**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ll long long
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define sdefine \
    string s;   \
    cin >> s;
#define vll vector<ll>
#define pll pair<ll, ll>
#define vvll vector<vll>
#define vpll vector<pll>
#define mll map<ll, ll>
#define pb push_back
#define ppb pop_back
#define ff first
#define ss second
#define set_bits(a) __builtin_popcountll(a)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define forl(i, a, b) for (int i = a; i < b; i++)
#define yesno(x, y)            \
    if ((x) == (y))            \
        cout << "YES" << endl; \
    else                       \
        cout << "NO" << endl;
#define endl "\n"
#define display(a, x, y)            \
    {                               \
        for (int i = x; i < y; i++) \
            cout << a[i] << " ";    \
        cout << endl;               \
    }
#define int long long

bool dfs(int val, int s, int pr, int op, vll(&v), vll G[])
{
    int curr = v[s] - op;
    if (s == 1)
        op += (val - curr);
    if (curr < 0)
        op += abs(curr);
    int ct = 0;
    bool ck = true;
    for (auto it : G[s])
    {
        if (it == pr)
            continue;
        ct++;
        ck &= dfs(val, it, s, op, v, G);
    }
    if ((ct == 0) && (curr < 0))
        ck = false;
    return ck;
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vll v(n + 1, 0);
        forl(i, 1, n + 1)
        {
            cin >> v[i];
        }
        vll par(n - 1, -1);
        vll G[n + 1];
        forl(i, 0, n - 1)
        {
            cin >> par[i];
            G[i + 2].push_back(par[i]);
            G[par[i]].push_back(i + 2);
        }
        int st = v[1], ed = 1e15;
        while ((ed - st) > 1)
        {
            int mid = (st + ed) / 2;
            bool ck = dfs(mid, 1, -1, 0, v, G);
            if (ck)
                st = mid;
            else
                ed = mid - 1;
        }
        int ans = st;
        if (dfs(ed, 1, -1, 0, v, G))
            ans = ed;
        cout << ans << endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases:**
   - Trees with a single child.
   - Trees with multiple levels of children.
   - Trees where the initial values at the root and its children are zero or very small.

2. **Scenarios:**
   - Trees where the root value is already very high and the children values are very low.
   - Trees where the root value is low and the children values are very high.
   - Trees with a mix of high and low values in the subtree.

3. **Constructing the Test Case:**
   - We will create a small tree with a root and a few children to ensure the test case is complete and valid.
   - The values at the nodes will be chosen to challenge the binary search and the DFS logic in the buggy code.

Given these considerations, let's construct a test input:

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

### Explanation:
- **Number of Test Cases:** 1
- **Number of Vertices:** 5
- **Initial Values:** [1, 10, 10, 10, 10]
- **Parent Array:** [1, 1, 1, 1] (indicating that vertices 2, 3, 4, and 5 are all children of vertex 1)

This test case is designed to:
- Ensure the root value is initially very low (1).
- Ensure the children values are very high (10).
- Test the binary search and DFS logic in the buggy code to see if it can correctly determine the maximum possible value at the root.

By using this test input, we can effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define pii pair<int, int>
#define all(a) a.begin(), a.end()
const int mod = 998244353, N = 500005;

void solve() {
    int n;
    cin >> n;
    vector <ll> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    vector <vector <int>> adj(n);
    for (int i = 1, p; i < n; ++i) {
        cin >> p, --p;
        adj[p].pb(i);
    }
    auto ask = [&](ll x) {
        vector <ll> b = a, lz(n, 0);
        b[0] -= x;
        auto dfs = [&](auto self, int v, ll op) -> void {
            b[v] -= op;
            if (adj[v].empty()) {
                return;
            }
            if (b[v] < 0) {
                op -= b[v], b[v] = 0;
            }
            for (int u : adj[v]) {
                self(self, u, op);
            }
        };
        dfs(dfs, 0, 0);
        return *min_element(all(b)) >= 0;
    };
    ll l = a[0], r = 1ll << 50;
    while (r - l > 1) {
        ll m = l + r >> 1;
        if (ask(m)) {
            l = m;
        } else {
            r = m;
        }
    }
    cout << l << '\n';
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3
1 0 0
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

Buggy Code:
if __name__ == "__main__":

    def solve():
        def dfs(v,k):
            if tree[v] == []:return a[v] >= k
            for u in tree[v]:
                if a[u] >= k:
                    sub = dfs(u,k)
                else:
                    sub = dfs(u,k + k - a[u])
                if not sub:
                    return False
            return True

        n = int(input())
        a = list(map(int, input().split()))
        parents = list(map(int, input().split()))
        tree = {i:[] for i in range(n)}

        for i in range(n - 1):
            parent = parents[i] - 1
            child = i + 1
            tree[parent].append(child)
        # print(tree, a)
        # @cache
        ans = 0
        l = 0
        r = 2* 10 **9+1
        while l<r:
            mid = (l+r)//2
            if a[0]>=mid or dfs(0,mid - a[0]):
                ans = mid
                l = mid +1
            else:
                r = mid

        print(ans)

    for _ in range(int(input())):
        solve()

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

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

Think step by step.

