**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

using namespace std;

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

#define ar array
#define ll long long
#define ld long double
#define sz(x) ((int)x.size())
#define all(a) (a).begin(), (a).end()

const int MAX_N = 1e5 + 5;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int tc;
    cin >> tc;
    while(tc--){
        int n;
        cin>>n;
        vector<int> a(n);
        for (int i = 0; i<n; i++){
                cin>>a[i];
        }
        int curr = 0;
        int max_diff = 0;
        int curr_x = 0;
        for (int i = 0; i<n; i++){
            if (a[i]>curr_x){
                curr_x+=1;
                curr+=1;
            }
            else if (a[i]<curr_x){
                curr_x-=1;
                curr-=1;
            }
            if (curr>=0){
                curr=0;
            }
            max_diff =max(max_diff,abs(curr));
        }
        if (max_diff ==0){
            max_diff = -1;
        }
        cout<<curr_x+max_diff<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios where it might fail.

### Analysis of the Buggy Code

1. **Rating Calculation**:
   - The code initializes `curr_x` to 0 and iterates through the contests.
   - It updates `curr_x` based on the performance rating `a[i]`:
     - If `a[i] > curr_x`, `curr_x` increases by 1.
     - If `a[i] < curr_x`, `curr_x` decreases by 1.
   - It also updates a `curr` value, which is used to calculate `max_diff`.

2. **Bug Identification**:
   - The code uses `curr` to track the difference between the current rating and the initial rating, but it resets `curr` to 0 whenever it becomes non-negative.
   - This reset can lead to incorrect calculations of `max_diff`, especially in cases where the rating fluctuates significantly.

### Challenging Test Input

To expose the bug, we need a test case where the rating fluctuates a lot, causing the `curr` value to be reset frequently. This can lead to an incorrect `max_diff` calculation.

### Example Test Input

Let's consider a test case with a sequence of ratings that fluctuates between high and low values:

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

### Explanation

- **Number of Test Cases**: 1
- **Number of Contests**: 10
- **Performance Ratings**: `1 10 1 10 1 10 1 10 1 10`

### Expected Behavior

- The rating starts at 0.
- The first contest (1) decreases the rating to -1.
- The second contest (10) increases the rating to 0.
- The third contest (1) decreases the rating to -1.
- The fourth contest (10) increases the rating to 0.
- This pattern continues.

The optimal interval to skip would be any interval that includes a high rating (10) followed by a low rating (1). For example, skipping the interval `[2, 3]` would result in the following rating changes:

- Initial rating: 0
- Contest 1 (1): Rating becomes -1
- Contest 2 (10): Rating becomes 0 (skipped)
- Contest 3 (1): Rating remains 0 (skipped)
- Contest 4 (10): Rating becomes 1
- Contest 5 (1): Rating becomes 0
- Contest 6 (10): Rating becomes 1
- Contest 7 (1): Rating becomes 0
- Contest 8 (10): Rating becomes 1
- Contest 9 (1): Rating becomes 0
- Contest 10 (10): Rating becomes 1

The final rating would be 1, but the code might incorrectly calculate a higher `max_diff` due to the frequent resets of `curr`.

### Output

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

This test input should expose the bug in the provided buggy code by causing it to incorrectly calculate the maximum possible rating.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
t = int(input())
for z in range (t):
    n = int(input())
    a = list(map(int, input().split()))
    dp = [0,-10**10,-10**10]
    for i in a:
        if i > dp[0]:
            dp[0] += 1
        if i < dp[0]:
            dp[0] -= 1
        dp[1] = max(dp[0], dp[1])
        dp[2] = max(dp[1], dp[2])
        if dp[2] < i:
            dp[2] += 1
        if dp[2] > i:
            dp[2] -= 1
    if dp[0] == dp[1] == dp[2]:
        print(dp[2]-1)
    else:
        print(max(dp[1], dp[2]))

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

### Explanation:
This test case is designed to expose the bug in the given buggy code. Here’s why:

1. **Sequence of Ratings**: The sequence `1 2 3 2 1 4` is carefully chosen to create a scenario where the rating changes in a non-monotonic way. This can help highlight issues in the logic that updates the `dp` array.

