**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

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

using namespace std;
#define pb push_back
#define ff first
#define ss second
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int, int>
#define ppi pair<pii, int>
#define vii vector<pii>
#define mii map<int, int>
#define miv map<int, vi>
#define setbits(n) __builtin_popcount(n)
#define msb(x) (31 - __builtin_clz(x))
#define lsb(x) __builtin_ctz(x)
#define sz(b) (int)b.size()
#define all(v) (v).begin(), (v).end()
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define endl "\n"
#define in(a, n) \
  for (int i = 0; i < n; i++) cin >> a[i];
#define show2(a, b) cout << a << ' ' << b << endl
#define show3(a, b, c) cout << a << ' ' << b << ' ' << c << endl
#define show(arr) \
  for (auto i : arr) cout << i << ' ';
#define Endl endl
#define dbg(x) cout << #x << ": " << x << endl;
const long long N = 500005;
const long long INF = 1000000009;
const long long mod = 1000000007;  // 998244353;
int n, q;
int a[N + 1];
vi v;

bool check() {
  int eq = 0, les = 1;
  for (int i = 2; i < sz(v); i++) {
    if (v[i] == v[i - 1] && les) {
      les--;
      eq++;
      i++;
    } else
      les++;
  }
  if (eq >= 2) return true;
  multiset<int> s;
  s.insert(v[0]);
  int cnt = 0;
  for (int i = 2; i < sz(v); i++) {
    int x = v[i] - v[i - 1] + 1;
    if (s.lower_bound(x) != s.end()) {
      cnt++;
      s.erase(s.lower_bound(x));
      i++;
    } else
      s.insert(v[i - 1]);
  }
  return cnt >= 2;
}

void cases() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (r - l >= 100) {
      yes;
      continue;
    }
    v.clear();
    for (int i = l; i <= r; i++) v.pb(a[i]);
    sort(all(v));
    if (check())
      yes;
    else
      no;
  }
}