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

1. **Tree Structure**: The tree structure should be such that it challenges the DFS logic in the code.
2. **Initial Values**: The initial values should be chosen to make the binary search and DFS checks fail in specific ways.
3. **Parent-Child Relationships**: Ensure that the parent-child relationships are set up in a way that can expose the logic errors in the DFS function.

Let's construct a small-scale test input that meets these criteria:

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

### Explanation
- **Test Case Count**: 1
- **Number of Vertices**: 5
- **Initial Values**: [1, 2, 3, 4, 5]
- **Parent-Child Relationships**:
  - Vertex 1 (root) has children 2 and 3.
  - Vertex 3 has children 4 and 5.

### Why This Input is Challenging
1. **Tree Structure**: The tree is not a simple linear or balanced tree. It has a root with two children, and one of those children has two children.
2. **Initial Values**: The values are increasing, which can make the DFS checks more complex. The root value is 1, and the values in the subtree are higher.
3. **Binary Search and DFS**: The binary search in the code might not handle the case where the root value is much smaller than the values in the subtree. The DFS function might not correctly propagate the values and constraints.

This input should help expose the bug in the provided code by challenging the logic of the DFS and binary search.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

Buggy Code:
# My competitive programming template - July 2024
# by bernie

from types import GeneratorType
from sys import stdin, stdout, setrecursionlimit
from collections import Counter, deque, defaultdict, OrderedDict
from math import ceil, comb, sqrt, factorial, perm, log, gcd, lcm, floor, inf
from functools import lru_cache, cache
from bisect import bisect_left, bisect_right
from heapq import heapify, heappush, heappop
from itertools import permutations, chain
import time

setrecursionlimit(100000)

def measure(func):
    def wrapper(*args, **kwargs):
        start_time = time.perf_counter()
        result = func(*args, **kwargs)
        end_time = time.perf_counter()
        print(f"Function '{func.__name__}' took {end_time - start_time} seconds to execute.")
        return result
    return wrapper

# pajenegod orz
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        to = f(*args, **kwargs)
        while True:
            if type(to) is GeneratorType:
                stack.append(to)
                to = next(to)
            else:
                stack.pop()
                if not stack:
                    break
                to = stack[-1].send(to)
        return to
    return wrappedfunc

def input():
    """ Faster input parsing """
    return stdin.readline().strip()

def print(string="", newline=True):
    """Prints a single string on a new line"""
    if newline:
        stdout.write(str(string) + "\n")
    else:
        stdout.write(str(string))

def print_array(array):
    """ Prints an array of values, the way CodeForces requires """
    stdout.write(" ".join(map(str, array)) + "\n")

def bin(x, width=None):
    """A replacement for the built-in bin function, without the 0x"""
    if width:
        return format(x, "b").zfill(width)
    return format(x, "b")

def find_primes(limit, return_all=True):
    """ Finds all the prime numbers from 1 to a specified limit (inclusive). Sieve of Erastothenes """
    is_prime = [1] * (limit + 1)
    is_prime[1] = 0

    for i in range(2, limit + 1):
        if i * i > limit:
            break
        if is_prime[i]:
            for j in range(i * i, limit + 1, i):
                is_prime[j] = 0
    if return_all:
        return is_prime
    else:
        return [i for i in range(2, limit + 1) if is_prime[i]]

def factorize(x):
    factors = set()
    for i in range(1, ceil(sqrt(x)) + 1):
        if x % i == 0:
            factors.add(i)
            factors.add(x // i)
    return factors

MOD = int(1e9 + 7)

def parity(x):
    return x >= 0

def is_even(x):
    return x % 2 == 0

def solve():
    def increase_min(node):
        min_child = float("inf")

        if tree[f"{node}"]:
            for child in tree[f"{node}"]:
                min_child = min(min_child, increase_min(child))

            if min_child <= a[node]:
                return min_child
            else:
                best = (min_child + a[node]) // 2
                return best
        else:
            return a[node]

    for t in range(int(input())):
        n = int(input())
        a = [int(_) for _ in input().split()]
        b = [int(_) for _ in input().split()]
        tree = {}

        for v in range(n):
            tree[f"{v}"] = []

        for i in range(len(b)):
            tree[f"{b[i] - 1}"].append(i + 1)

        ans = a[0]
        min_child = float("inf")

        for child in tree["0"]:
            min_child = min(min_child, increase_min(child))

        if min_child == float("inf"):
            min_child = 0

        print(ans + min_child)

if __name__ == "__main__":
    try:
        solve()
    except Exception as error:
        print(error)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases that can challenge the logic of the algorithm. The main issues to consider are:

1. **Handling of subtrees with very low values**: The code might not correctly handle situations where the values in the subtrees are very low, leading to incorrect minimum values being propagated.
2. **Handling of large values at the root**: The code might not correctly handle situations where the root has a large initial value.
3. **Handling of deep trees**: The code might not correctly handle deep trees where the operations need to be performed multiple times to maximize the root value.

Given these considerations, let's construct a small-scale test input that could expose these issues:

### Test Input
```plaintext
1
5
10 0 0 0 0
1 1 3 3
```

### Explanation
1. **Initial Values**:
   - The root (vertex 1) has an initial value of 10.
   - The other vertices (2, 3, 4, 5) have initial values of 0.

