            c.append(int(data[idx]))
            idx += 1
        # Create list of (cost, month index)
        months = [(c[i], i+1) for i in range(m)]
        # Sort by cost ascending
        months.sort(key=lambda x: x[0])
        balance = 0
        pointer = 1
        happiness = 0
        for cost, i in months:
            if i < pointer:
                continue  # Cannot move back
            # Accumulate earnings from pointer to i
            earnings_to_add = x * (i - pointer)
            balance += earnings_to_add
            pointer = i
            if balance >= cost:
                balance -= cost
                happiness += 1
        outputs.append(str(happiness))
    print('\n'.join(outputs))

if __name__ == '__main__':
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

public class G_Money_Buys_Less_Happiness_Now {
    static Scanner in = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int testCases = in.nextInt();
        for (int t = 0; t < testCases; ++t) {
            int n = in.nextInt();
            long monthlyEarnings = in.nextLong();
            long[] costs = new long[n];
            for (int i = 0; i < n; ++i) {
                costs[i] = in.nextLong();
            }
            if (t != 2) {
                out.println(maxHappiness(n, monthlyEarnings, costs));
            } else {
                int res = maxHappiness(n, monthlyEarnings, costs);
                out.println(res+1);
            }
        }
        out.close();
    }

    private static int maxHappiness(int months, long monthlyEarnings, long[] costs) {
        PriorityQueue<Long> minHeap = new PriorityQueue<>();
        long availableMoney = 0; // Money available for spending on happiness.
        int count = 0; // Counter for happiness units bought.

        for (int i = 0; i < months - 1; i++) {
            availableMoney += monthlyEarnings; // Add salary to available funds.
            if (availableMoney >= costs[i + 1]) {
                minHeap.add(costs[i]);
                availableMoney -= costs[i + 1];
                count++;
            }
        }

        return count;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        // Modified nextLine() to handle empty lines
        public String nextLine() {
            String str = "";
            while (str.isEmpty()) {
                try {
                    str = br.readLine().trim();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static void main(String[] args) {
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();

            int testCases = in.nextInt();
            while (testCases-- > 0) {
                int m = in.nextInt();
                int x = in.nextInt();
                int[] c = new int[m];
                PriorityQueue<int[]> Q = new PriorityQueue<>(Comparator.comparingInt(a -> a[1]));
                for (int i = 0; i < m; i++) {
                    c[i] = in.nextInt();
                    if (i > 0) {
                        Q.offer(new int[]{i, c[i]});
                    }
                }
                int val = solve(m, x, Q);
                out.println(val);
            }
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static int solve(int n, int x, PriorityQueue<int[]> Q) {
        int i = 1;
        int cnt = 0;
        int amt = x;
        while (i < n) {
            while (!Q.isEmpty() && amt >= Q.peek()[1]) {
                int[] pop = Q.poll();
                if (pop[0] < i) {
                    continue;
                }
                amt -= pop[1];
                cnt++;
            }
            amt += x;
            i++;
        }
        return cnt;
    }
}

**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, we need to consider edge cases and scenarios that might cause the algorithm to fail. The code has a few potential issues:

1. **Early Termination**: The code might terminate early if it doesn't handle the case where the queue is empty correctly.
2. **Incorrect Priority Queue Usage**: The priority queue might not be used correctly to ensure that the smallest cost is always considered first.
3. **Incorrect Money Management**: The code might not correctly manage the money over time, especially when the monthly salary is very small.

Let's create a test input that challenges these aspects:

1. **Small Monthly Salary**: This can cause the code to run out of money quickly, making it harder to buy happiness.
2. **High Costs Early**: This can cause the code to skip opportunities early, which might be crucial for later months.
3. **Low Costs Later**: This can cause the code to miss out on buying happiness when it has more money.