int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  for (int i = 0; i < tc; i++) {
    // cout<<"Case #"<<i+1<<": ";
    cases();
  }
  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
10 5
1 1 1 1 1 1 10 10 10 10
1 6
2 7
3 8
4 9
5 10
```

### Explanation:
This test input is designed to challenge the buggy code in the following ways:

1. **High Frequency of Equal Lengths**: The sticks have lengths that are highly repetitive (many 1s and 10s). This can make it difficult to form non-degenerate triangles if the algorithm does not handle equal lengths correctly.

2. ** Queries with Small Ranges**: The queries are designed to cover small ranges within the array. This forces the algorithm to handle the specific conditions for forming triangles with a limited set of stick lengths.

3. **Edge Cases**: The queries are chosen to test the boundary conditions where the number of sticks is just enough to potentially form two triangles (6 sticks) or slightly more.

### Expected Behavior:
- **Query 1 (1 6)**: The lengths are `[1, 1, 1, 1, 1, 1]`. It is impossible to form any non-degenerate triangle, so the expected output is "NO".
- **Query 2 (2 7)**: The lengths are `[1, 1, 1, 1, 1, 10]`. It is impossible to form any non-degenerate triangle, so the expected output is "NO".
- **Query 3 (3 8)**: The lengths are `[1, 1, 1, 1, 10, 10]`. It is impossible to form any non-degenerate triangle, so the expected output is "NO".
- **Query 4 (4 9)**: The lengths are `[1, 1, 1, 10, 10, 10]`. It is impossible to form any non-degenerate triangle, so the expected output is "NO".
- **Query 5 (5 10)**: The lengths are `[1, 1, 10, 10, 10, 10]`. It is impossible to form any non-degenerate triangle, so the expected output is "NO".

This test input should expose the bug in the algorithm, particularly in the `check` function, which may not handle the case of multiple equal lengths correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

Buggy Code:
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f3f3f3f3fll
#define debug(x) cerr<<#x<<"="<<x<<endl
using namespace std;
using ll=long long;
using ld=long double;
using pli=pair<ll,int>;
using pi=pair<int,int>;
template<typename A>
using vc=vector<A>;
inline int read()
{
    int s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
inline ll lread()
{
    ll s=0,w=1;char ch;
    while((ch=getchar())>'9'||ch<'0') if(ch=='-') w=-1;
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    return s*w;
}
struct node
{
    int v,x,y;
    node(int v=0,int x=0,int y=0):v(v),x(x),y(y){}
    bool operator < (node b) const
    {
        if(v!=b.v) return v<b.v;
        if(x!=b.x) return x<b.x;
        return y<b.y;
    }
};
set<node>S1,S2;
node x[11];
int a[100001];
int ans[100001];
int n,m;
inline void add(int w)
{
    // printf("add %d\n",w);
    auto it=S1.lower_bound(node(a[w],w,0));
    int lt=-1,nx=-1;

    if(it!=S1.begin())
    {
        auto I=it;I--;lt=I->x;
        S2.insert(node(a[w]-a[lt],lt,w));
    }
    if(it!=S1.end())
    {
        nx=it->x;
        S2.insert(node(a[nx]-a[w],w,nx));
    }
    if(lt!=-1&&nx!=-1) S2.erase(node(a[nx]-a[lt],lt,nx));
    S1.insert(node(a[w],w,0));

    // printf("S1 : \n");
    // for(auto i:S1) printf("(%d,%d) ",i.v,i.x);;putchar('\n');

    // printf("S2 : \n");
    // for(auto i:S2) printf("(%d,%d,%d) ",i.v,i.x,i.y);;putchar('\n');
}
inline void del(int w)
{
    // printf("del %d\n",w);
    S1.erase(node(a[w],w,0));
    int lt=-1,nx=-1;
    auto it=S1.lower_bound(node(a[w],w,0));

    if(it!=S1.begin())
    {
        auto I=it;I--;lt=I->x;
        S2.erase(node(a[w]-a[lt],lt,w));
    }
    if(it!=S1.end())
    {
        nx=it->x;
        S2.erase(node(a[nx]-a[w],w,nx));
    }
    if(lt!=-1&&nx!=-1) S2.insert(node(a[nx]-a[lt],lt,nx));
}
inline bool no(int i,int w)
{
    return x[i].x==w||x[i].y==w;
}
inline bool check(int i,int w)
{
    if(!w) return false;
    assert(node(a[w],w,0)<node(a[x[i].x],x[i].x,0));
    return a[w]>x[i].v;
}
inline bool check()
{
    // printf("in check\n");
    auto it=S2.begin();int c=0;
    for(int i=1;i<=8;i++)
    {
        if(it==S2.end()) break;
        c++,x[i]=*it;it++;
    }
    // printf("c=%d\n",c);

    for(int i=1;i<=c;i++) for(int j=i+1;j<=c;j++)
    {
        // printf("i=%d(%d,%d) j=%d(%d,%d)\n",i,x[i].x,x[i].y,j,x[j].x,x[j].y);
        if(x[i].x==x[j].x) continue;
        if(x[i].x==x[j].y) continue;
        if(x[i].y==x[j].x) continue;
        if(x[i].y==x[j].y) continue;

        int f1=0,f2=0;
        auto it=S1.lower_bound(node(a[x[j].x],x[j].x,0));
        while(it!=S1.begin())
        {
            it--;int w=it->x;
            if(no(i,w)||no(j,w)) continue;
            f1=w;break;
        }
        it=S1.lower_bound(node(a[x[i].x],x[i].x,0));
        while(it!=S1.begin())
        {
            it--;int w=it->x;
            if(no(i,w)||no(j,w)||f1==w) continue;
            f2=w;break;
        }
        // printf("f1=%d f2=%d\n",f1,f2);
        if(check(j,f1)&&check(i,f2))
        {
            // printf("ok\n");
            return true;
        }
    }
    // printf("nope\n");
    return false;
}
int main()
{
    n=read(),m=read();
    for(int i=1;i<=n;i++) a[i]=read();

    int r=0;
    for(int i=1;i<=n;i++)
    {
        // printf("i=%d\n",i);
        while(r<=n&&!check())
        {
            r++;
            if(r<=n) add(r);
        }
        ans[i]=r;
        del(i);
    }

    for(int i=1;i<=m;i++)
    {
        int l=read(),r=read();
        if(r>=ans[l]) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider scenarios where the code might fail. The code is designed to check if it's possible to form two non-degenerate triangles from a subset of sticks. The main issues in the code could arise from:

1. **Edge cases with small or large stick lengths.**
2. **Queries where the stick lengths are very close to each other.**
3. **Queries where the stick lengths are very different.**
4. **Queries where the number of sticks is exactly 6.**

Given these considerations, let's create a test input that might expose these issues:

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

### Explanation:
- **Stick Lengths:** The stick lengths are chosen to be small and close to each other, which can make it difficult to form non-degenerate triangles.
- **Queries:**
  - **1 6:** The lengths are [1, 2, 2, 3, 3, 4]. This should be a "YES" because we can form two non-degenerate triangles.
  - **2 7:** The lengths are [2, 2, 3, 3, 4, 4]. This should be a "NO" because it's impossible to form two non-degenerate triangles.
  - **2 8:** The lengths are [2, 2, 3, 3, 4, 4, 5]. This should be a "YES" because we can form two non-degenerate triangles.
  - **5 10:** The lengths are [3, 3, 4, 4, 5, 5, 6]. This should be a "NO" because it's impossible to form two non-degenerate triangles.
  - **4 10:** The lengths are [2, 3, 3, 4, 4, 5, 5, 6]. This should be a "YES" because we can form two non-degenerate triangles.

This test input is designed to be small and valid, while also challenging the code to handle edge cases and specific stick length distributions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

Buggy Code:
def can_form_two_triangles(sticks):
    n = len(sticks)
    if n < 6:
        return "NO"
    sticks.sort()
    for i in range(n - 5):
        if sticks[i] + sticks[i + 1] > sticks[i + 2] and \
           sticks[i + 1] + sticks[i + 2] > sticks[i] and \
           sticks[i] + sticks[i + 2] > sticks[i + 1]:
            for j in range(i + 3, n - 2):
                if sticks[j] + sticks[j + 1] > sticks[j + 2] and \
                   sticks[j + 1] + sticks[j + 2] > sticks[j] and \
                   sticks[j] + sticks[j + 2] > sticks[j + 1]:
                    return "YES"
    return "NO"
n,q  = [int(x) for x in input().split()]
sticks = [int(x) for x in input().split()]
for _ in range(q):
        l,r = [int(x) for x in input().split()]
        subarr = sticks[l-1:r]
        res = can_form_two_triangles(subarr)
        if l==1:
            print("YES")
        else:
            print(res)

**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 1
1 2 2 3 3 4
1 6
```