2. **Tree Structure**:
   - Vertex 1 is the root.
   - Vertex 2 and 3 are children of vertex 1.
   - Vertex 4 and 5 are children of vertex 3.

### Expected Behavior
- The code should attempt to move the values from the children (2, 3, 4, 5) to the root (1) to maximize the value at the root.
- Since the initial values of the children are 0, the only way to increase the root value is to use the initial value of the root itself.
- The expected output should be 10, as no additional operations can be performed to increase the root value.

### Why This Test Input is Challenging
- **Initial Value at Root**: The root has a large initial value (10), which should not change unless the code incorrectly modifies it.
- **Subtree Values**: The subtrees have very low values (0), which should not contribute to the root value.
- **Tree Depth**: The tree is relatively shallow, but the structure ensures that the code must correctly handle the propagation of values through the tree.

This test input should help expose any issues in the logic of the buggy code, particularly in how it handles initial values and subtree operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

Buggy Code:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
  static final int LIMIT = 1_000_000_000;

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

    int t = sc.nextInt();
    for (int tc = 0; tc < t; ++tc) {
      int n = sc.nextInt();
      int[] a = new int[n];
      for (int i = 0; i < a.length; ++i) {
        a[i] = sc.nextInt();
      }
      int[] p = new int[n - 1];
      for (int i = 0; i < p.length; ++i) {
        p[i] = sc.nextInt();
      }

      System.out.println(solve(a, p));
    }

    sc.close();
  }

  static int solve(int[] a, int[] p) {
    int n = a.length;

    @SuppressWarnings("unchecked")
    List<Integer>[] childLists = new List[n];
    for (int i = 0; i < childLists.length; ++i) {
      childLists[i] = new ArrayList<>();
    }
    for (int i = 0; i < p.length; ++i) {
      childLists[p[i] - 1].add(i + 1);
    }

    int base = a[0];

    int delta = -1;
    int lower = 0;
    int upper = LIMIT;
    while (lower <= upper) {
      int middle = (lower + upper) / 2;

      a[0] = middle;
      if (search(a, childLists, middle, 0)) {
        delta = middle;
        lower = middle + 1;
      } else {
        upper = middle - 1;
      }
    }

    return base + delta;
  }

  static boolean search(int[] a, List<Integer>[] childLists, long minValue, int node) {
    if (childLists[node].isEmpty()) {
      return a[node] >= minValue;
    }

    for (int child : childLists[node]) {
      if (!search(a, childLists, minValue + Math.max(0, minValue - a[node]), child)) {
        return false;
      }
    }

    return true;
  }
}

**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
1 0 0 0
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Root
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices. The vertices in the tree are numbered from $1$ to $n$, and the root is the vertex $1$. The value $a_i$ is written at the $i$-th vertex.

You can perform the following operation any number of times (possibly zero): choose a vertex $v$ which has at least one child; increase $a_v$ by $1$; and decrease $a_u$ by $1$ for all vertices $u$ that are in the subtree of $v$ (except $v$ itself). However, after each operation, the values on all vertices should be non-negative.

Your task is to calculate the maximum possible value written at the root using the aforementioned operation.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^9$) — the initial values written at vertices.

The third line contains $n-1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i \le n$), where $p_i$ is the parent of the $i$-th vertex in the tree. Vertex $1$ is the root.

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible value written at the root using the aforementioned operation.

Examples:
input:
3
4
0 1 0 2
1 1 3
2
3 0
1
5
2 5 3 9 6
3 1 5 2
output:
1
3
6

Note: In the first test case, the following sequence of operations is possible:

  * perform the operation on $v=3$, then the values on the vertices will be $[0, 1, 1, 1]$;    * perform the operation on $v=1$, then the values on the vertices will be $[1, 0, 0, 0]$.

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static void solve() throws IOException {
        int n = sc.nextInt();
        int a [] = sc.nextIntArray(n);
        ArrayList<Integer> adjl []  = new ArrayList [n + 1];
        for(int i = 0; i <= n ; i++)
            adjl[i] = new ArrayList<>();
        for(int i = 2 ; i <= n ; i++){
            int u = sc.nextInt(), v = i;
            adjl[u].add(v);
        }
        long low = a[0], high = (long) 4e9;
        while(low <= high) {
            long mid = (low + high) / 2;
            Queue<long []> qu= new LinkedList<>();
            qu.add(new long[]{1, mid - a[0]});
            boolean f = true;
            while (!qu.isEmpty()) {
                long [] arr = qu.poll();
//                pw.println(Arrays.toString(arr));
                int idx = (int) arr[0];
                long req = arr[1];
                if(adjl[idx].isEmpty()){
                    f &= a[idx - 1] >= req;
                }
                for(int i = 0; i < adjl[idx].size() ; i++){
                    long val = a[adjl[idx].get(i) - 1] - req;
                    qu.add(new long[]{adjl[idx].get(i), req - Math.min(val, 0)});
                }
            }
            if(f)
                low = mid + 1;
            else
                high = mid - 1;
        }
        pw.println(high);
    }

    public static void main(String[] args) throws IOException {
        int t = sc.nextInt();
        while(t -- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

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

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

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

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

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

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

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

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

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some steps to consider:

1. **Test Case with a Single Child**: This can help verify if the code correctly handles the simplest non-trivial tree structure.
2. **Test Case with a Long Chain**: This can help verify if the code correctly handles deep trees where the root has a long chain of children.
3. **Test Case with Multiple Children**: This can help verify if the code correctly handles trees with multiple branches.
4. **Test Case with Negative Initial Values**: Although the problem states that values are non-negative, it's good to test if the code handles negative values gracefully (though this should not be a valid input).
5. **Test Case with Large Values**: This can help verify if the code handles large values and the operations correctly.

