**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
//  _____   _   _____    _____   _____   _      __    __
// |  ___| | | |  _  \  | ____| |  ___| | |     \ \  / /
// | |__   | | | |_| |  | |__   | |__   | |      \ \/ /
// |  __|  | | |  _  /  |  __|  |  __|  | |       \  /
// | |     | | | | \ \  | |___  | |     | |___    / /
// |_|     |_| |_|  \_\ |_____| |_|     |_____|  /_/
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define lowbit(i) ((i) & -(i))
#define ull unsigned long long
// #define int long long
// #define i64 long long
#define endl '\n'
using namespace std;
int cmax(int &a, int b) { return a < b ? a = b : a; }
int cmin(int &a, int b) { return a > b ? a = b : a; }
const int mod = 998244353;
const int N = 1e4 + 1;
bitset<N> d[32], q, h;
int a[200005];
int b[200005];
int c[200005];
int ct = 0, kkkk, tes = -1;
void solve()
{
    ct += kkkk;
    int n, m, cnt = 0, res = 0;
    cin >> n >> m;
    if (ct == tes)
        cout << n << ' ' << m << endl;
    for (int i = 0; i < 32; i++)
        d[i].reset();
    int f = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        if (ct == tes)
            cout << a[i] << ' ';
        if (a[i] == 0)
            f = i;
    }
    if (f)
    {
        c[0] = 0;
        d[0][0] = 1;
        q.reset(), h.reset();
        h[0] = 1;
        int kk = 0;
        for (int i = 1; i <= n; i++)
        {
            if (a[i] == 0)
            {
                if (i == 1 || a[i - 1] != 0)
                {
                    q = h;
                    h.reset();
                }
                kk = 1;
                continue;
            }
            if (a[i] == 1)
            {
                if (a[i - 1] == 1)
                    b[i] = b[i - 1];
                else
                    b[i] = i - 1;
                cnt++;
                if (i == n || a[i + 1] != 1)
                {
                    res++;
                    if (res == 32)
                        res = 0;
                    c[i] = res;
                    d[res].reset();
                    for (int j = 0; j <= cnt; j++)
                    {
                        if (kk == 0)
                            d[res] |= (d[c[b[i]]] << j);
                        d[res] |= (q << j);
                    }
                    d[res] |= q;
                    h |= d[res];
                    cnt = 0;
                    // cout << i << endl;
                    // for (int i = 0; i <= 20; i++)
                    //     cout << d[res][i] << ' ';
                    // cout << endl;
                }
                continue;
            }
            kk = 0;
            b[i] = i - 1;
            res++;
            if (res == 32)
                res = 0;
            c[i] = res;
            d[res].reset();
            int now = 1, k = i;
            while (k != 0 && now)
            {
                now *= a[k];
                if (now > m)
                    break;
                k = b[k];
                d[res] |= (d[c[k]] << now);
                d[res] |= (q << now);
                // if (a[i] == 3)
                    // cout << "now "<<now << ' ';
            }
            d[res] |= q;
            // cout << i << endl;
            // for (int i = 0; i <= 20; i++)
            //     cout << d[res][i] << ' ';
            // cout << endl;
            h |= d[res];
            // for (int i = 0; i <= 20; i++)
            //     cout << h[i] << ' ';
            // cout << endl;
        }
        if (a[n] == 0)
            d[res] |= q;
        // for (int i = 0; i <= 20; i++)
        //     cout << d[res][i] << ' ';
        // cout << endl;
        // for (int i = 0; i <= 20; i++)
        //     cout << q[i] << ' ';
        // cout << endl;
        if (!ct)
            cout << (d[res][m] ? "YES\n" : "NO\n");
        return;
    }
    else
    {
        c[0] = 0;
        d[0][0] = 1;
        for (int i = 1; i <= n; i++)
        {
            if (a[i] == 1)
            {
                if (a[i - 1] == 1)
                    b[i] = b[i - 1];
                else
                    b[i] = i - 1;
                cnt++;
                if (i == n || a[i + 1] != 1)
                {
                    res++;
                    if (res == 32)
                        res = 0;
                    c[i] = res;
                    d[res].reset();
                    for (int j = 0; j <= cnt; j++)
                    {
                        d[res] |= (d[c[b[i]]] << j);
                    }
                }
                continue;
            }
            cnt = 0;
            b[i] = i - 1;
            res++;
            if (res == 32)
                res = 0;
            c[i] = res;
            d[res].reset();
            int now = 1, k = i;
            while (k != 0)
            {
                now *= a[k];
                if (now > m)
                    break;
                k = b[k];
                d[res] |= (d[c[k]] << now);
            }
        }
        if (!ct)
            cout << (d[res][m] ? "YES\n" : "NO\n");
        return;
    }
    return;
}
signed main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int tt = 1;
    cin >> tt;
    if (tt > 10)
        kkkk = tes != -1;
    while (tt--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
#pragma GCC optimize("O3")

#include <algorithm>
#include <bitset>
#include <iostream>
#include <map>
#include <vector>

using namespace std;

const int N = 1e4 + 28;

void solve() {
    int n, m; cin >> n >> m;
    vector <int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    map <int, bitset<N>> mp;
    mp[1][0] = true;
    bitset<N> all; all[0] = true;
    for (int i = 0; i < n; ++i) {
        int x = a[i];
        map <int, bitset<N>> mp1;
        if (x) {
            for (auto [last, res] : mp) {
                if (last * x <= m && last * x - last >= 0) {
                    mp1[last * x] |= (res << (last * x - last));
                    all |= (res << (last * x - last));
                }
                mp1[x] |= (res << x);
                all |= (res << x);
            }
        } else {
            mp1[0] |= all;
        }
        mp = mp1;
    }
    for (auto [last, res] : mp) {
        if (res[m]) {
            cout << "YES" << '\n'; return;
        }
    }
    cout << "NO" << '\n';
}