**Explanation:**

- **Test Case Design:**
  - **Number of Sticks (n = 6):** This is the minimum number of sticks required to form two triangles, as per the problem constraints.
  - **Number of Queries (q = 1):** A single query to focus on the edge case.
  - **Stick Lengths:** The stick lengths are `[1, 2, 2, 3, 3, 4]`.
  - **Query (1 6):** The query covers the entire range of sticks.

- **Reasoning:**
  - **Buggy Code Analysis:** The buggy code has a flaw where it always prints "YES" if the query starts from the first stick (`l == 1`). This is incorrect because it does not consider whether the sticks can actually form two non-degenerate triangles.
  - **Challenging Input:** The input `[1, 2, 2, 3, 3, 4]` is designed to test the scenario where the first stick is included in the query. The correct answer should be "NO" because the sticks `[1, 2, 2]` cannot form a non-degenerate triangle (the sum of the two smallest sticks is not greater than the third stick).

This test input should expose the bug in the buggy code, as it will print "YES" incorrectly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

Buggy Code:
from itertools import combinations

def is_valid_triangle(sides):
    """ Проверка, может ли тройка чисел образовать невырожденный треугольник """
    a, b, c = sorted(sides)
    return a + b > c

def check_query(array, l, r):
    """ Проверка одного запроса """
    subarray = array[l-1:r]  # Извлекаем подмассив от l-1 до r-1
    seen_sextets = {}  # Словарь для хранения уникальных шестерок

    # Проверяем все возможные комбинации шестерок
    for sextet in combinations(range(len(subarray)), 6):
        sextet_numbers = tuple(sorted(subarray[i] for i in sextet))  # Сортируем числа в шестерке для уникальности
        if sextet_numbers in seen_sextets:
            continue  # Пропускаем уже обработанную шестерку
        seen_sextets[sextet_numbers] = True

        seen_triangles = {}  # Словарь для хранения уникальных троек
        for triangle1_indices in combinations(range(6), 3):
            triangle1 = tuple(sorted(sextet_numbers[i] for i in triangle1_indices))
            if triangle1 in seen_triangles:
                continue  # Пропускаем уже обработанную тройку
            seen_triangles[triangle1] = True

            if is_valid_triangle(triangle1):
                remaining_indices = [i for i in range(6) if i not in triangle1_indices]
                triangle2 = tuple(sorted(sextet_numbers[i] for i in remaining_indices))
                if triangle2 in seen_triangles:
                    continue  # Пропускаем уже обработанную тройку
                seen_triangles[triangle2] = True

                if is_valid_triangle(triangle2):
                    return "YES"

    return "NO"

def process_queries(n, q, array, queries):
    """ Основная функция для обработки всех запросов """
    results = []
    for l, r in queries:
        result = check_query(array, l, r)
        results.append(result)
    return results

# Ввод данных
def main():
    # Пример ввода данных
    n, q = 10, 5
    array = [5, 2, 2, 10, 4, 10, 6, 1, 5, 3]
    queries = [(1, 6), (2, 7), (2, 8), (5, 10), (4, 10)]

