**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import java.util.*;

public class dyp {
    public static void main(String[] args)
     {
    Scanner sc=new Scanner((System.in));
    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[3][n+1];
    dp[0][0]=0;
    dp[1][0]=0;
    dp[2][0]=0;

    for(int i=0;i<3;i++){
        for(int j=1;j<n+1;j++){
             dp[i][j]=-1;
    }
}
    if(n>0)
    {
        dp[0][1]=2*a[0]+1;
        dp[1][1]=a[0];
        dp[2][1]=a[0];

    }
    if(n>1)
    {
        dp[0][2]=2*Math.max(a[0],a[1])+1;
        if(a[0]==a[1])
        dp[0][2]=a[0]+a[1]+2;
        dp[1][2]=Math.max(a[0],a[1]);
        dp[2][2]=Math.max(a[0],a[1]);
    }
    if(n>2&&((2*a[1]+1)<(a[0]+a[2]+2)))
    {

        dp[0][3]=a[0]+a[2]+2;
        dp[1][3]=Math.min(a[0],a[2]);
        dp[2][3]=Math.max(a[0],a[2]);

    }
    else if(n>2&&((2*a[1]+1)>=(a[0]+a[2]+2)))
    {

        dp[0][3]=2*a[1]+1;dp[1][3]=a[1];dp[2][3]=a[1];

    }
    if(n>2&&2*a[0]+1>=dp[0][3]){
        dp[0][3]=2*a[0]+1;dp[1][3]=a[0];dp[2][3]=a[0];

    }if(n>2&&2*a[2]+1>dp[0][3]){
        dp[0][3]=2*a[2]+1;dp[1][3]=a[2];dp[2][3]=a[2];

    }

        if(n<4) System.out.println(dp[0][n]);
        else
    System.out.println(count(a,dp,n));
    //System.out.println(Arrays.deepToString(dp));
}   sc.close();
}
    static int count(int a[],int dp[][],int n){

       for(int i=4;i<n+1;i++)
       {
            if(dp[0][i]==-1)
            {
                if(a[i-1]<=dp[1][i-2]-1)
                {
                    dp[1][i]=dp[1][i-2];
                    dp[2][i]=dp[2][i-2];
                    dp[0][i]=dp[0][i-2];
            }
                else if(a[i-1]>dp[2][i-2])
                {
                    dp[1][i]=dp[1][i-2];dp[2][i]=a[i-1];
                    dp[0][i]=dp[0][i-2]-dp[2][i-2]+dp[2][i]+1;
            }else if(a[i-1]==dp[1][i-2])
            {
                dp[1][i]=a[i-1];
                dp[2][i]=dp[2][i-2];
                dp[0][i]=dp[0][i-2]+1;

            }else if(a[i-1]==dp[2][i-2])
                {
                    dp[1][i]=dp[1][i-2];
                dp[2][i]=dp[2][i-2];
                dp[0][i]=dp[0][i-2]+1;}
                else
                {
                    dp[1][i]=dp[1][i-2];
                    dp[2][i]=dp[2][i-2];
                    dp[0][i]=dp[0][i-2]-dp[1][i-2]+dp[1][i]+1;

                }if(dp[0][i-1]>dp[0][i]){   dp[1][i]=dp[1][i-1];
                    dp[2][i]=dp[2][i-1];
                    dp[0][i]=dp[0][i-1];}            }
       }
        return dp[0][n];

}
}

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

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

Think step by step.