2. **Optimal Skip Interval**: The optimal interval to skip is `[3, 5]`. Here’s the detailed breakdown:
   - Initial rating: `0`
   - Contest 1: `a_1 = 1` → Rating: `1`
   - Contest 2: `a_2 = 2` → Rating: `2`
   - Contest 3 (skipped): `a_3 = 3` → Rating: `2`
   - Contest 4 (skipped): `a_4 = 2` → Rating: `2`
   - Contest 5 (skipped): `a_5 = 1` → Rating: `2`
   - Contest 6: `a_6 = 4` → Rating: `3`

   The final rating after skipping the interval `[3, 5]` is `3`.

3. **Buggy Code Analysis**:
   - The buggy code updates the `dp` array in a way that does not correctly handle the skip intervals. Specifically, the logic for updating `dp[2]` and the final output logic are flawed.
   - The `dp[2]` value should ideally represent the maximum possible rating after considering the skip intervals, but the given code does not handle this correctly.

4. **Expected Output**:
   - The correct output for this test case is `3`.

By using this test case, the bug in the buggy code can be more easily identified and corrected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
import sys

def main():
    import sys
    import threading

    def solve():
        import sys

        input = sys.stdin.read().split()
        ptr = 0
        t = int(input[ptr])
        ptr +=1
        for _ in range(t):
            n = int(input[ptr])
            ptr +=1
            a = list(map(int, input[ptr:ptr+n]))
            ptr +=n

            delta = []
            x =0
            for ai in a:
                if ai > x:
                    delta_i =1
                elif ai ==x:
                    delta_i =0
                else:
                    delta_i =-1
                delta.append(delta_i)
                x += delta_i
            x_n = x

            # Find the minimal subarray sum using Kadane's algorithm
            min_subarray_sum = delta[0]
            current_min = delta[0]
            for i in range(1,n):
                current_min = min(delta[i], current_min + delta[i])
                if current_min < min_subarray_sum:
                    min_subarray_sum = current_min

            # The final answer is x_n - min_subarray_sum
            result = x_n - min_subarray_sum
            print(result)

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

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The key points to focus on are:

1. **Edge Cases:**
   - Single contest.
   - All contests have the same rating.
   - All contests have ratings in a strictly increasing or decreasing order.
   - A mix of increasing and decreasing ratings with a large range.

2. **Specific Scenarios:**
   - Skipped intervals that affect the final rating significantly.
   - Scenarios where the minimal subarray sum is not correctly calculated.

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

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `1\n1`
   - **Output:** `0`
   - **Explanation:** Kevin must skip the only contest, so his rating remains at 0.

2. **Test Case 2:**
   - **Input:** `5\n5 4 3 2 1`
   - **Output:** `4`
   - **Explanation:** The ratings are in a strictly decreasing order. The optimal choice is to skip the last contest, resulting in a final rating of 4.

3. **Test Case 3:**
   - **Input:** `7\n1 3 2 5 4 6 7`
   - **Output:** `5`
   - **Explanation:** The ratings are a mix of increasing and decreasing. The optimal choice is to skip the interval [3,5], resulting in a final rating of 5.

This test input is designed to cover a range of scenarios and edge cases that can help expose potential bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
#!/usr/bin/env python
from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

def solve(ratings):
    n = len(ratings)
    dp = [[-10**9]*3 for _ in range(n+1)]
    dp[0][0] = 0

    for i in range(n):
        curr_pos = i + 1

        if dp[i][0] != -10**9:
            rating = dp[i][0]
            dp[curr_pos][1] = max(dp[curr_pos][1], rating)
            new_rating = rating
            if ratings[i] > rating:
                new_rating += 1
            elif ratings[i] < rating:
                new_rating -= 1
            dp[curr_pos][0] = max(dp[curr_pos][0], new_rating)

        if dp[i][1] != -10**9:
            rating = dp[i][1]
            dp[curr_pos][1] = max(dp[curr_pos][1], rating)
            new_rating = rating
            if ratings[i] > rating:
                new_rating += 1
            elif ratings[i] < rating:
                new_rating -= 1
            dp[curr_pos][2] = max(dp[curr_pos][2], new_rating)

        if dp[i][2] != -10**9:
            rating = dp[i][2]
            new_rating = rating
            if ratings[i] > rating:
                new_rating += 1
            elif ratings[i] < rating:
                new_rating -= 1
            dp[curr_pos][2] = max(dp[curr_pos][2], new_rating)

    return dp[n][2]

def main():
    for _ in range(int(input())):
        n = int(input())
        ratings = list(map(int, input().split()))
        print(max(0, solve(ratings)))

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

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

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The main issue in the code seems to be related to how it handles the states and transitions, especially when skipping contests.