    # Обработка запросов
    results = process_queries(n, q, array, queries)

    # Вывод результатов
    for result in results:
        print(result)

# Запуск программы
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
6 1
1 1 1 1 1 1
1 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        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;
    }

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

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

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n, q;
    long[] a;
    boolean find(List<Long> l, int i){
        return l.get(i-1) + l.get(i-5) > l.get(i) && l.get(i-4) + l.get(i-3) > l.get(i-2);
    }
    public void solve() throws IOException {
        t = 1;
        while (t-- > 0) {
            n =i();
            q = i();
            a = new long[n+1];
            for(int i = 1;i<=n;i++){
                a[i] = i();
            }
            while(q-->0){
                int l = i(), r = i();
                if(r-l+1>50){
                    out.print("YES");
                }
                else{
                    List<Long> list = new ArrayList<>();
                    int[] dp = new int[r-l+1];
                    for(int i = l;i<=r;i++) list.add(a[i]);
                    Collections.sort(list);
                    dp[2] = list.get(0)+list.get(1)>list.get(2)?1:0;
                    for(int i = 3;i<list.size();i++){
                         long x = list.get(i-2) , y = list.get(i-1), z = list.get(i);
                         dp[i] =max(dp[i-1], dp[i-3] + (x+y>z?1:0));
                         if(i==5){
                            dp[i] = max(dp[i], find(list, i)?2:0);
                         } else if(i>5){
                            dp[i] = max(dp[i], dp[i-6] + (find(list , i)?2:0));
                         }
                    }
                    out.print(dp[list.size()-1]>1?"YES":"NO");
                }
                out.print("\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:
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 are likely to fail. The buggy code has a specific condition where it assumes that if the range size is greater than 50, it always outputs "YES". This is a significant flaw because it bypasses the actual logic for checking triangle formation.

### Step-by-Step Reasoning:

1. **Edge Case with Small Range:**
   - Test a small range where the number of sticks is exactly 6. This is the minimum number of sticks required to form two triangles. If the sticks are not properly chosen, it should return "NO".

2. **Edge Case with Large Range:**
   - Test a large range where the number of sticks is greater than 50. The buggy code will always output "YES" for such ranges, which might be incorrect. We need to ensure that the sticks in the range cannot form two non-degenerate triangles.

3. **Mixed Case:**
   - Test a range where the number of sticks is between 6 and 50. This will ensure that the actual logic for checking triangle formation is used.

### Test Input:

```plaintext
10 3
1 1 1 1 1 1 1 1 1 1
1 6
1 51
1 10
```

### Explanation:

1. **First Query (1 6):**
   - The lengths of the sticks are `[1, 1, 1, 1, 1, 1]`.
   - It is impossible to form two non-degenerate triangles with these lengths because all sides are equal, and the sum of any two sides is not greater than the third side.
   - Expected output: "NO".

2. **Second Query (1 51):**
   - The lengths of the sticks are `[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]`.
   - The buggy code will output "YES" because the range size is greater than 50, but it is impossible to form two non-degenerate triangles with these lengths.
   - Expected output: "NO".

3. **Third Query (1 10):**
   - The lengths of the sticks are `[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]`.
   - It is impossible to form two non-degenerate triangles with these lengths.
   - Expected output: "NO".

This test input will help expose the bug in the buggy code by providing scenarios where the actual logic for checking triangle formation should be used, but the buggy code fails to do so.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
10 5
5 2 2 10 4 10 6 1 5 3
1 6
2 7
2 8
5 10
4 10
output:
YES
NO
YES
NO
YES

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

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

public class Main {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args) throws IOException {
        int t = 1;
//        t = Integer.parseInt(reader.readLine());
        while (--t >= 0) solve();
        writer.flush();
        writer.close();
    }

    static void solve() throws IOException {
        String[] tokens = reader.readLine().split(" ");
        int n = Integer.parseInt(tokens[0]), q = Integer.parseInt(tokens[1]);
        tokens = reader.readLine().split(" ");
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = Long.parseLong(tokens[i]);
        int[] p = new int[n];
        int[] tmp = new int[]{-1, -1};
        for (int i = 0; i < n - 6; i++) {
            int r = i;
            while (r < n && tmp[0] < i) {
                tmp = valid(a, i, r + 1);
                r++;
            }
            p[i] = r;
        }
        for (int i = n - 6; i < n; i++) p[i] = 0x3f3f3f3f;
        while (--q >= 0) {
            tokens = reader.readLine().split(" ");
            int l = Integer.parseInt(tokens[0]) - 1, r = Integer.parseInt(tokens[1]);
            writer.write(r >= p[l] ? "Yes\n" : "No\n");
        }
    }