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
import math
t = int(input())
while (t>0):
  n,m = map(int,input().split())
  a = list(map(int,input().split()))
  zero = []
  for i in range(n):
      if a[i]==0:
          zero.append(i)
  zero.append(n)
  prevget = set()
  prevget.add((a[0],a[0]))
  currget = set()
  lateadd = set()
  next = 0
  for i in range(1,n):
    if i == zero[next]:
        next += 1
        prevget = prevget.union(lateadd)
        lateadd.clear()
    for j in prevget:

      if(j[0]<=(m)):
        currget.add((j[0] + a[i],a[i]))
        currget.add((j[0] - j[1] + j[1] * a[i],j[1] * a[i]))
      if(j[0]-j[1]<=m and a[i] != 0):
        lateadd.add((j[0]-j[1],0))
      elif(j[0]-j[1]<=m):
        currget.add((j[0]-j[1],0))
    prevget = currget
    currget = set()
  prevget = prevget.union(lateadd)
  ans = [tup[0] for tup in prevget]
  if(m in ans):
    print("YES")
  else:
    print("NO")
  t -= 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
import math
t = int(input())
while (t>0):
  n,m = map(int,input().split())
  a = list(map(int,input().split()))
  prevget = set()
  prevget.add((a[0],a[0]))
  currget = set()
  for i in range(1,n):
    for j in prevget:
      if(j[0] + a[i]<=m):
        currget.add((j[0] + a[i],a[i]))
      if(j[0] - j[1] + j[1] * a[i]<=m):
        currget.add((j[0] - j[1] + j[1] * a[i],j[1] * a[i]))
    prevget = currget
    currget = set()
  ans = [tup[0] for tup in prevget]
  if(m in ans):
    print("YES")
  else:
    print("NO")
  t -= 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

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

public class WonderlandEscape {
    static final int MOD = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();

        int t = Integer.parseInt(reader.readLine());
        while (t-- > 0) {
            StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
            int n = Integer.parseInt(tokenizer.nextToken());
            int m = Integer.parseInt(tokenizer.nextToken());

            tokenizer = new StringTokenizer(reader.readLine());
            int[] a = new int[n];
            int sum = 0;
            int max = 0;
            boolean hasZero = false;

            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(tokenizer.nextToken());
                sum += a[i];
                max = Math.max(max, a[i]);
                if (a[i] == 0) hasZero = true;
            }

            if (sum == m) {
                output.append("YES\n");
            } else if (sum < m) {
                output.append("NO\n");
            } else if (sum > m) {
                output.append(checkCombinations(a, n, m) ? "YES\n" : "NO\n");
            }
        }

        System.out.print(output);
    }

    private static boolean checkCombinations(int[] a, int n, int target) {
        Set<Integer> dp = new HashSet<>();
        dp.add(0);

        for (int num : a) {
            Set<Integer> newDp = new HashSet<>(dp);
            for (int value : dp) {
                if (value + num == target || value * num == target) return true;
                newDp.add(value + num);
                newDp.add(value * num);
            }
            dp = newDp;
        }

        return dp.contains(target);
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 10000
10000
2 10000
5000 5000
3 10000
1 9999 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Objects;
import java.util.Queue;
import java.util.Scanner;

public class CombinationProSum
{
	private static class CombinationHolder implements Cloneable
	{
		private Integer first;
		private Integer second;

		private boolean isSecondPresent()
		{
			return Objects.nonNull(second);
		}

		@Override public CombinationHolder clone()
		{
			try
			{
				CombinationHolder clone = (CombinationHolder) super.clone();
				return clone;
			}
			catch(CloneNotSupportedException e)
			{
				throw new AssertionError();
			}
		}
	}
	public static void main(String[] args)
	{
		Scanner scanner = new Scanner(System.in);
		int testCase = scanner.nextInt();
		for(int i = 0; i < testCase; i++)
		{
			int size = scanner.nextInt();
			int target = scanner.nextInt();
			scanner.nextLine();

			// Initialize the array
			int[] arr = new int[size];

			// Read the second line into the array
			for (int j = 0; j < size; j++) {
				arr[j] = scanner.nextInt();
			}

			System.out.println(isCombinationPresent(target, arr)?"yes":"no");
		}
	}
	private static boolean isCombinationPresent(int target, int[] numbers)
	{
		if(0 == target && Arrays.stream(numbers).anyMatch(number -> 0==number)) return true;
		Queue<CombinationHolder> queue = new LinkedList<>();
		CombinationHolder combinationHolder = new CombinationHolder();
		combinationHolder.first = numbers[0];
		queue.offer(combinationHolder);
		int temp, size;
		CombinationHolder element;
		for(int i = 1; i < numbers.length; i++)
		{
			temp = numbers[i];
			size = queue.size();
			for(int j = 0; j < size; j++)
			{
				element = queue.poll();
				proNumber(queue, element.clone(), temp, target);
				addNumber(queue, element.clone(), temp, target);
			}
		}

		return queue.stream().anyMatch(answer -> target == value(answer));
	}

	private static void addNumber(Queue<CombinationHolder> queue, CombinationHolder holder, int number, int target)
	{
		if(holder.isSecondPresent())
		{
			holder.first+=holder.second;
		}
		holder.second = number;
		addQueue(queue, holder, target);
	}

	private static void proNumber(Queue<CombinationHolder> queue, CombinationHolder holder, int number, int target)
	{
		if(holder.isSecondPresent()) holder.second*=number;
		else holder.first*=number;
		addQueue(queue, holder, target);
	}

	private static int value(CombinationHolder holder)
	{
		return takeValueIfNotNull(holder.first)+takeValueIfNotNull(holder.second);
	}

	private static int takeValueIfNotNull(Integer value)
	{
		return Objects.isNull(value)?0:value;
	}

	private static void addQueue(Queue<CombinationHolder> queue, CombinationHolder holder, int target)
	{
		if(value(holder)<=target) queue.offer(holder);
	}
}

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

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

Think step by step.