Response:
```plaintext
4
3
1 1 1
4
1 2 3 4
5
1000000000 1000000000 1000000000 1000000000 1000000000
6
1 1000000000 1 1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class dyp {
    public static void main(String[] args)
     {
    Scanner sc=new Scanner((System.in));
    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[3][n+1];
    for(int i=0;i<3;i++){
        for(int j=1;j<n+1;j++){
             dp[i][j]=-1;
    }
}
    if(n>0)
    {
        dp[0][1]=2*a[0]+1;
        dp[1][1]=a[0];
        dp[2][1]=a[0];

    }
    if(n>1)
    {
        dp[0][2]=2*Math.max(a[0],a[1])+1;
        dp[1][2]=Math.max(a[0],a[1]);
        dp[2][2]=Math.max(a[0],a[1]);
    }
    if(n>2&&((2*a[1]+1)<(a[0]+a[2]+2)))
    {

        dp[0][3]=a[0]+a[2]+2;
        dp[1][3]=Math.min(a[0],a[2]);
        dp[2][3]=Math.max(a[0],a[2]);

    }
    else if(n>2&&((2*a[1]+1)>=(a[0]+a[2]+2)))
    {

        dp[0][3]=2*a[1]+1;dp[1][3]=a[1];dp[2][3]=a[1];

    }
    if(2*a[0]+1>=dp[0][3]){
        dp[0][3]=2*a[0]+1;dp[1][3]=a[0];dp[2][3]=a[0];

    }if(2*a[2]+1>dp[0][3]){
        dp[0][3]=2*a[2]+1;dp[1][3]=a[2];dp[2][3]=a[2];

    }
    System.out.println(count(a,dp,n));
}
}
    static int count(int a[],int dp[][],int n){
        if(n==0) return 0;
        if(n==1) return dp[0][1];
        if(n==2) return dp[0][2];
        if(n==3) return dp[0][3];
       for(int i=4;i<n+1;i++)
       {long subans=0;

            if(dp[0][i]==-1)
            {
                if(a[i-1]<=dp[1][i-2]-1)
                {
                    dp[1][i]=dp[1][i-2];
                    dp[2][i]=dp[2][i-2];
                    dp[0][i]=dp[0][i-2];
            }
                else if(a[i-1]>dp[2][i-2])
                {
                    dp[1][i]=dp[1][i-2];dp[2][i]=a[i-1];
                    dp[0][i]=dp[0][i-2]-dp[2][i-2]+dp[2][i]+1;
            }else if(a[i-1]==dp[1][i-2])
            {
                dp[1][i]=a[i-1];
                dp[2][i]=dp[2][i-2];
                dp[0][i]=dp[0][i-2]+1;

            }else if(a[i-1]==dp[2][i-2])
                {
                    dp[1][i]=dp[1][i-2];
                dp[2][i]=dp[2][i-2];
                dp[0][i]=dp[0][i-2]+1;}
                else
                {
                    dp[1][i]=dp[1][i-2];
                    dp[2][i]=dp[2][i-2];
                    dp[0][i]=dp[0][i-2]-dp[1][i-2]+dp[1][i]+1;

                }if(dp[0][i-1]>dp[0][i]){   dp[1][i]=dp[1][i-1];
                    dp[2][i]=dp[2][i-1];
                    dp[0][i]=dp[0][i-1];}            }
       }
        return dp[0][n];

}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
for ______ in range(fg()):
    n = fg()
    a = fgh()
    if n == 1:
        print(2 * a[0] + 1)
        continue
    d = {}
    for i in range(n):
        if a[i] not in d:
            d[a[i]] = []
        d[a[i]].append(i)
    z = []
    for i in range(n):
        if 0 < i < n - 1:
            z.append((a[i], -max(a[i - 1], a[i + 1])))
        else:
            z.append((a[i], 3293293293))
    t = z.index(max(z))
    cnt = 1
    ans = 0
    G = [0] * n
    l = [i for i in range(n)]
    r = [i for i in range(n)]
    mx = max(a)
    for x in range(mx, mx - n - 2, -1):
        if x in d:
            for j in d[x]:
                if abs(j - t) > 1:
                    G[j] = 1
                    i = j
                    if j == 0:
                        if G[1] == 0:
                            r[0] = 1
                            cnt += 1
                        else:
                            if (r[1]) % 2:
                                cnt += 1
                            l[r[1] - 1] = 0
                            r[0] = r[1]
                    elif j == n - 1:
                        if G[n - 2] == 0:
                            r[n - 1] = n
                            cnt += 1
                        else:
                            if (n - l[n - 2]) % 2:
                                cnt += 1
                            r[l[n - 2]] = n
                            l[n - 1] = l[n - 2]
                    else:
                        i = j
                        if G[i - 1] == 0 and G[i + 1] == 0:
                            cnt += 1
                            r[i] = i + 1
                        elif G[i - 1] == 0:
                            if (r[i + 1] - i) % 2:
                                cnt += 1
                            r[i] = r[i + 1]
                            l[r[i] - 1] = i
                        elif G[i + 1] == 0:
                            if (i + 1 - l[i - 1]) % 2:
                                cnt += 1
                            l[i] = l[i - 1]
                            r[l[i]] = i + 1
                        else:
                            if (r[i + 1] - i - 1) % 2 == 0 and (i - l[i - 1]) % 2 == 0:
                                cnt += 1
                            l[r[i + 1] - 1] = l[i - 1]
                            r[l[i - 1]] = r[i + 1]
            ans = max(ans, cnt + mx + x)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
4
3
1 2 1
3
2 1 2
10
1 2 3 4 5 6 7 8 9 10
10
10 9 8 7 6 5 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: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
for ______ in range(fg()):
    n = fg()
    a = fgh()
    if n == 1:
        print(2 * a[0] + 1)
        continue
    d = {}
    for i in range(n):
        if a[i] not in d:
            d[a[i]] = []
        d[a[i]].append(i)
    z = []
    for i in range(n):
        if 0 < i < n - 1:
            z.append((a[i], -min(a[i - 1], a[i + 1])))
        else:
            z.append((a[i], 3293293293293293))
    t = z.index(max(z))
    cnt = 1
    ans = 0
    G = [0] * n
    l = [i for i in range(n)]
    r = [i for i in range(n)]
    mx = max(a)
    for x in range(mx, mx - n - 1, -1):
        if x in d:
            for j in d[x]:
                if abs(j - t) > 1:
                    G[j] = 1
                    i = j
                    if j == 0:
                        if G[1] == 0:
                            r[0] = 1
                            cnt += 1
                        else:
                            if (r[1] - 1) % 2 == 0:
                                cnt += 1
                            l[r[1] - 1] = 0
                            r[0] = r[1]
                    elif j == n - 1:
                        if G[n - 2] == 0:
                            r[n - 1] = n
                            cnt += 1
                        else:
                            if (n - l[n - 2]) % 2 == 0:
                                cnt += 1
                            r[l[n - 2]] = n
                            l[n - 1] = l[n - 2]
                    else:
                        i = j
                        if G[i - 1] == 0 and G[i + 1] == 0:
                            cnt += 1
                            r[i] = i + 1
                        elif G[i - 1] == 0:
                            if (r[i + 1] - i - 1) % 2 == 0:
                                cnt += 1
                            r[i] = r[i + 1]
                            l[r[i] - 1] = i
                        elif G[i + 1] == 0:
                            if (i + 1 - l[i - 1] - 1) % 2 == 0:
                                cnt += 1
                            l[i] = l[i - 1]
                            r[l[i]] = i + 1
                        else:
                            if (r[i + 1] - i - 1) % 2 == 0 and (i - l[i - 1]) % 2 == 0:
                                cnt += 1

                            l[r[i + 1] - 1] = l[i - 1]
                            r[l[i - 1]] = r[i + 1]
            #print(cnt, x, mx)
            ans = max(ans, cnt + mx + x)
    print(ans)

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

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

Think step by step.

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

1. **Single Element Array**: This is a straightforward edge case where the array has only one element. The expected output is straightforward.
2. **Array with All Elements the Same**: This can cause issues with the logic that handles the maximum and minimum values.
3. **Array with Alternating Large and Small Values**: This can test the logic that handles the adjacency constraint.
4. **Array with a Single Large Value in the Middle**: This can test the logic that handles the maximum value and the adjacency constraint.
5. **Array with a Single Small Value in the Middle**: This can test the logic that handles the minimum value and the adjacency constraint.

Given these considerations, let's create a test input that combines some of these scenarios:

```plaintext
5
1
10
3
10 10 10
5
1 1000000000 1 1000000000 1
7
1 2 3 1000000000 3 2 1
9
10 1 2 3 1000000000 3 2 1 10
```

This test input includes:
1. A single element array.
2. An array with all elements the same.
3. An array with alternating large and small values.
4. An array with a single large value in the middle.
5. An array with a single small value in the middle and large values at the ends.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
for ______ in range(fg()):
    n = fg()
    a = fgh()
    if n == 1:
        print(2 * a[0] + 1)
        continue
    d = {}
    for i in range(n):
        if a[i] not in d:
            d[a[i]] = []
        d[a[i]].append(i)
    z = []
    for i in range(n):
        if 0 < i < n - 1:
            z.append((a[i], -max(a[i - 1], a[i + 1])))
        else:
            z.append((a[i], 3293293293293293))
    t = z.index(max(z))
    cnt = 1
    ans = 0
    G = [0] * n
    l = [i for i in range(n)]
    r = [i for i in range(n)]
    mx = max(a)
    for x in range(mx, mx - n - 1, -1):
        if x in d:
            for j in d[x]:
                if abs(j - t) > 1:
                    G[j] = 1
                    i = j
                    if j == 0:
                        if G[1] == 0:
                            r[0] = 1
                            cnt += 1
                        else:
                            if (r[1] - 1) % 2 == 0:
                                cnt += 1
                            l[r[1] - 1] = 0
                            r[0] = r[1]
                    elif j == n - 1:
                        if G[n - 2] == 0:
                            r[n - 1] = n
                            cnt += 1
                        else:
                            if (n - l[n - 2]) % 2 == 0:
                                cnt += 1
                            r[l[n - 2]] = n
                            l[n - 1] = l[n - 2]
                    else:
                        i = j
                        if G[i - 1] == 0 and G[i + 1] == 0:
                            cnt += 1
                            r[i] = i + 1
                        elif G[i - 1] == 0:
                            if (r[i + 1] - i - 1) % 2 == 0:
                                cnt += 1
                            r[i] = r[i + 1]
                            l[r[i] - 1] = i
                        elif G[i + 1] == 0:
                            if (i + 1 - l[i - 1] - 1) % 2 == 0:
                                cnt += 1
                            l[i] = l[i - 1]
                            r[l[i]] = i + 1
                        else:
                            if (r[i + 1] - i - 1) % 2 == 0 and (i - l[i - 1]) % 2 == 0:
                                cnt += 1

                            l[r[i + 1] - 1] = l[i - 1]
                            r[l[i - 1]] = r[i + 1]
            #print(cnt, x, mx)
            ans = max(ans, cnt + mx + x)
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
// Problem: D. Max Plus Min Plus Size
// Contest: Codeforces - Codeforces Round 975 (Div. 1)
// URL: https://codeforces.com/contest/2018/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Author: @socho
// Parsed at: 27-09-2024 09:35:12

#include <bits/stdc++.h>
using namespace std;
#define int LL
typedef long long LL;
typedef double D;
typedef vector<int> VI;
typedef set<int> SI;
typedef map<int, int> MII;
typedef pair<int, int> PII;

#define A first
#define B second
#define SZ(x) int(x.size())
#define PB push_back
#define FR(i, a, b) for (int i = (a); i < (b); i++)
#define FOR(i, n) FR(i, 0, n)
#define RF(i, a, b) for (int i = (a); i >= (b); i--)
#define FRA(a, x) for (auto a: (x))

template <typename T> inline void set_min(T &a, T b) {if(b < a) a = b;}
template <typename T> inline void set_max(T &a, T b) {if(b > a) a = b;}

const int K = 3; // K best max values per segment

struct segd {

	PII datasets[2][2][K]; // pairs are: max, count (maxes in decreasing order)
	// ltaken, rtaken, rank

	void reset() {
		FOR(j, 2) FOR(k, 2) FOR(i, K)  datasets[j][k][i] = {LLONG_MIN / 8, LLONG_MIN / 8};
	}

	segd() {
		reset();
	}

	segd(int v) {
		reset();
		datasets[0][0][0] = {LLONG_MIN / 8, 0};
		datasets[1][1][0] = {v, 1};
	}

	segd(bool t) {
		// drop
		reset();
		datasets[0][0][0] = {LLONG_MIN / 8, 0};
	}

	segd(segd& lef, segd& rig) {
		reset();
		for (int l=0; l<2; l++) {
			for (int r=0; r<2; r++) {
				// solve for l, r in final answer

				vector<PII> candidates;
				map<int, int> frm;

				for (int ml=0; ml<2; ml++) {
					for (int mr=0; mr<2; mr++) {
						if (ml == 1 && mr == 1) continue; // can't combine

						// lef[l][ml] and rig[mr][r] combine
						for (int LK=0; LK<K; LK++) {
							for (int RK=0; RK<K; RK++) {
								PII& f = lef.datasets[l][ml][LK];
								PII& s = rig.datasets[mr][r][RK];
								set_max(frm[max(f.first, s.first)], f.second + s.second);
							}
						}

					}
				}

				int ptr = 0;

				for (auto xe = frm.rbegin(); xe != frm.rend(); xe++) {
					if (ptr == K) break;
					datasets[l][r][ptr] = *xe;
					ptr++;
				}

			}
		}
	}

};

int n;
const int MXN = 200005;
int arr[MXN];
segd seg[MXN * 4];

void build(int ind, int l, int r) {
	if (l == r) {
		seg[ind] = segd(arr[l]);
		return;
	}
	int m = (l + r) / 2;
	build(ind*2, l, m);
	build(ind*2+1, m+1, r);
	seg[ind] = segd(seg[ind*2], seg[ind*2+1]);
}

void update(int ind, int l, int r, int p, int v) {
	if (l == r) {
		seg[ind] = segd(true);
		return;
	}
	int m = (l + r) / 2;
	if (p <= m) update(ind*2, l, m, p, v);
	else update(ind*2+1, m+1, r, p, v);
	seg[ind] = segd(seg[ind*2], seg[ind*2+1]);
}

int get_value() {
	int ans = LLONG_MIN / 2;
	FOR(a, 2) FOR(b, 2) FOR(k, K) {
		set_max(ans, seg[1].datasets[a][b][k].A + seg[1].datasets[a][b][k].B);
	}
	return ans;
}

void trav(int ind, int l, int r) {
	if (l != r) {
		int m = (l + r) / 2;
		trav(ind*2, l, m);
		trav(ind*2+1, m+1, r);
	}
	// cout << l << " to " << r << endl;
	FOR(i, 2) {
		FOR(j, 2) {
			cout << i << ' ' << j << endl;
			FOR(k, K) {
				cout << seg[ind].datasets[i][j][k].A << ' ' << seg[ind].datasets[i][j][k].B << endl;
			}
		}
	}
	cout << endl;
}

void solve() {

	vector<PII> so;

	cin >> n;
	for (int i=1; i<=n; i++) {
		cin >> arr[i];
		so.PB({arr[i], i});
	}

	sort(so.begin(), so.end());

	build(1, 1, n);
	// trav(1, 1, n);

	int result = get_value() + so[0].A;
	int rems = 0;
	// cout << "aft rem " << rems << ' ' << result << endl;

	FRA(x, so) {
		rems++;
		if (rems == n) break;
		update(1, 1, n, x.second, LLONG_MIN / 4);
		int here = get_value() + so[rems].A;
		// trav(1, 1, n);
		// cout << "aft rem " << rems << " (" << x.B << ")" << ' ' << here << ' ' << get_value() << ' ' << so[rems].A << endl;
		set_max(result, here);
	}

	cout << result << endl;

}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

	int t = 1;
	cin >> t;
	while (t--) solve();

}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The code involves a segment tree to manage and query the maximum, minimum, and count of red elements. Here are some potential edge cases to consider:

1. **Single Element Array**: The simplest case where the array has only one element.
2. **All Elements are the Same**: An array where all elements are the same, which could lead to issues in managing the segment tree.
3. **Alternating Elements**: An array where elements alternate between two values, which could test the logic for non-adjacent red elements.
4. **Large Values with Small Differences**: An array with large values but small differences between them, which could test the precision and logic of the segment tree.

Let's generate a small-scale test input that covers these cases:

```plaintext
4
1
10
3
1 1 1
5
1 2 1 2 1
5
999999999 1000000000 999999999 1000000000 999999999
```

### Explanation:
1. **Single Element Array**:
   - `n = 1`, `a = [10]`
   - The only element can be colored red, resulting in a score of `10 + 10 + 1 = 21`.

2. **All Elements are the Same**:
   - `n = 3`, `a = [1, 1, 1]`
   - All elements can be colored red, resulting in a score of `1 + 1 + 3 = 5`.

3. **Alternating Elements**:
   - `n = 5`, `a = [1, 2, 1, 2, 1]`
   - The optimal coloring is `[1, 2, 1, 2, 1]` with red elements at positions 1, 3, and 5, resulting in a score of `2 + 1 + 3 = 6`.

4. **Large Values with Small Differences**:
   - `n = 5`, `a = [999999999, 1000000000, 999999999, 1000000000, 999999999]`
   - The optimal coloring is `[999999999, 1000000000, 999999999, 1000000000, 999999999]` with red elements at positions 1, 3, and 5, resulting in a score of `1000000000 + 999999999 + 3 = 2000000002`.

This test input should help in identifying any issues in the segment tree management and query logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;

public class MaxPlusMinPlusSize {
    static int answer=0;
    public static void main(String[] args) throws IOException {
        BufferedReader reader= new BufferedReader( new InputStreamReader(System.in));

        int t =Integer.parseInt(reader.readLine());
        while(t-->0) {

            int n = Integer.parseInt(reader.readLine());

            String[] arrEl = reader.readLine().split(" ");
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(arrEl[i]);
            }
            HashMap<String, Long> mp = new HashMap<>();
            long ans = helper(n - 1, 0, 0, 0, arr, mp);

            System.out.println(ans);
        }
    }

    private static long helper(int ind, int max, int min, int size, int[] arr,HashMap<String,Long> mp){
            if (ind < 0) return 0;
            String dp = ind + "," + max + "," + min + "," + size;
            if (mp.containsKey(dp)) return mp.get(dp);
            int mini = min != 0 ? Math.min(arr[ind], min) : arr[ind];
            int maxi = Math.max(arr[ind], max);
            int fSize = size + 1;
            long sum = mini + maxi + fSize;
            long pick = Math.max(sum, helper(ind - 2, maxi, mini, fSize, arr, mp));
            long notPick = Math.max(max + min + size, helper(ind - 1, max, min, size, arr, mp));
            long maxSum = Math.max(pick, notPick);
            mp.put(dp, maxSum);
            return maxSum;

    }
}

**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
4
3
1 1 1
5
1 2 3 4 5
6
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
10
1 1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include <iostream>
#include <string>
#include <math.h>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <map>
#include <array>
#include <set>
#include <stack>
#include <limits>
#include <queue>
#include <list>
#include <cstring>
using namespace std;

const int C = 3;

struct node2 {
  // l unused, l used, r used, r unused, max
  long long lr[4];
  node2(long long val) {
    for (int i = 0; i < 4; i++) {
      lr[i] = 0;
    }
    lr[3] = val;
  }

};

template<class T> struct Seg2 {
	const T ID = node2(0);
	T comb(const T & a, const T & b) {
    node2 ret = node2(0);
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < 4; j++) {
        int llu = i >> 1;
        int lru = i & 1;
        int rlu = j >> 1;
        int rru = j & 1;
        if (rlu && lru) continue;
        long long nn = a.lr[i] + b.lr[j];
        int nu = llu * 2 + rru;
        ret.lr[nu] = max(ret.lr[nu], nn);
      }
    }
    return ret;
	}
	int n;
	vector<T> seg;
	void init(int _n) {
		n = _n;
		seg.assign(2*n, ID);
	}
	void pull(int p) {
		seg[p] = comb(seg[2*p], seg[2*p+1]);
	}

	void upd(int p, T val) {
		seg[p += n] = val;
		for (p /= 2; p; p/= 2) {
			pull(p);
		}
	}
	T query(int l, int r) {
		if (l > r) return ID;
		T ra = ID, rb = ID;
		for (l += n, r += n+1; l < r; l /= 2, r /= 2) {
			if (l&1) ra = comb(ra, seg[l++]);
			if (r&1) rb = comb(seg[--r], rb);
		}
		return comb(ra,rb);
	}
};

struct node {
  // l unused, l used, r used, r unused, max
  array<array<long long, 2>, C> lr[4];
  node(long long val) {
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < C; j++) {
        lr[i][j] = {0, 0};
      }
    }
    lr[3][0] = {val + 1, val};
  }

  node() {
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < C; j++) {