Here are the steps to think about:

1. **Initial State Handling**: Ensure the initial state is correctly set up.
2. **Skipping Contests**: Verify that the code correctly handles the skipping of contests.
3. **Rating Updates**: Check if the rating updates are correctly applied based on the contest performance.
4. **Edge Cases**: Consider cases where the ratings are all the same, or where the ratings are in a specific pattern that might cause the algorithm to fail.

Given these considerations, a good test case to expose the bug would be one where the ratings are in a pattern that challenges the skipping logic and the rating updates.

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

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

Explanation:
- This test case has 10 contests, and all ratings are the same (5).
- The optimal strategy would be to skip any single contest, as skipping any contest will not change the rating.
- The expected output should be 5, but if the code incorrectly handles the skipping logic, it might produce a different result.

This test case is simple but effective in exposing issues with the state transitions and skipping logic in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
import java.util.*;

public class C_New_Rating {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tests = sc.nextInt();

        for(int t = 0; t < tests; t++){
            int n = sc.nextInt();
            int[] a = new int[n + 1];
            for(int i = 1; i <= n; i++){
                a[i] = sc.nextInt();
            }
            int x = 0;

            int dp[] = new int[n + 1];
            for(int i = 1; i <= n; i++){
                dp[i] = dp[i - 1];
                if(a[i] > dp[i])dp[i]++;
                else if(a[i] < dp[i])dp[i]--;
            }

            Pair[] mxs = new Pair[n + 2];

            for(int i = 0; i <= n; i++){
                mxs[i] = new Pair();
            }

            for(int i = 1; i <= n; i++){
                mxs[i] = mxs[i - 1];
                if(dp[i] > mxs[i].a){
                    mxs[i].a = dp[i];
                    mxs[i].b = i;
                }
                //System.out.print(mxs[i].a + " " + mxs[i].b + " ");
            }
            //System.out.println("");
            dp[0] = 0;

            int diff = 0;
            for(int i = 1; i <= n; i++){
                dp[i] = dp[i - 1];
                diff = 0;
                if(a[i] > dp[i])diff = 1;
                else if(a[i] < dp[i])diff = -1;
                dp[i] = Math.max(dp[i - 1] + diff, mxs[i].a);
            }

            if(dp[n] == n)dp[n]--;
            System.out.println(dp[n]);

            // for(int i = 1; i <= n; i++){
            //     System.out.print(dp[i] + " ");
            // }
            // System.out.println("");
        }
    }
}

class Pair {
    int a = -1, b = -1;
    Pair(){
        a = -1;
        b = -1;
    }
    Pair(int a, int b){
        this.a = a;
        this.b = b;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has issues with how it handles the dynamic programming (DP) array and the optimal interval selection. Here are the steps to create a test input:

1. **Edge Case: Single Contest**: Ensure the code handles the case where there is only one contest.
2. **Edge Case: All Contests Have the Same Rating**: Test the scenario where all contests have the same rating.
3. **Edge Case: Sequential Increasing Ratings**: Test the scenario where the ratings are strictly increasing.
4. **Edge Case: Sequential Decreasing Ratings**: Test the scenario where the ratings are strictly decreasing.
5. **Edge Case: Mixed Ratings**: Test the scenario with a mix of increasing and decreasing ratings.

Let's create a small-scale test input that covers these scenarios:

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

This test input includes:
- A single contest with a rating of 1.
- Five contests with the same rating of 1.
- Five contests with strictly increasing ratings.
- Five contests with strictly decreasing ratings.
- Ten contests with a mix of increasing and decreasing ratings.

These cases should help expose any issues in the buggy code's handling of different scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

public class C {
    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int[][] dp = new int[n][3];//before in after
            dp[0][0] = 1;
            dp[0][1] = 0;
            dp[0][2] = -1 << 30;
            for (int i = 1; i < n; i++) {
                dp[i][0] = dp[i - 1][0] + f(dp[i - 1][0], a[i]);
                dp[i][1] = Math.max(dp[i - 1][0], dp[i - 1][1]);
                dp[i][2] = Math.max(dp[i - 1][1], dp[i - 1][2] + f(dp[i - 1][2], a[i]));
            }
            out.println(Math.max(dp[n - 1][1], dp[n - 1][2]));
        }
        out.close();
    }

    static int f(int a, int b) {
        int v = 0;
        if (a < b) v = 1;
        else if (a > b) v = -1;
        return v;
    }

    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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios where the logic in the code might fail. The key points to focus on are:

1. **Edge Cases:**
   - Single contest.
   - All contests have the same rating.
   - Contests with a strictly increasing or decreasing sequence of ratings.
   - Contests with a mix of increasing and decreasing ratings.