Response:
```plaintext
3
1 10000
10000
1 1
0
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: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
#include <stack>
#include <bitset>
#include <string>
#include <cstring>
#include <iterator>
#include <random>
#define ll long long
#define ld long double
#define inf (ll)(2*1e18)
#define sort(a) sort(a.begin(), a.end())
#define reverse(a) reverse(a.begin(), a.end())
#define pb push_back
// #define endl "\n"
using namespace std;

vector<bitset<10007>> dp;
bitset<10007> zero, all, buf;

void solve() {
    ll n, m, i, j, k, x;
    cin>>n>>m;
    vector<ll> a(n+1);
    vector<ll> pref(n+1);
    dp.clear();
    dp.resize(n+1);
    for(i=1;i<=n;++i){
        cin>>a[i];
        if(a[i] == 1){
            pref[i] = pref[i-1];
        }else{
            pref[i] = i;
        }
    }
    dp[0][0] = true;
    all[0] = true;
    for(i=1;i<=n;++i){
        if(a[i] == 0){
            zero = all;
            dp[i] = zero;
            continue;
        }
        x = 1;
        for(j=i;j>0;--j){
            x *= a[j];
            if(x > m+7) break;
            dp[i] |= zero;
            buf = dp[j-1];
            buf <<= x;
            dp[i] |= buf;
//            for(k=0;k<=m;++k){
//                if(k >= x){
//                    if(dp[j-1][k-x]){
//                        dp[i][k] = true;
//                    }
//                }
//            }
            if(x == 0) break;
        }
        all |= dp[i];
    }
//    for(i=0;i<=n;++i) {
//        for(j=0;j<=m;++j) {
//            cout<<dp[i][j]<<" ";
//        }
//        cout<<endl;
//    }
    for(i=0;i<10007;++i){
        zero[i] = false;
        all[i] = false;
    }
    if(dp.back()[m]){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    srand(time(nullptr));
    ll t=1;
    cin>>t;
    for(;t>0;--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
1
10 10000
0 0 0 0 0 0 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

// Evaluate expression following order of operations
long long evaluate(const vector<int>& nums, long long mask) {
    int n = nums.size();
    vector<long long> values(n);
    vector<bool> used(n, false);

    // Copy initial numbers
    for(int i = 0; i < n; i++) {
        values[i] = nums[i];
    }

    // First handle multiplications
    for(int i = 0; i < n-1; i++) {
        if(!(mask & (1LL << i))) { // if multiplication
            values[i+1] = values[i] * values[i+1];
            used[i] = true;
        }
    }

    // Then handle additions
    long long result = 0;
    for(int i = 0; i < n; i++) {
        if(!used[i]) {
            result += values[i];
        }
    }

    return result;
}

bool solve(const vector<int>& nums, int target) {
    int n = nums.size();

    // Try all possible combinations
    long long maxMask = 1LL << (n-1);
    for(long long mask = 0; mask < maxMask; mask++) {
        if(evaluate(nums, mask) == target) {
            return true;
        }
    }

    return false;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

    while(t--) {
        int n, m;
        cin >> n >> m;

        vector<int> nums(n);
        for(int i = 0; i < n; i++) {
            cin >> nums[i];
        }

        bool possible = solve(nums, m);
        cout << (possible ? "YES\n" : "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:
```plaintext
1
3 10
5 0 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
def can_make_target(arr, target):
    n = len(arr)
    if n == 1:
        return arr[0] == target

    # Try all possible combinations of operations
    def solve(idx, expr):
        if idx == n:
            # Evaluate expression following order of operations
            # First, split by '+' and evaluate each multiplication group
            terms = expr.split('+')
            total = 0
            for term in terms:
                product = 1
                for num in term.split('*'):
                    product *= int(num)
                total += product
            return total == target

        # Try multiplication
        if idx > 0:
            # Add multiplication
            if solve(idx + 1, expr + '*' + str(arr[idx])):
                return True

        # Try addition
        if solve(idx + 1, expr + '+' + str(arr[idx])):
            return True

        return False

    # Start with first number
    return solve(1, str(arr[0]))

def solve_test_case():
    n, m = map(int, input().split())
    arr = list(map(int, input().split()))

    result = can_make_target(arr, m)
    print("YES" if result else "NO")

# Handle multiple test cases
t = int(input())
for _ in range(t):
    solve_test_case()

**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 10
1 2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
from itertools import product

def evaluarCombinaciones(nums, target):
    operadores = ['+', '*']

    for ops in product(operadores, repeat=len(nums)-1):
        expr = str(nums[0])

        for i in range(1, len(nums)):
            expr += ' ' + ops[i-1] + ' ' + str(nums[i])

        if eval(expr) == target:
            return True
    return False

def main():
    t = int(input())
    for _ in range(t):
        n, m = map(int, input().split())
        resultados = list(map(int, input().split()))
        if evaluarCombinaciones(resultados, m):
            print("yes")
        else:
            print("no")

main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
def can_make_target(arr, m):
    if not arr:
        return False

    n = len(arr)
    sett = set(arr)

    if m == 0 and 0 in sett:
        return True

    stack = [(1, arr[0], arr[0])]
    while stack:
        index, current_value, previous_operand = stack.pop()

        if index == n:
            if current_value == m:
                return True
            continue

        next_operand = arr[index]

        stack.append((index + 1, current_value + next_operand, next_operand))

        new_value = current_value - previous_operand + previous_operand * next_operand
        stack.append((index + 1, new_value, previous_operand * next_operand))

    return False

t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    arr = list(map(int, input().split()))
    if can_make_target(arr, m):
        print("YES")
    else:
        print("NO")

**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 10
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in Addition
time_limit_ms: 3000
memory_limit_mb: 32
Description: Note that the memory limit is unusual.

The Cheshire Cat has a riddle for Alice: given $n$ integers $a_1, a_2, \ldots, a_n$ and a target $m$, is there a way to insert $+$ and $\times$ into the circles of the expression $$a_1 \circ a_2 \circ \cdots \circ a_n = m$$ to make it true? We follow the usual order of operations: $\times$ is done before $+$.