        lr[i][j] = {0, 0};
      }
    }
  }

};

template<class T> struct Seg {
	const T ID = node();
	T comb(const T & a, const T & b) {
    node ret = node();
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < 4; j++) {
        for (int x = 0; x < C; x++) {
          for (int y = 0; y < C; y++) {
            int llu = i >> 1;
            int lru = i & 1;
            int rlu = j >> 1;
            int rru = j & 1;
            if (rlu && lru) continue;
            long long lm = a.lr[i][x][1];
            long long ln = a.lr[i][x][0] - lm;
            long long rm = b.lr[j][y][1];
            long long rn = b.lr[j][y][0] - rm;
            long long nm = max(lm, rm);
            long long nn = ln + rn;
            int nu = llu * 2 + rru;
            //cout << nu << " " << lm << " " << ln << " " << rm << " " << rn << " " << nm << " " << nn << endl;
            for (int k = 0; k < C; k++) {
              if (ret.lr[nu][k][1] == nm) {
                ret.lr[nu][k][0] = max(nn + nm, ret.lr[nu][k][0]);
                break;
              } else if (nn + nm > ret.lr[nu][k][0]) {
                for (int l = C - 2; l >= k; l--) {
                  swap(ret.lr[nu][l][0], ret.lr[nu][l + 1][0]);
                  swap(ret.lr[nu][l][1], ret.lr[nu][l + 1][1]);
                }
                ret.lr[nu][k][0] = nn + nm;
                ret.lr[nu][k][1] = nm;
                break;
              }
            }
          }
        }
      }
    }
    return ret;
	}
	int n;
	vector<T> seg;
	void init(int _n) {
		n = _n;
		seg.assign(2*n, ID);
	}
	void pull(int p) {
		seg[p] = comb(seg[2*p], seg[2*p+1]);
	}

	void upd(int p, T val) {
		seg[p += n] = val;
		for (p /= 2; p; p/= 2) {
			pull(p);
		}
	}
	T query(int l, int r) {
		if (l > r) return ID;
		T ra = ID, rb = ID;
		for (l += n, r += n+1; l < r; l /= 2, r /= 2) {
			if (l&1) ra = comb(ra, seg[l++]);
			if (r&1) rb = comb(seg[--r], rb);
		}
		return comb(ra,rb);
	}
};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tests;
  cin >> tests;
  for (int test = 1; test <= tests; test++) {
    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<pair<long long, int> > v(n);
    for (int i = 0; i < n; i++) {
      v[i] = {a[i], i};
    }
    Seg<node> seg;
    Seg2<node2> seg2;
    seg.init(n);
    seg2.init(n);
    for (int i = 0; i < n; i++) {
      seg.upd(i, node());
      seg2.upd(i, node2(0));
    }
    sort(v.begin(), v.end());

    int p = n - 1;
    long long res = 0;
    for (int i = n - 1; i >= 0; i--) {
      while (p >= 0 && v[p].first >= v[i].first) {
        seg.upd(v[p].second, v[p].first);
        seg2.upd(v[p].second, 1);
        p--;
      }
      {
        auto l = seg.query(0, v[i].second);
        auto r = seg2.query(v[i].second, n - 1);
        long long lm = 0, rm = 0;
        for (int j = 0; j < 2; j++) {
          for (int k = 0; k < C; k++) {
            lm = max(lm, l.lr[j * 2 + 1][k][0]);
          }
          rm = max(rm, r.lr[2 + j]);
        }
        res = max(res, lm + rm - 1 + v[i].first);
      }
      {
        // rev
        auto l = seg2.query(0, v[i].second);
        auto r = seg.query(v[i].second, n - 1);
        long long lm = 0, rm = 0;
        for (int j = 0; j < 2; j++) {
          for (int k = 0; k < C; k++) {
            rm = max(rm, r.lr[2 + j][k][0]);
          }
          lm = max(lm, l.lr[2 * j + 1]);
        }
        //cout << lm << " " << rm << " " << v[i].first << " " << lm + rm - 1 + v[i].first << endl;
        res = max(res, lm + rm - 1 + v[i].first);
      }
    }
    cout << res << endl;

  }

  // IF STUCK:
    // DIV CONQUER?
    // CONSIDER SMALL CASES
    // INDUCTION
  return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are a few potential edge cases to consider:

1. **Single Element Array**: This is a simple case but important to check.
2. **Array with All Elements the Same**: This can help identify issues with handling uniform data.
3. **Array with Alternating Values**: This can test the logic for non-adjacent red elements.
4. **Array with a Large Difference in Values**: This can test the logic for handling extreme values.
5. **Array with a Small and a Large Value at the Ends**: This can test the logic for handling boundary conditions.

Let's generate a small-scale test input that combines these edge cases to maximize the failure probability in the faulty solution.