2. **Optimal Skipping:**
   - Ensure that the optimal interval for skipping is not trivial (e.g., skipping the first or last contest alone).

3. **Initialization and State Transitions:**
   - Check if the initialization of the `dp` array and the state transitions are correct.

Given these considerations, let's construct a test case that might expose the bug in the code:

### Test Case:
```plaintext
5
1
1
3
1 1 1
6
1 2 3 4 5 6
6
6 5 4 3 2 1
9
1 2 3 1 2 3 1 2 3
```

### Explanation:
1. **Single Contest:**
   - This test case checks if the code handles a single contest correctly. The rating should remain 0 since Kevin must skip at least one contest.

2. **All Contests with the Same Rating:**
   - This test case checks if the code can handle a sequence of the same ratings. The optimal rating should be 0 because skipping any interval will result in no change.

3. **Strictly Increasing Sequence:**
   - This test case checks if the code can handle a strictly increasing sequence of ratings. The optimal rating should be 5, as skipping any single contest will not change the final rating.

4. **Strictly Decreasing Sequence:**
   - This test case checks if the code can handle a strictly decreasing sequence of ratings. The optimal rating should be 1, as skipping the first few contests will maximize the final rating.

5. **Mixed Sequence with Repeating Patterns:**
   - This test case checks if the code can handle a sequence with repeating patterns. The optimal rating should be 3, as skipping the middle part of the sequence will maximize the final rating.

This test case should help in identifying potential issues with the state transitions and the logic in the `dp` array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

int main() {
	cin.tie(0)->sync_with_stdio(0);

	int T;
	cin >> T;

	while (T--) {
		int N;
		cin >> N;

		vector<int> A(N);
		for (auto &Ai : A) cin >> Ai;

		vector<int> pfx(N);
		int x = 0;
		for (int i = 0; i < N; i++) {
			pfx[i] = x;
			if (A[i] > x) {
				x++;
			} else if (A[i] < x) {
				x--;
			}
		}

		auto check = [&](int X) -> bool {
			set<int> S;
			for (int i = 0; i < N; i++) {
				S.insert(pfx[i]);
			}
			bool rv = false;
			for (int i = N - 1; i >= 0; i--) {
				if (A[i] < X) {
					X++;
				} else {
					X--;
				}
				rv = rv or S.lower_bound(X) != S.end();
			}
			return rv;
		};
		int l = 0, r = N - 1;
		while (l < r) {
			int m = l + (r - l + 1) / 2;
			check(m) ? l = m : r = m - 1;
		}
		cout << l << '\n';
	}
}
/*
1. Think big picture!

2. Don't waste time coding before thinking it through!

3. Think fast! Don't be afraid to churn through ideas!

4. Try modelling the problem with as few moving parts as possible!
	- What does the problem look like?
	- What might the solution look like?

5. Try listing properties of components of the problem!
	- All ____ have ____
	- All ____ can be represented as ____
	- There is at most ____ ____'s

6. Try solving an easier version of the problem!
	- What if the problem didn't have this constraint?
	- How would I solve a more general version of the problem?
	- Remember, what is presented to you isn't all you can use!

7. Ask questions!
	- What has similar behavior?
	- I've assumed this, what now?

*/

**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, let's consider the following steps:

1. **Understand the Bug**: The buggy code uses a binary search to find the maximum possible rating. However, it does not correctly account for the intervals where the rating can be maximized. The `check` function is flawed because it uses a set to store prefix ratings and then checks if a certain value can be reached. This approach may not correctly simulate the behavior of skipping an interval and recalculating the rating.

2. **Identify Edge Cases**:
   - **All ratings are the same**: This can cause issues with the binary search.
   - **Ratings that form a strictly increasing or decreasing sequence**: This can challenge the prefix sum approach.
   - **Ratings that have a large gap in the middle**: This can expose issues with the interval selection.
   - **Ratings that have a peak and then drop**: This can test the recalculation logic.