    static int[] valid(long[] num, int l, int r) { // [st, ed]
        int n = r - l + 1;
        if (n < 6) return new int[] {-1, -1};
        for (int a = l; a < r; a++) {
            for (int b = l; b < r; b++) {
                for (int c = l; c < r; c++) {
                    for (int d = l; d < r; d++) {
                        for (int e = l; e < r; e++) {
                            for (int f = l; f < r; f++) {
                                if (a == b || a == c || a == d || a == e || a == f ||
                                        b == c || b == d || b == e || b == f ||
                                        c == d || c == e || c == f ||
                                        d == e || d == f ||
                                        e == f) continue;
                                if (isTriangle(num[a], num[b], num[c]) && isTriangle(num[d], num[e], num[f])) return new int[] {a, f};
                            }
                        }
                    }
                }
            }
        }
        return new int[]{-1, -1};
    }

    static boolean isTriangle(long a, long b, long c) {
        return a + b > c && a + c > b && b + c > 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
6 1
1 2 3 4 5 6
1 6
```

### Explanation:
- **Objective:** To test the edge case where the number of sticks is exactly 6 (the minimum required to form two triangles).
- **Input:**
  - `6 1`: There are 6 sticks and 1 query.
  - `1 2 3 4 5 6`: The lengths of the sticks.
  - `1 6`: The query to check if it's possible to form two triangles using sticks from index 1 to index 6.
- **Expected Output:**
  - `YES`: It is possible to form two non-degenerate triangles with the given stick lengths. For example, the triangles could be `[1, 2, 3]` and `[4, 5, 6]`.

This test case is designed to check if the algorithm can handle the minimum number of sticks required to form two triangles and correctly identify that it is possible to form them. If the code fails to handle this simple case, it indicates a significant issue with the logic.


**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 endl '\n'
using namespace std;

long long t, n, k, i, l, r, mid, ans, a[200001];

long long cnt(long long kol) {
    long long x = 0;
    for (i = 1; i <= n; i++) {
        if (kol / n + (kol % n >= i) > a[i]) {
            kol -= a[i]+1;
            x += a[i];
        } else {
            x += kol;
            break;
        }
    }
    return x;
}

void solve() {
    cin >> n >> k;
    for (i = 1; i <= n; i++) {
        cin >> a[i];
    }
    sort(a + 1, a + n + 1);
    l = k;
    r = k + n;
    ans = LONG_MAX;
    while(l <= r) {
        mid = (l + r) / 2;
        if (cnt(mid) > k) {
            r = mid - 1;
        } else if (cnt(mid) < k) {
            l = mid + 1;
        } else {
            r = mid - 1;
            ans = min(mid, ans);
        }
    }
    cout << ans << endl;
}

int main() {

    cin.tie(0);
    ios::sync_with_stdio(0);

    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 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:
#include<bits/stdc++.h>
using namespace std;
#define int long long int
#define pb push_back
#define all(v) (v.begin(), v.end())
#define nl <<endl
#define pii             pair<int,int>
#define all(v) v.begin(),v.end()
#define vi      vector<int>
#define loop(lcv,lower,upper) for (int lcv = lower; lcv < upper; lcv++)
#define rloop(lcv,upper,lower) for(int lcv = upper;lcv>=lower;lcv--)

#define in(arr,n)   vector<int>arr(n); for(int x = 0; x < n; ++x) cin>>arr[x]
class DisjointSet {
vector<int> rank, parent, size;
public:
DisjointSet(int n) {
rank.resize(n+1, 0);
parent.resize(n+1);
size.resize(n+1);
for(int i = 0;i<=n;i++) {
parent[i] = i;
size[i] = 1;
}
}
int findUPar(int node) {
if(node == parent[node])
return node;
return parent[node] = findUPar(parent[node]);
}
void unionByRank(int u, int v) {
int ulp_u = findUPar(u);
int ulp_v = findUPar(v);
if(ulp_u == ulp_v) return;
if(rank[ulp_u] < rank[ulp_v]) {
parent[ulp_u] = ulp_v;
}
else if(rank[ulp_v] < rank[ulp_u]) {
parent[ulp_v] = ulp_u;
}
else {
parent[ulp_v] = ulp_u;
rank[ulp_u]++;
}
}
void unionBySize(int u, int v) {
int ulp_u = findUPar(u);
int ulp_v = findUPar(v);
if(ulp_u == ulp_v) return;
if(size[ulp_u] < size[ulp_v]) {
parent[ulp_u] = ulp_v;
size[ulp_v] += size[ulp_u];
}
else {
parent[ulp_v] = ulp_u;
size[ulp_u] += size[ulp_v];
}
}
};

unsigned long long power(unsigned long long x,
                     int y, int p)
{
unsigned long long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
while (y > 0)
{
if (y & 1)
res = (res * x) % p;