Although Alice is excellent at chess, she is not good at math. Please help her so she can find a way out of Wonderland!

Input: Each test contains multiple test cases. The first line of input 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 $n, m$ ($1\le n\le 2\cdot 10^5$; $1\le m\le 10^4$) — the number of integers and the target, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0\le a_i\le 10^4$) — the elements of the array $a$.

The sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output "YES" without quotes if it is possible to get the target by inserting $+$ or $\times$ and "NO" otherwise.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
5 4
2 1 1 1 2
5 5
2 1 1 1 2
5 6
2 1 1 1 2
5 7
2 1 1 1 2
5 8
2 1 1 1 2
5 6
2 0 2 2 3
output:
YES
YES
YES
YES
NO
YES

Note: Possible solutions for the first four test cases are shown below. $$\begin{align*} 2 \times 1 + 1 \times 1 \times 2 &= 4 \\\ 2 \times 1 + 1 + 1 \times 2 &= 5 \\\ 2 \times 1 + 1 + 1 \times 2 &= 6 \\\ 2 + 1 + 1 + 1 + 2 &= 7 \\\ \end{align*}$$ It is impossible to get a result of $8$ in the fifth test case.

Buggy Code:
// coached by rainboy
import java.io.*;
import java.util.*;

public class CF2028F extends PrintWriter {
	CF2028F() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF2028F o = new CF2028F(); o.main(); o.flush();
	}

	static final int M = 10000, K = 13;	// K = floor(log(M))

	boolean test(long[] aa, int o) {
		return (aa[o >> 6] & 1L << (o & 0x3f)) != 0;
	}

	void set(long[] aa, int o) {
		aa[o >> 6] |= 1L << (o & 0x3f);
	}

	int l;

	void set(long[] aa, long[] bb) {
		for (int i = 0; i < l; i++)
			aa[i] = bb[i];
	}

	void or(long[] aa, long[] bb) {
		for (int i = 0; i < l; i++)
			aa[i] |= bb[i];
	}

	void or(long[] aa, long[] bb, int o) {
		int p = o >> 6, q = o & 0x3f;
		for (int i = l - 1, j = l - 1 - p; j >= 0; i--, j--) {
			long b = bb[j] << q;
			if (j > 0 && q > 0)
				b |= bb[j - 1] >> 0x40 - q;
			aa[i] |= b;
		}
	}

	void main() {
		for (int t = sc.nextInt(); t-- > 0; ) {
			int n = sc.nextInt();
			int m = sc.nextInt();
			int[] aa = new int[n];
			for (int i = 0; i < n; i++)
				aa[i] = sc.nextInt();

			l = (m >> 6) + 1;
			int[] qu = new int[K]; Arrays.fill(qu, -1);
			int tail = 0;
			long[][] q1_ = new long[K][l];
			long[] bs = new long[l];
			long[] p0 = new long[l];
			long[] q1 = new long[l];
			long[] pp = new long[l];
			set(q1, 0);
			set(pp, 0);
			for (int i0 = -1, i = 0; i < n; i++) {
				int a = aa[i];
				if (a == 0) {
					i0 = i;
					set(p0, pp);
					set(bs, pp);
					set(q1, bs);
				} else if (a == 1) {
					if (i == 0)
						set(bs, 1);
					else
						or(bs, bs, 1);
					or(q1, bs);
				} else {
					set(q1_[tail], q1);
					qu[tail] = i;
					if (++tail == K)
						tail = 0;
					a = 1;
					set(bs, p0);
					for (int g = 1; g <= K; g++) {
						int h = tail - g;
						if (h < 0)
							h += K;
						int j = qu[h];
						if (j == -1)
							break;
						if (j < i0)
							break;
						if ((a *= aa[j]) > m)
							break;
						or(bs, q1_[h], a);
					}
					set(q1, bs);
				}
				or(pp, bs);
			}
			println(test(bs, m) ? "YES" : "NO");
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
5 10000
10000 0 0 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
from collections import deque
import  math

class FENWICK_TREE():
    def __init__(self,L):
        pre=[0]
        pre.extend(L)
        n=len(pre)
        for i in range(1,n):
            pre[i]+=pre[i-1]
        self.tree=[0]*n
        for i in range(1,n):
            self.tree[i]=pre[i]-pre[i-(i&-i)]
    def update(self,m,n):
        ind=m
        while ind<=len(self.tree):
            self.tree[ind]+=n
            ind+=ind&-ind
    def su(self,l,r):
        ans1=0
        l-=1
        while r>0:
            ans1+=self.tree[r]
            r-=r&-r
        ans2=0
        while l>0:
            ans2+=self.tree[l]
            l-=l&-l
        return ans1-ans2

import bisect
# from functools import cache
# @cache
for _ in range(1):
    n,q=MII()
    L=LMI()
    D={}
    for i in range(1,n+1):
        D[i]=0
    ind={}
    for i in range(n):
        if L[i] not in ind:
            ind[L[i]]=[]
        ind[L[i]].append(i)
    ans=[[] for _ in range(n+1)]
    X=FENWICK_TREE([1]*n)
    for i in range(n):
        if i in ind:
            for j in ind[i]:
                X.update(j+1,-1)
        Y=set()
        for x in D:
            if X.su(D[x]+1,n)<x:
                Y.add(x)
                continue
            l=D[x]+1
            h=n
            while l<=h:
                mid=(l+h)//2
                if X.su(D[x]+1,mid)>=x:
                    res=mid
                    h=mid-1
                else:
                    l=mid+1
            ans[x].append(res)
            D[x]=res
        for j in Y:
            del D[j]

    for _ in range(q):
        i,x=MII()
        cc=bisect.bisect_left(ans[x],i)-1
        if cc+1>=L[i-1]:
            print("NO")
        else:
            print("YES")

