**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

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

// --------------------------- Defines -------------------------------------  //
template<typename A, typename B>
ostream& operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
}

template <typename Tc,
          typename T = typename enable_if<!is_same<Tc, string>::value,
                                          typename Tc::value_type>::type>
ostream &operator<<(ostream &os, const Tc &v) {
    os << '{';
    for (const T &x : v) os << x << ',';
    return os << '}';
}

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }

#ifdef DEBUG
    #define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
    #define dbg(...)
#endif

#define all(x) (x).begin(), (x).end()
#define forn(i, n) for(int i = 0; i < n; i++)
#define MOD(n) ( ( ((n) % mod) + mod ) % mod)

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vp;
// --------------------------- Constants ----------------------------------- //
const ll LN = 0,
         LM = 0,
         mod = (ll)(1e9) + 7;
// ------------------------- Your code ---------------------------------- //
// clang-format on

void Solve() {
    ll n, k; cin >> n >> k;
    vl vec(n); for (auto& it:vec) cin >> it;
    sort(all(vec));
    ll ans = 0;
    for(ll i = 0; i < n; i++) {
        if (i > 0) vec[i] -= vec[i-1];
        if (k > vec[i] * (n-i)) ans += vec[i] * (n-i) + 1;
        else ans += k;
        k -= vec[i] * (n-i);
        if (k <= 0) break;
    }
    cout << ans << '\n';
}

// clang-format off
// --------------------------------------------------------------------- //
int main() {
    #ifdef DEBUG
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #else
        ios_base::sync_with_stdio(0); cin.tie(0);
    #endif

    int Tc = 1;
    cin >> Tc;

    forn(i, Tc)
        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 focus on edge cases that can highlight the incorrect logic. The buggy code has an issue with how it calculates the number of presses required to guarantee receiving at least `k` cans of lemonade. Specifically, it incorrectly handles the distribution of cans across slots and the subtraction of cans from the total required.

Here are the steps to generate a challenging test input:

1. **Small-scale test case**: Ensure the input is small enough to fit within the output length limit but still complex enough to expose the bug.
2. **Edge cases**: Consider cases where the distribution of cans is uneven, or where the required number of cans `k` is close to the total number of cans available.
3. **Optimal strategy**: The input should force the algorithm to make suboptimal choices if it doesn't handle the logic correctly.

Let's create a test case where the distribution of cans is such that the buggy code's logic will fail to correctly calculate the number of presses.

### Challenging Test Input
```plaintext
1
3 5
1 2 3
```

### Explanation
- **Test case description**: There are 3 slots with 1, 2, and 3 cans of lemonade respectively. We need to guarantee receiving at least 5 cans of lemonade.
- **Correct solution**: The optimal strategy is to press the button corresponding to the slot with 3 cans first, then press the button corresponding to the slot with 2 cans, and finally press the button corresponding to the slot with 1 can. This will ensure we get 5 cans in 3 presses.
- **Buggy code behavior**: The buggy code incorrectly handles the subtraction and distribution of cans, potentially leading to an incorrect number of presses.

This test case should 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: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define lld  long double
#define fn(n) for (ll i = 0; i < n; i++)
#define f(i, x, n) for (ll i = x; i < n; i++)
#define fr(i, x, n) for (ll i = x; i >= n; i--)
#define pb push_back
#define ff first
#define ss second
#define all(v) (v).begin(), (v).end()
#define cyes cout<<"YES"<<"\n"
#define cno cout<<"NO"<<"\n"
#define vi vector<int>
#define vll vector<ll>
#define inf 2147483647
#define msdfan ios_base::sync_with_stdio(false);cin.tie(NULL);
const ll mod = 1e9 + 7;
const ll N = 1e3 + 5;
template <typename T>
void printv(const T &t){{std::copy(t.cbegin(), t.cend(), std::ostream_iterator<typename T::value_type>(std::cout, " "));  }  cout << "\n";}
template <class T>
void cc(T a) { cout << a << "\n"; }
template <class T>
void kk(T a) { cout << a << " "; }

void solve(){
int n,k;
cin >> n >> k;
vi v(n); fn(n) cin >> v[i];

map<int,int> mp;

fn(n) mp[v[i]]++;

int t = k;
int extra = 0;
int prev = 0;

int tot = n;

for(auto i : mp){
    t -= (i.ff - prev)*tot;
    if(t <= 0) break;
    prev += i.ff;
    tot -= i.ss;
    extra += i.ss;
}

cc(k + extra);