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

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

Explanation:
- **First Test Case**:
  - Monthly salary: 1
  - Costs: 1, 2, 3, 4, 5
  - Charlie can only buy happiness in the first month, as he will not have enough money to buy it in any subsequent months.
  - Expected output: 1

- **Second Test Case**:
  - Monthly salary: 2
  - Costs: 5, 4, 3, 2, 1
  - Charlie can buy happiness in the last three months (3, 2, 1), as he will have enough money by then.
  - Expected output: 3

This test input should help identify if the code correctly manages the money and prioritizes the costs correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

#define int long long

void init(){
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
    #endif //
}

int LIS(int n, vector<int>&cost, vector<int>&sal){
    vector <int> lis;
    lis.push_back(0);
    for(int i=1; i<=n; i++){
        if((lis.back()+cost[i])<=sal[i]){
            lis.push_back(lis.back()+cost[i]);
        }
        else{
            auto it = upper_bound(lis.begin(), lis.end(), sal[i]-cost[i]);
            if(it==lis.begin()){
                // cout << lis.back() << "\n";
                continue;
            }
            else{
                auto it2 = it;
                it2--;
                while(it!=lis.begin()){
                    auto it2 = it;
                    it2--;
                    if((*it2)+cost[i] < (*it)){
                        *it = (*it2)+cost[i];
                        break;
                    }
                    it = it2;
                }
            }
        }
        // cout << lis.back() << "\n";
    }
    return lis.size()-1;
}

void solve(){
    int n,x;
    cin >> n >> x;
    vector <int> v(n+1);
    for(int i=1; i<=n; i++){
        cin >> v[i];
    }
    vector <int> sal(n+1,0);
    for(int i=1; i<=n; i++){
        sal[i] = x*(i-1);
    }
    cout << LIS(n,v,sal) << "\n";
}

signed main(){
    ios::sync_with_stdio(0);cin.tie(0); init();

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

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code seems to be related to the logic for finding the longest increasing subsequence (LIS) of happiness costs that can be afforded. The code incorrectly handles the case where the cost of happiness exceeds the available salary in a way that doesn't always find the optimal solution.

2. **Edge Cases**:
   - **Small Salary**: If the monthly salary is very small, it will be hard to afford the happiness costs, and the algorithm might fail to find the optimal solution.
   - **Large Cost Spikes**: If there are large cost spikes in the middle of the months, the algorithm might incorrectly skip over cheaper opportunities that could be combined to buy more happiness units.
   - **Costs in Decreasing Order**: If the costs are in decreasing order, the algorithm might not correctly handle the upper bound check and miss out on buying happiness units.

3. **Test Input**:
   - We will create a test case with a small salary and costs that have a large spike in the middle, followed by cheaper costs.
   - This will test the algorithm's ability to handle large cost spikes and find the optimal solution.

Here is the test input:

```plaintext
1
6 3
2 2 10 2 2 2
```

**Explanation**:
- **t = 1**: One test case.
- **m = 6, x = 3**: 6 months with a monthly salary of 3 pounds.
- **Costs = [2, 2, 10, 2, 2, 2]**: The costs of happiness units for each month.

**Expected Output**:
- The maximum reachable units of happiness should be 4.

This test case is designed to challenge the algorithm's ability to handle a large cost spike (10) and still find the optimal solution by buying happiness units in the cheaper months.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

*******************************************************************************/

#include <bits/stdc++.h>
using namespace std;
void solve(){
    int m, x;
    cin>>m>>x;
    vector <int> a(m);
    for (int i=0; i<m; i++) cin>>a[i];
    map <int, int> ma;
    ma[0]=0;
    int maxi=0;
    for (int i=0; i<m; i++){
        vector <pair<int, int>> vec;
        for (auto e: ma){
            auto [mon, kol]=e;
            maxi=max(kol, maxi);
            vec.push_back({mon+x, kol});
            if (mon>=a[i]){
                vec.push_back({mon+x-a[i], kol+1});
            }
        }
        ma.clear();
        for (int i=0; i<vec.size(); i++){
            ma[vec[i].first]=max(vec[i].second, ma[vec[i].first]);
        }
    }
    for (auto e: ma){
        auto [mon, kol]=e;
        maxi=max(kol, maxi);
    }
    cout<<maxi;
    /*vector <long long> a(n);
    for (int i=0; i<n; i++) cin>>a[i];*/
    cout<<"\n";
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while (t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases that can cause the algorithm to fail. The code uses a map to keep track of the maximum units of happiness that can be achieved with a given amount of money. The main issues to focus on are:

1. **Initial Conditions**: Ensure the initial conditions are set correctly.
2. **Money Accumulation and Spending**: Check how the code handles the accumulation of money and spending it on happiness.
3. **Edge Cases**: Test with the minimum and maximum values for `m`, `x`, and `c_i`.
4. **Repeating Costs**: Test with repeated costs to see if the algorithm can handle it correctly.

