**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>
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define MOD 1000000007LL
#define INF 987654321987654321LL
#define v(x) x.begin(),x.end()
#define fi(x) for(i=0;i<x;i++)
#define fi1(x) for(i=1;i<=x;i++)
#define fj(x) for(j=0;j<x;j++)
#define fj1(x) for(j=1;j<=x;j++)
#define fk(x) for(k=0;k<x;k++)
#define fk1(x) for(k=1;k<=x;k++)
#define fastio() {ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define pq priority_queue
#define vvi vector<vector<int> >
#define X first
#define Y second
#define endl '\n'
#define int long long
#define pbds tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>

using namespace __gnu_pbds;
using namespace std;
typedef pair<int, int> P;

int t;
int n,k;
int a[400000];
int seg[2000000];
int maxseg[2000000];
int sz;

void put(int now=1, int l=0, int r=sz-1)
{
    if (l==r) {
        seg[now] = a[l];
        return;
    }
    int mid = (l+r)/2;
    put(now * 2, l, mid);
    put(now * 2 + 1, mid + 1, r);
    seg[now] = seg[now * 2] + seg[now * 2 + 1];
}

void putMax(int now=1, int l=0, int r= sz - 1)
{
    if (l==r) {
        maxseg[now] = a[l];
        return;
    }
    int mid = (l+r)/2;
    putMax(now * 2, l, mid);
    putMax(now * 2 + 1, mid + 1, r);
    maxseg[now] = max(maxseg[now * 2], maxseg[now * 2 + 1]);
}
void init()
{
    sz = 1;
    while (sz < n) sz *= 2;
    int i;
    fi(sz+5) {
        maxseg[i] = 0;
        seg[i] = 0;
    }
}

int sum(int x, int y, int now=1, int l=0, int r=sz-1)
{
    if (r < x || y < l) return 0;
    if (x <= l && r <= y) return seg[now];

    int mid = (l+r)/2;
    return sum(x,y,now*2,l,mid)+sum(x,y,now*2+1,mid+1,r);
}

P findrange(int x, int y) // x보다 작은 것 중에서 최초로 y보다 큰것
{
    int now = x + sz;
    int past = now;

    int lo = -1, hi = sz; // 못찾을 수도 있으니
    while (now) {
        now /= 2;

        if (now*2 != past && maxseg[now*2] > y && lo == -1) {
            lo = now * 2;
            while (lo < sz) {
                if (maxseg[lo*2+1] > y) lo = lo * 2 + 1;
                else lo *= 2;
            }
            lo -= sz;
        }

        if (now*2+1 != past && maxseg[now*2+1] > y && hi == sz) { // 오른쪽을 보면서 올라가다 y보다 작은거 찾으면
            hi = now * 2 + 1;
            while (hi < sz) { // 리프에 도달하기 전까지
                if (maxseg[hi * 2] > y) hi *= 2; // 오른쪽은 크던 말던 무시, 최대한 왼쪽으로
                else hi = hi * 2 + 1;
            }
            hi -= sz;
        }
        past = now;
    }
    //cout<<lo<<" "<<hi<<endl;
    return {lo + 1, hi - 1};
}

signed main()
{
    int i,j;
    fastio();

    cin>>t;

    while (t--) {
        cin >> n >> k;
        fi(n) cin >> a[i];
        init();
        put();
        putMax();

        int ans = 0;
        fi(n) {
            int pastl = -1, pastr = -1;
            int now = a[i];
//            cout<<"doing "<<a[i]<<endl;
            while (1) {
                auto [l, r] = findrange(i, now);
//                cout<<now<<":"<<l<<" "<<r<<endl;
                if (l == pastl && r == pastr) break;
                now = sum(l,r);
                pastl = l;
                pastr = r;
            }
            if (pastl == 0 && pastr == sz-1) {
//                cout<<i<<"possible"<<endl;
                ans++;
            }
        }
        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:
```plaintext
1
4 4
2 2 2 2
```


**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
import threading
from collections import defaultdict

def main():
    import sys
    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    for _ in range(t):
        n, x = map(int, sys.stdin.readline().split())
        a = list(map(int, sys.stdin.readline().split()))
        # Since x =n in this version
        if n > 20:
            # Check if all elements are the same
            all_same = all(ai == a[0] for ai in a)
            if all_same:
                print(n)
            else:
                print(1)
        else:
            # Perform exact simulation
            from functools import lru_cache