}

int32_t main(){
msdfan;
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif

int tt = 1;
cin >> tt;

fn(tt){
    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:
```plaintext
1
3 3
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
r=[]
for i in range(int(input())):
    c=0
    b=0
    d=0
    s,req=map(int,input().split())
    l1=list(map(int,input().split()))
    l1.sort()
    x=len(l1)
    while req>=0 and d<x:
        a=l1[d]
        y=((a-b)*(x-d))
        req-=y
        if req>0:
            c+=(y+1)
        else:
            c+=y
        if d!=0:
            b=l1[d-1]
        d+=1
    if req<0:
        c-=(abs(req))
    r.append(c)
for i in r:
    print(i)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that highlights the issues in the logic. The code seems to have problems with handling the distribution of lemonade cans and the calculation of the required button presses.

### Step-by-Step Analysis:
1. **Initial Setup**: The code initializes variables and reads the input.
2. **Sorting**: The list of lemonade cans in each slot is sorted.
3. **Loop**: The code iterates through the sorted list, calculating the number of button presses needed to get the required number of cans.
4. **Calculation**: The code calculates the number of cans that can be obtained from the current slot and the remaining slots, and adjusts the required number of cans accordingly.
5. **Adjustment**: If the required number of cans becomes negative, the code adjusts the count of button presses.

### Potential Bugs:
1. **Incorrect Calculation of Button Presses**: The code might not correctly calculate the number of button presses needed when the distribution of cans is uneven.
2. **Handling of Zero Cans**: The code might not handle cases where some slots have zero cans correctly.
3. **Edge Cases**: The code might not handle edge cases where the required number of cans is very close to the total number of cans or where the number of slots is very small.

### Test Case to Expose the Bug:
Let's create a test case with a small number of slots and a specific distribution of cans that could expose the bug.

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

### Explanation:
- **Number of Test Cases**: 1
- **Number of Slots (n)**: 4
- **Required Cans (k)**: 6
- **Cans in Slots (a1, a2, a3, a4)**: 1, 2, 2, 3

### Why This Test Case:
- **Uneven Distribution**: The distribution of cans (1, 2, 2, 3) is uneven, which can challenge the logic of the code.
- **Specific Requirement**: The required number of cans (6) is exactly the sum of the cans in the slots, which can test how the code handles the case where all cans need to be taken.
- **Small Scale**: The test case is small enough to be easily analyzed and understood, but complex enough to potentially expose bugs in the code.

This test case should help identify any issues with the logic used to calculate the minimum number of button presses.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
for _ in range(int(input())):
    count,need = list(map(int,input().split()))
    massive = list(map(int,input().split()))
    massive.sort()
    answer = 0
    percent = 0
    for i in range(count):
        if need > 0:
            if massive[i]-percent > 0:
                need = need - (massive[i]-percent)*(count-i)
                answer = answer + (massive[i]-percent)*(count-i)
                percent += (massive[i]-percent)
                if need > 0:
                    answer = answer + 1
            elif massive[i]-percent == 0:
                answer = answer + 1
        else:
            print(answer-abs(need))
            break

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

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

import static java.lang.Math.*;

@SuppressWarnings("unused")
public class Tej {
    public static void main(String[] args) {
        FastIO sc = new FastIO();
        long t = sc.nextLong();
        while (t-- > 0) {
            long n = sc.nextLong();
            long k = sc.nextLong();
            long[] a = new long[(int) n];
            for (int i=0; i<n; i++) a[i] = sc.nextLong();
            Arrays.sort(a);
            LinkedHashMap<Long, Long> mp = new LinkedHashMap<>();
            for (int i=0; i<n; i++) {
                mp.put(a[i], mp.getOrDefault(a[i], 0L) + 1);
            }

            long perpress = n, press = 0, cans = 0, diff = 0;
            Iterator<Long> it = mp.keySet().iterator();
            long ptr = it.next();
            long p=mp.get(ptr);
            for(; true;) {
                press += (ptr)*perpress;
                cans += (ptr)*perpress;

                if(cans>=k) {
                    press -= (cans-k);
                    break;
                }

                press += perpress;
                perpress -= mp.get(ptr);
                cans += perpress;

                if(cans>=k) {
                    press -= (cans-k);
                    break;
                }

                if(it.hasNext()) {
                    ptr = it.next();
                    p += mp.get(ptr);
                } else {
                    break;
                }
            }
            System.out.println(press);
        }
        sc.close();
    }
}

class FastIO extends PrintWriter {
    private InputStream stream;
    private byte[] buf = new byte[1 << 16];
    private int pChar;
    private int numChars;

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

    public FastIO(InputStream i, OutputStream o) {
        super(o);
        stream = i;
    }

    public FastIO(String i, String o) throws IOException {
        super(new FileWriter(o));
        stream = new FileInputStream(i);
    }

    private int nextByte() {
        if (numChars == -1) {
            throw new InputMismatchException();
        }
        if (pChar >= numChars) {
            pChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars == -1) {
                return -1;
            }
        }
        return buf[pChar++];
    }

    public String next() {
        int c;
        do {
            c = nextByte();
        } while (c <= ' ');

        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = nextByte();
        } while (c > ' ');
        return res.toString();
    }

    public int nextInt() {
        int c;
        do {
            c = nextByte();
        } while (c <= ' ');

        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = nextByte();
        }

        int res = 0;
        do {
            if (c < '0' || c > '9') {
                throw new InputMismatchException();
            }
            res = 10 * res + c - '0';
            c = nextByte();
        } while (c > ' ');
        return res * sgn;
    }

    public long nextLong() {
        long c;
        do {
            c = nextByte();
        } while (c <= ' ');

        long sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = nextByte();
        }

        long res = 0;
        do {
            if (c < '0' || c > '9') {
                throw new InputMismatchException();
            }
            res = 10 * res + c - '0';
            c = nextByte();
        } while (c > ' ');
        return res * sgn;
    }

    public double nextDouble() {
        return Double.parseDouble(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:
```plaintext
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            long k = sc.nextLong();
            int no = n;
            long[] a = new long[n];
            long zc = 0;
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextLong();
            }
            Arrays.sort(a);
            for (int j = 0; j < n; j++) {
                long d = ((k - 1) / n) + 1;
                if(a[j] == 1) {
                    a[j] = 0;
                    zc++;
                } else {
                    a[j] = a[j] / d;
                    if(a[j] == 0) {
                        zc++;
                    }
                }
            }
            if(k <= n) System.out.println(k);
            else {
                System.out.println(k + zc);
            }
        }
    }
}