# if __name__=="__main__":

#     # for _ in range(II()):
#     #     t()
#     t()

**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 10
2 1 3 2 1
1 1
2 1
3 1
4 1
5 1
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
#define ll  long long
#define F_PS ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define sz(s) (int) (s.size())
#define all(vect) vect.begin(), vect.end()
struct segment_tree {
    int sz;
    vector<vector<int>> seg;
    vector<int> v;

    segment_tree(vector<int> &arr, int n) {
        v = arr;
        sz = n;
        seg = vector<vector<int>>(n * 4);
        build(1, 1, n);
    }
    vector<int>merge(auto &a,auto &b) {
        vector<int> res;
        int i = 0, j = 0;
        while (i < sz(a) and j < sz(b)) {
            if (a[i] < b[j])
                res.push_back(a[i++]);
            else
                res.push_back(b[j++]);
        }
        while (i < sz(a))
            res.push_back(a[i++]);
        while (j < sz(b))
            res.push_back(b[j++]);
        return res;
    }
    void build(int n, int l, int r) {
        if (l == r) {
            seg[n].push_back(v[l]);
            return;
        }
        int mid = (l + r) / 2;
        build(n * 2, l, mid);
        build(n * 2 + 1, mid + 1, r);
        seg[n] = merge(seg[n * 2], seg[n * 2 + 1]);
    }
    int get(int n,int l,int r,int st,int end,int x) {
        if (l > end or r < st)return 0;
        if (l >= st and r <= end) {
            int i = lower_bound(all(seg[n]), x) - seg[n].begin();
            return (r - l + 1) - i;
        }
        int mid = (l + r) / 2;
        auto a = get(n * 2, l, mid, st, end, x);
        auto b = get(n * 2 + 1, mid + 1, r, st, end, x);
        return a + b;
    }

};
struct query {
    int idx, val, input;

    bool operator<(const query &sec) const {
        return val > sec.val;
    }
};
int main() {
    F_PS;
    int n, q;
    cin >> n >> q;
    vector<int> v(n + 1);
    for (int g = 1; g <= n; g++)cin >> v[g];
    segment_tree sgt(v, n);
    vector<bool> query_ans(q + 1, 1);
    vector<vector<query>> queries(n + 1);
    for (int g = 1; g <= q; g++) {
        int i, x;
        cin >> i >> x;
        queries[x].push_back({i, v[i], g});
    }
    for (int g = 1; g <= n; g++)
        sort(all(queries[g]));

    vector<int> go(n + 1, 0);
    for (int buff_rate = n; buff_rate >= 1; buff_rate--) {
        for (int step = 1, ptr = 1, power = 1; ptr <= n and !queries[buff_rate].empty(); step++, power++) {
            int st = max(ptr, go[step]), end = n, mid, ans = n;
            while (st <= end) {
                mid = (st + end) / 2;
                int fights = sgt.get(1, 1, n, ptr, mid, power);
                if (fights < buff_rate) {
                    st = mid + 1;
                    continue;
                }
                if (fights > buff_rate) {
                    end = mid - 1;
                    continue;
                }
                if (fights == buff_rate) {
                    ans = mid;
                    end = mid - 1;
                }
            }
            go[step] = ans - 5;
            ptr = ans + 1;
            while (!queries[buff_rate].empty() and queries[buff_rate].back().val == power) {
                if (queries[buff_rate].back().idx >= ptr)
                    query_ans[queries[buff_rate].back().input] = 0;
                queries[buff_rate].pop_back();
            }
        }
    }
    for (int g = 1; g <= q; g++)
        cout << (query_ans[g] ? "YES\n" : "NO\n");
}
/*
10 1
3 1 4 4 3 1 1 4 4 4
9 1
 */

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

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

Think step by step.