```plaintext
4
1
5
2
5 5
5
1 2 3 2 1
6
1 1000000000 1 1000000000 1 1
```

### Explanation:
1. **Single Element Array**:
   - Test case: `1 5`
   - Expected output: `6` (The only element is colored red, and the score is `5 + 5 + 1 = 11`).

2. **Array with All Elements the Same**:
   - Test case: `2 5 5`
   - Expected output: `11` (The array can be colored as `[5, 5]` or `[5, 5]`, and the score is `5 + 5 + 1 = 11`).

3. **Array with Alternating Values**:
   - Test case: `5 1 2 3 2 1`
   - Expected output: `12` (The array can be colored as `[1, 2, 3, 2, 1]`, and the score is `3 + 1 + 5 = 9`).

4. **Array with a Large Difference in Values**:
   - Test case: `6 1 1000000000 1 1000000000 1 1`
   - Expected output: `2000000005` (The array can be colored as `[1, 1000000000, 1, 1000000000, 1, 1]`, and the score is `1000000000 + 1 + 5 = 2000000005`).

This test input combines various edge cases to ensure the algorithm handles different scenarios correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

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

void solve(int l, int r, vector <int> &a, vector <pair <int, int>> &ans) {
    if (l > r) return;

    for (int i = l; i <= r; i++) {
        if (a[i] == i - l) continue;
        ans.emplace_back(l, i);
        solve(l, i - 1, a, ans);
    }
}

void testcase() {
    int n;
    cin >> n;

    vector <int> a(n);
    for (auto &x : a) cin >> x;

    int ans = 0, ans_mask = 0;
    for (int mask = 0; mask < (1<<n); mask++) {
        int sum = 0, lt = 0;
        for (int i = 0; i < n; i++) {
            if (!(mask & (1<<i))) continue;
            sum += a[i] + (i - lt) * (i - lt);
            lt = i + 1;
        }
        sum += (n - lt) * (n - lt);
        if (ans < sum) ans = sum, ans_mask = mask;
    }

    int lt = 0;
    vector <pair <int, int>> ans_op;
    for (int i = 0; i < n; i++) {
        if (!(ans_mask & (1<<i))) continue;
        solve(lt, i - 1, a, ans_op);
        ans_op.emplace_back(lt, i - 1);
        for (int j = lt; j < i; j++) a[j] = i - lt;
        lt = i + 1;
    }
    solve(lt, n - 1, a, ans_op);
    ans_op.emplace_back(lt, n - 1);
    for (int j = lt; j < n; j++) a[j] = n - lt;

    cout << ans << ' ' << ans_op.size() << '\n';
    for (auto [l, r] : ans_op) cout << l + 1 << ' ' << r + 1 << '\n';
}

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

    int t = 1;
    // cin >> t;

    while (t--) testcase();
    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
4
0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int maxn=20;
typedef long long ll;
int n;
ll a[maxn];
ll dp[maxn];
int from[maxn];// 记录每个位置的 dp 是从哪里转移过来的
// from[i] 代表第 i 个为结尾的 mex 操作区间的上一个区间结尾是 from[i]
// 若 from[i] = -1，代表第 i 个数不操作
/*
把长度为 len 的数组变成 len-1,len-2,...,0 的操作方法：
f(len):
1. 把自己点成 0
2. 对后面的数执行 f(len-1)
3. 然后，把所有数点成 len-1
4. 对后面的数执行 f(len-1)
*/
vector<pair<int,int>> ans;
void op(int len,int offset){
    // 变成 len-1,len-2,...,0
    // 假设操作序列以 1 开始，下标在 [1,len]，但是实际上操作区间应该右移 offset
    if(a[1+offset]!=0){
   //     printf("1+offs = %d len = %d push: %d\n",offset+1,len,1+offset);
        ans.emplace_back(make_pair(1+offset,1+offset));
    }
    a[1+offset]=0;
    if(len!=1)op(len-1,offset+1);
   // printf("1+offs = %d len = %d push: %d %d\n",offset+1,len,1+offset,len+offset);
    if(len!=1)ans.emplace_back(make_pair(1+offset,len+offset));
    for(int i=1;i<=len;i++){
        a[i+offset]=len-1;
    }
    if(len!=1)op(len-1,offset+1);
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }
    for(int i=1;i<=n;i++){
        dp[i]=dp[i-1]+a[i];
        from[i]=-1;
        for(int j=1;j<=i;j++){
            if(dp[j-1]+ll(i-j+1)*ll(i-j+1)>dp[i]){
                from[i]=j-1;
                dp[i]=dp[j-1]+ll(i-j+1)*ll(i-j+1);
            }
        }
    }
    int nowr=n;
    while(nowr>=1){
        int nowl=from[nowr]+1;
        if(from[nowr]==-1){
            nowl=nowr;
        }
        for(int i=nowl;i<=nowr;i++){
            if(a[i]!=0){
                ans.emplace_back(make_pair(i,i));
                a[i]=0;
            }
        }
        op(nowr-nowl+1,nowl-1);
        nowr=nowl-1;
    }
    ans.emplace_back(make_pair(1,n));
    printf("%lld %d\n",dp[n],int(ans.size()));
    for(auto [l,r]:ans){
        printf("%d %d\n",l,r);
    }
    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
4
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

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

#pragma GCC target("popcnt")
#define ll long long
#define pii pair<int, int>

const int MOD = 1e9 + 7;
const int MAXN = 1e1 + 5;

void getRec(vector<pii> &result, int l, int r){
    if (l >= r)
        return;

    result.push_back(make_pair(l, r));
    result.push_back(make_pair(l, r - 1));
    getRec(result, l, r - 1);
}

int getAns(vector<pii> &result, vector<int> &a, int l, int r){
    int cnt = 0;
    for (int i = l - 1; i < r; i++){
        cnt += a[i] == 0;
    }

    if (!cnt){
        result.push_back(make_pair(l, r));
    }
    else if (cnt < r - l + 1){
        result.push_back(make_pair(l, r));
        result.push_back(make_pair(l, r));
    }

    getRec(result, l, r);

    result.push_back(make_pair(l, r));
    return (r - l + 1) * (r - l + 1);
}

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

    vector<pii> result;
    vector<int> b(n, 1);
    priority_queue<pii> pq;
    for (int i = 0; i < n; i++){
        pq.push(make_pair(-a[i], i));
    }

    while(!pq.empty()){
        auto top = pq.top();
        pq.pop();

        int l, r;
        for (int i = top.second; i < n; i++){
            if (a[i] <= abs(top.first)){
                r = i;
            }
            else{
                break;
            }
        }

        for (int i = top.second; i >= 0; i--){
            if (a[i] <= abs(top.first)){
                l = i;
            }
            else{
                break;
            }
        }

        int no_update = a[top.second] + (top.second - l) * (top.second - l) + (r - top.second) * (r - top.second);
        if (no_update <= (r - l + 1) * (r - l + 1)){
            for (int i = l; i <= r; i++){
                b[i] = 0;
            }
        }
    }

    int l = 0, r = 0, ans = 0;
    for (int i = 0; i < n; i++){
        if (b[i]){
            ans += getAns(result, a, l + 1, r + 1);
            ans += a[i];
            l = i + 1; r = -1;
        }
        else{
            r = i;
        }
    }

    if (r != 0){
        ans += getAns(result, a, l + 1, r + 1);
    }

    cout << ans << " " << result.size() << "\n";
    for (auto &it : result)
        cout << it.first << " " << it.second << "\n";

}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    ll numTests = 1;
    // cin >> numTests;
    while(numTests-- ){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
def apply_operations(a, l, r, ops):
    if l == r:
        if a[l] != 0:
            ops.append((l, l))
            a[l] = 0
        return

    apply_operations(a, l + 1, r, ops)
    if a[l] != r - l + 1:
        ops.append((l, r))
        for i in range(l, r + 1):
            a[i] = r - l + 1

        apply_operations(a, l + 1, r, ops)

def maximize_sum(a):
    n = len(a)
    ops = []

    def recursive_maximize_sum(l, r):
        s = sum(a[l:r+1])
        if s <= (r - l + 1) * (r - l + 1):
            apply_operations(a, l, r, ops)
            ops.append((l, r))
            for i in range(l, r + 1):
                a[i] = r - l + 1
        else:
            mx = max(a[l:r+1])
            pos = a[l:r+1].index(mx) + l
            if pos != l:
                recursive_maximize_sum(l, pos - 1)
            if pos != r:
                recursive_maximize_sum(pos + 1, r)

    recursive_maximize_sum(0, n - 1)
    return sum(a), len(ops), ops

def main():
    n = int(input())
    a = list(map(int, input().split()))
    s, m, ops = maximize_sum(a)
    print(s, m)
    for l, r in ops:
        print(l + 1, r + 1)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
4
0 0 0 0
```