Let's create a small-scale test input that combines some of these elements:

```plaintext
3
4
0 1 0 2
1 1 3
5
2 5 3 9 6
3 1 5 2
6
1 2 3 4 5 6
1 2 2 3 3
```

### Explanation:
1. **First Test Case (n=4)**:
   - Values: `0 1 0 2`
   - Parent array: `1 1 3`
   - This is a simple tree with a root and a few children. The root has two children, and one of those children has another child.

2. **Second Test Case (n=5)**:
   - Values: `2 5 3 9 6`
   - Parent array: `3 1 5 2`
   - This is a more complex tree with multiple branches. The root has two children, and one of those children has another child, and so on.

3. **Third Test Case (n=6)**:
   - Values: `1 2 3 4 5 6`
   - Parent array: `1 2 2 3 3`
   - This is a tree with a root and multiple children, some of which have their own children. This can help verify if the code correctly handles multiple levels of children.

This test input should help expose any bugs in the provided code by covering a range of tree structures and value distributions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define s(v) (int)v.size()
#define m1 cout << "-1" << endl
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), r.rend()
#define _lb(v, x) lower_bound(all(v), x)
#define lb(S, x) S.lower_bound(x)
#define _ub(v, x) lower_bound(all(v), x)
#define ub(S, x) S.lower_bound(x)
#define mAx(v) *max_element(v.begin(), v.end())
#define Sum(v) accumulate(v.begin(), v.end(), 0LL)
#define mIn(v) *min_element(v.begin(), v.end())
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define asort(v) sort(v.begin(), v.end())
#define dsort(v) sort(v.begin(), v.end(), greater<int>())
#define rep(i, a, b) for (int i = a; i < b; i++)
#define pb push_back
#define ff first
#define ss second
#define cnf(p) cout << "@@" << p << endl;
vector<vector<int>> adj;
vector<int> vis;
const int nax = 2e5 + 10;
void Display(vector<int> &v)
{
    for (auto &A : v)
        cout << A << " ";
    cout << endl;
}
void Scan(vector<int> &v)
{
    for (auto &A : v)
        cin >> A;
}
vector<int> get_primes(int n)
{
    vector<int> ans;
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
            ans.pb(p);
    return ans;
}
int binexp(int a, int b, int MOD)
{
    int res = 1;
    while (b > 0)
    {
        if (b & 1)
            res = (res * 1LL * a) % MOD;
        a = (a * 1LL * a) % MOD;
        b >>= 1;
    }
    return res;
}
void runcase()
{
    int n, x;
    cin >> n >> x;
    vector<int> a(n);
    Scan(a);
    int ans = 0;
    int max_nahi = 0, max_haan = 0;
    vector<int> pref(n + 1);
    for (int i = 1; i <= n; i++)
        pref[i] = pref[i - 1] + a[i - 1];
    int min_haan = mAx(a);
    int l = 0;
    for (int i = 0; i < n - 1; i++)
    {
        if (!ans)
        {
            if (a[i] <= max_nahi)
                continue;
            int sum = pref[i + 1];
            int j = i;
            while (j < n - 1 && sum >= a[j + 1])
            {
                sum += a[j + 1], j++;
                if (sum >= min_haan)
                {
                    ans++;
                    min_haan = a[i];
                    break;
                }
            }
            if (sum < min_haan)
                max_nahi = a[i];
            else
                l = i;
        }
        else
        {
            int sum;
            if (a[i] >= a[i - 1])
                sum = pref[i + 1] - pref[l + 1];
            else
                sum = a[i];
            int j = i;
            while (j < n - 1 && sum >= a[j + 1])
            {
                sum += a[j + 1], j++;
                if (sum >= min_haan)
                {
                    ans++;
                    min_haan = min(min_haan, a[i]);
                    break;
                }
            }
        }
        // cout << i << " " << ans << endl;
    }
    int sum = a.back();
    int ind = n - 1;
    while (ind > 0 && sum >= a[ind - 1])
        sum += a[ind - 1], ind--;
    if (ind == 0)
        ans++;
    cout << ans << "\n";
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--)
        runcase();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

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

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

/***********************************************/
/* dear online judge:
 * i've read the problem, and tried to solve it.
 * even if you don't accept my solution, you should respect my effort.
 * i hope my code compile and get accepted.
 *      ____________
 *     /            \
 *    /  /\      /\  \
 *   /  /  \    /  \  \
 *   \                /
 *    \     \___/    /
 *     \____________/
 */

typedef tree<int, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> oset;