Response:
```plaintext
10 1
3 1 4 4 3 1 1 4 4 4
9 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

#define ll long long
#define ull unsigned long long

int read()
{
    int x = 0; bool f = false; char c = getchar();
    while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
    while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int n, q, a[N];
vector<int> pos[N];
vector<int> dp[N]; // 设dp[i][j]表示k=i时，第一次达到j时的位置

#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)

int sum[N << 2];

void update(int k, int l, int r, int pos, int val)
{
    sum[k] += val;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(pos <= mid) update(ls(k), l, mid, pos, val);
    else update(rs(k), mid + 1, r, pos, val);
}

int getpos(int k, int l, int r, int Size)
{
    if(l == r) return l;
    int mid = (l + r) >> 1;
    if(sum[ls(k)] >= Size) return getpos(ls(k), l, mid, Size);
    else return getpos(rs(k), mid + 1, r, Size - sum[ls(k)]);
}

int getsum(int k, int l, int r, int L, int R)
{
    if(L <= l && r <= R) return sum[k];
    int mid = (l + r) >> 1;
    if(R <= mid) return getsum(ls(k), l, mid, L, R);
    if(L > mid) return getsum(rs(k), mid + 1, r, L, R);
    return getsum(ls(k), l, mid, L, R) + getsum(rs(k), mid + 1, r, L, R);
}

int main()
{
    n = read(), q = read();
    for(int i = 1; i <= n; ++i)
    {
        a[i] = read();
        pos[a[i]].emplace_back(i);
        dp[i].emplace_back(0), dp[i].emplace_back(i);
        update(1, 1, n, i, 1);
    }

    for(int j = 1; j <= n; ++j)
    {
        for(auto x : pos[j]) update(1, 1, n, x, -1);
        for(int i = 1; i * j <= n; ++i)
        {
            if(dp[i].size() <= j) continue;
            if(dp[i][j] == n + 1) continue;
            int sum1 = getsum(1, 1, n, 1, dp[i][j]);
            int sum2 = getsum(1, 1, n, 1, n);
            if(sum2 - sum1 < i){ dp[i].emplace_back(n + 1); continue; }
            int pos = getpos(1, 1, n, sum1 + i);
            dp[i].emplace_back(pos);
        }
    }

    while(q--)
    {
        int i = read(), x = read();
        if(dp[x].size() <= a[i]) printf("NO\n");
        else if(i > dp[x][a[i]]) printf("NO\n");
        else printf("YES\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 provided buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. Here are the steps to consider:

1. **Edge Cases for `k` and `n`**:
   - Test with `k = 1` and `k = n` to see how the code handles the extreme values.
   - Test with `n` and `q` at their maximum limits to check for performance issues.

2. **Monster Levels**:
   - Create a sequence of monster levels that are increasing, decreasing, or mixed to see how the code handles different patterns.
   - Include levels that are equal to the number of fights required for a level up (`k`).

3. **Queries**:
   - Include queries for the first and last monsters to ensure the code handles the boundary conditions correctly.
   - Include queries for monsters that are likely to cause the code to fail, such as those with levels equal to `k`.

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

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

This test input includes:
- 5 monsters with levels 1, 2, 3, 4, 5.
- 10 queries that test different values of `k` for each monster, including the edge cases where `k = 1` and `k = 5`.

This should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

const int N = 2e5+5;
int a[N], bit[N], loc[N], nextloc[N], lvl = 1;
vector<int> positions[N];

void add(int i, int x) {
  for (; i < N; i += i & -i) {
    bit[i] += x;
  }
}

int query(int i) {
  int ans = 0;
  for (; i; i -= i & -i) {
    ans += bit[i];
  }
  return ans;
}

int find(int s) {
  int l = 0;
  int cur = 0;
  for (int i = __lg(N); i >= 0; i--) {
    int k = l+(1<<i);
    if (k >= N) continue;
    if (cur+bit[k] < s) {
      l = k;
      cur += bit[k];
    }
  }
  return l+1;
}

void increment_lvl() {
  for (int i = 1; i < N && loc[i] < N; i++) {
    int s = query(loc[i]);
    loc[i] = find(s+i);
  }
  for (int i: positions[lvl]) add(i, -1);
  lvl++;
}

int main () {
  ios_base::sync_with_stdio(0); cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    add(i, 1);
    cin >> a[i];
    positions[a[i]].push_back(i);
  }

  vector<bool> ans(q);
  vector<vector<tuple<int, int, int>>> queries_by_lvl(N);
  for (int i = 0; i < q; i++) {
    int j, x;
    cin >> j >> x;
    queries_by_lvl[a[j]].emplace_back(i, j, x);
  }
  for (; lvl < 4; increment_lvl()) {
    for (auto& [qid, j, x]: queries_by_lvl[lvl-1]) {
      ans[qid] = j <= loc[x];
    }
  }

  for (bool b: ans) {
    cout << (b ? "YES" : "NO") << '\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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code seems to have issues with level increments and the logic for determining when Monocarp fights a monster. Specifically, the `increment_lvl` function and the way levels are updated might be problematic.

2. **Edge Cases**:
   - **Monocarp's initial level**: Ensure the initial level of 1 is handled correctly.
   - **Level increments**: Test scenarios where Monocarp's level increments after a certain number of fights.
   - **Monsters with the same level**: Test cases where multiple monsters have the same level.
   - **Monsters with varying levels**: Test cases where the levels of monsters are not in a simple increasing or decreasing order.

3. **Small-Scale Test Input**: Create a test input that is small enough to be manageable but complex enough to expose the bug.

Given these considerations, here is a small-scale test input:

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

### Explanation:
- **Number of Monsters (n) and Queries (q)**: 5 monsters and 10 queries.
- **Monster Levels**: The levels of the monsters are [2, 1, 3, 1, 2].
- **Queries**:
  - Queries for `k = 1` and `k = 2` for each monster.
  - This setup ensures that the code needs to handle level increments and the logic for fighting or fleeing correctly.

This test input is designed to be small and manageable while still being complex enough to expose any issues in the level increment and fight/flee logic of the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase

class LazySegmentTree:
    def __init__(self, data, default=0, func=max):
        """initialize the lazy segment tree with data"""
        self._default = default
        self._func = func

        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()
        self._lazy = [0] * (2 * _size)

        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(range(_size)):
            self.data[i] = func(self.data[i + i], self.data[i + i + 1])

    def __len__(self):
        return self._len

    def _push(self, idx):
        """push query on idx to its children"""
        # Let the children know of the queries
        q, self._lazy[idx] = self._lazy[idx], 0

        self._lazy[2 * idx] += q
        self._lazy[2 * idx + 1] += q
        self.data[2 * idx] += q
        self.data[2 * idx + 1] += q

    def _update(self, idx):
        """updates the node idx to know of all queries applied to it via its ancestors"""
        for i in reversed(range(1, idx.bit_length())):
            self._push(idx >> i)

    def _build(self, idx):
        """make the changes to idx be known to its ancestors"""
        idx >>= 1
        while idx:
            self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) + self._lazy[idx]
            idx >>= 1

    def add(self, start, stop, value):