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
unsigned long long modInverse(unsigned long long n,
int p)
{
return power(n, p - 2, p);
}
unsigned long long nCr(unsigned long long n,
int r, int p)
{
if (n < r)
return 0;
if (r == 0)
return 1;
unsigned long long fac[n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = (fac[i - 1] * i) % p;
return (fac[n] * modInverse(fac[r], p) % p
* modInverse(fac[n - r], p) % p)
% p;
}

//MEX
int findmex(vector<int>v){
set<int>s;
loop(i,0,v.size()){s.insert(v[i]);}
 int MEX{},i{};
for(auto j:s){
if(j!=i)break;
else i++;
}
MEX = i;
return MEX;
 }

//Check prime
bool is_prime(int n){
if(n==1){
return false;
}
for(int i=2; i*i<=n; i++){
if(n%i==0){
return false;
}
}
return true;
}

long long binpow(long long a, long long b) {
if (b == 0)
return 1;
long long res = binpow(a, b / 2);
if (b % 2)
return res * res * a;
else
return res * res;
}

void build(vector<int>&arr,int v,int l,int r,vector<int>&seg){
if(l==r){
seg[v]=arr[l];
}
else {
int mid=(l+r)/2;
build(arr,v*2,l,mid,seg);
build(arr,v*2+1,mid+1,r,seg);
seg[v]=seg[v*2]+seg[v*2 + 1];
}
}
int query(int v, int segl,int segr,int l,int r, vector<int>&seg){
if(l>r)return 0;
if(segl>r || segr<l) return 0;
if(l==segl && r==segr) return seg[v];
int mid=(segl+segr)/2;
return  query(v*2,segl,mid,l,min(r,mid),seg)+query(v*2+1,mid+1,segr,max(l,mid+1),r,seg);
}
void update(int v,int segl,int segr,int pos,int new_val,vector<int>&seg){
if(segl == segr){
seg[v] = new_val;
}
else{
int mid= (segl+segr)/2;
if(pos <= mid)update(v*2,segl,mid,pos,new_val,seg);
else  update(v*2+1,mid+1,segr,pos,new_val,seg);
seg[v] = seg[v*2] + seg [v*2+1];
}
}

void solve(){
        int n,k;cin>>n>>k;
        vector<int>arr(n);
        for(int i=0;i<n;i++)cin>>arr[i];
        //in(arr,n);
        int ans=0,sum=0;
        sort(all(arr));
        int a=k;
        map<int,int>mp;
        for (int i = 0; i < n; i++)
        {
            mp[arr[i]]++;
        }
        int val=0,prev=0,count=0;
        for(auto x:mp){
            if((x.first-val)*(n-count)>=k)break;
            else{
               ans+=x.second;

               k-=(x.first+1-prev)*(n-count);
               prev=val;
               val=x.first+1;
               count+=x.second;

            }
        }
        cout<<ans+a<<endl;
        // for(int i=0;i<n;i++){
        //     if(arr[i]==sum
        //     )continue;
        //     if(arr[i]*(n-i)>=k)break;
        //     else{
        //         ans+=mp[arr[i]];
        //         k-=mp[arr[i]];
        //         sum=arr[i];
        //     }
        // }
        // cout<<ans+val<<endl;

}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Identify Common Bugs**:
   - The code might have issues with how it handles the distribution of cans across slots.
   - The logic for calculating the minimum number of presses might be flawed, especially in cases where the distribution is uneven.

2. **Edge Cases**:
   - **Single Slot**: Test with a single slot to ensure the code handles it correctly.
   - **All Slots Equal**: Test with all slots having the same number of cans.
   - **Uneven Distribution**: Test with a highly uneven distribution of cans.
   - **Large k**: Test with k close to the total number of cans.
   - **Small k**: Test with k much smaller than the total number of cans.

3. **Generate Test Input**:
   - We will create a test case with a small number of slots and a specific distribution of cans to ensure it is a valid and complete test input.

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

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

### Explanation of the Test Input:
1. **Test Case 1**:
   - **n = 1, k = 1, a = [1]**
   - There is only one slot with one can. The minimum number of presses needed to get one can is 1.

2. **Test Case 2**:
   - **n = 2, k = 2, a = [1, 1]**
   - There are two slots, each with one can. The minimum number of presses needed to get two cans is 2.

3. **Test Case 3**:
   - **n = 3, k = 4, a = [1, 2, 3]**
   - There are three slots with 1, 2, and 3 cans respectively. The minimum number of presses needed to get four cans is 5. This is because you can press the button corresponding to the slot with 3 cans twice, and then press the other two buttons once each.

This test input covers a range of scenarios, including single slots, equal distribution, and uneven distribution, and should help expose any bugs in the given 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;
using ll = long long;
#define TryAndCry cin.tie(nullptr), cout.tie(nullptr), ios::sync_with_stdio(false)
#define log(x) (31 ^ __builtin_clz(x))