**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
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
for i in range(int(input())):

    n, ned = [int(x) for x in input().split()]
    data = [int(x) for x in input().split()]
    do = 0
    now = 0

    while now < ned:

        ls = len(data)

        k = (ned - now + ls - 1) // ls

        for i in range(ls):

            plas = min(ned - now, k)

            if not plas:
                break

            if data[i] >= plas:
                data[i] -= plas
                do += plas
                now += plas
            else:
                do += data[i] + 1
                now += data[i]
                data[i] = 0

        data = [x for x in data if x]

    if ((1 + n) / 2 * n) == ned and ned >= 6:
        do += 1

    print(do)

**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
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
def rem_zero(q):
    return [x for x in q if x != 0]
def num_presses(n, k, a):
    a.sort()
    if k <= a[0]:
        return k
    elif k <= len(a):
        return k
    else:
        c = 0
        b = a[0]
        while k > 0:
            for i in range(len(a)):
                a[i] -= b
                c += b
                k -= b
                if k == 0:
                    return c
                elif k < 0:
                    return c + k
            c += a.count(0)
            a = rem_zero(a)
        return c
t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    print(num_presses(n, k, a))

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

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

Think step by step.

Response:
```plaintext
1
3 6
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void solve() {
    long long n, k;
    long long ans = 0;
    cin >> n >> k;
    vector<int> a(n);
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    //依次按下每个按钮
    int pre_index = 0;
    int index = 0;
    int wrong_count = 0;
    while(k > 0) {
        ans += wrong_count;
        if(a[index] * (n - index) >= k) {
            ans += k;
            k = 0;
            break;
        }
        for(int i = 0; i < n; i++) {
            a[i]--;
        }

        k -= (n - index);
        ans += (n - index);
        index = upper_bound(a.begin(), a.end(), 0) - a.begin();
        wrong_count = index - pre_index;
        pre_index = index;
    }
    cout << ans << endl;
}
int main() {
    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:
```plaintext
5
2 2
1 1
3 4
1 1 2
4 7
1 2 2 3
10 50
1 1 1 1 1 1 1 1 1 40
2 100
50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
void solve() {
	int n,k;
	cin>>n>>k;
	vector<int>a(n);
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a.begin(),a.end());
	ll l,r;
	l=0,r=1e9;
	auto check=[&](int val)->bool {
		int tmp=0;
		for(int i=0;i<n;i++) {
			tmp+=min(val,a[i]);
		}
		if(tmp<k) return false;
		return true;
	};
	while(l<=r) {
	//	cout<<l<<" "<<r<<'\n';
		ll mid=(l+r)>>1;
		if(check(mid)) {r=mid-1;}
		else {l=mid+1;}
	}
	ll res=r+1,ans=0,m=0;
	for(int i=0;i<n;i++) {
		if(a[i]<res) ans+=a[i]+1,m+=a[i];
		else ans+=res,m+=res;
	}
	cout<<ans-(m-k)<<'\n';
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	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 10
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

Buggy Code:
import java.util.*;
public class ran2 {
   public static void main(String args[]){
      Scanner sc=new Scanner(System.in);
      int t=sc.nextInt();
      while(t-->0){
      int n=sc.nextInt();
      int k=sc.nextInt();
      int arr[]=new int[n];
      for(int i=0;i<n;i++){
         arr[i]=sc.nextInt();
      }
      Arrays.sort(arr);
      int ans=0,count=0;
      for(int i=0;i<n && count!=k; i++){
         int c=arr[i];
         for(int j=i;j<n;j++){