        """lazily add value to [start, stop)"""
        start = start_copy = start + self._size
        stop = stop_copy = stop + self._size
        while start < stop:
            if start & 1:
                self._lazy[start] += value
                self.data[start] += value
                start += 1
            if stop & 1:
                stop -= 1
                self._lazy[stop] += value
                self.data[stop] += value
            start >>= 1
            stop >>= 1

        # Tell all nodes above of the updated area of the updates
        self._build(start_copy)
        self._build(stop_copy - 1)

    def query(self, start, stop, default=0):
        """func of data[start, stop)"""
        start += self._size
        stop += self._size

        # Apply all the lazily stored queries
        self._update(start)
        self._update(stop - 1)

        res = default
        while start < stop:
            if start & 1:
                res = self._func(res, self.data[start])
                start += 1
            if stop & 1:
                stop -= 1
                res = self._func(res, self.data[stop])
            start >>= 1
            stop >>= 1
        return res

    def __repr__(self):
        return "LazySegmentTree({0})".format(self.data)

def discrete_binary_search(func, lo, hi):
    """ Locate the first value x s.t. func(x) = True within [lo, hi] """
    while lo < hi:
        mi = lo + (hi - lo) // 2
        if func(mi):
            hi = mi
        else:
            lo = mi + 1

    return lo

def main():
    n, q = map(int, input().split())
    a = [int(ai) for ai in input().split()]

    mask_set = 0x40000
    mask_val = 0x3ffff
    def func(a, b):
        if a & mask_set == 0 and b & mask_set == 0:
            return ((a > 0) + (b > 0)) + mask_set
        elif a & mask_set == 0:
            return (a > 0) + b
        elif b & mask_set == 0:
            return (b > 0) + a
        return a + b - mask_set

    seg = LazySegmentTree(a, func=func)

    queries = [[int(j) for j in input().split()] for i in range(q)]
    s = [[] for _ in range(n + 1)]
    for m, (i, x) in enumerate(queries):
        s[x].append([i, m])

    sol = [False] * q
    for x in reversed(range(1, n + 1)):
        start, ops = x, []
        while start < n:
            seg.add(start, n, -1)
            ops.append(start)
            start = discrete_binary_search(lambda y: seg.query(start, y) - mask_set > x, start + 1, n)

        for i, m in s[x]:
            sol[m] = seg.query(i - 1, i) & mask_val == 1

        for start in ops:
            seg.add(start, n, 1)

    for i in sol:
        print(["NO", "YES"][i])

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on scenarios that might cause the code to behave incorrectly. Here are the steps to think through this:

1. **Understanding the Bug**: The provided code uses a lazy segment tree to manage the levels and queries. A common issue in such problems is incorrect handling of updates and queries, especially when dealing with lazy propagation.

2. **Edge Cases**:
   - **Monsters with the same level**: This can cause issues if the segment tree doesn't handle repeated levels correctly.
   - **Queries with small \( k \)**: Small \( k \) values can cause frequent level ups, which might not be handled correctly.
   - **Queries with large \( k \)**: Large \( k \) values can cause fewer level ups, which might also expose issues.
   - **Monsters with levels close to the initial level**: This can cause the segment tree to incorrectly determine whether a monster flees or is fought.

3. **Small-Scale Test Input**:
   - We will use a small number of monsters and queries to keep the input manageable.
   - We will include a mix of monsters with levels close to the initial level and some with higher levels.
   - We will include queries with both small and large \( k \) values.

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

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

This input includes:
- 5 monsters with levels [2, 1, 3, 2, 1].
- 10 queries:
  - 5 queries with \( k = 1 \).
  - 5 queries with \( k = 2 \).

This setup should challenge the code to correctly handle level ups and determine whether the monsters flee or are fought.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
class SegmentTree:
    def __init__(self, data):
        self.n = len(data)
        self.tree = [None] * (2 * self.n)
        self.build(data)

    def build(self, data):
        for i in range(self.n):
            self.tree[self.n + i] = sorted([(data[i], i)])
        for i in range(self.n - 1, 0, -1):
            self.tree[i] = merge(self.tree[2 * i], self.tree[2 * i + 1])

    def query(self, left, right, t, k):
        left += self.n
        right += self.n
        result = []
        while left < right:
            if left % 2 == 1:
                result += self._get_elements_ge(self.tree[left], t)
                left += 1
            if right % 2 == 1:
                right -= 1
                result += self._get_elements_ge(self.tree[right], t)
            left //= 2
            right //= 2
        result.sort()
        return result[k - 1][1] if k <= len(result) else float('inf')

    def _get_elements_ge(self, arr, t):
        left, right = 0, len(arr)
        while left < right:
            mid = (left + right) // 2
            if arr[mid][0] < t:
                left = mid + 1
            else:
                right = mid
        return arr[left:]

def merge(left, right):
    result = []
    i = j = 0
    while i < len(left) and j < len(right):
        if left[i] < right[j]:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            j += 1
    result.extend(left[i:])
    result.extend(right[j:])
    return result

n, q = map(int, input().split())
a = list(map(int, input().split()))

w = {k: [] for k in range(1, n + 1)}
st = SegmentTree(a)

for k in w:
    last = k - 1
    cur = 2
    while last < float('inf'):
        w[k].append(last)
        last = st.query(last + 1, n, k, cur)
        cur += 1

for _ in range(q):
    i, k = map(int, input().split())
    p = a[i - 1]
    if len(w[k]) < p or w[k][p - 1] >= i - 1:
        print('yEs')
    else:
        print('nO')

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

public class Main {

    static  long mod=998244353;
    static int MOD = (int)(1e9 + 7);
    static boolean MULTI_CASE = false;
    static Integer dp[][];
    static boolean print = false;

    public static void solve() throws IOException {

       int n = sc.nextInt();
       int q = sc.nextInt();

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

       ArrayList<Integer> arr = new ArrayList<>();
       arr.add(a[0]);
       int pre[] = new int[n];

       for(int i = 1; i<n; i++) {
            int ind = bs(a[i], arr);
            pre[i] = ind;
       }

       while(q-- > 0) {
        int i = sc.nextInt() - 1;
        int k = sc.nextInt();

        int div = pre[i] / k;
        int nlvl = 1 + div;
        if(nlvl <= a[i]) {
            System.out.println("YES");
        }
        else {
            System.out.println("NO");
        }
       }
    }

        private static int bs(int val, ArrayList<Integer> arr) {
        int low = 0;
        int high = arr.size() - 1;
        int count = 0;

        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr.get(mid) >= val) {