#define f(i,j,k) for(long long i=j;i<k;i++)
#define fb(i,j,k) for(long long i=j;i>=k;i--)
#define fs(i,j,k,p) for(long long i=j;i<k;i+=p)
#define fbs(i,j,k,p) for(long long i=j;i>=k;i-=p)
#define cond(a) if(a)cout<<"yes"<<endl;else cout<<"no"<<endl
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define test int t;cin>>t;while(t--)
#define endl "\n"
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define ppb pop_back
//#define mp make_pair
#define ff first
#define ss second
#define pi 3.141592653589793238462
//#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<int>> vvi;

ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}

struct SegmentTree{
private:
    vector<int>seg;int sz,skip=0;
    int merge(int a,int b)
    {
        return (a+b);
    }
    void update(int l,int r,int node,int idx,int val)
    {
        if(l==r)
        {
            seg[node]=val;
            return;
        }
        int mid=l+r>>1;
        if(mid<idx)
        {
            update(mid+1,r,2*node+2,idx,val);
        }
        else
        {
            update(l,mid,2*node+1,idx,val);
        }
        seg[node]=merge(seg[2*node+1],seg[2*node+2]);
    }
    int query(int l,int r,int node,int lx,int rx)
    {
        if(l>rx||r<lx)return skip;
        if(l>=lx&&r<=rx)return seg[node];
        int mid=l+r>>1;
        int a=query(l,mid,2*node+1,lx,rx);
        int b=query(mid+1,r,2*node+2,lx,rx);
        return merge(a,b);
    }
public:
    SegmentTree(int n)
    {
        sz=1;
        while(sz<=n)sz<<=1;
        seg=vector<int>(sz<<1,skip);
    }
    void update(int idx,int val)
    {
        update(0,sz-1,0,idx,val);
    }
    int query(int l,int r)
    {
        return query(0,sz-1,0,l,r);
    }
};

int tt = 1;

void solve()
{
    ll n , x;
    cin>>n>>x;
    vector<pair<ll,ll>>v(n);
    vll a(n);
    f(i,0,n){
        cin>>v[i].first;
        a[i]= v[i].first;
        v[i].second = i+1;
    }
    vll can(n+1,0);
    vll pref(n+1,0);
    f(i,1,n+1){
        pref[i]=pref[i-1]+v[i-1].first;
    }
    sort(rall(v));
    set<ll>s;
    can[v[0].second]=1;
    s.insert(v[0].second);
    f(i,1,n){
        s.insert(v[i].second);

        ll ind = v[i].second;
        ll sum = v[i].first;
        auto it = s.upper_bound(ind);
        if(it==s.end()){
            sum+=pref[n]-pref[ind];
        }
        else {
            sum+=pref[*it-1]-pref[ind];
        }
        auto itt = s.lower_bound(ind);

        if(itt==s.begin()){
            sum+=pref[ind-1];
        }
        else {
            itt--;
            sum+=pref[ind-1]-pref[*itt];

        }

        if(can[*it] && sum>=a[*it-1])can[ind]=1;
        if(can[*itt] && sum>=a[*itt-1])can[ind]=1;
    }
    ll cnt =0;
    f(i,1,n+1){
        if(can[i])cnt++;
    }
    if(tt==9619)cout<<a[1]<<endl;
    cout<<cnt<<endl;
}

int main()
{
    fastio();
    //combinatorics::init(1,1e9+7);
    // std::cout << std::setprecision(11) << a << '\n';
    //freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
    test
    {
        solve();
        tt++;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

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

#define int long long
using namespace std;
int arr[200009];
int st[200009][30];
int lg[2000009];
int sum[2000009];
int maxr[2000009];
int ans[2000009];

int n;
int get_max(int l,int r)
{
    if(r==n+1) r=n;
  int k=lg[r-l+1];
  return max(st[l][k],st[r-(1<<k)+1][k]);
}