            if(c+count<=k){
               ans+=c;
               count+=c;
               arr[j]=arr[j]-c;
            }
            else{
               int d=k-count;
               ans+=d;
               count+=d;
               break;
            }
         }
         if(arr[i]==0 && count!=k){
            ans+=1;
         }
      }
      System.out.println(ans);
      }
   }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Buying Lemonade
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a vending machine that sells lemonade. The machine has a total of $n$ slots. You know that initially, the $i$-th slot contains $a_i$ cans of lemonade. There are also $n$ buttons on the machine, each button corresponds to a slot, with exactly one button corresponding to each slot. Unfortunately, the labels on the buttons have worn off, so you do not know which button corresponds to which slot.

When you press the button corresponding to the $i$-th slot, one of two events occurs:

  * If there is a can of lemonade in the $i$-th slot, it will drop out and you will take it. At this point, the number of cans in the $i$-th slot decreases by $1$.   * If there are no cans of lemonade left in the $i$-th slot, nothing will drop out.

After pressing, the can drops out so quickly that it is impossible to track from which slot it fell. The contents of the slots are hidden from your view, so you cannot see how many cans are left in each slot. The only thing you know is the initial number of cans in the slots: $a_1, a_2, \ldots, a_n$.

Determine the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Note that you can adapt your strategy during the button presses based on whether you received a can or not. It is guaranteed that there are at least $k$ cans of lemonade in total in the machine. In other words, $k \leq a_1 + a_2 + \ldots + a_n$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $1 \leq k \leq 10^9$) — the number of slots in the machine and the required number of cans of lemonade.

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

It is guaranteed that $k \leq a_1 + a_2 + \ldots + a_n$, meaning there are at least $k$ cans of lemonade in the machine.

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

Output: For each test case, output a single integer — the minimum number of button presses needed to guarantee that you receive at least $k$ cans of lemonade.

Examples:
input:
5
2 1
1 1
2 2
1 2
3 4
2 1 3
10 50
1 1 3 8 8 9 12 13 27 27
2 1000000000
1000000000 500000000
output:
1
2
5
53
1000000000

Note: In the first test case, we can simply press the first button and receive one can of lemonade.

In the second test case, we can press each button once and guarantee that we receive $2$ cans of lemonade. Note that if we simply press one button twice, we might not be lucky, and that button could correspond to the first slot, in which case we would only receive $1$ can of lemonade for two presses.

In the third test case, one of the optimal strategies is as follows:

Press the first button twice. After the first press, a can of lemonade will definitely drop out. Then there are two options:

  * If no can of lemonade drops after the second press, we know that this button must correspond to the second slot, since $a_2 = 1$ and $a_1, a_3 > 1$. Then we can press the second button twice and the third button once. Since $a_1, a_3 \geq 2$, we will definitely receive three cans of lemonade for these three presses. Thus, after $5$ presses, we will have $4$ cans of lemonade.   * If a can of lemonade drops after the second press, we can make one press on the second button and one press on the third button. After each of these presses, we will definitely receive a can of lemonade. Thus, after $4$ presses, we will have $4$ cans of lemonade.