3. **Construct the Test Case**: We will create a test case that includes a mix of these edge cases to ensure the buggy code fails.

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

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

This test input includes:
- A strictly increasing sequence.
- A sequence with a peak and then a drop.
- A single contest.
- A sequence with repeated values and a large gap in the middle.
- A sequence with a mix of increasing and decreasing values.

This should challenge the buggy code and expose its flaws in handling different types of intervals and ratings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
#include <bits/stdc++.h>
#include <vector>
using namespace std;
using ll = long long;
using sll = signed long long;
using ull = unsigned long long;
#define ff first
#define ss second
#define pb push_back
typedef pair<int,int> pai;
const ll MOD = 1e9 + 7;
const ll inf = 1e18;
const int maxN = 4e4 + 10;

void f(){
    int n;cin >> n;
    vector<int>v(n + 1), dp(n + 1), dat(n + 1), dat2(n + 2);
    for(int i = 1; i <= n; i++){
        cin >> v[i];
        if(v[i] < dat[i - 1]) dat[i] = dat[i - 1] - 1;
        else if(v[i] > dat[i - 1]) dat[i] = dat[i - 1] + 1;
        else dat[i] = dat[i - 1];
        dp[i] = max(dp[i - 1],dat[i]);
    }
    if(n == 1){
        cout << 0 << '\n';
        return;
    }
    if(is_sorted(v.begin(),v.end())){
        map<int,int>mp;
        bool b = false;
        for(int i = 1; i <= n; i++) mp[v[i]]++, b |= (mp[v[i]] > 1);
        if(b) cout << dp[n] << "\n";
        else cout << dp[n - 1] << "\n";
        return;
    }
    if(v[n]  < v[n - 1]) dat2[n] = -1;
    else dat2[n] = 1;
    for(int i = n - 1; i >= 1; i--){
        if(v[i] > v[i + 1]) dat2[i] = dat2[i + 1] + 1;
        else if(v[i] < v[i - 1]) dat2[i] = dat2[i + 1] - 1;
        else dat2[i] = dat2[i + 1];
    }
    int ans = dp[n];
    for(int i = 1; i <= n; i++){
        int idx = -1;
        if(v[i] > dp[i - 1]) idx = min(v[i],dp[i - 1] + 1) + dat2[i + 1];
        ans = max(ans,idx);
    }
    cout << ans << "\n";

}

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int t = 1;
    cin >> t;
    while(t--) {
        f();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Single Contest**: The code should handle the case where there is only one contest.
2. **Sorted Contests**: The code should handle the case where the ratings are sorted.
3. **Identical Ratings**: The code should handle the case where there are multiple identical ratings.
4. **Optimal Skip Interval**: The code should correctly identify the optimal interval to skip.

Given these points, let's create a test input that combines these aspects:

1. **Single Contest**:
   - This should return 0 because Kevin must skip at least one contest.
2. **Sorted Contests**:
   - This should handle both strictly increasing and strictly decreasing sequences.
3. **Identical Ratings**:
   - This should handle multiple identical ratings in the sequence.
4. **Optimal Skip Interval**:
   - This should test the code's ability to find the optimal interval to skip.

Here is a small-scale test input that covers these scenarios:

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

Explanation:
1. **Test Case 1**: Single contest with rating 1. The output should be 0.
2. **Test Case 2**: Strictly increasing sequence. The output should be 6.
3. **Test Case 3**: Strictly decreasing sequence. The output should be 6.
4. **Test Case 4**: All ratings are identical. The output should be 0.
5. **Test Case 5**: Increasing sequence. The output should be 9.

This test input should help expose any issues in the buggy code, particularly in handling edge cases and identifying the optimal interval to skip.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
f = int(input())
for _ in range(f):
    d = int(input())
    a = list(map(int, input().split()))
    rtg = 0
    m_r = float('-inf')
    prefix = [0] * (d + 1)
    for i in range(d):
        if a[i] > rtg:
            rtg += 1
        elif a[i] < rtg:
            rtg -= 1
        prefix[i + 1] = rtg
    for l in range(d):
        for r in range(l + 1, d + 1):
            rtg_skipped = prefix[l] + (prefix[d] - prefix[r])
            m_r = max(m_r, rtg_skipped)
    print(m_r)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code has a nested loop to calculate the maximum possible rating after skipping an interval, which is inefficient and may not handle certain cases correctly.