signed main()
{
    int tt=0;
    lg[1]=0;
    for(int i=2;i<=200009;i++)
    {
       lg[i]=lg[i/2]+1;
    }
    scanf("%lld",&tt);
    while(tt--)
    {
       int k;
       scanf("%lld%lld",&n,&k);
       for(int i=1;i<=n;i++) ans[i]=0;
        for(int i=1;i<=n;i++)
       {
         scanf("%lld",&arr[i]);
         sum[i]=sum[i-1]+arr[i];
       }
       for(int i=1;i<=n;i++)
       {
          st[i][0]=arr[i];
       }
       for(int i=1;(1<<i)<=n;i++)
       {
          for(int j=1;j+(1<<i)-1<=n;j++)
          {
             st[j][i]=max(st[j][i-1],st[j+(1<<(i-1))][i-1]);
          }
       }
        for(int i=1;i<=n;i++)
        {
           int cur=sum[i];
           int r=i+1;
           while(r<=n)
           {
               ///bool flag=1;
               if(get_max(r,n)<=cur)
               {
                  r=n+1;
                  break;
               }
               if(arr[r]>cur) break;
               int ll=r,rr=n,answer;
               while(ll<=rr)
               {
                 int mid=(ll+rr)>>1;
                 if(get_max(ll,mid)>cur)
                 {
                     answer=mid;
                     rr=mid-1;
                 }
                 else ll=mid+1;
               }
               cur+=sum[answer-1]-sum[r-1];
               r=answer;
           }
           maxr[i]=r-1;
        }
        for(int i=1;i<=n;i++)///看这个地方能不能扩展到1的位置以及扩展到1位置最小的右端点是多少;
        {
          int cur=arr[i],l=i,r=i;
          int flag=1;
          int t=0;
          while(flag)
          {
              t++;
              if(t>1000000) break;
              //printf("l,ri,:%d %d %d %d\n",i,l,r,cur);
              flag=0;
              if(l!=1&&arr[l-1]<=cur)///暴力向左扩展,能扩展则一直扩展知道无法扩展,再在右边二分直到能找到一段区间扩展;
              {
                 int ll=1,rr=l-1,answer;
                 while(ll<=rr)
                 {
                    int mid=(ll+rr)>>1;
                    if(get_max(mid,l)<=cur)
                    {
                        answer=mid;
                        rr=mid-1;
                    }
                    else ll=mid+1;
                 }
                 cur+=sum[l-1]-sum[answer-1];
                 l=answer;
                 flag=1;
              }
              if(l!=1&&arr[l-1]>arr[l])
              {
                 if(r!=n&&arr[r+1]<=cur)
                 {
                     int ll=r+1,rr=n+1,answer;
                    // printf("%d\n",get_max());
                     int xuqiu=arr[l-1]-cur;
                     while(ll<=rr)
                     {
                       int mid=(ll+rr)>>1;
                       if( (get_max(r,mid)>cur)||sum[mid-1]-sum[r]>=xuqiu)
                       {
                          answer=mid;
                          rr=mid-1;
                       }
                       else ll=mid+1;
                     }
                     flag=1;
                     //r=answer-1;
                     cur+=sum[answer-1]-sum[r];
                     r=answer-1;
                     //printf("iid:%d %d %d\n",i,r,arr[r+1]);
                     //_sleep(1000);
                 }
              }
              if(!flag) break;
          }
           if(l==1)
           {
              ans[r]++;
              ans[maxr[i]+1]--;
           }
        }
        for(int i=1;i<=n;i++)
        {
           ans[i]+=ans[i-1];
           //printf("%d ",ans[i]);
        }
        for(int i=k;i<=n;i++)
        {
           printf("%lld ",ans[i]);
        }
        //printf("%lld\n",ans[n]);
       // for(int i=1;i<=n;i++) ans[i]=0;
        printf("\n");
    }
    return 0;
}

/*
3
7 1
4 5 1 2 1 4 5
*/

**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
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

Buggy Code:
import sys
input = sys.stdin.read

MAX_SIZE = 200005
values = [0] * MAX_SIZE
cumulative_sum = [0] * MAX_SIZE

def process_test_case(num_elements, unused, value_with_index):
    global cumulative_sum
    cumulative_sum[0] = 0
    for i in range(1, num_elements + 1):
        cumulative_sum[i] = cumulative_sum[i - 1] + value_with_index[i]

    sorted_indices = list(range(1, num_elements + 1))
    sorted_indices.sort(key=lambda x: value_with_index[x])

    valid_indices = set()
    valid_indices.add(0)
    valid_indices.add(num_elements + 1)
    valid_indices.add(sorted_indices[num_elements - 1])

    for j in range(num_elements - 1, 0, -1):
        current_index = sorted_indices[j]
        right_boundary = next((i for i in sorted(valid_indices) if i > current_index), num_elements + 1)
        left_boundary = next((i for i in reversed(sorted(valid_indices)) if i < current_index), 0)

        current_range_sum = cumulative_sum[right_boundary - 1] - cumulative_sum[left_boundary]
        can_include = False

        if right_boundary <= num_elements and current_range_sum >= cumulative_sum[right_boundary] - cumulative_sum[right_boundary - 1]:
            can_include = True
        if left_boundary >= 1 and current_range_sum >= cumulative_sum[left_boundary] - cumulative_sum[left_boundary - 1]:
            can_include = True
        if can_include:
            valid_indices.add(current_index)

    print(len(valid_indices) - 2)

def main():
    data = input().split()
    index = 0
    test_case_count = int(data[index])
    index += 1

    while test_case_count > 0:
        test_case_count -= 1
        num_elements = int(data[index])
        unused = int(data[index + 1])
        index += 2

        value_with_index = [0] * (num_elements + 1)
        for i in range(1, num_elements + 1):
            value_with_index[i] = int(data[index])
            index += 1

        process_test_case(num_elements, unused, value_with_index)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

Buggy Code:
import sys
input = sys.stdin.readline
from math import inf
from bisect import bisect_left

def cal():
    n, _ = map(int, input().split())
    nums = [inf] + list(map(int, input().split()))

    preS = [0]
    for i in range(1, n + 1):
        preS.append(preS[i - 1] + nums[i])

    preT = preS.copy()
    for i in range(1, n):
        preT[i] -= nums[i + 1]

    for i in range(n - 1, -1, -1):
        preT[i] = min(preT[i], preT[i + 1])

    judge = [True] * (n + 1)
    for i in range(1, n + 1):
        if judge[i] is False: continue

        j = min(bisect_left(preS, preS[i - 1] + nums[i - 1]),
                bisect_left(preT, preS[i - 1]))

        for k in range(i, j):
            judge[k] = False

    print(sum(judge) - 1 - (nums[-1] < nums[-2]))

for _ in range(int(input())):
    cal()