// freopen('', 'r', stdin);
// freopen('', 'w', stdout);

void solve()
{
    ll n, k, ans = 0, have = 0;
    cin >> n >> k;
    ll v[n];
    for (auto &i : v)
        cin >> i;
    ll sz = (k / n) + (k % n != 0);
    for (ll i = 0; i < n; i++)
    {
        if (v[i] < sz)
            ans++;
    }
    cout << ans + k << '\n';
}

int main()
{
    TryAndCry;
    ll t = 1;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the code's logic fails to correctly determine the minimum number of button presses required to guarantee receiving at least \( k \) cans of lemonade. The code's logic seems to be incorrect in handling cases where the distribution of cans across slots is uneven, especially when the number of presses required is not directly proportional to the number of slots and the number of cans.

### Step-by-Step Analysis:

1. **Understanding the Bug**:
   - The code calculates a `sz` value which is the average number of cans that should be in each slot to meet the requirement of \( k \) cans.
   - It then counts how many slots have fewer cans than `sz` and adds this count to \( k \) to determine the total number of presses.
   - This logic fails to account for the possibility of pressing the same slot multiple times and the need to adapt the strategy based on the outcome of each press.

2. **Edge Cases**:
   - **Uneven Distribution of Cans**: If the cans are distributed unevenly across the slots, the code might overestimate or underestimate the number of presses needed.
   - **Large \( k \) with Few Slots**: If \( k \) is large and the number of slots \( n \) is small, the code might not correctly handle the required number of presses.
   - **Small \( k \) with Large \( n \)**: If \( k \) is small and the number of slots \( n \) is large, the code might not correctly handle the minimum number of presses needed.

3. **Generating a Test Case**:
   - **Uneven Distribution**: Consider a case where one slot has a large number of cans and the others have very few.
   - **Large \( k \) with Few Slots**: Consider a case where \( k \) is large and \( n \) is small.

### Test Case:
```plaintext
1
3 5
1 1 5
```

### Explanation:
- **Input**:
  - \( n = 3 \) (3 slots)
  - \( k = 5 \) (5 cans needed)
  - Initial number of cans in slots: \( 1, 1, 5 \)

- **Expected Output**:
  - The minimum number of button presses needed to guarantee receiving at least 5 cans of lemonade is 5.
  - **Correct Strategy**:
    - Press the button corresponding to the slot with 5 cans (the third slot) 5 times. This will guarantee that you receive 5 cans of lemonade.

- **Why This Test Case Exposes the Bug**:
  - The code's logic will calculate `sz` as 2 (since \( \text{ceil}(5 / 3) = 2 \)).
  - It will then count how many slots have fewer than 2 cans, which is 2 (the first and second slots).
  - The code will output `2 + 5 = 7`, which is incorrect.
  - The correct answer is 5, as pressing the button corresponding to the slot with 5 cans 5 times will guarantee the required 5 cans.

This test case will expose the bug in the given code by demonstrating that the logic for determining the minimum number of presses is flawed.


**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 math
t = int(input())
ans = []

for i in range(t):
    n,k = map(int, input().split())
    a = list(map(int, input().split()))
    a.sort()
    p = k/n
    count = 0
    for x in a:
        if x<p:
            count+=1
            p = (k-x)/(n-count)
    ans.append(k+count)
for i in range(t):
    print(ans[i])

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

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

Think step by step.

Response:
```plaintext
3
3 4
1 2 3
4 6
1 1 2 2
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:
from bisect import bisect_left as bl
from bisect import bisect_right as br
from math import inf,log2,ceil,sqrt,gcd
from collections import defaultdict,deque,Counter
from bisect import bisect_left as bl
import sys; input=sys.stdin.readline
tt=1
tt = int(input())
for _ in range(tt):
    #n=int(input())
    n,k=map(int,input().split())
    a=list(map(int,input().split()))
    #s=input().strip()
    #s=list(input())
    cnt=Counter(a)
    s=sorted(set(a))
    if n>=k:
        print(k)
        continue
    ans=0;lst=0;lst0=0;mn=min(a)
    for x in s:
        if k>x*(n-lst):
            ans+=(x-lst0)*(n-lst)
            k-=(x-lst0)*(n-lst)
            lst+=cnt[x]
            lst0=x
        else:
            if k<=cnt[x]:
                ans+=k
            else:
                ans+=(k+cnt[x]-1)//cnt[x]*cnt[x]
            break
        ans+=cnt[x]
    print(ans)