### Explanation:
This input is designed to challenge the buggy code by providing an array where all elements are zero. The correct solution should transform the array into `[4, 4, 4, 4]` with a sum of 16, and it should require only one operation to achieve this.

However, the buggy code in the `apply_operations` function has a flaw where it does not correctly handle the case where all elements are zero. Specifically, the condition `if a[l] != r - l + 1` will always be true for the first call, but it will incorrectly set the elements to `r - l + 1` and then recursively call itself, potentially leading to incorrect or redundant operations.

This input will help expose the issue where the code does not efficiently handle an array of all zeros and may perform unnecessary operations or fail to achieve the correct maximum sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#import time
#start_time = time.perf_counter()
global operations

def maximize_array_sum_optimized(a):
    global operations
    n = len(a)
    operations = []
    # Инициализация массива для отслеживания MEX значений
    mex_values = [0] * (19)
    for i in range(n):
        mex_values[a[i]] = 1
    # Находим первое незанятое значение MEX
    mex_value = 0
    while mex_values[mex_value] == 1:
        mex_value += 1
    # Применяем замены
    for i in range(n):
        if a[i] != mex_value:
            operations.append((i, i, mex_value))
            a[i] = mex_value
    return sum(a), len(operations), operations

def mex(arr):
    mex_value = 0
    while mex_value in arr:
        mex_value += 1
    return mex_value

a=[]
n=int(input())
sa=input()
na=sa.split()
for ai in na:
  a.append(int(ai))

max_sum, num_operations, operations = maximize_array_sum_optimized(a)
print(max_sum, num_operations-1)
for operation in operations:
    print(operation[0]+1, end=" ")
print()

#end_time = time.perf_counter()
#execution_time = end_time - start_time
#print(f"Time: {execution_time:.5f} sec")

**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
4
0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
n = int(input())

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

rec = [([(0, [])] * (n + 1)) for _ in range(n + 1)]

for i in range(1, n + 1):
    for j in range(0, n - i + 1):
        l = j
        r = j + i
        if i == 1:
            rec[l][r] = (max(a[l], 1), [[l, r]])
        else:
            s = (r - l) ** 2
            p = [[l, r]]
            for k in range(l, r):
                if rec[l][k][0] + a[k] + rec[k + 1][r][0] > s:
                    s = rec[l][k][0] + a[k] + rec[k + 1][r][0]
                    p = rec[l][k][1] + [[k]] + rec[k + 1][r][1]
            rec[l][r] = (s, p)

def do_len_x(l, r):
    if r - l == 0:
        return
    elif r - l == 1:
        print(l, l)
        print(l, r)
    else:
        do_len_x(l, r - 1)
        print(l, r - 1)
        do_len_x(l + 1, r)
        print(l, r)

cnt = 0
for ops in rec[0][n][1]:
    if len(ops) == 1:
        if a[ops[0]] == 0:
            cnt += 1
    else:
        if sum(a[ops[0]: ops[1]]) == 0:
            cnt += 0
        else:
            if min(a[ops[0]: ops[1]]) == 0:
                cnt += 2
            else:
                cnt += 1
        cnt += (2 ** (ops[1] - ops[0])) - 2

# print(rec[0][n])
print(rec[0][n][0], cnt)

for ops in rec[0][n][1]:
    if len(ops) == 1:
        if a[ops[0]] == 0:
            print(ops[0] + 1, ops[0] + 1)
    else:
        if sum(a[ops[0]: ops[1]]) == 0:
            pass
        else:
            if min(a[ops[0]: ops[1]]) == 0:
                print(ops[0] + 1, ops[1])
                print(ops[0] + 1, ops[1])
            else:
                print(ops[0] + 1, ops[1])
        do_len_x(ops[0] + 1, ops[1])

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

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

Think step by step.

Response:
```plaintext
4
0 1 2 3
```

**Reasoning:**

1. **Edge Case with Initial MEX Values:**
   - The array `[0, 1, 2, 3]` is a sequence of the first four non-negative integers. The MEX of this array is 4, which is the smallest non-negative integer not present in the array.
   - This input is designed to test the algorithm's ability to correctly compute the MEX and apply the operation to maximize the sum.

2. **Small-Scale Test:**
   - The array length is 4, which is within the constraints and manageable for testing.
   - This size allows for a detailed examination of the algorithm's behavior without overwhelming the input.

3. **Expected Behavior:**
   - The optimal strategy is to apply the MEX operation to the entire array, which will change all elements to 4.
   - The resulting array will be `[4, 4, 4, 4]`, with a sum of 16.
   - The number of operations required is 1, as applying the MEX operation to the entire array once is sufficient.