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

Buggy Code:
def count_distinct_balls(test_cases):
    results = []

    for n, x, a in test_cases:
        if n == 11:
            # Directly output 4 for the special case where n == 11
            results.append([4])
        else:
            # Implement the normal logic for other cases
            f_values = [0] * (n - x + 1)

            for size in range(n, x - 1, -1):
                possible_values = set()
                for i in range(size):
                    possible_values.add(a[i])
                f_values[size - x] = len(possible_values)

            results.append(f_values)

    return results

# Reading input and preparing the test cases
import sys
input = sys.stdin.read
data = input().split()

t = int(data[0])
index = 1
test_cases = []

for _ in range(t):
    n = int(data[index])
    x = int(data[index + 1])
    a = list(map(int, data[index + 2: index + 2 + n]))
    test_cases.append((n, x, a))
    index += 2 + n

results = count_distinct_balls(test_cases)

# Output the results
for res in results:
    print(" ".join(map(str, res)))

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The bug in the code is that it incorrectly handles the special case where \( n = 11 \) by directly outputting `[4]` without considering the actual values of the array \( a \). This means that for any input where \( n = 11 \), the code will always output `[4]`, regardless of the actual values in \( a \).

To expose this bug, we can create a test case where \( n = 11 \) and the values in \( a \) are such that the correct output should not be `[4]`.

Here's a step-by-step approach to generating such a test case:

1. **Choose \( n = 11 \)**: This is the special case that the buggy code handles incorrectly.
2. **Choose values for \( a \) such that the correct output is not `[4]`**:
   - We need to ensure that the number of distinct values that can be left after performing the operations is not 4.
   - One way to do this is to create an array where the number of distinct values that can be left is different from 4.

Let's create an array \( a \) where the number of distinct values that can be left is 5. This will ensure that the correct output is `[5]`, which will differ from the buggy code's output of `[4]`.

Here's a valid and small-scale test input that meets these criteria:

```plaintext
1
11 11
1 2 3 4 5 6 7 8 9 10 11
```

### Explanation:
- **Test Case 1**:
  - \( n = 11 \)
  - \( x = 11 \)
  - \( a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] \)

In this test case, the correct output should be `[11]` because there are 11 distinct values, and each value can be left after performing the operations. The buggy code will output `[4]`, which is incorrect.

This test input will effectively expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

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

public class E_1_Eliminating_Balls_With_Merging_Easy_Version {
    public static void main(String[] args){
        // Read the number of test cases
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();

        // Process each test case
        for (int k = 0; k < t; k++) {
            int n = scanner.nextInt();
            scanner.nextInt();
            long[] arr = new long[n];
            for(int i=0;i<n;i++) arr[i]=scanner.nextLong();
            int result = helper(arr, n);
            System.out.println(result);
        }

        scanner.close();
    }

    public static boolean bSearch(ArrayList<Integer> list, int val){
        int n=list.size();
        int lo=0;
        int hi=n-1;
        while(hi>=lo){
            int mid=(hi+lo)/2;
            if(list.get(mid)>val) hi=mid-1;
            else if(list.get(mid)<val) lo=mid+1;
            else return true;
        }
        return false;
    }

    // Define the helper function
    public static int helper(long[] arr, int n) {
        ArrayList<Integer> indices = new ArrayList<>();
        boolean flag=true;
        while(true){
            int idx=0;
            int index=indices.isEmpty()? 0 : indices.get(indices.size()-1);
            if(!indices.isEmpty()){
                if(index==n-1) break;
                else idx=index+1;
            }
            long sum=arr[idx];
            for(int i=idx+1;i<n;i++){
                if(arr[i]>sum) idx=i;
                sum+=arr[i];
            }
            for(int i=idx-1;i>=0;i--){
                if(arr[i]>sum) idx=i;
                sum+=arr[i];
            }
            if(bSearch(indices, idx)) break;
            indices.add(idx);
        }
        return indices.size();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

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

public class Main {
    static final int N = (int) 2e5 + 5;
    static int n, x;
    static int[] a;
    static int[] dp;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int t = Integer.parseInt(br.readLine());

        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            n = Integer.parseInt(st.nextToken());
            x = Integer.parseInt(st.nextToken());

            a = new int[n + 1];
            st = new StringTokenizer(br.readLine());
            for (int i = 1; i <= n; i++) {
                a[i] = Integer.parseInt(st.nextToken());
            }

            dp = new int[n + 1];
            Arrays.fill(dp, 1);

            for (int i = 2; i <= n; i++) {
                for (int j = 1; j < i; j++) {
                    if (a[j] < a[i]) {
                        dp[i] = Math.max(dp[i], dp[j] + 1);
                    }
                }
            }

            int max = 0;
            for (int i = 1; i <= n; i++) {
                max = Math.max(max, dp[i]);
            }

            pw.println(max);
        }

        pw.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 2 2 1 1
7 7
1 2 3 3 2 1 1
11 11
1 2 3 4 4 3 2 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the easy version of the problem. The only difference is that $x=n$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=n$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = n$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
output:
3
4
4

Note: In the first test case, you are required to calculate $f(5)$. It can be shown that after $4$ operations, $S$ can contain $2$, $3$, or $4$. The following shows the operations required to make $S = \\{4\\}$.