Given these considerations, let's create a small-scale test input that can expose potential bugs:

```plaintext
3
1 1000
1000
2 1000
1000 1000
3 1
1 1 1
```

### Explanation:
1. **Test Case 1**:
   - `m = 1`, `x = 1000`, `c_1 = 1000`
   - Charlie earns 1000 pounds in the first month and can buy one unit of happiness for 1000 pounds.
   - Expected output: 1

2. **Test Case 2**:
   - `m = 2`, `x = 1000`, `c_1 = 1000`, `c_2 = 1000`
   - Charlie earns 1000 pounds in the first month and can buy one unit of happiness for 1000 pounds. In the second month, he earns another 1000 pounds and can buy another unit of happiness for 1000 pounds.
   - Expected output: 2

3. **Test Case 3**:
   - `m = 3`, `x = 1`, `c_1 = 1`, `c_2 = 1`, `c_3 = 1`
   - Charlie earns 1 pound per month. In the first month, he can buy one unit of happiness for 1 pound. In the second month, he earns another pound and can buy another unit of happiness. In the third month, he earns another pound and can buy another unit of happiness.
   - Expected output: 3

These test cases cover a range of scenarios, including the maximum and minimum values for `x` and `c_i`, and repeated costs. They should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
t = int(input())
for i in range(t):
    m,x = map(int,input().split(" "))
    D = list(map(int,input().split(" ")))
    def DP(i,t):
        if i == m:
            return 0
        if t>=D[i]:
            return max(DP(i+1,t+x),DP(i+1,t+x-D[i])+1)
        return DP(i+1,t+x)
    print(DP(0,0))

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

1. **Initial Conditions and Edge Cases:**
   - Test cases with very small values for `m` and `x`.
   - Test cases where `x` is very large compared to the costs `c_i`.
   - Test cases where `x` is very small compared to the costs `c_i`.

2. **Distribution of Costs:**
   - Costs that are uniformly distributed.
   - Costs that are all the same.
   - Costs that increase or decrease linearly.
   - Costs that have a few very high values and a few very low values.

3. **Dynamic Programming Issues:**
   - The provided code does not handle memoization, which can lead to performance issues for large `m`.
   - The code does not consider the possibility of saving money for future months where the cost might be lower.

Given these considerations, let's create a test input that will likely expose the bug:

```plaintext
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
```

This input includes:
- A small number of months with a moderate salary (`3 3`).
- A larger number of months with a moderate salary (`6 5`).
- A larger number of months with a higher salary (`6 4`).
- A small number of months with a very low salary (`2 1`).
- A larger number of months with a very low salary (`4 1`).
- A larger number of months with a moderate salary (`4 2`).

Each of these test cases is designed to challenge different aspects of the algorithm, such as handling small and large values of `m` and `x`, and varying distributions of costs. The provided code might fail to handle these cases efficiently due to the lack of memoization and the recursive nature of the dynamic programming approach.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

Buggy Code:
t = int(input())

for _ in range(t):
    m, x = list(map(int, input().split()))
    c = list(map(int, input().split()))

    def f(start_month=0, happiness_=0, money=0):
        for i in range(start_month, m):
            if c[i] <= money:
                return max(f(i+1, happiness_+1, money+x-c[i]), f(i+1, happiness_, money+x))
            money += x
        return happiness_
    print(f())

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

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

Think step by step.

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

1. **Edge Cases**: Test cases with extreme values of `m` and `x` to see how the code handles large inputs.
2. **Cost Distribution**: Test cases where the cost of happiness is distributed in a way that makes the greedy approach fail.
3. **Boundary Conditions**: Test cases where the cost of happiness is just slightly more than the available money to see if the code correctly handles such scenarios.

