**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

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

    using namespace std;

    #define ll long long
    #define endl "\n"

    int main(){

        cin.tie(0)->sync_with_stdio(0);

        ll ntc; cin >> ntc;

        while (ntc--){
            ll n, k; cin >> n >> k;

            ll arr[n] = {0};
            ll media = 0;

            for(ll i = 0; i < n; i++){
                ll a; cin >> a;
                arr[i] = a;
            }
            sort(arr, arr+n);

            ll l = 0, r = 10e16+10;
            ll ans = 0;
            ll tmpBom = 0;
            while(l <= r){
                ll mid = (l+r)/2;
                ll tmp = 0;
                for(ll i = 0; i < n; i++){
                    if(arr[i] < mid){
                        tmp += mid-arr[i];
                    }
                }
                //cout << tmp << endl;
                if(tmp <= k){
                    l = mid+1;
                    ans = mid;
                    tmpBom = tmp;
                }
                else{
                    r = mid-1;
                }
                //cout << mid << endl;
            }

            //cout << ans << endl;
            ll resposta = 0;
            for(ll i = 0; i < n ;i++){
                arr[i] = max(0LL, arr[i] - ans);
                if(arr[i] != 0) resposta++;
            }

            //for(auto item: arr){ cout << item << endl;}
            ll aux = n;
            ll valor = (ans*n);
            while(aux--){
                resposta += (valor/n);
                valor--;
            }
            cout << resposta + k-tmpBom << endl;
        }
        return 0;
    }

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

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

Think step by step.

Response:
```plaintext
3
1 10
1
2 4
8 4
3 4
6 1 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
/***
*   سبحان الله والحمد لله ولا إله إلا الله والله أكبر
*         لا حول ولا قوة إلا بالله,
***/

#include "bits/stdc++.h"

using namespace std;

using ll = long long;

#ifdef BuGs
#include "Debug.h"
#else
#define debug(...) 1.618033
#endif
//#define int long  long

/// 20% ==> 80%
///-------------------------------------------------------------------------------+

signed main()
{

    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int tc=1;
    cin>>tc;
    while(tc--)
    {

        ll n,k;
        cin>>n>>k;
        vector<ll>a(n);
        for(ll&I:a)cin>>I;
        auto check=[&](ll m)
        {
            ll sum=0;
            for(int i=0; i<n; i++)
            {
                if(a[i]<m)
                {
                    sum+=(m-a[i]);
                }
            }
            return k>=sum;
        };
        ll l=0,r=1e18;
        while(l<r)
        {
            ll mid=l+(r-l+1)/2;
            if(check(mid))
            {
                l=mid;
            }
            else r=mid-1;
        }
        ll answer=((n)*(l)-n)+1;
       //  debug(l);
        for(int i=0; i<n; i++)
        {
            if(a[i]<l)
            {
                k-=(l-a[i]);
                a[i]=l;
            }
        }
        //debug(a);
       // debug(k);
        int cnt=0;
        for(int i=0; i<n; i++)
        {
            if(a[i]>l)
            {
                answer++;
            }
            else if(k>0)
            {
                answer++;
                k--;
            }
        }
        cout<<answer<<"\n";

    }

}

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

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

Think step by step.

Response:
```plaintext
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: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import sys

def main():
    mod = 1000000007
    t = int(input().strip())
    for _ in range(t):
        n, k = map(int, input().strip().split())
        arr = list(map(int, input().strip().split()))
        arr.sort()

        if n == 1:
            print(arr[0] + k)
            continue

        i = 0
        while k > 0 and i <= n - 2:
            diff = arr[i + 1] - arr[i]
            if diff > 0:
                distribute = min(k, diff * (i + 1))
                if distribute == k:
                    coins_per_element = k // (i + 1)
                    arr[:i + 1] = [x + coins_per_element for x in arr[:i + 1]]
                    remaining_coins = k % (i + 1)
                    arr[:remaining_coins] = [x + 1 for x in arr[:remaining_coins]]
                    k = 0
                else:
                    arr[:i + 1] = [x + distribute // (i + 1) for x in arr[:i + 1]]
                    k -= distribute
            i += 1

        if k > 0:
            coins_per_element = k // n
            arr = [x + coins_per_element for x in arr]
            remaining_coins = k % n
            arr[:remaining_coins] = [x + 1 for x in arr[:remaining_coins]]
            k = 0

        arr.sort()
        c = 0
        for i in range(1, n):
            if arr[i] == arr[0]:
                c += 1
            else:
                break

        ans = (n * arr[0]) - c
        print(ans)

if __name__ == "__main__":
        main()

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

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

Think step by step.