It can be shown that it is impossible to guarantee receiving $4$ cans of lemonade with only $4$ presses, so the answer is $5$.

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

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        // Uncomment these lines to read input from a file and write output to a file
        // BufferedReader br = new BufferedReader(new FileReader("input.txt"));
        // PrintWriter out = new PrintWriter(new FileWriter("output.txt"));

        int tt = Integer.parseInt(br.readLine());

        while (tt-- > 0) {
            String[] input = br.readLine().split(" ");
            int n = Integer.parseInt(input[0]);
            int k = Integer.parseInt(input[1]);
            int subtract = 0;
            int answer = k;

            int[] a = new int[n];
            String[] arrInput = br.readLine().split(" ");
            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(arrInput[i]);
            }
            Arrays.sort(a);

            for (int i = 0; i < n; i++) {
                a[i] -= subtract; // update the number of cans we have
                k -= Math.min(k, (n - i) * a[i]); // subtract the number of cans we can get

                if (k == 0) break;
                else answer++; // we waste a move discarding this slot, that is this slot is empty and we want to know

                subtract += a[i];
            }

            out.println(answer);
        }

        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:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can cause the code to fail. The code seems to have issues with handling the distribution of cans in slots and the number of button presses needed to guarantee receiving a certain number of cans.

Here are some key points to consider:
1. **Slots with varying numbers of cans**: The code might not handle slots with very different numbers of cans correctly.
2. **High number of required cans**: The code might fail when the required number of cans is close to the total number of cans.
3. **Multiple slots with the same number of cans**: The code might not handle multiple slots with the same number of cans efficiently.

Given these considerations, let's create a test input that includes these edge cases.

### Test Input
```plaintext
1
5 10
1 2 3 4 5
```

### Explanation
- **Number of test cases**: 1
- **Number of slots (n)**: 5
- **Required cans (k)**: 10
- **Initial number of cans in each slot**: 1, 2, 3, 4, 5

### Why This Test Input is Challenging
1. **Varying number of cans**: The slots have different numbers of cans, which can test how the code handles distributing presses across slots.
2. **High number of required cans**: The required number of cans (10) is close to the total number of cans (15), which can test the code's ability to ensure all required cans are obtained.
3. **Optimal strategy**: The optimal strategy involves pressing buttons in a way that maximizes the number of cans obtained with the minimum number of presses. This can expose any inefficiencies in the code's logic.

By using this test input, we can observe if the code correctly calculates the minimum number of button presses needed to guarantee receiving at least 10 cans of lemonade.


**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:
// #pragma GCC target( "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <queue>
#include <numeric>
#include <bitset>
#include <stdio.h>
#include <time.h>
#include <assert.h>
#include <random>
#include <chrono>
// #include <bits/stdc++.h>

// #define int long long
#define pb push_back
#define f first
#define s second
#define ndl '\n'
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ryoiki_tenkai ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define muryokusho srand(time(0));

const double pi = 3.14159265359;
const int mod = 998244353;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const long long inf = 1ll<<62;
const int P = 31, P1 = 37;
const int maxn = 200200;

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int a[200200];
int t[800800];
int n, k;
void upd(int pos, int x, int v = 1, int tl = 1, int tr = n) {
    if(tl == tr) {
        t[v] = x;
    } else {
        int mid = (tl + tr)>>1;
        if(pos <= mid) {
            upd(pos, x, v * 2, tl, mid);
        } else {
            upd(pos, x, v * 2 + 1, mid + 1, tr);
        }
        t[v] = t[v * 2] + t[v * 2 + 1];
    }
}
int get(int r, int l = 1, int v = 1, int tl = 1, int tr = n) {
    if(tl > r || tr < l) return 0;
    if(tl >= l && tr <= r) return t[v];
    int mid = (tl + tr)>>1;
    return get(r, l, v * 2, tl, mid) + get(r, l, v * 2 + 1, mid + 1, tr);
}
void infinite() {
    cin>>n>>k;
    for(int i=0;i<=n*4;i++) t[i] = 0;
    multiset<pair<int, pair<int, int> > > st;
    int sum = 0;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
        st.insert({a[i], {sum, i}});
        sum += k;
    }
    int ans = 0;
    set<pair<int, int> > check;
    check.insert({n+1, 2e9});
    while(st.size()) {
        pair<int, pair<int, int> > dx = *st.begin();
        st.erase(st.begin());
        pair<int, int> tek = *(check.lower_bound({dx.s.s, 0}));
        if(get(dx.s.s) + dx.f <= min(tek.s, dx.s.f)) {
            ans++;
            upd(dx.s.s, dx.f);
            check.insert({dx.s.s, (min(tek.s, dx.s.f)) - (get(dx.s.s) + dx.f)});
        }
    } cout<<ans<<ndl;
}
signed main() {
    ryoiki_tenkai
    muryokusho
    #ifndef ONLINE_JUDGE