Given the constraints:
- \(1 \le m \le 2 \cdot 10^5\)
- \(1 \le x \le 10^3\)
- \(1 \le c_i \le 10^3\)

A small-scale test input that can expose the bug in the provided code is:

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

### Explanation:
1. **Test Case 1:**
   - \(m = 3\), \(x = 3\)
   - \(c = [1, 2, 3]\)
   - This is a simple case where the cost increases linearly. The correct output should be 3, but the buggy code might fail if it doesn't correctly manage the money over the months.

2. **Test Case 2:**
   - \(m = 4\), \(x = 2\)
   - \(c = [1, 3, 2, 4]\)
   - This case has a non-linear cost distribution. The correct output should be 2, but the buggy code might fail if it doesn't consider the optimal sequence of purchases.

3. **Test Case 3:**
   - \(m = 5\), \(x = 1\)
   - \(c = [1, 2, 3, 4, 5]\)
   - This case has a linear cost distribution with a low monthly salary. The correct output should be 1, but the buggy code might fail if it doesn't handle the low salary correctly.

4. **Test Case 4:**
   - \(m = 6\), \(x = 4\)
   - \(c = [1, 10, 1, 10, 1, 10]\)
   - This case has a high cost with low-frequency low-cost opportunities. The correct output should be 3, but the buggy code might fail if it doesn't correctly manage the high cost and low frequency.

These test cases cover a range of scenarios that can help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

public class Main {
    static final int N = 510;
    static long mod = (long) 1e9 + 7;
    static int[] a;

    public static void main(String[] args) throws IOException {
        FastScanner f = new FastScanner();
        PrintWriter w = new PrintWriter(System.out);
        int T = f.nextInt();
        while (T-- > 0) {
            int n = f.nextInt();
            int x = f.nextInt();
            a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = f.nextInt();
            PriorityQueue<Integer> q = new PriorityQueue<>((o1, o2) -> o2 - o1);
            long sum = x, ans = 0;
            for (int i = 2; i <= n; i++) {
                if (sum >= a[i]) {
                    sum -= a[i];
                    q.add(a[i] + x);
                    ans++;
                } else {
                    if (!q.isEmpty() && q.peek() >= a[i] + x) {
                        sum += q.poll() - x - a[i];
                    }
                }
                sum += x;
            }
            w.println(ans);
        }
        w.flush();
    }

    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios where the code's logic might fail. The key aspects to focus on are:

1. **Initial Conditions**: Ensure the initial conditions (like the first month's cost) are set up in a way that can challenge the code.
2. **Edge Cases**: Test with edge cases such as very low or very high costs, and scenarios where the costs are just slightly above or below the available money.
3. **Priority Queue Behavior**: The priority queue is used to manage the costs, so we need to ensure that the code correctly handles the addition and removal of costs.

Given the code's logic, a particularly challenging scenario would be where the costs are not sorted and the code might incorrectly manage the priority queue, leading to suboptimal results.

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

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

Explanation:
1. **Test Case 1**:
   - **m = 5, x = 10**
   - **Costs = [10, 1, 10, 1, 10]**
   - The costs are not sorted, and there are multiple opportunities to buy happiness at a cost of 1. The code should be able to buy happiness in the 2nd and 4th months, but it might fail if it incorrectly manages the priority queue.

2. **Test Case 2**:
   - **m = 5, x = 5**
   - **Costs = [1, 10, 1, 10, 1]**
   - Similar to the first test case, but with a lower monthly salary. The code should be able to buy happiness in the 1st, 3rd, and 5th months, but it might fail if it incorrectly handles the costs.

3. **Test Case 3**:
   - **m = 5, x = 3**
   - **Costs = [3, 1, 3, 1, 3]**
   - The costs are not sorted, and the monthly salary is just enough to buy happiness in the 2nd and 4th months. The code should be able to buy happiness in these months, but it might fail if it incorrectly manages the priority queue.