Response:
```plaintext
1
5 10
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import heapq

output = []

t = int(input())

for _ in range(t):
    # number of card types, number of coins
    n, k = map(int, input().split())
    # number of card types you have initially
    a = list(map(int, input().split()))
    a.sort()

    if n == 1:
        output.append(n + k)
        continue

    make_all = sum(a[-1]-x for x in a)

    if k >= make_all:
        q, r = divmod(k - make_all, n)
        out = (max(a) + q) * n + r - (n-1)
        output.append(out)
        continue

    heap = [x for x in a]
    heapq.heapify(heap)
    while k > 0:
        freq1 = heapq.heappop(heap)
        freq2 = heap[0]

        diff = freq2 - freq1
        if diff == 0:
            temp = 1
        else:
            temp = min(k, diff)

        k -= temp
        heapq.heappush(heap, freq1 + temp)

    out = heap[0] * n + sum(x > heap[0] for x in heap[1:])
    output.append(out-(n-1))

for line in output:
    print(line)

**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
1 10
1
2 4
8 4
3 4
6 1 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
t = int(input())
for _ in range(t):
  n, k = map(int, input().split())
  a = list(map(int, input().split()))

  a = sorted(a)
  a.reverse()

  count = 0
  i = 0
  while True:

    if a[i] > 0:
      count += 1
      a[i] -= 1

    elif k:
      count += 1
      k -= 1

    else:
      break

    ###print(a[i], count, k)

    i += 1
    i = i % n

  count -= n - 1
  print(count)

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

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

Think step by step.

Response:
```plaintext
1
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: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

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

public class Main{

   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));

      int t = sc.nextInt();

      while(t-- != 0) {
        int n = sc.nextInt();
        int k = sc.nextInt();

        List<Integer> cards = new ArrayList();
        for(int i = 0 ; i < n ; i++) {
            cards.add(sc.nextInt());
        }

        out.println(new Solution().solve(n,k,cards));
      }

      out.close();
   }

    public static class Solution {
        public Solution() {}
        public int solve(int n,  int k, List<Integer> cards) {
            Collections.sort(cards);

            int i = 0;
            long count = 0, current = cards.get(0);;

            while(i < n && current == cards.get(i)) {
                count++;
                i++;
            }

            while(i < n) {
                if(current != cards.get(i)) {
                    long diff = cards.get(i) - current;
                    if(diff * count > k) {
                        break;
                    }
                    k -= diff * count;
                    current = cards.get(i);
                }
                count++;
                i++;
            }

            current += k / count;
            k %= count;
            count -= k;

            return (int) (n * current - count + 1);
        }
    }

   public static PrintWriter out;
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

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

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

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

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

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

      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
   //--------------------------------------------------------
}

**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 focus on edge cases and scenarios that can cause the algorithm to fail. The bug in the code is related to how it handles the distribution of cards and the calculation of the maximum score. Here are the steps to create such a test input:

1. **Small-scale Test Case**: Ensure the test case is small enough to be manageable but large enough to highlight the issue.
2. **Edge Cases**: Consider cases where the number of coins is exactly enough to balance the distribution of cards, or where the number of cards is already perfectly balanced.
3. **Maximal Failure Probability**: Create a scenario where the algorithm's logic for balancing the cards and calculating the score fails.

Given the constraints and the nature of the problem, a test case with a small number of distinct card types and a small number of coins can be effective. Here's a test input that could expose the bug:

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

### Explanation:
- **Test Case**: There is one test case.
- **n = 3**: There are 3 distinct types of cards.
- **k = 4**: You have 4 coins to buy additional cards.
- **a_1 = 1, a_2 = 2, a_3 = 3**: You have 1 card of type 1, 2 cards of type 2, and 3 cards of type 3.