            @lru_cache(None)
            def dfs(state, current_a_tuple):
                if len(state) == 1:
                    return {state[0]}
                results = set()
                state_list = list(state)
                current_a = list(current_a_tuple)
                for idx in range(len(state_list)-1):
                    l = state_list[idx]
                    r = state_list[idx+1]
                    a_l = current_a[idx]
                    a_r = current_a[idx+1]
                    if a_l > a_r:
                        # Keep l, update a_l
                        new_a = current_a.copy()
                        new_a[idx] = a_l + a_r
                        new_a.pop(idx+1)
                        new_state = tuple(state_list[:idx+1] + state_list[idx+2:])
                        results |= dfs(new_state, tuple(new_a))
                    elif a_l < a_r:
                        # Keep r, update a_r
                        new_a = current_a.copy()
                        new_a[idx+1] = a_l + a_r
                        new_a.pop(idx)
                        new_state = tuple(state_list[:idx] + state_list[idx+1:])
                        results |= dfs(new_state, tuple(new_a))
                    else:
                        # a_l ==a_r, two choices
                        # Choice 1: keep l, remove r
                        new_a1 = current_a.copy()
                        new_a1[idx] = a_l + a_r
                        new_a1.pop(idx+1)
                        new_state1 = tuple(state_list[:idx+1] + state_list[idx+2:])
                        res1 = dfs(new_state1, tuple(new_a1))
                        results |= res1
                        # Choice 2: keep r, remove l
                        new_a2 = current_a.copy()
                        new_a2[idx+1] = a_l + a_r
                        new_a2.pop(idx)
                        new_state2 = tuple(state_list[:idx] + state_list[idx+1:])
                        res2 = dfs(new_state2, tuple(new_a2))
                        results |= res2
                return results

            initial_state = tuple(range(1, n+1))
            initial_a = tuple(a)
            possible_js = dfs(initial_state, initial_a)
            print(len(possible_js))

threading.Thread(target=main,).start()

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

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

Think step by step.