        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);

    #endif
    srand(time(0));
    int tt=1;
    cin>>tt;
    for(int i=1;i<=tt;i++) {
        // cout<<"Case "<<i<<": ";
        infinite();
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider scenarios that might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Months with very low costs compared to the monthly salary**: This can cause the algorithm to make suboptimal choices if it doesn't handle the cost effectively.
2. **Months with costs exactly equal to the monthly salary**: This can test the boundary conditions and ensure the algorithm correctly handles the exact match.
3. **Months with costs that are multiples of the monthly salary**: This can test the algorithm's ability to handle large values and ensure it doesn't miss opportunities.
4. **Months with costs that are just above the monthly salary**: This can test the algorithm's ability to handle near-miss scenarios and ensure it doesn't waste opportunities.

Given these considerations, let's create a small-scale test input that includes these edge cases:

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

### Explanation:
1. **Test Case 1**:
   - **Months**: 5
   - **Monthly Salary**: 3
   - **Costs**: 1, 2, 3, 4, 5
   - **Expected Output**: 3
   - **Reason**: Charlie can buy happiness in the 1st, 2nd, and 3rd months, but not in the 4th and 5th months because he won't have enough money by then.

2. **Test Case 2**:
   - **Months**: 5
   - **Monthly Salary**: 3
   - **Costs**: 3, 3, 3, 3, 3
   - **Expected Output**: 5
   - **Reason**: Charlie can buy happiness in every month because the cost is exactly equal to his monthly salary.

3. **Test Case 3**:
   - **Months**: 5
   - **Monthly Salary**: 3
   - **Costs**: 4, 4, 4, 4, 4
   - **Expected Output**: 2
   - **Reason**: Charlie can only buy happiness in the 2nd and 5th months because he will have 6 pounds by the 5th month, which is enough to buy happiness in the 5th month.

This test input should help expose any issues in the buggy code by covering a range of cost scenarios and boundary conditions.


**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;

template <typename T>
void debug_print(const T& value) {
    if (std::is_same<T, std::string>::value) {
        std::cout << '"' << value << '"';
    } else if (std::is_same<T, char>::value) {
        std::cout << '\'' << value << '\'';
    } else {
        std::cout << value;
    }
}
template <typename T>
void debug_print(const std::vector<T>& vec) {
    std::cout << "[";
    for (size_t i = 0; i < vec.size(); ++i) {
        debug_print(vec[i]);
        if (i < vec.size() - 1) std::cout << ", ";
    }
    std::cout << "]";
}
template <typename K, typename V>
void debug_print(const std::map<K, V>& m) {
    std::cout << "{";
    for (auto it = m.begin(); it != m.end(); ++it) {
        debug_print(it->first);
        std::cout << ": ";
        debug_print(it->second);
        if (std::next(it) != m.end()) std::cout << ", ";
    }
    std::cout << "}";
}
template <typename T>
void debug_print(const std::set<T>& s) {
    std::cout << "{";
    for (auto it = s.begin(); it != s.end(); ++it) {
        debug_print(*it);
        if (std::next(it) != s.end()) std::cout << ", ";
    }
    std::cout << "}";
}
template <typename T, typename... Args>
void debug_print(const T& first, const Args&... args) {
    debug_print(first);
    if (sizeof...(args) > 0) {
        std::cout << ", ";
        debug_print(args...);
    }
    std::cout << "\n";
}
#define onandon(...) do {                          \
    std::cout << "[" << #__VA_ARGS__ << "]: ";   \
    debug_print(__VA_ARGS__);                    \
    std::cout << std::endl;                      \
} while (0)