### Expected Output:
The expected output for this test case is 3. Here's why:
- Initially, you have the cards: [1, 2, 2, 3, 3, 3].
- You can buy 4 new cards. To maximize the score, you should try to balance the distribution of cards to form as many permutations of [1, 2, 3] as possible.
- One optimal way to do this is to buy 1 card of type 1, 1 card of type 2, and 2 cards of type 3. This results in the cards: [1, 1, 2, 2, 2, 3, 3, 3, 3].
- You can form 3 subarrays of length 3 that are permutations of [1, 2, 3]:
  - [1, 1, 2, 2, 2, 3, 3, 3, 3] (subarray [1, 2, 3])
  - [1, 1, 2, 2, 2, 3, 3, 3, 3] (subarray [2, 2, 3])
  - [1, 1, 2, 2, 2, 3, 3, 3, 3] (subarray [2, 3, 3])

This test input should expose the bug in the provided code by challenging its logic for balancing the cards and calculating the score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

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

public class C {
    public static void main(String[] args) throws IOException {
        FastReader s = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int tt = s.nextInt();
        outer: while(tt-->0){
            int n = s.nextInt();
            long k = s.nextInt();
            long[] arr = s.nextArrayLong(n);
            TreeMap<Long, Long> map = new TreeMap<>();
            map.put(Long.MAX_VALUE, 1L);
            for(int i = 0; i < n; i++){
                map.put(arr[i], map.getOrDefault(arr[i], 0L) + 1);
            }
            while(k > 0){
                Map.Entry<Long, Long> en = map.pollFirstEntry();
                long need = (map.firstKey() - en.getKey()) * en.getValue();
                if(need <= k){
                    k -= need;
                    map.put(map.firstKey(), map.get(map.firstKey()) + en.getValue());
                }
                else{
                    long poss = k / en.getValue();
                    k -= poss * en.getValue();
                    map.put(en.getKey() + poss, en.getValue() - k);
                    if(k > 0){
                        map.put(en.getKey() + poss + 1, map.getOrDefault(en.getKey() + poss + 1, 0L) + k);
                        k = 0;
                    }
                }
            }
            long ans = n * map.pollFirstEntry().getKey();
            for(long i: map.values()){
                ans += i;
            }
            out.println((ans - 1) - (n-1));
        }
        out.flush();
    }

    public static class Pair<T, E> {
        T p1;
        E p2;
        Pair(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }
        void setValue(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }

        void setA(T a){
            this.p1 = a;
        }
        void setB(E b){
            this.p2 = b;
        }
        Pair getValue() {
            return this;
        }
        @Override
        public int hashCode() {
            return Objects.hash(p1, p2);
        }

        @Override
        public boolean equals(Object startObject) {
            if(this == startObject){
                return true;
            }
            if(startObject == null || getClass() != startObject.getClass()){
                return false;
            }
            Pair<T, E> pair = (Pair<T, E>) startObject;
            return p1.equals(pair.p1) && p2.equals(pair.p2);
        }
    }

    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

        double[] nextArrayDouble(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }

        String[] nextArrayString(int n) throws IOException {
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = next();
            }
            return arr;
        }

        char[] nextArrayChar() throws IOException {
            return next().toCharArray();
        }

        static void shuffleArray(int[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                int a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }

        static void shuffleArray(long[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                long a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }

        public int[] nextSortedArrayInt(int n) throws IOException {
            int[] arr = nextArrayInt(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }

        public long[] nextSortedArrayLong(int n) throws IOException {
            long[] arr = nextArrayLong(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some steps to think through the problem:

1. **Edge Cases for Small `n` and `k`:**
   - Test with `n = 1` and various values of `k`.
   - Test with `n = 2` and various values of `k`.

2. **Large `k` and Small `n`:**
   - Test with a small `n` and a very large `k`.

3. **Unbalanced Distribution of `a_i`:**
   - Test with a highly unbalanced distribution of `a_i` values.

4. **Equal Distribution of `a_i`:**
   - Test with all `a_i` values being equal.

5. **Mixed Distribution of `a_i`:**
   - Test with a mix of high and low values for `a_i`.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

This test input includes:
- A test case with `n = 1` and `k = 10`.
- A test case with `n = 2` and `k = 4`.
- A test case with `n = 3` and `k = 4` with an unbalanced distribution of `a_i`.
- A test case with `n = 3` and `k = 9` with a mixed distribution of `a_i`.
- A test case with `n = 5` and `k = 3` with a mixed distribution of `a_i`.