Response:
```plaintext
1
5 5
1 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:
import sys
from bisect import bisect_left, bisect_right

def process_test_case(num_elements, values):
    value_with_index = [(values[i], i + 1) for i in range(num_elements)]
    cumulative_sum = [0] * (num_elements + 1)

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

    value_with_index.sort()

    valid_indices = {0, num_elements + 1, value_with_index[-1][1]}

    for j in range(num_elements - 1, 0, -1):
        current_index = value_with_index[j - 1][1]
        sorted_indices = sorted(valid_indices)

        pos = bisect_right(sorted_indices, current_index)
        right_boundary = sorted_indices[pos] if pos < len(sorted_indices) else num_elements + 1
        left_boundary = sorted_indices[pos - 1] if pos > 0 else 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():
    input = sys.stdin.read().split()
    index = 0
    test_case_count = int(input[index])
    index += 1

    results = []

    while test_case_count > 0:
        test_case_count -= 1

        num_elements = int(input[index])
        _unused = int(input[index + 1])  # unused value
        index += 2

        values = list(map(int, input[index:index + num_elements]))
        index += num_elements

        process_test_case(num_elements, values)

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 1 1 1 1
7 7
1 2 3 4 5 6 7
11 11
1 1 1 1 1 1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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.util.*;
import java.io.*;

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

    static class segmentTree {

        int N;
        int[] sgTree;

        public segmentTree(int arr []) {
            int n = arr.length;
            N = 1;
            while (N < n) {
                N <<= 1;
            }
            sgTree = new int[N << 1];
            build(1, 1, N, arr);
        }

        int merge(int left, int right) {
            return Math.max(left, right);
        }

        void build(int node, int l, int r, int[] arr) { // O(n)
            if (l == r) {
                if (l <= arr.length) {
                    sgTree[node] = arr[l - 1];
                }
                return;
            }
            int mid = (l + r) >> 1;
            int leftChild = 2 * node, rightChild = leftChild + 1;
            build(leftChild, l, mid, arr);
            build(rightChild, mid + 1, r, arr);
            sgTree[node] = merge(sgTree[leftChild], sgTree[rightChild]);
        }

        void updatePoint(int idx, int val) {
            idx += N - 1;
            sgTree[idx] = val;
            while (idx > 1) {
                idx >>= 1;
                sgTree[idx] = merge(sgTree[idx << 1], sgTree[(idx << 1) | 1]);
            }
        }

        int query(int start, int end) {
            return query(1, 1, N, start, end);
        }

        int query(int node, int l, int r, int start, int end) {
            if (l > end || r < start) {
                return 0;
            }
            if (l >= start && r <= end) {
                return sgTree[node];
            }
            int mid = (l + r) >> 1;
            int left = node << 1, right = left | 1;
            return merge(query(left, l, mid, start, end), query(right, mid + 1, r, start, end));
        }

        public String toString(){
            return  Arrays.toString(sgTree);
        }
    }

    public static void solve() throws IOException {
        int n = sc.nextInt(), x = sc.nextInt();
        int a[] = sc.nextIntArray(n);
        segmentTree st = new segmentTree(a);
        long prefix[] = new long[n + 1];
        for(int i = 1 ; i <= n ; i++)
            prefix[i] = prefix[i - 1] + a[i - 1];
        int ans = 0;
        for(int i = 0; i < n ; i++){
            int l = i + 1, r = i + 1;
            for(int j = 0; j < 40 ; j++){
                int low = r + 1, high = n;
                while(low <= high){
                    int mid = (low + high) / 2;
                    int max = st.query(r + 1, mid);
                    if(max <= prefix[r] - prefix[l - 1])
                        low = mid + 1;
                    else
                        high = mid - 1;
                }
                if(high <= n) {
                    if (prefix[r] - prefix[l - 1] >= a[high - 1]) {
                        r = high;
                    }
                }
                low = 1; high = l - 1;
                while(low <= high){
                    int mid = (low + high) / 2;
                    int max = st.query(mid, l - 1);
                    if(max <= prefix[r] - prefix[l - 1])
                        high = mid - 1;
                    else
                        low = mid + 1;
                }
                if(low > 0) {
                    if (prefix[r] - prefix[l - 1] >= a[low - 1]) {
                        l = low;
                    }
                }
            }
            if(l == 1 && r == n)
                ans++;
        }
        pw.println(ans);
    }

    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 buggy code, we need to consider the following aspects:

1. **Edge Cases**: The code should handle edge cases such as the smallest and largest possible values of \( n \) and \( x \).
2. **Special Sequences**: Sequences that might cause the segment tree or the merging logic to fail.
3. **Large Values**: Values that might cause overflow or incorrect behavior in the segment tree operations.
4. **Complex Merging Scenarios**: Sequences that require multiple operations to merge correctly.

Given the constraints and the nature of the problem, a small-scale test input that is likely to expose the bug could be:

- A sequence where the values are such that the merging operations are non-trivial and might lead to incorrect segment tree updates or queries.
- A sequence that tests the boundary conditions of the segment tree.

Here is a small-scale test input that meets these criteria:

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

Explanation:
- **Test Case Count**: 1
- **Number of Balls (n)**: 5
- **Value of x**: 5 (as \( x = n \))
- **Initial Values**: [1, 3, 2, 4, 1]

This input is designed to test the merging logic and the segment tree operations. The sequence [1, 3, 2, 4, 1] requires multiple steps to merge and might expose issues in the segment tree updates and queries. The segment tree should correctly handle the maximum values and the updates as the operations progress.


**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 PrintWriter out;
	static Reader in;
	public static void main(String[] args) throws Exception {
		input_output();
		Main solver = new Main();
		solver.solve();
		out.close();

		out.flush();
	}

	static int MOD = 1_000_000_000 + 7;
	static int MAX = 1_000_000_000;
	static int MAXN = 2_00_000 + 1;

	void solve() throws Exception {
		int t = in.nextInt();

		while (t --> 0) {
			int n = in.nextInt(),
				x = in.nextInt();

			int[][] sorted = new int[n][2];
			int[] arr = new int[n];
			int[] sum = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = in.nextInt();
				sorted[i][0] = arr[i];
				sorted[i][1] = i;
				if (i == 0) sum[i] = sorted[i][0];
				else sum[i] = sum[i-1] + sorted[i][0];
			}

			Arrays.sort(sorted, Comparator.comparingInt(a -> -a[0]));

			TreeSet<Integer> set = new TreeSet<>();
			set.add(-1); set.add(n);
			set.add(sorted[0][1]);
			boolean[] ans = new boolean[n];
			ans[sorted[0][1]] = true;
			for (int i = 1; i < n; i++) {
				int l = set.lower(sorted[i][1]),
					r = set.higher(sorted[i][1]);

				if (l != -1 && r != n) {
					if (!ans[l] || !ans[r]) continue;
					int val = sum[r-1] - sum[l];
					if (val >= arr[l] || val >= arr[r]) ans[sorted[i][1]] = true;
				} else if (l == -1) {
					if (!ans[r]) continue;
					int val = sum[r-1];
					if (val >= arr[r]) ans[sorted[i][1]] = true;
				} else {
					if (!ans[l]) continue;
					int val = sum[n-1] - sum[l];
					if (val >= arr[l]) ans[sorted[i][1]] = true;
				}

				set.add(sorted[i][1]);
			}

			int total = 0;
			for (int i = 0; i < n; i++)
				if (ans[i]) total++;
			out.println(total);
		}
	}

	static class Reader {
		private InputStream mIs;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public Reader() {
			this(System.in);
		}

		public Reader(InputStream is) {
			mIs = is;
		}

		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();

			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = mIs.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

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

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}
	static void input_output() throws IOException {
		File f = new File("in.txt");
		if (f.exists() && !f.isDirectory()) {
			in = new Reader(new FileInputStream("in.txt"));
		} else in = new Reader();
		f = new File("out.txt");
		if (f.exists() && !f.isDirectory()) {
			out = new PrintWriter(new File("out.txt"));
		} else out = new PrintWriter(System.out);
	}
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are some considerations:

1. **Edge Cases:**
   - Small values of `n` (e.g., `n = 1` or `n = 2`).
   - Large values of `n` (up to the limit, but we will keep it small for this example).
   - Cases where all values in the array are the same.
   - Cases where the array is sorted in ascending or descending order.
   - Cases with a mix of large and small values.

2. **Scenarios to Maximize Failure Probability:**
   - Cases where the operations lead to a specific pattern that the code might not handle correctly.
   - Cases where the merging process should result in a specific final set but the code fails to recognize it.

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

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

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `x = 3`
   - Array: `[1, 2, 3]`
   - This is a small, simple case to check if the code can handle basic merging correctly. The expected output is `1` because the only way to end up with a single element is to merge in a specific order.

2. **Test Case 2:**
   - `n = 5`, `x = 5`
   - Array: `[3, 3, 3, 3, 3]`
   - This case checks how the code handles an array with all identical values. The expected output is `5` because any single element can be the final result.

3. **Test Case 3:**
   - `n = 7`, `x = 7`
   - Array: `[1, 2, 3, 4, 5, 6, 7]`
   - This case checks how the code handles a sorted array. The expected output is `7` because any single element can be the final result.

This test input should help expose any bugs in the given code, particularly in how it handles identical values and sorted arrays.


**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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#define rint int
#define mem(a,b) memset(a,b,sizeof a)
#define lowbit(x) (x&-x)
#define debug(a,b) cout<<"line"<<__LINE__<<":"<<a<<" = "<<b<<endl;
#define eb emplace_back
#define pb push_back
#define read raed
#define int long long
#define ls (p<<1)
#define rs (p<<1|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const ll INF=1e18;
const int N=2e5+8;
template<typename T>
void read(T &x){
    x=0;int f=1;char c=getchar();
    while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    x*=f;
}
template<typename T,typename ... Args>
void read(T &x, Args &... y){read(x);read(y...);}
int t,n,mx;
int a[N];
int b[N],pre[N],nxt[N];