Let's consider the following scenarios:
1. **All ratings are the same**: This can cause the code to skip the entire range and not update the rating correctly.
2. **Ratings are in a strictly increasing or decreasing order**: This can help in testing the logic for rating updates.
3. **Ratings with a small range but a large number of contests**: This can test the efficiency and correctness of the interval skipping logic.
4. **Ratings with a large range but a small number of contests**: This can test the edge cases where the rating changes significantly.

Given the constraints and the need for a small-scale test input, let's create a test case with a mix of these scenarios.

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

### Explanation:
1. **Test Case 1**: All ratings are in strictly increasing order.
2. **Test Case 2**: All ratings are the same.
3. **Test Case 3**: Ratings form a pyramid shape, which can test the logic for both increasing and decreasing ratings.
4. **Test Case 4**: All ratings are in strictly decreasing order.
5. **Test Case 5**: Ratings with repeated values in a pattern, testing the logic for multiple intervals with the same ratings.

This test input should help expose any issues in the buggy code, particularly in handling different types of rating sequences and ensuring the correct interval is chosen for maximum rating.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
def max_final_rating(cases):
    results = []
    for n, a in cases:
        total = 0
        impacts = []
        for p in a:
            if p > total:
                impact = 1
            elif p < total:
                impact = -1
            else:
                impact = 0
            impacts.append(impact)
            total += impact

        maxImp = float('-inf')
        curr = 0
        for impact in impacts:
            curr += -impact

            maxImp = max(maxImp, curr)
            if curr < 0:
                curr = 0
        results.append(total + maxImp)

    return results

T = int(input())
cases = []
for _ in range(T):
    N = int(input())
    nums = list(map(int, input().split()))
    cases.append((N, nums))

results = max_final_rating(cases)
for x in results:
    print(x)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
import java.io.*;
import java.util.*;
public class C {
	public static PrintWriter out;
	public static void main(String[] args)throws IOException{
        JS sc=new JS();
        out=new PrintWriter(System.out);
        int t=sc.nextInt();
        outer: while(t-->0) {
        	int n=sc.nextInt();
        	int a[]=new int[n];
        	for(int i=0;i<n;i++) {
        		a[i]=sc.nextInt();
        	}
        	if(n==1) {
        		out.println(0);
        		continue outer;
        	}
        	int dp[][]=new int[n+1][2];
        	for(int i=0;i<=n;i++) {
        		Arrays.fill(dp[i], (int)-1e9);
        	}
        	dp[0][0]=0;
        	int max=0;
        	for(int i=1;i<=n;i++) {
        		if(a[i-1]>dp[i-1][0]) {
        			dp[i][0]=dp[i-1][0]+1;
        		}else if(a[i-1]==dp[i-1][0]){
        			dp[i][0]=dp[i-1][0];
        		}else {
        			dp[i][0]=dp[i-1][0]-1;
        		}
        		if(a[i-1]>dp[i-1][1]) {
        			dp[i][1]=dp[i-1][1]+1;
        		}else if(a[i-1]==dp[i-1][1]){
        			dp[i][1]=dp[i-1][1];
        		}else {
        			dp[i][1]=dp[i-1][1]-1;
        		}
        		dp[i][1]=Math.max(dp[i][1], max);
        		max=Math.max(dp[i-1][0], max);
        	}
        	out.println(dp[n][1]);
        }
        out.close();
	}
	static class Pair implements Comparable<Pair>{
		long x;
		int y;
		public Pair(long x, int y) {
			this.x=x;this.y=y;
		}
		public int compareTo(Pair p) {
			return Long.compare(x, p.x);
		}
	}
	public static long power(long a, long b, int mod) {
		long res=1;
		while(b!=0) {
			if(b%2==1) {
				res=(res*a)%mod;
			}
			a=(a*a)%mod;
			b/=2;
		}
		return res;
	}
	static class JS {
	    public int BS = 1<<16;
	    public char NC = (char)0;
	    byte[] buf = new byte[BS];
	    int bId = 0, size = 0;
	    char c = NC;
	    double num = 1;
	    BufferedInputStream in;

	    public JS() {
	        in = new BufferedInputStream(System.in, BS);
	    }

	    public JS(String s) throws FileNotFoundException {
	        in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
	    }