These cases should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
from sys import stdin, stdout

input = stdin.readline
print = lambda s: stdout.write(s+'\n')

out = []
for _ in range(int(input())):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    mx = max(a)
    cnt = [0]*(mx+1)
    m = 0
    for val in a:
        cnt[val] += 1
        m += mx - val
    if k >= m:
        out.append(str(n*(mx-1)+1+k-m))
        continue
    for i in range(mx):
        if not cnt[i]:
            continue
        if k > cnt[i]:
            k -= cnt[i]
            cnt[i+1] += cnt[i]
            cnt[i] = 0
        else:
            cnt[i+1] += k
            cnt[i] -= k
            i += cnt[i] == 0
            break
    out.append(str(n*i+1-cnt[i]))
print('\n'.join(out))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import sys

def find(a, i):
    res = 0
    return res

def solve():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    sum = 0
    while k:
        min_index = a.index(min(a))
        a[min_index] += 1
        k -= 1
    a.sort()
    num = a[0]
    for i in range(n):
        sum += num
        a[i] -= num
    j = 0
    i = 0
    while a[(i - 1 + n) % n]:
        a[(i - 1 + n) % n] -= 1
        sum += 1
        i = (i - 1 + n) % n
    while a[(j + 1 + n) % n]:
        a[(j + 1 + n) % n] -= 1
        sum += 1
        j = (j + 1 + n) % n
    print(sum - n + 1)

def main():
    t = int(input())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The buggy code has several potential issues, including:

1. **Incorrect handling of the minimum element**: The code uses `a.index(min(a))` to find the index of the minimum element, but this can be inefficient and may not handle cases where multiple elements are the same minimum value correctly.
2. **Incorrect redistribution of cards**: The code attempts to redistribute the cards to form valid permutations, but the logic for doing so is flawed, especially in the loops that adjust the array.

To expose these issues, we can create a test case where the initial distribution of cards is such that the minimum element is not unique, and the number of coins is large enough to significantly alter the distribution. This will test the code's ability to handle multiple minimum elements and large redistributions.

Here is a small-scale test input that should challenge the buggy code:

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

### Explanation:
- **Number of test cases**: 1
- **n**: 5 (number of distinct types of cards)
- **k**: 10 (number of coins)
- **a**: [1, 1, 1, 1, 1] (number of cards of each type)

### Why this test input is challenging:
1. **Multiple minimum elements**: All elements in the array `a` are the same (1), so the code must correctly handle the situation where multiple elements are the minimum.
2. **Large number of coins**: With 10 coins, the code will need to distribute these coins effectively to maximize the score. The code's logic for finding and updating the minimum element will be tested.
3. **Expected behavior**: The optimal strategy is to distribute the 10 coins such that each type of card has 3 cards, resulting in a total of 3 cards for each type. This should yield a score of 3, as there will be 3 valid subarrays of length 5 that are permutations of [1, 2, 3, 4, 5].

This test input should help identify issues with the code's handling of multiple minimum elements and its strategy for distributing coins.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <unordered_set>
#include <string>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
typedef long long ll;

ll good(vector<ll> a, ll cnt){ //returns -1
    ll res = 0;
    for (int i = 0; i < a.size(); i++){
        if (cnt > a[i]){
            res += cnt - a[i];
        }
    }
    return res;
}
ll ans(vector<ll> a, ll k){
    ll cnt = 0;
    for (int i = 0; i < a.size(); i++) if (a[i] > k) cnt++;
    return cnt;
}
int main(){
    ll tc;
    cin >> tc;

    for (int t = 0; t < tc; t++){
        ll n,k;
        cin >> n >> k;
        vector<ll> a;
        for (int i = 0; i < n; ++i){
            ll curr;
            cin >> curr;
            a.push_back(curr);
        }

        ll hi = pow(10,15);
        ll lo = 0;
        while (lo + 1 < hi){
            ll mid = (lo + hi) >> 1;
            if (good(a,mid) <= k){
                lo = mid;
            }
            else{
                hi = mid;
            }
        }
        if (good(a,hi) <= k) cout << 1 + (hi-1)*n + k - good(a,hi) + ans(a,hi)  << endl;
        else cout << 1 + (lo-1)*n + k - good(a,lo) + ans(a,lo)<< endl;

    }

}

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

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

Think step by step.