using vi = vector<int>;
using vl = vector<int64_t>;
#define pb push_back
#define F first
#define S second
#define pii pair<int, int>
#define pll pair<int64_t, int64_t>
#define all(x) (x).begin(), (x).end()
#define sz(a) int((a).size())
#define uni(a) (a).erase(unique(all((a))), (a).end())

const int64_t mod = (int)1e9+7;
const int MAXN = 1e5+5;
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
const int LOG = 31;
const int INF = 1e9;
const int NINF = -1e9;

void solve () {
    int M, X; cin >> M >> X;
    vector<pii> A(M);
    for (int i = 0; i < M; i++) {
        cin >> A[i].F;
        A[i].S = i;
    }
    sort(all(A));
    int ans = 0, tot = 0;
    for (int i = 0; i < M; i++) {
        if (A[i].F + tot > X * A[i].S) continue;
        tot += A[i].F;
        ans++;
    }
    cout << ans << "\n";
}

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

    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }

    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases that might cause the algorithm to fail. The key issues in the code are related to the sorting and the way the algorithm checks if the cost can be afforded. Specifically, the code sorts the happiness costs but doesn't consider the time constraint (i.e., money earned in month \(i\) can only be spent in month \(j\) where \(j > i\)).

Here are some potential edge cases to consider:
1. **High Costs in Early Months**: If the costs in the early months are very high, the algorithm might skip them and fail to maximize the total happiness.
2. **Low Costs in Late Months**: If the costs in the late months are very low, the algorithm might not have enough money to buy happiness in those months.
3. **Mixed Costs**: A mix of high and low costs in different months can confuse the algorithm.

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

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

### Explanation:
1. **First Test Case**:
   - \(m = 5\), \(x = 5\)
   - Costs: \(10, 10, 10, 10, 1\)
   - The algorithm should recognize that the only affordable cost is in the 5th month, but it might fail to do so because it sorts the costs and might skip the early high costs.

2. **Second Test Case**:
   - \(m = 5\), \(x = 5\)
   - Costs: \(1, 10, 1, 10, 1\)
   - The algorithm should buy happiness in the 1st, 3rd, and 5th months, but it might fail to do so if it doesn't correctly manage the time constraint.

3. **Third Test Case**:
   - \(m = 5\), \(x = 5\)
   - Costs: \(10, 1, 10, 1, 10\)
   - The algorithm should buy happiness in the 2nd and 4th months, but it might fail to do so if it doesn't correctly manage the time constraint.

This test input is designed to challenge the algorithm by including scenarios where the costs are not uniformly distributed and where the time constraint plays a crucial role.


**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 <iostream>
#include <stdio.h>
#include <vector>
#include <array>
#include <string>
#include <set>
#include <algorithm>
#include <utility>
#include <functional>
#include <fstream>
#include <istream>
#include <climits>
#include <map>
#include <list>
#include <cmath>
#include <deque>
#include <numeric>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <cstdint>
#include <iomanip>
#include <queue>
//recursion, dp -> matrix expo
using namespace std::chrono;
using namespace std;
typedef long long int ll;
typedef vector<ll> vll;
ll mod = 1e9 + 7;
ll MAXN = 1000001;
vector<ll> sieveerato(MAXN, 1);
vector<ll> spf(MAXN, 1);
vector<ll> primes;

void sievespf()
{
	spf[0] = 0;
	for (ll i = 2; i <= MAXN - 1; i++)
	{
		if (spf[i] == 1)
		{
			for (ll j = i; j <= MAXN - 1; j += i)
			{
				if (spf[j] == 1)
					spf[j] = i;
			}
		}
	}
}

vector<ll> getFactorization(ll x)
{
	vector<ll> ret;
	while (x != 1) {
		ret.push_back(spf[x]);
		x = x / spf[x];
	}
	return ret;
}

void sieve()
{
	sieveerato[0] = 0;
	sieveerato[1] = 0;
	for (ll i = 2; i * i <= MAXN - 1; ++i)
	{
		if (sieveerato[i])
		{
			for (ll j = i * i; j <= MAXN - 1; j += i)
			{
				sieveerato[j] = 0;
			}
		}
	}
}

void prime()
{
	sieve();
	for (ll i = 0; i <= MAXN - 1; ++i)
	{
		if (sieveerato[i])
		{
			primes.push_back(i);
		}
	}
	sort(primes.begin(), primes.end());
}

bool isPrime(ll n)
{
	if (sieveerato[n] == 1)
	{
		return true;
	}
	return false;
}