This test input is designed to challenge the code's ability to correctly manage the priority queue and handle unsorted costs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Money Buys Less Happiness Now
time_limit_ms: 2000
memory_limit_mb: 256
Description: You can never buy enough happiness, so here we go again! In this version, you can only buy $h_i = 1$ unit of happiness each month, but the number of months is hugely increased. We are in the realm of quantum happiness and time dilation.

Being a physicist, Charlie likes to plan his life in simple and precise terms.

For the next $m$ months, starting with no money, Charlie will work hard and earn $x$ pounds per month. For the $i$-th month $(1 \le i \le m)$, there'll be a single opportunity of paying cost $c_i$ pounds to obtain one unit of happiness. You cannot buy more than one unit each month.

Borrowing is not allowed. Money earned in the $i$-th month can only be spent in a later $j$-th month ($j>i$).

Since physicists don't code, help Charlie find the maximum reachable units of happiness.

Input: The first line of the input contains $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains two integers, $m$ and $x$ ($1 \le m \le 2 \cdot 10^5$, $1 \le x \le 10^3$) — the total number of months and the monthly salary.

The second line of each test case contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \leq c_i \leq 10^3$) — the cost of one unit of happiness for each month.

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

Output: For each test case, output one integer — the maximal amount of happiness Charlie can get.

Examples:
input:
6
3 3
2 2 2
6 5
2 2 8 2 6 8
6 4
4 10 3 8 6 10
2 1
1 1
4 1
4 1 3 1
4 2
1 3 4 3
output:
2
4
3
1
2
1

Note:

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

public class Main {
    static final int N = 510;
    static long mod = (long) 1e9 + 7;
    static int[] a;

    public static void main(String[] args) throws IOException {
        FastScanner f = new FastScanner();
        PrintWriter w = new PrintWriter(System.out);
        int T = f.nextInt();
        while (T-- > 0) {
            int n = f.nextInt();
            int x = f.nextInt();
            a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = f.nextInt();
            PriorityQueue<Integer> q = new PriorityQueue<>((o1, o2) -> o2 - o1);
            long sum = x, ans = 0;
            for (int i = 2; i <= n; i++) {
                if (sum >= a[i]) {
                    sum -= a[i];
                    q.add(a[i] + x);
                    ans++;
                } else {
                    if (!q.isEmpty() && q.peek() > a[i] + x) {
                        sum += q.peek() - x - a[i];
                        q.poll();
                    }
                }
                sum += x;
            }
            w.println(ans);
        }
        w.flush();
    }

    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can maximize the failure probability. The key issues in the buggy code are:

1. **Incorrect Initialization of `sum`**: The `sum` variable is initialized to `x` instead of `0`.
2. **Incorrect Handling of `PriorityQueue`**: The `PriorityQueue` is not used correctly to manage the costs and the available money.

Given these issues, we can create a test case that:

- Has a small number of months to make it easy to verify.
- Has costs that are either very high or very low, to test the edge cases of the `PriorityQueue` and the `sum` logic.
- Ensures that the initial `sum` value is not sufficient to buy happiness in the first month, which will highlight the incorrect initialization.

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

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

### Explanation:
- **Test Case**: 1
- **Months**: 3
- **Monthly Salary**: 1
- **Costs**: 2, 1, 1

### Expected Output:
- The correct output should be `2`.

### Why This Test Case Exposes the Bug:
1. **Initial `sum` Value**: The `sum` variable is initialized to `1` (monthly salary), but it should be `0` since Charlie starts with no money.
2. **First Month**: The cost is `2`, which is greater than the initial `sum` of `1`. The code should not be able to buy happiness in the first month, but due to the incorrect initialization, it might incorrectly assume that `sum` is `1` and try to buy happiness.
3. **Second Month**: The cost is `1`, which should be affordable with the new `sum` value (after adding the monthly salary of `1`).
4. **Third Month**: The cost is `1` again, which should also be affordable.

By using this test case, we can see if the buggy code incorrectly buys happiness in the first month, which it should not. This will help identify the issue with the `sum` initialization and the incorrect handling of the `PriorityQueue`.