Response:
```plaintext
3
1 0
1
2 1
1 1
3 10
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define double long double
#define pb push_back
#define mod 1000000007
#define mod1 998244353
#define inf 1e18
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

#define print(v) for (auto x: v)cout<<x<<' ';cout<<endl;
#define vin(v) for(auto&x:v)cin>>x;
#define cno {cout<<"NO"<<endl;return;}
#define cyes {cout<<"YES"<<endl;return;}

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<pair<int,int>> vpi;
typedef pair<int,int> pi;
typedef priority_queue<int> prq;

//if the thing you are calucalating is increaseing with m(1) else (2)

//(1)
// if(sum<=m1)
// return 1;
// return 0;

//(2)
// if(sum<=m1)
// return 0;
// return 1;

// be carefull whether u want to the last 0 or the last 1

int fun(vi& a,int m,int k,int n)    //any other required values for the fun
{
    int sum=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]<m)
        {
            sum+=m-a[i];
        }
    }
    if(sum>k)
    return 1;
    return 0;
}

void code()
{
    int n,k;
    cin>>n>>k;
    vi a(n);
    vin(a);
    int l=0;
    int r=1000000000000000;
    // cout<<r<<"\n";
    int m=(l+r)/2;
    while(l<r-1)
    {
        if(m<0)
        {
            l=m;
            continue;
        }
        if(fun(a,m,k,n)==0)
        {
          l=m;
        }
        else
        {
          r=m;
        }
        m=(l+r)/2;
    }
    if(l<=0)
    {
        cout<<"0\n";
        return;
    }
    // l=r-1;
    for(int i=0;i<n;i++)
    {
        if(a[i]<l)
        {
            k-=l-a[i];
            a[i]=l;
        }
    }
    for(int i=0;i<n;i++)
    {
        if(k==0)
        break;
        if(a[i]<=l)
        {
            a[i]++;
            k--;
        }
    }
    int count=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]>l)
        count++;
    }
    // print(a);
    // cout<<count<<" "<<k<<" ";
    int ans=(l-1)*n+1+count;
    cout<<ans<<"\n";
}
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    cin>>t;
    while(t--) code();
    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
2 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import java.math.BigInteger;
import java.sql.SQLOutput;
import java.util.*;

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

        int t;
        t = in.nextInt();
        while(t-- > 0) {
            int n;
            n = in.nextInt();
            long k;
            k = in.nextLong();
            long[] a = new long[n];
            for(int i = 0; i < n; i++) {
                a[i] = in.nextLong();
            }
//            System.out.println(n);
//            System.out.println(k);
            long l = 0;
            long r = (long)2e15;
            long ind = 0;
            while(l <= r) {
                long mid = (l + r) / 2;
//                System.out.println(mid);
                if(check(a, mid, k)) {
                    l = mid + 1;
                    ind = mid;
                }
                else {
                    r = mid - 1;
                }
            }
//            System.out.println(ind);
            for(int i = 0; i < n; i++) {
                if(a[i] < ind) {
                    k -= ind - a[i];
                    a[i] = 0;
                }
                else {
                    a[i] -= ind;
                }
            }
//            System.out.println(k);
            BigInteger bind= new BigInteger(String.valueOf(ind));
            BigInteger bn= new BigInteger(String.valueOf(n));
            BigInteger res = bind.multiply(bn);

            for(int i = 0; i < n; i++) {
                if(a[i] > 0) {
                    k++;
                }
            }
            k -= n - 1;
            BigInteger bk = new BigInteger(String.valueOf(k));
            String ans = res.add(bk).toString();
            System.out.println(ans);

        }

    }
    static boolean check(long[] b, long m, long k) {

        for(int i = 0; i < b.length; i++) {
            if(b[i] < m) {
                k -= m - b[i];
            }
        }
        if(k >= 0) {
            return true;
        }
        else {
            return false;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the algorithm to fail. The main issues in the buggy code are related to the binary search and the way it handles the distribution of cards. Here are the steps to generate a suitable test input:

1. **Binary Search Range**: The binary search range is set from `0` to `2e15`, which is very large. This can lead to precision issues or incorrect termination conditions.
2. **Card Distribution**: The way the code distributes cards after the binary search can be problematic, especially if the `k` value is not used correctly.
3. **Edge Cases**: We need to test with small values of `n` and `k` to ensure the binary search and distribution logic are correct.

Let's create a test input that includes:
- A small value of `n` to ensure the binary search range is manageable.
- A value of `k` that is just enough to make the distribution tricky.
- A distribution of cards that will challenge the binary search and the final card distribution logic.

Here is the test input:

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

Explanation:
- **Test Case 1**: `n = 1, k = 10, a = [1]`
  - This is a simple case where the only card is 1. The expected output is 11, as we can buy 10 more cards of type 1.
- **Test Case 2**: `n = 2, k = 4, a = [8, 4]`
  - This case tests the distribution of cards when the number of cards of each type is different. The expected output is 15, as we can buy 4 more cards of type 2.
- **Test Case 3**: `n = 3, k = 4, a = [6, 1, 8]`
  - This case tests the distribution of cards when the number of cards of each type is different and `k` is small. The expected output is 15, as we can buy 4 more cards to optimize the score.

This test input should help expose any issues in the binary search and card distribution logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import java.util.*;

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

        int t;
        t = in.nextInt();
        while(t-- > 0) {
            int n;
            n = in.nextInt();
            long k;
            k = in.nextLong();
            long[] a = new long[n];
            for(int i = 0; i < n; i++) {
                a[i] = in.nextLong();
            }
//            System.out.println(n);
//            System.out.println(k);
            long l = 0;
            long r = (long)1e15;
            long ind = 0;
            while(l <= r) {
                long mid = (l + r) / 2;
//                System.out.println(mid);
                if(check(a, mid, k)) {
                    l = mid + 1;
                    ind = mid;
                }
                else {
                    r = mid - 1;
                }
            }
//            System.out.println(ind);
            for(int i = 0; i < n; i++) {
                if(a[i] < ind) {
                    k -= ind - a[i];
                    a[i] = 0;
                }
                else {
                    a[i] -= ind;
                }
            }
//            System.out.println(k);
            long res = ind * n + k;
            for(int i = 0; i < n; i++) {
                if(a[i] > 0) {
                    res++;
                }
            }
            res -= n - 1;
            System.out.println(res);

        }

    }
    static boolean check(long[] b, long m, long k) {

        for(int i = 0; i < b.length; i++) {
            if(b[i] < m) {
                k -= m - b[i];
            }
        }
        if(k >= 0) {
            return true;
        }
        else {
            return false;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The code has a binary search to determine the maximum number of complete permutations of length `n` that can be formed, and then it calculates the remaining score based on the leftover cards.

### Key Points to Consider:
1. **Large `k` with small `n` and `a` values**: This can expose issues with the binary search and the calculation of the remaining score.
2. **Small `k` with large `n` and `a` values**: This can test the code's ability to handle cases where the number of coins is insufficient to form even one complete permutation.
3. **Equal or nearly equal `a` values**: This can test the code's ability to handle balanced distributions of card types.
4. **Very unbalanced `a` values**: This can test the code's ability to handle extreme cases where one card type is significantly more or less frequent than others.

### Chosen Test Input:
```plaintext
5
1 10
1
2 1
8 4
3 10
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `k = 10`, `a = [1]`
   - This tests the case where `n` is 1, which should result in a score of 11 (since we can buy 10 more cards of type 1).
   - Expected output: `11`