bool ff[N];
inline int work(int p){
	for(rint i=1;i<=n;i++) b[i]=a[i],pre[i]=i-1,nxt[i]=i+1;
	while(b[p]<mx){
		bool f=0;
		if(b[pre[p]]<=b[p]&&ff[pre[p]]&&pre[p]>=1) return 1;
		if(b[nxt[p]]<=b[p]&&nxt[p]<=n) b[p]+=b[nxt[p]],nxt[p]=nxt[nxt[p]],f=1;
		if(b[pre[p]]<=b[p]&&pre[p]>=1) b[p]+=b[pre[p]],pre[p]=pre[pre[p]],f=1;
		if(!f) return 0;
	}
	return 1;
}
signed main(){
	read(t);
	while(t--){
		read(n,n);mx=0;int ans=0;
		for(rint i=1;i<=n;i++) read(a[i]),mx=max(mx,a[i]),ff[i]=0;
		for(rint i=1;i<=n;i++) if(work(i)) ans++,ff[i]=1;
		printf("%lld\n",ans);
	}
	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
5 5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 maxn = 1e6 + 10;
int a[maxn], b[maxn];
int n, x;
int dp[maxn];
void solve()
{
    cin >> n >> x;
    int sum = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        dp[i] = 0;

        sum += a[i];
    }
    if (n == 1)
    {
        cout << 1 << '\n';
        return;
    }
    a[0] = 1e18;
    a[n + 1] = 1e18;
    dp[0] = 1;
    dp[n + 1] = 1;
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        int l = i - 1;
        int r = i + 1;
        int flag = 1;
        int oddsum = 0;
        int now = a[i];
        if (a[i] < a[i + 1] && a[i] < a[i - 1])
            continue;
        if (a[i] >= a[i + 1])
        {
            while (now >= a[r] && r <= n)
            {
                now += a[r];
                r++;
            }
            flag = 1;
        }
        else if (a[i] >= a[i - 1])
        {
            int ff = 0;
            while (now >= a[l] && l >= 1)
            {
                now += a[l];
                if (dp[l])
                {
                    dp[i] = 1;
                    ff = 1;
                    break;
                    // continue;
                }
                l--;
            }
            flag = 0;
            if (ff)
            {
                // cout << i << ' ';
                ans++;
                continue;
            }
        }
        while (1)
        {
            if (flag)
            {
                if (now >= a[l] && l >= 1)
                {
                    while (now >= a[l] && l >= 1)
                    {
                        now += a[l];
                        if (dp[l])
                        {
                            now = sum;
                            break;
                        }
                        l--;
                    }
                }
                else
                {
                    break;
                }
            }
            else
            {
                if (now >= a[r] && r <= n)
                {
                    while (now >= a[r] && r <= n)
                    {
                        now += a[r];
                        r++;
                    }
                }
                else
                    break;
            }
            flag = flag ^ 1;
        }
        if (now == sum)
        {
            // cout << i << ' ';
            dp[i] = 1;
            ans++;
        }
    }
    // cout << '\n';
    cout << ans << '\n';
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `5 5` with the sequence `1 1 1 1 1`.
   - **Output:** `1`
   - **Reason:** All elements are the same, so the only way to end up with a single element is to start with the middle element. The middle element is 3, and it is the only valid starting point.

2. **Test Case 2:**
   - **Input:** `7 7` with the sequence `1 1 2 2 1 1 1`.
   - **Output:** `2`
   - **Reason:** The possible valid starting points are 3 and 5. Starting with 3, you can merge the sequence to end up with a single element. Similarly, starting with 5 also leads to a single element.

3. **Test Case 3:**
   - **Input:** `11 11` with the sequence `1 2 2 1 1 3 3 1 1 2 2`.
   - **Output:** `3`
   - **Reason:** The possible valid starting points are 3, 6, and 8. Starting with these points, you can merge the sequence to end up with a single element.

These test cases are designed to expose issues in the buggy code by focusing on edge cases where the elements are the same or have symmetric patterns, which can cause the algorithm to fail if it does not handle these scenarios correctly.


**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:
for _ in range(int(input())):
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    ans=0
    for i in range(n):
        p=i-1;q=i+1;x=l[i]
        while True:
            f=0
            if p>=0 and l[p]<=x:x+=l[p];p-=1;f+=1
            if q<n and l[q]<=x:x+=l[q];q+=1;f+=1
            if p<0 and q>=n:ans+=1;break
            if f==0:break
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 solve():
    def find_max_index(a, l, r):
        max_value = max(a[l:r+1])
        max_index = a.index(max_value, l, r+1)
        return max_index

    x, n = map(int, input().split())
    a = [0] + list(map(int, input().split()))
    pre = [0] * (n + 1)
    for i in range(1, n + 1):
        pre[i] = a[i] + pre[i - 1]

    cnt = 1
    stack = [(1, n)]
    while stack:
        l, r = stack.pop()
        if l >= r:
            continue

        idx = find_max_index(a, l, r)
        if pre[r] - pre[idx] >= a[idx]:
            cnt += 1
            stack.append((idx + 1, r))
        if pre[idx - 1] - pre[l - 1] >= a[idx]:
            cnt += 1
            stack.append((l, idx - 1))

    print(cnt)

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:
```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:
//package com.aakash.Contest.C965;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class E1 {
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));
    public static void main(String[] args) throws IOException {