	    public char nextChar(){
	        while(bId==size) {
	            try {
	                size = in.read(buf);
	            }catch(Exception e) {
	                return NC;
	            }
	            if(size==-1)return NC;
	            bId=0;
	        }
	        return (char)buf[bId++];
	    }

	    public int nextInt() {
	        return (int)nextLong();
	    }

	    public long nextLong() {
	        num=1;
	        boolean neg = false;
	        if(c==NC)c=nextChar();
	        for(;(c<'0' || c>'9'); c = nextChar()) {
	            if(c=='-')neg=true;
	        }
	        long res = 0;
	        for(; c>='0' && c <='9'; c=nextChar()) {
	            res = (res<<3)+(res<<1)+c-'0';
	            num*=10;
	        }
	        return neg?-res:res;
	    }

	    public double nextDouble() {
	        double cur = nextLong();
	        return c!='.' ? cur:cur+nextLong()/num;
	    }

	    public String next() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c>32) {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public String nextLine() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c!='\n') {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public boolean hasNext() {
	        if(c>32)return true;
	        while(true) {
	            c=nextChar();
	            if(c==NC)return false;
	            else if(c>32)return true;
	        }
	    }
	}
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Single contest (n = 1)
   - All contests have the same rating
   - Contests with ratings in increasing or decreasing order
   - Contests with ratings that fluctuate significantly

2. **Optimal Interval Selection:**
   - Ensure that the optimal interval selection is not straightforward and requires careful consideration.

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

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

### Explanation:
1. **Single Contest (n = 1):**
   - This tests the edge case where Kevin must skip the only contest, resulting in a rating of 0.

2. **Increasing Order (n = 5):**
   - This tests the scenario where the ratings are in increasing order. The optimal interval to skip would be any single contest, resulting in a rating of 4.

3. **Decreasing Order (n = 5):**
   - This tests the scenario where the ratings are in decreasing order. The optimal interval to skip would be any single contest, resulting in a rating of 4.

This test input covers a range of scenarios that can help identify issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

public class C_New_Rating {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            int testCases = vt.nextInt();
            // int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int[] ar = new int[n];
        for (int i = 0; i < n; i++) {
            ar[i] = vt.nextInt();
        }
        if (n == 1) {
            out.println(0);
            return;
        }
        int[] br = new int[n];
        int x = 0;
        for (int i = 0; i < n; i++) {
            if (ar[i] > x) {
                x++;
            } else if (ar[i] < x) {
                x--;
            }
            br[i] = x;
        }
        x = 1;
        boolean ok = false;
        int prev = -1;
        for (int i = 1; i < n; i++) {
            if (br[i] > x) {
                x = br[i];
                ok = false;
            } else if (ar[i] > x) {
                x++;
            } else {
                if (ok && prev != i - 1) {
                    x--;
                } else {
                    ok = true;
                    prev = i;
                }
            }
        }
        if (x == n) {
            x--;
        }
        out.println(x);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
f = int(input())

for _ in range(f):
    d = int(input())
    a = list(map(int, input().split()))

    rtg = 0
    m_r = float('-inf')
    prefix = [0] * (d + 1)

    for i in range(d):
        if a[i] > rtg:
            rtg += 1
        elif a[i] < rtg:
            rtg -= 1
        prefix[i + 1] = rtg

    for l in range(d):
        for r in range(l + 1, d + 1):
            rtg_skipped = prefix[l] + (prefix[d] - prefix[r])
            m_r = max(m_r, rtg_skipped)

    print(m_r)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    A = list(map(int, input().split()))
    x = [0] * (n + 1)
    for i in range(1, n + 1):
        if A[i - 1] > x[i - 1]:
            x[i] = x[i-1] + 1
        elif A[i - 1] == x[i - 1]:
            x[i] = x[i - 1]
        else:
            x[i] = x[i - 1] - 1

    min_xr = [0] * (n + 1) + [float('inf')]
    for l in range(n, 0, -1):
        min_xr[l] = min(x[l], min_xr[l + 1])

    ans = -float('inf')
    for l in range(1, n + 1):
        ans = max(ans, x[l - 1] - min_xr[l])

    print(x[n] + ans if ans > 0 else x[n] - 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 that can maximize the failure probability. The buggy code has a few potential issues, such as incorrect handling of the interval selection and the calculation of the minimum values.