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

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

Think step by step.

Response:
```plaintext
5
2 3
2 2
3 6
3 3 3
4 10
5 5 5 5
5 15
1 2 3 4 5
10 50
1 1 3 8 8 9 12 13 27 27
```

### Explanation:

1. **Test Case 1:**
   - **Input:** `2 3`
   - **Array:** `[2, 2]`
   - **Expected Output:** `3`
   - **Reason:** Each slot has 2 cans. To get 3 cans, you need to press each button twice (since the second press on each button will definitely get a can).

2. **Test Case 2:**
   - **Input:** `3 6`
   - **Array:** `[3, 3, 3]`
   - **Expected Output:** `6`
   - **Reason:** Each slot has 3 cans. To get 6 cans, you need to press each button twice (since the second press on each button will definitely get a can).

3. **Test Case 3:**
   - **Input:** `4 10`
   - **Array:** `[5, 5, 5, 5]`
   - **Expected Output:** `10`
   - **Reason:** Each slot has 5 cans. To get 10 cans, you need to press each button twice (since the second press on each button will definitely get a can).

4. **Test Case 4:**
   - **Input:** `5 15`
   - **Array:** `[1, 2, 3, 4, 5]`
   - **Expected Output:** `15`
   - **Reason:** The total number of cans is 15. To get all 15 cans, you need to press each button a number of times equal to the number of cans in that slot.

5. **Test Case 5:**
   - **Input:** `10 50`
   - **Array:** `[1, 1, 3, 8, 8, 9, 12, 13, 27, 27]`
   - **Expected Output:** `53`
   - **Reason:** The total number of cans is 107. To get 50 cans, you need to press buttons in a strategic manner to ensure you get the required cans. The expected minimum number of button presses is 53, which is a challenging case for the buggy code to handle correctly.

These test cases are designed to expose potential issues in the buggy code, especially in handling scenarios where the number of cans in each slot is uniform and where the total number of cans is significantly larger than the required number of cans.


**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 count(a:list,x:int):
    c = 0
    cc = 1
    k = 0
    u = 0
    while a and x>k and cc:

        p=((x-k)/len(a))+u
        if p == int(p):
            q = exception(a,int(p-1))
            b,cc = q[0],q[1]
        else:
            q = exception(a,int(p))
            b,cc = q[0],q[1]
        k += sum(a[:cc]) - u*(len(a)-cc) + (a[cc-1]-u)*(len(a)-cc)
        u = a[cc-1]
        a = b
        c += cc
    return x+c

def exception(a,x):
    i = 0
    while i<len(a) and a[i]<=x:
        i+=1
    return a[i:],i

for i in range(int(input())):
    m,x = map(int,input().split(' '))
    a = sorted(list(map(int,input().split(' '))))
    print(count(a,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
1
3 5
1 2 3
```


**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 B_Buying_Lemonade {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int req = sc.nextInt();
            int[] arr = new int[n];
            for(int i=0;i<n;i++){
                arr[i] = sc.nextInt();
            }
            double min = req / (double)n;

            int count = 0;
            Arrays.sort(arr);
            for(int i=0;i<n;i++){
                if(arr[i] < min){
                    count++;
                }
            }

            long ans = 0;
            for(int i=0;i<n;i++){
                if(req > 0){
                    ans+= Math.min(req,arr[i]);
                    req -= arr[i];
                }
                if(count > 0){
                    ans++;
                    count--;
                }
            }

            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
5
2 2
1 1
2 3
1 2
3 6
2 1 3
10 50
1 1 1 1 1 1 1 1 1 1
2 3
1 2
```


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

        int t = sc.nextInt();

        for (int test = 0; test < t; test++) {
            int n = sc.nextInt();
            int k = sc.nextInt();

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

            Arrays.sort(lis);

            int[] sortedLis = new int[n + 1];
            sortedLis[0] = 0;
            System.arraycopy(lis, 0, sortedLis, 1, n);

            long val = 0;
            long ops = 0;

            for (int i = 1; i < n; i++) {
                long adj = sortedLis[i] - sortedLis[i - 1];
                long remainingSlots = n - i + 1;

                if (val + remainingSlots * adj >= k) {
                    ops += (k - val);
                    break;
                }

                long x = remainingSlots * adj;
                ops += x + 1;
                val += x;
            }

            System.out.println(ops);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
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
```