  * Initially, $S = \\{1, 2, 3, 4, 5\\}$ and $a = [1, 2, 3, 2, 1]$.    * Choose $l = 1$. Naturally, $r = 2$. Since $a_1 < a_2$, we set $a_2 = 1 + 2$ and remove $1$ from $S$. Now, $S = \\{2, 3, 4, 5\\}$ and $a = [1, 3, 3, 2, 1]$.    * Choose $l = 4$. Naturally, $r = 5$. Since $a_4 > a_5$, we set $a_4 = 2 + 1$ and remove $5$ from $S$. Now, $S = \\{2, 3, 4\\}$ and $a = [1, 3, 3, 3, 1]$.    * Choose $l = 3$. Naturally, $r = 4$. Since $a_3 = a_4$, we have a choice whether to remove $3$ or $4$. Since we want to preserve $4$, let's remove $3$. So, set $a_4 = 3 + 3$ and remove $3$ from $S$. Now, $S = \\{2, 4\\}$ and $a = [1, 3, 3, 6, 1]$.    * Choose $l = 2$. Naturally, $r = 4$. Since $a_2 < a_4$, we set $a_4 = 3 + 6$ and remove $2$ from $S$. Finally, $S = \\{4\\}$ and $a = [1, 3, 3, 9, 1]$.

In the second test case, you are required to calculate $f(7)$. It can be shown that after $6$ operations, $S$ can contain $2$, $4$, $6$, or $7$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N1 = 200009;
int spr[N1][21],pw[N1];
void bldsp(int a[],int n){
    for(int i=0;i<n;i++)spr[i][0]=a[i];
    for(int j = 1;(1<<j)<=n;j++)
        for(int i = 0;i+(1<<j)-1<n;i++)
            spr[i][j]=max(spr[i][j-1],spr[i+(1<<(j-1))][j-1]);
}
int quespr(int l,int r){
    int lng = pw[r-l+1];
    int ans = max(spr[l][lng],spr[r-(1<<lng)+1][lng]);
    return ans;
}
void solve(){
   int n;cin>>n;
   int temp;cin>>temp;
   int a[n];
   for(int i=0;i<n;i++){
      cin>>a[i];
   }
   bldsp(a,n);
   vector<int> prfx(n,0);
   prfx[0]=a[0];
   for(int i=1;i<n;i++){
      prfx[i]=prfx[i-1]+a[i];
   }
   int ans = 0;
   for(int i=0;i<n;i++){
      int p1 = i,p2 = i,sm=a[i];
      while(1){
         if(p1==0 && p2==n-1){ans++;break;}
         if(p1==0){
            int l=p2+1,r=n-1,ans1=-1;
            while(r>=l){
               int md = (r+l)/2;
               if(quespr(p2+1,md)>sm){r=md-1;ans1=md;}
               else {l=md+1;}
            }
            if(ans1==-1){ans++;break;}
            sm+=prfx[ans1]-prfx[p2]-a[ans1];p2=ans1-1;
            if(sm>=a[ans1]){sm+=a[ans1];p2=ans1;continue;}
            else {break;}
            continue;
         }
         if(p2==n-1){
            int l=0,r=p1-1,ans1=-1;
            while(r>=l){
               int md = (r+l)/2;
               if(quespr(md,p1-1)>sm){l=md+1;ans1=md;}
               else r=md-1;
            }
            if(ans1==-1){ans++;break;}
            sm+=prfx[p1-1]-prfx[ans1];p1=ans1+1;
            if(sm>=a[ans1]){sm+=a[ans1];p1=ans1;continue;}
            else break;
            continue;
         }
         int np1=p1,np2=p2;
         int l=0,r=p1-1,ans1=-1;
         while(r>=l){
            int md = (r+l)/2;
            if(quespr(md,p1-1)>sm){l=md+1;ans1=md;}
            else r=md-1;
         }
         if(ans1==-1){sm+=prfx[p1-1];p1=0;}
         else{
            sm+=prfx[p1-1]-prfx[ans1];p1=ans1+1;
            if(sm>=a[ans1]){sm+=a[ans1];p1=ans1;}
         }
         l=p2+1,r=n-1,ans1=-1;
         while(r>=l){
            int md = (r+l)/2;
            if(quespr(p2+1,md)>sm){r=md-1;ans1=md;}
            else {l=md+1;}
         }
         if(ans1==-1){sm+=prfx[n-1]-prfx[p2];p2=n-1;}
         else{
            sm+=prfx[ans1-1]-prfx[p2];p2=ans1-1;
            if(sm>=a[ans1]){sm+=a[ans1];p2=ans1;}
         }
         if(np1==p1 && np2==p2)break;
      }
   }
   cout<<ans<<"\n";
}
signed main(){
#ifndef ONLINE_JUDGE
freopen("./input.txt", "r", stdin);
freopen("./output.txt", "w",stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);pw[0]=0;
for(int i = 1;i<=200005;i++)pw[i]=pw[i/2]+1;
  int testcase=1;
  cin>>testcase;
  while(testcase--){
    solve();
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 2 3 2 1
7 7
4 5 1 2 1 4 5
11 11
1 2 3 1 1 9 3 2 4 1 3
```