Here are some key points to consider:
1. **Single Contest Case**: Ensure the code handles the case where there is only one contest.
2. **All Contests Skipped**: Test the case where all contests are skipped.
3. **Monotonic Sequences**: Test sequences where the ratings are strictly increasing or decreasing.
4. **Mixed Ratings**: Test sequences with a mix of increasing and decreasing ratings.
5. **Edge Cases**: Test cases where the optimal interval is at the beginning, middle, or end of the sequence.

Based on these considerations, let's create a test input that covers these scenarios:

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

This test input includes:
1. A single contest case.
2. A strictly increasing sequence.
3. A strictly decreasing sequence.
4. A mixed sequence with a peak in the middle.
5. A mixed sequence with a peak at the end.

These cases should help expose any issues in the buggy code's logic for interval selection and rating recalculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define nin int n; cin >> n;
#define nil ll n; cin >> n;
#define tit int t = 1; cin >> t;
#define til ll t = 1; cin >> t;
#define vi vector<int>
#define vvi vector<vi>
#define vb vector<bool>
#define vl vector<long>
#define vvl vector<long long>
#define all(v) v.begin(), v.end()
#define srt(v) sort(v.begin(), v.end())         // sort
#define mxe(v) *max_element(v.begin(), v.end()) // find max element in vector
#define mne(v) *min_element(v.begin(), v.end()) // find min element in vector
#define unq(v) v.resize(distance(v.begin(), unique(v.begin(), v.end())));
#define uq(b) b.erase(unique(b.begin(),b.end()),b.end());
#define chk(x,y) count(x.begin(), x.end(), y);
#define posr1(k) int d=1; while((k&1)==0){k>>=1;d++;}

// make sure to sort before applying unique // else only consecutive duplicates would be removed

#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define f0(lcv, hi) for (long long int lcv = 0; lcv < hi; lcv++)
#define fb(lcv, hi) for (int lcv = hi; lcv >= 0; lcv--)
#define f1(lcv, hi) for (int lcv = 1; lcv <= hi; lcv++)
#define pisec(u) sort(u.begin(), u.end(), [](auto &left, auto &right){return left.second < right.second;});

int MOD = 1e9 + 7;
const int MX_SZ = 1e5 + 5;
int par[MX_SZ];

void tests()
{
    long long int a;cin>>a;
    vector<ll>b(a+1);
    vector<vector<ll>>x(a+1,vector<ll>(3));
    for(auto& it:x){
        fill(it.begin(),it.end(),-1000000000);
    }
    f1(i,a)cin>>b[i];
    x[0][0]=0;
    f1(i,a){
        if(x[i-1][0]>b[i]){x[i][0]=x[i-1][0]-1;}
        else if(x[i-1][0]<b[i]){x[i][0]=x[i-1][0]+1;}
        else if(x[i-1][0]==b[i]){x[i][0]=x[i-1][0]+0;}
        x[i][1]=max(x[i-1][0],x[i-1][1]);
        if(x[i-1][1]>b[i]&&x[i-1][2]>b[i]){x[i][2]=max(x[i-1][1]-1,x[i-1][2]-1);}
        else if(x[i-1][1]<b[i]&&x[i-1][2]<b[i]){x[i][2]=max(x[i-1][1]+1,x[i-1][2]+1);}
        else if(x[i-1][1]>b[i]&&x[i-1][2]==b[i]){x[i][2]=max(x[i-1][1]-1,x[i-1][2]);}
        else if(x[i-1][1]<b[i]&&x[i-1][2]==b[i]){x[i][2]=max(x[i-1][1]+1,x[i-1][2]);}
        else if(x[i-1][1]==b[i]&&x[i-1][2]>b[i]){x[i][2]=max(x[i-1][1],x[i-1][2]-1);}
        else if(x[i-1][1]==b[i]&&x[i-1][2]<b[i]){x[i][2]=max(x[i-1][1],x[i-1][2]+1);}
        else if(x[i-1][1]==b[i]&&x[i-1][2]==b[i]){x[i][2]=max(x[i-1][1],x[i-1][2]);}
    }
    cout<<max(x[a][1],x[a][2])<<'\n';
}

int32_t main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(NULL);

    tit
    while (t--){

        tests();

    }
    return 0;
}

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

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

Think step by step.

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