ll power(ll x, ll y)
{
	ll res = 1;
	while (y)
	{
		if (y % 2)
		{
			res = res * x;
			--y;
		}
		else
		{
			x = x * x;
			y /= 2;
		}
	}
	return res;
}

vector<vector<ll>> matrixMultiply(vector<vector<ll>> a, vector<vector<ll>> b)
{
	vector<vector<ll>> res(a.size(), vector<ll>(b[0].size()));
	for (ll i = 0; i < a.size(); ++i)
	{
		for (ll j = 0; j < b[0].size(); ++j)
		{
			res[i][j] = 0;
			for (ll k = 0; k < a[0].size(); ++k)
			{
				res[i][j] = (res[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
			}
		}
	}
	return res;
}

vector<vector<ll>> matrixPower(vector<vector<ll>> matrix, ll n)
{
	vector<vector<ll>> res(matrix.size(), vector<ll>(matrix.size()));
	for (ll i = 0; i < matrix.size(); ++i)
	{
		for (ll j = 0; j < matrix.size(); ++j)
		{
			if (i == j)
			{
				res[i][j] = 1;
			}
			else
			{
				res[i][j] = 0;
			}
		}
	}
	while (n)
	{
		if (n % 2)
		{
			res = matrixMultiply(res, matrix);
			--n;
		}
		else
		{
			matrix = matrixMultiply(matrix, matrix);
			n = n / 2;
		}
	}
	return res;
}

ll check_kth_bit(ll x, ll k) {
	return (x >> k) & 1;
}

void print_on_bits(ll x) {
	for (ll k = 0; k < 32; k++) {
		if (check_kth_bit(x, k)) {
			cout << k << ' ';
		}
	}
	cout << '\n';
}

ll count_on_bits(ll x) {
	ll ans = 0;
	for (ll k = 0; k < 32; k++) {
		if (check_kth_bit(x, k)) {
			ans++;
		}
	}
	return ans;
}

bool is_even(ll x) {
	if (x & 1) {
		return false;
	}
	else {
		return true;
	}
}

ll set_kth_bit(ll x, ll k) {
	return x | (1LL << k);
}

ll unset_kth_bit(ll x, ll k) {
	return x & (~(1LL << k));
}

ll toggle_kth_bit(ll x, ll k) {
	return x ^ (1LL << k);
}

bool check_power_of_2(ll x) {
	return count_on_bits(x) == 1;
}

vector<ll> get_divisors(ll n)
{
	vector<ll> divs;
	for (ll i = 2; 1LL * i * i <= n; i++)
	{
		if (n % i == 0)
		{
			divs.push_back(i);
			if (i != n / i)
			{
				divs.push_back(n / i);
			}
		}
	}
	sort(divs.begin(), divs.end());
	return divs;
}

void solve()
{
	ll m, x;
	cin >> m >> x;
	vector<ll> c(m);
	multiset<ll> cc;
	for (ll i = 0; i < m; ++i)
	{
		cin >> c[i];
		cc.insert(c[i]);
	}
	vector<pair<ll, bool>> isRemoved;
	for (ll i = 0; i < m; ++i)
	{
		isRemoved.push_back({c[i], false});
	}
	ll money = 0, ans = 0;
	for (ll i = 1; i < m; ++i)
	{
		if (!isRemoved[i-1].second)
		{
			cc.erase(cc.find(c[i - 1]));
			isRemoved[i - 1].second = 1;
		}
		if (cc.size()>0)
		{
			money += x;
			ll mini = *cc.begin();
			if (mini <= money)
			{
				money -= mini;
				++ans;
				cc.erase(cc.begin());
				auto it = find(c.begin() + i, c.end(), mini) - c.begin();
				isRemoved[it].second = 1;
			}
		}
	}
	cout << ans << endl;

}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	auto start = high_resolution_clock::now();
	long long t = 1, n = 200000;
	//prime();  // uncomment to get prime numbers till 1e6 and isPrime functionality
	//sievespf(); // uncomment for getFactorization functionality
	cin >> t; // uncomment to use test cases
	while (t--)
	{
		solve();
	}
	auto stop = high_resolution_clock::now();
	duration<double> duration = stop - start;
#ifndef ONLINE_JUDGE
	cout << duration.count() << " seconds" << endl; // uncomment to get the time taken by the whole code
#endif

	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
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
```