                count = arr.size() - mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        int pos = low;
        arr.add(pos, val);

        return count;
    }

    private static void dfs(char[][] grid, boolean[][] visited, int[][] regions, int x, int y, int n, int regionId) {
        int[] dx = {-1, 1, 0, 0};
        int[] dy = {0, 0, -1, 1};
        Stack<int[]> stack = new Stack<>();
        stack.push(new int[]{x, y});
        visited[x][y] = true;
        regions[x][y] = regionId;

        while (!stack.isEmpty()) {
            int[] cell = stack.pop();
            int cx = cell[0];
            int cy = cell[1];

            for (int i = 0; i < 4; i++) {
                int nx = cx + dx[i];
                int ny = cy + dy[i];

                if (nx >= 0 && nx < 2 && ny >= 0 && ny < n && grid[nx][ny] == '.' && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    regions[nx][ny] = regionId;
                    stack.push(new int[]{nx, ny});
                }
            }
        }
    }

    public static String DecimalToBase(long number, int base) {

        StringBuilder conv = new StringBuilder();
        while (number > 0) {
            conv.append(number % base);
            number /= base;
        }
        return conv.reverse().toString();
    }

    public static long BaseToDecimal(String base) {
        long result = 0;
        for (char digit : base.toCharArray()) {
            result *= 10;
            int baseDigit = digit - '0';
            result += (baseDigit >= 4) ? (baseDigit + 1) : baseDigit;
        }
        return result;
    }

    public static boolean isNonDecreasing(long[] arr) {
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] < arr[i - 1]) {
                return false;
            }
        }
        return true;
    }

    static long gcd(long a, long b) {
    if (a == 0)
      return b;
    return gcd(b % a, a);
  }

        static long ncr(int n, int r) {

        long sum = 1l;

        for (int i = 1; i <= r; i++) {
            sum = sum * (n - r + i) / i;
        }

        return sum;
    }

    static boolean isPrime(int num) {
        int rev = 0;
        int tmp = num;
        while(tmp > 0) {
            rev = rev * 10 + tmp % 10;
            tmp = tmp / 10;
        }
        return rev == num;
    }

    static long fact(int n) {
    return (n == 1 || n == 0) ? 1l : 1l * n * fact(n - 1);
  }

    static long fast_pow(long a, long b) {
        if(b == 0)
            return 1L;

        long val = fast_pow(a, b / 2);

        if(b % 2 == 0)
            return val * val % mod;
        else
            return val * val % mod * a % mod;
    }

    static boolean[] sieveOfEratosthenes(int n)
    {
        boolean prime[] = new boolean[n+1];
        for(int i=0;i<=n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++)
        {
            if(prime[p] == true)
            {
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;

    }

    static long lcm_of_array(List<Long> arr)
    {
        long lcm = arr.get(0);
        for (int i = 1; i < arr.size(); i++) {
            long num1 = lcm;
            long num2 = arr.get(i);
            long gcd_val = gcdl(num1, num2);
            lcm = (lcm * arr.get(i)) / gcd_val;
        }
        return lcm;
    }

    static int gcdi(int a,int b)
    {
        if(b==0) return a;

        return gcdi(b,a%b);
    }

    static long gcdl(long a,long b)
    {
        if(b==0) return a;

        return gcdl(b,a%b);
    }

    static boolean isprime(int n)
    {
        if (n <= 1)
            return false;

        if (n == 2 || n == 3)
            return true;

        if (n % 2 == 0 || n % 3 == 0)
            return false;

        for (int i = 5; i < Math.sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;

    }

  public static void main(String[] args) throws IOException {
    if (MULTI_CASE) {
      int T = sc.nextInt();
      for (int i = 0; i < T; ++i) {
        // if(i == 494) print = true;
        solve();
      }
    } else {
      solve();
    }
    out.close();
  }

  static InputReader sc = new InputReader();
  static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

  static class InputReader {
    private StringTokenizer st;
    private BufferedReader bf;

    public InputReader() {
      bf = new BufferedReader(new InputStreamReader(System.in));
      st = null;
    }

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

    public String nextLine() throws IOException {
      return bf.readLine();
    }

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

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

    public double nextDouble() throws IOException {
      return Double.parseDouble(next());
    }
  }
}
class Pair {
    int first;
    int second;
    Pair(int a, int b) {
        first = a;
        second = b;
    }
}
class Node {
    long val;
    List<Node> child;

    Node(long a) {
        val = a;
        child = new ArrayList<>();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to create a test case that highlights the issue with the way the code handles the levels and the binary search. The code incorrectly calculates the number of monsters that Monocarp has to fight to level up, and it also has issues with the binary search logic.

### Step-by-Step Analysis:

1. **Binary Search Issue:**
   - The binary search function `bs` is supposed to find the position where the current monster's level should be inserted in the list of levels. However, it incorrectly calculates the number of monsters that are less than or equal to the current monster's level.

2. **Level Calculation Issue:**
   - The code uses the `pre` array to store the number of monsters that are less than or equal to the current monster's level. However, this is not the correct way to determine how many fights Monocarp has had to level up.

### Test Case Design:

To create a test case that exposes these issues, we can use a scenario where the levels of the monsters are such that the binary search will incorrectly place them, and the level calculation will be off.

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

### Explanation:
- **Monsters' Levels:** `[2, 1, 3, 1, 2]`
  - Monster 1: Level 2
  - Monster 2: Level 1
  - Monster 3: Level 3
  - Monster 4: Level 1
  - Monster 5: Level 2

- **Queries:**
  - Query 1: `(3, 2)` - Will Monocarp fight the 3rd monster if `k` is 2?
  - Query 2: `(4, 2)` - Will Monocarp fight the 4th monster if `k` is 2?
  - Query 3: `(3, 3)` - Will Monocarp fight the 3rd monster if `k` is 3?
  - Query 4: `(4, 3)` - Will Monocarp fight the 4th monster if `k` is 3?

### Expected Output:
- For Query 1: `(3, 2)` - **YES**
  - Monocarp will have fought 2 monsters by the time he reaches the 3rd monster, so his level will be 2, and he will fight the 3rd monster (level 3).

- For Query 2: `(4, 2)` - **NO**
  - Monocarp will have fought 3 monsters by the time he reaches the 4th monster, so his level will be 2, and he will not fight the 4th monster (level 1).

- For Query 3: `(3, 3)` - **YES**
  - Monocarp will have fought 3 monsters by the time he reaches the 3rd monster, so his level will be 2, and he will fight the 3rd monster (level 3).

- For Query 4: `(4, 3)` - **NO**
  - Monocarp will have fought 4 monsters by the time he reaches the 4th monster, so his level will be 2, and he will not fight the 4th monster (level 1).

This test case should expose the issues with the binary search and the level calculation in the provided buggy code.