2. **Test Case 2**:
   - `n = 2`, `k = 1`, `a = [8, 4]`
   - This tests the case where `k` is very small, and we can only buy one card. The optimal solution is to buy one card of type 2.
   - Expected output: `12` (8 subarrays of [1, 2] and 4 subarrays of [2, 1])

3. **Test Case 3**:
   - `n = 3`, `k = 10`, `a = [6, 1, 8]`
   - This tests a case with a moderate number of coins and a mix of card types.
   - Expected output: `15` (5 subarrays of [1, 2, 3] and 5 subarrays of [3, 1, 2] and 5 subarrays of [2, 3, 1])

4. **Test Case 4**:
   - `n = 3`, `k = 9`, `a = [7, 6, 2]`
   - This tests a case with a moderate number of coins and a mix of card types.
   - Expected output: `22` (7 subarrays of [1, 2, 3] and 7 subarrays of [3, 1, 2] and 7 subarrays of [2, 3, 1])

5. **Test Case 5**:
   - `n = 5`, `k = 3`, `a = [6, 6, 7, 4, 6]`
   - This tests a case with a larger number of card types and a moderate number of coins.
   - Expected output: `28` (6 subarrays of [1, 2, 3, 4, 5] and 6 subarrays of [5, 1, 2, 3, 4] and 6 subarrays of [4, 5, 1, 2, 3] and 6 subarrays of [3, 4, 5, 1, 2] and 6 subarrays of [2, 3, 4, 5, 1])

This test input should expose the bug in the binary search and the calculation of the remaining score in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

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

using namespace std;
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		long long n, k;
		cin >> n >> k;
		long long q[n + 1];
		long long pos = 0;
		for(int i = 1;i <= n;i++)
		{
			cin >> q[i];
		}
		long long l = 0,r = 1e13;
		while(l < r)
		{
			int mid = (l + r + 1) / 2;
			long long sum = 0;
			for(int i = 1;i <= n;i++)
			{
				if(q[i] < mid) sum += mid - q[i];
			}
			if(sum <= k)
			{
				l = mid;
			}
			else r = mid - 1;
		}
		long long sum = 0;
		int p = 0;
		for(int i = 1;i <= n;i++)
		{
			if(q[i] <= l)
			{
				sum +=  l - q[i];
				p++;
			}
 		}
 		long long cns = p - (k - sum);

		cout << n * l - n + 1 + n - cns << endl;
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. Here are the steps to identify such a test input:

1. **Understand the Bug**: The buggy code uses a binary search to find the maximum number of cards of each type that can be achieved. It then calculates the number of subarrays that are permutations of `[1, 2, ..., n]`. The bug seems to be in the calculation of the final score, particularly in how it handles the number of cards that need to be adjusted.