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        while (t-- > 0) {
            StringTokenizer s2
                    = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(s2.nextToken());
            int[] arr = new int[n];
            StringTokenizer s3
                    = new StringTokenizer(br.readLine());
            for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(s3.nextToken());
            solve(arr, n);
        }

    }

    static void solve(int[] arr, int n) {
        int res=dnq(arr,0,n-1);
        System.out.println(res);
    }
    static int dnq(int[] arr,int s,int e) {
        if(s>e) ;
        if(s==e) return 1;
        int max=-1,ind=-1;
        for(int i=s;i<=e;i++) {
            if(arr[i]>max) {
                ind=i;
                max=arr[i];
            }
        }
        int c=1;
        long left=0,right=0;
        for(int i=s;i<ind;i++) left+=arr[i];
        for(int i=ind+1;i<=e;i++) right+=arr[i];
        if(left>=arr[ind]) c+=dnq(arr,s,ind-1);
        if(right>=arr[ind]) c+=dnq(arr,ind+1,e);
        return c;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
2 2 2 2 2
```


**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 {
    /**
     * 给你一个数组，每次操作可以选择相邻两个元素，大的元素吃掉小的元素，值 += 小的元素值
     * 问有多少个元素最终可以活下来
     *
     * @param args
     */
    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt(), x = sc.nextInt();
            int[] a = new int[n];
            int max = 0;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                max = Math.max(max, a[i]);
            }
            long[] ps = new long[n + 1];
            for (int i = 0; i < n; i++) {
                ps[i + 1] = ps[i] + a[i];
            }
            int[] l = new int[n], r = new int[n];
            Stack<Integer> stack = new Stack<>();
            for (int i = 0; i < n; i++) {
                while (!stack.isEmpty() && a[i] > a[stack.peek()]) {
                    r[stack.pop()] = i;
                }
                stack.push(i);
            }

            while (!stack.isEmpty()) r[stack.pop()] = n;

            for (int i = n - 1; i >= 0; i--) {
                while (!stack.isEmpty() && a[i] > a[stack.peek()]) {
                    l[stack.pop()] = i;
                }
                stack.push(i);
            }
            while (!stack.isEmpty()) l[stack.pop()] = -1;
            int ans = 0;
            //[l + 1,r - 1]
            for (int i = 0; i < n; i++) {
                long v = ps[r[i]] - ps[l[i] + 1];
                if (v >= max) {
                    ans++;
                    continue;
                }
                int le = l[i], ri = r[i];
                while (v < max && le >= 0 && ri < n) {
                    if (a[le] < a[ri]) {
                        if (v < a[le]) {
                            break;
                        } else {
                            v += a[le];
                            le--;
                        }
                    } else {
                        if (v < a[ri]) {
                            break;
                        } else {
                            v += a[ri];
                            ri++;
                        }
                    }
                }
                if (v >= max) {
                    ans++;
                    continue;
                }
                while (v < max && le >= 0) {
                    if (v < a[le]) {
                        break;
                    } else {
                        v += a[le];
                        le--;
                    }
                }
                while (v < max && ri < n) {
                    if (v < a[ri]) {
                        break;
                    } else {
                        v += a[ri];
                        ri++;
                    }
                }
                if (v >= max) {
                    ans++;
                }
            }
            out.println(ans);

        }
        out.close();
    }

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

    static class Kattio {
        static BufferedReader r;
        static StringTokenizer st;

        public Kattio() {
            r = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(r.readLine());
                }
                return st.nextToken();
            } catch (Exception e) {
                return null;
            }
        }

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

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

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include<stdio.h>
int main()
{
	int t, i, j, flag = 0;
	char a[100] = {0};
	scanf("%d", &t);
	if (t < 1 || t>100)
		return 0;
	for (i = 0; i < t; i++)
	{
		for (j = 0; j < 100; j++)
		{
			scanf("%c", &a[j]);
			if (j!=0 && a[j] == '\n')
			{
				flag = j ;
				break;
			}
		}
		for (j = flag-1 ; j>=0; j--)
		{
			if ((a[j] == 'p') || (a[j] == 'q') || (a[j] == 'w'))
			{
				if (a[j] == 'p')
					printf("q");
				else if (a[j] == 'q')
					printf("p");
				else
					printf("w");
			}
		}
		printf("\n");
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
pq
qw
wp
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
int main()
{
    int test;
    scanf("%d", &test);
    for(int t=1; t<=test; t++)
    {
        int i;
        char a[100], b[100];
        scanf("%s", a);
        for(int i=0; a[i]; i++)
        {
            if(a[i]=='q') a[i]='p';
            else if(a[i]=='p') a[i]='q';
        }
        for(i=0; i<strlen(a); i++)
        {
            b[strlen(a)-i-1] = a[i];
        }
        b[i] = '\0';
        printf("%s\n", b);
    }
    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
w
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

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

int main(){
 int n;
 cin>>n;
 string store[n];
 string rev[n];
 for (int i = 0; i < n; i++)
 {
    cin>>store[i];

 }
//  cout<<store[0][store[0].length()-1];

for (int i = 0; i < n; i++)
{
    for (int j = 0; j < store[i].length(); j++)
    {
        rev[i][j]=store[i][store[i].length()-1-j];
    }

}

for (int i = 0; i < n; i++)
{
    for (int j = 0; j < store[i].length();  j++)
    {
        if (rev[i][j]=='p')
        {
            cout<<"q";
        }
        else if(rev[i][j]=='q'){
            cout<<"p";
        }
        else{
             cout<<rev[i][j];
        }

    }
    cout<<endl;
}

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the following aspects:

1. **Array Initialization**: The `rev` array is not properly initialized, which can lead to undefined behavior when accessing its elements.
2. **Character Swapping**: The code correctly swaps 'p' and 'q', but it does not handle the case where the string is empty or has a single character.
3. **Edge Cases**: We should consider strings with single characters, empty strings, and strings with only one type of character.

Given these points, let's create a small-scale test input that can expose the bug:

```plaintext
5
p
q
w
pp
qq
```

This test input includes:
- Single character strings ('p', 'q', 'w') to check if the code handles them correctly.
- Two-character strings ('pp', 'qq') to check if the code handles repeated characters correctly.

This should help identify issues with the array initialization and character swapping logic 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: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
def mirror_character(ch):
    if ch == 'p':
        return 'q'
    elif ch == 'q':
        return 'p'
    else:  # ch == 'w'
        return 'w'

def transform_string(s):
    # Reverse the string
    reversed_s = s[::-1]
    # Replace each character with its mirrored character
    transformed_s = ''.join(mirror_character(ch) for ch in reversed_s)
    return transformed_s

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    t = int(data[0])
    results = []

    for i in range(1, t + 1):
        a = data[i]
        b = transform_string(a)
        results.append(b)

    for result in results:
        print(result)

# Example usage with provided sample input
if __name__ == "__main__":
    import sys
    from io import StringIO

    sample_input = """5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq"""

    sys.stdin = StringIO(sample_input)
    main()

# aNHMA7qwEjWbiK065O1

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

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

Think step by step.

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

1. **Empty String**: Although the problem guarantees that the string will have at least one character, it's good to test this boundary condition.
2. **Single Character Strings**: Test with strings containing only one character to ensure the transformation works correctly.
3. **Reversal and Mirroring**: Ensure the code correctly handles the reversal and mirroring of characters.
4. **Special Characters**: Since the string can only contain 'p', 'q', and 'w', we need to test combinations of these characters to ensure the transformation is correct.

Given these points, let's create a small-scale test input:

```plaintext
6
p
q
w
pq
wp
pqp
```

This test input includes:
- Single character strings: 'p', 'q', 'w'
- A two-character string with 'p' and 'q': 'pq'
- A two-character string with 'w' and 'p': 'wp'
- A three-character string with 'p', 'q', and 'p': 'pqp'

These test cases will help ensure that the code correctly handles different lengths and combinations of characters.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
t = int(input())
while t != 0:
    a = str(input())
    l = list(a)
    l1=[]
    n1=l.count('p')
    n2=l.count('q')
    if n1==n2:
        print(a)
    else:
        for i in range(len(l)):
            if l[i]=='p':
                l1.append('q')
            elif l[i]=='q':
                l1.append('p')
            else:
                l1.append('w')
    l1.reverse()

    for i in l1:
        print(i,end='')
    print()

    t -= 1

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

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

Think step by step.

Response:
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 code has a logic issue where it only reverses the string and swaps 'p' and 'q' if the counts of 'p' and 'q' are not equal. This logic is flawed because the reflection should always reverse the string and swap 'p' and 'q' regardless of their counts.

Let's consider the following scenarios:
1. **Equal counts of 'p' and 'q'**: The code will print the original string without any modifications, which is incorrect.
2. **Unequal counts of 'p' and 'q'**: The code will reverse the string and swap 'p' and 'q', which is the correct behavior.

A small-scale test input that can expose this bug is a string where the counts of 'p' and 'q' are equal. For example, the string "pq".

Here is the test input:
```plaintext
1
pq
```

When this input is provided to the buggy code:
- The code will count the 'p' and 'q' in the string "pq".
- Since the counts are equal (1 'p' and 1 'q'), the code will print the original string "pq" without any modifications.
- The correct output should be "qp" (reversed and swapped).

This test input will clearly show that the code is not handling the reflection correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
for _ in range(int(input())):
    a = list(input())
    for i in range(len(a)):
        if a[i] == 'p':
            a[i] = 'q'
        elif a[i] == 'q':
            a[i] = 'p'
        a.reverse()
    print(''.join(a))

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

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

Think step by step.

Response:
```plaintext
3
p
q
w
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    int t;
    cin>>t;
    while(t--){
        string s;
        cin>>s;
        int sl=s.size();
        if((s[0]=='q' && s[sl-1]=='q') || (s[0]=='p' && s[sl-1]=='p')){
            for(int i=0;i<sl;i++){
                if(s[i]=='q') s[i]='p';
                else if(s[i]=='p') s[i]='q';
            }
        }
        else if((s[0]=='w' && s[sl-1]=='p' || s[sl-1]=='q') ||
                (s[sl-1]=='w' && s[0]=='p' || s[0]=='q')){
                    for(int i=0;i<(sl/2);i++){
                        char temp=s[i];
                        s[i]=s[sl-i-1];
                        s[sl-i-1]=temp;
                        if(s[i]=='p') s[i]='q';
                        else if(s[i]=='q') s[i]='p';
                        if(sl%2!=0 && s[(sl/2)]=='p') s[(sl/2)]='q';
                        else if(sl%2!=0 && s[(sl/2)]=='q') s[(sl/2)]='p';
                        if(s[sl-i-1]=='p') s[sl-i-1]='q';
                        else if(s[sl-i-1]=='q') s[sl-i-1]='p';
                    }
                }

        cout<<s<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
3
pwp
qwp
wqq
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main(){
    int t;
    cin>>t;
    vector<string>b;
    while(t--){
        char p,q;
        string a;
        cin>>a;
        for(int i =0;i<a.size();i++){
            reverse(a.begin(),a.end());
            if(a[i]=='p'){
                a[i]='q';
            }
            else if(a[i]=='q'){
                a[i]='p';
            }
            else{
                continue;
            }
        }
        b.push_back(a);
    }
    for(string i:b){
        cout<<i<<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 given buggy code, we need to focus on the following issues in the code:

1. **Reverse Function Misplacement**: The `reverse` function is called inside the loop for each character, which is incorrect. It should be called once after the loop to reverse the entire string.
2. **Character Transformation**: The transformation of 'p' to 'q' and 'q' to 'p' is correct, but the `continue` statement is unnecessary and can be removed.

Given these issues, we can create a test case that highlights the problem with the incorrect placement of the `reverse` function. A string that is not symmetric and has a mix of 'p', 'q', and 'w' characters would be ideal.

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

```plaintext
3
pqp
wqw
pqw
```

Explanation:
- For the string "pqp", the correct output should be "pqp" because it is symmetric.
- For the string "wqw", the correct output should be "wqw" because it is also symmetric.
- For the string "pqw", the correct output should be "wqp" because it is not symmetric and the characters need to be transformed and reversed.

This test input will help identify the issue with the `reverse` function being called incorrectly inside the loop.