2. **Edge Cases**:
   - **Small `n` and `k`**: Test with small values of `n` and `k` to ensure the binary search and final score calculation are correct.
   - **Large `n` and `k`**: Test with large values to ensure the binary search and final score calculation handle large numbers correctly.
   - **Uneven distribution of `a_i`**: Test with a highly uneven distribution of `a_i` to see how the code handles cases where some card types are much more frequent than others.
   - **Zero `k`**: Test with `k = 0` to ensure the code correctly handles the case where no new cards can be bought.

3. **Challenging Input**:
   - **Test Case 1**: Small `n` and `k` with an uneven distribution.
   - **Test Case 2**: Large `n` and `k` with an even distribution.
   - **Test Case 3**: Zero `k` with an even distribution.
   - **Test Case 4**: Small `n` and large `k` with an uneven distribution.

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
4
3 5
1 2 3
5 10
1 1 1 1 1
3 0
1 2 3
2 20
1 10
```

This test input includes:
- A small `n` and `k` with an uneven distribution.
- A large `n` and `k` with an even distribution.
- Zero `k` with an even distribution.
- A small `n` and large `k` with an uneven distribution.

These cases should help expose any issues in the binary search and final score calculation in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

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

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>

typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int>pii;
typedef unsigned long long ull;
priority_queue<int, vector<int>,greater<int>>rev_pq; //min heap
// Mathematical functions
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;    //__gcd
    return gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
    return (a/gcd(a,b)*b);
}
ll moduloMultiplication(ll a,ll b,ll mod)
{
    ll res = 0;
    a %= mod;
    while (b)
    {
        if (b & 1)
            res = (res + a) % mod;
        b >>= 1;
    }
    return res;
}
ll powermod(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
    if (x == 0)
        return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
//Sorting
bool sorta(const pair<ll,ll> &a,const pair<ll,ll> &b)
{
    return (a.second < b.second);   //sort pair second element a
}
bool sortd(const pair<ll,ll> &a,const pair<ll,ll> &b)
{
    return (a.second > b.second);   //sort pair second element desc
}

#define dbg          cout<<"Freak"<<endl;
#define dbgg(i)      cout<<i<<endl;
#define N            1e9+7
#define pb           push_back
#define pf           push_front
#define pi           acos(-1)
#define For(n)       for(int i=0;i<n;i++)
#define all(x)       x.begin(),x.end()
#define fast         ios_base::sync_with_stdio(0);cin.tie(0);
#define INF          0x3f3f3f3f
#define inputfile    freopen("input.txt", "r", stdin)
#define outputfile   freopen("output.txt", "w", stdout)
#define appendfile   freopen("output.txt", "a", stdout)
#define  endl        '\n'
int dx[]= {-1,1,0,0};
int dy[]= {0,0,-1,1};
int dxx[]= {0,0,1,-1,1,1,-1,-1};
int dyy[]= {1,-1,0,0,1,-1,1,-1};

void solve()
{
    ll i,j,n,m,k;
    cin>>n>>k;
    ll a[n+5];
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    a[i]=1e18;
    a[0]=0;
    sort(a,a+n+1);
    ll lo=1,hi=n;
    ll mx=0;
    while(lo<=hi)
    {
        ll mid=(lo+hi)/2;
        ll lagbe=0;
        for(i=1;i<mid;i++)
        {
            lagbe+=a[mid]-a[i];
        }
        if(lagbe>k)
        {
            hi=mid-1;
        }
        else{
            ll baki=k-lagbe;
            ll pls=baki/mid;
            ll tem=min(a[mid]+pls,a[mid+1]);
            ll extra=baki-(pls*mid)+(n-mid);
            if(tem==a[mid+1])extra--;
            ll cal=tem+(n-1)*(tem-1)+extra;
           // cout<<"lagbe--->"<<lagbe<<endl;
            //cout<<"baki-->"<<baki<<endl;
            //cout<<"tem-->"<<tem<<endl;
            //cout<<"extra--->"<<extra<<endl;
            //cout<<lo<<' '<<hi<<"---"<<cal<<endl<<endl;
            mx=max(mx,cal);
            lo=mid+1;
        }
    }
    cout<<mx<<endl;

}
int main()
{
    fast;
    int t=1;
    cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
t = int(input())
p = []
for _ in range(t):
    n,k = map(int,input().split())
    l = list(map(int,input().split()))
    l.sort()

    def manque(target):
        i = 0
        s = 0
        while i < n and l[i] < target:
            s += (target - l[i])
            i += 1
        return s

    i,j = 0,2*10**12
    while j-i > 0:
        m = (i+j+1)//2
        manq = manque(m)
        if manq > k:
            j = m-1
        elif manq == k:
            i = m
            j = m
        else:
            i = m
    assert manque(i) <= k and manque(i+1) > k
    target = i
    for i in range(n):
        if l[i] < target:
            k -= (target - l[i])
            l[i] = target
    count_target = 0
    for x in l:
        if x == target:
            count_target += 1
    count_target -= k
    k = 0
    p.append((target-1) * n + 1 + (n-count_target))

for x in p:
    print(x)

**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
1 0
1
2 1
1 1
3 2
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy $k$ new cards in total, and the cards you buy can contain any integer between $1$ and $n$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $n$ which are a permutation of $[1, 2, \ldots, n]$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t\ (1\le t\le 100)$. The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1\le n \le 2 \cdot 10^5$, $0\le k \le 10^{12}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$) — the number of cards of type $i$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$ (including $11$ single $1$s), which contains $11$ subarrays consisting of a permutation of $[1]$.

In the second test case, we can buy $0$ cards of type $1$ and $4$ cards of type $2$, and then we rearrange the cards as following: $[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$. There are $8$ subarrays equal to $[1, 2]$ and $7$ subarrays equal to $[2, 1]$, which make a total of $15$ subarrays which are a permutation of $[1, 2]$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$.

Buggy Code:
import collections
import math

def solve():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))

    range_start = 1
    range_end = sum(a) + k

    min_val = None
    min_cost = 0

    while range_start <= range_end:
        mid = (range_start + range_end) // 2
        cnt = sum(max(0, mid - v) for v in a)
        if cnt <= k:
            min_val = mid
            range_start = mid + 1
        else:
            range_end = mid - 1

    for i in range(n):
        if a[i] < min_val:
            min_cost += min_val - a[i]
            a[i] = min_val

    min_count = a.count(min_val)

    ans = min_val * n - (n - 1)
    if min_cost > (k - min_cost):
        ans += n - min_count + k - min_cost
    else:
        ans += n

    print(ans)

t = 1
t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

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

1. **Edge Cases:**
   - When `n` is very small (e.g., 1 or 2).
   - When `k` is 0, meaning no additional cards can be bought.
   - When the number of cards for each type is very high or very low.

2. **Critical Scenarios:**
   - When the number of cards for each type is already sufficient to form permutations without buying any cards.
   - When the number of cards for each type is insufficient, but the number of coins is large enough to buy the necessary cards.

3. **Behavior of the Buggy Code:**
   - The code uses a binary search to find the minimum value `min_val` such that the total cost of buying cards to reach this value is less than or equal to `k`.
   - It then calculates the number of times `min_val` appears and adjusts the answer based on this count and the remaining coins.

Given these points, a small-scale test input that can expose the bug is:

```plaintext
3
1 0
1
2 5
1 1
3 10
1 1 1
```

Explanation:
- **Test Case 1:**
  - `n = 1`, `k = 0`, `a = [1]`
  - The only possible array is `[1]`, which contains 1 subarray that is a permutation of `[1]`.
  - Expected output: 1

- **Test Case 2:**
  - `n = 2`, `k = 5`, `a = [1, 1]`
  - We can buy 4 cards of type 2, and the array can be rearranged as `[1, 2, 1, 2, 1, 2]`.
  - Expected output: 4 (since there are 4 subarrays that are permutations of `[1, 2]`).

- **Test Case 3:**
  - `n = 3`, `k = 10`, `a = [1, 1, 1]`
  - We can buy 7 cards of type 3, and the array can be rearranged as `[1, 2, 3, 1, 2, 3, 1, 2, 3]`.
  - Expected output: 7 (since there are 7 subarrays that are permutations of `[1, 2, 3]`).

This test input should help identify any issues in the binary search and the calculation of the final score in the buggy code.