4. **Faulty Code Exposure:**
   - The buggy code might fail to correctly compute the MEX or might incorrectly update the array elements.
   - This test case should highlight issues in the recursive computation and the logic for determining the operations.

By using this test input, you can effectively evaluate whether the buggy code correctly computes the MEX and applies the operations to maximize the sum of the array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import javax.print.FlavorException;
import javax.swing.text.AbstractDocument.Content;

public class Main {

  public static boolean isprime(long x) {
    if (x == 2 || x == 3 || x == 5) {
      return true;
    }
    for (long i = 2; i <= Math.sqrt(x); i++) {
      if (x % i == 0) {
        return false;
      }
    }
    return true;
  }

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

  static void sort(long[] a) {
    ArrayList<Long> l = new ArrayList<>();
    for (long i : a) l.add(i);
    Collections.sort(l);
    for (int i = 0; i < a.length; i++) a[i] = l.get(i);
  }

  static void sortInt(int[] a) {
    ArrayList<Integer> l = new ArrayList<>();
    for (int i : a) l.add(i);
    Collections.sort(l);
    for (int i = 0; i < a.length; i++) a[i] = l.get(i);
  }

  public static class pair implements Comparable<pair> {

    long x;
    int y;

    public pair(long x, int y) {
      this.x = x;
      this.y = y;
    }

    @Override
    public int compareTo(pair a) {
      return Long.compare(this.x, a.x); // Corrected the order of comparison
    }
  }

  public static class pp implements Comparable<pp> {

    long x;
    int y;

    public pp(long x, int y) {
      this.x = x;
      this.y = y;
    }

    @Override
    public int compareTo(pp a) {
      return this.y - a.y; // Corrected the order of comparison
    }
  }

  static Comparator<int[]> customComparator = new Comparator<int[]>() {
    @Override
    public int compare(int[] a, int[] b) {
      // First compare based on the values at index 0
      if (a[0] != b[0]) {
        return a[0] - b[0];
      } else {
        // If values at index 0 are equal, compare based on a[i][1] - a[i][0]
        return a[1] - a[0] - (b[1] - b[0]);
      }
    }
  };

  static long mod = 1_000_000_000 + 7;

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

    int n = sc.nextInt();

    int arr[] = sc.readArray(n);

    long sum = 0;

    for (int i = 0; i < n; i++) {
      sum += arr[i];
    }

    long max = sum;

    int dp[] = new int[n + 1];

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

    for (int i = n - 1; i >= 0; i--) {
      dp[i] = Math.max(arr[i] + dp[i + 1], dp[i]);
      for (int j = i; j < n; j++) {
        int y = dp[i];
        dp[i] = Math.max((j - i + 1) * (j - i + 1) + dp[j + 1], dp[i]);
        if (y != dp[i]) {
          index[i] = j;
        }
      }
    }

    if (max == dp[0]) {
      System.out.println(max + " " + 0);
    }

    int op = 0;
    StringBuilder sb = new StringBuilder("");

    int id = 0;

    while (id != n) {
      if (index[id] == -1) {
        id++;
      } else {
        int left = id;
        int right = index[id];
        for (int j = left; j <= right; j++) {
          if (arr[j] != 0) {
            sb.append((j + 1) + " " + (j + 1));
            op++;
            sb.append("\n");
          }
        }

        for (int j = left; j <= right; j++) {
          for (int i = left; i < j; i++) {
            sb.append((left + 1) + " " + (i + 1) + "\n");
            op++;
          }
          sb.append((left + 1) + " " + (right + 1) + "\n");
          op++;
        }
        id = index[id] + 1;
      }
    }

    System.out.println(dp[0]+" "+op);
    System.out.print(sb);
  }

  static class Scanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
      while (!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());
    }

    int[] readArray(int n) {
      int[] arr = new int[n];
      for (int i = 0; i < n; i++) arr[i] = nextInt();
      return arr;
    }

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

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on scenarios that can highlight the issues in the logic. The code seems to have issues with correctly calculating the MEX and applying the operations optimally. Here are the steps to generate a challenging test input:

1. **Small Array Size**: Since the problem constraints allow \( n \leq 18 \), we can use a small array to make the test input manageable and easy to verify.

2. **Edge Cases**:
   - **Array with All Zeros**: This can test if the code correctly handles the simplest case where the MEX is 0 and all elements should be set to 0.
   - **Array with Consecutive Integers**: This can test if the code correctly identifies the MEX and applies the operations to maximize the sum.
   - **Array with Large Values**: This can test if the code correctly handles large values and calculates the MEX correctly.

3. **Mixed Values**: An array with a mix of small and large values can test the code's ability to handle a variety of inputs and correctly identify the MEX.

Based on these considerations, a challenging test input that can expose the bug is:

```plaintext
4
0 1 2 3
```

This input is small but contains a mix of values that should challenge the code to correctly identify the MEX and apply the operations to maximize the sum. The expected output for this input should be:

```plaintext
10 1
1 4
```

Explanation:
- The MEX of the array \([0, 1, 2, 3]\) is 4.
- By applying the operation with \( l = 1 \) and \( r = 4 \), the array becomes \([4, 4, 4, 4]\).
- The sum of the array is \( 4 + 4 + 4 + 4 = 16 \).

However, the buggy code might not correctly handle this case, leading to a suboptimal solution. This input should help in identifying the issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

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

public class Main {
	public static void main(String[] args) throws IOException {
		FastIO io = new FastIO();
		int N = io.nextInt();

		int[] arr = new int[N + 1];
		for (int i = 1; i <= N; i++) {
			arr[i] = io.nextInt();
		}

		int sum = N * N;
		int prev = 0;
		ArrayList<int[]> ops = new ArrayList<>();
		for (int i = 1; i <= N; i++) {
			if (prev == 0 && arr[i] == 0) continue;
			if (prev == 0) prev = i;
			if (arr[i] == 0) {
				ops.add(new int[] {prev, i - 1});
				prev = 0;
			}
		}

		if (prev != 0) ops.add(new int[] {prev, N});
		int start = ops.size();

		for (int i = 1; i < N; i++) {
			ops.add(new int[] {1, i + 1});
			ops.add(new int[] {1, i});
			int size = ops.size();
			for (int j = start; j < size - 2; j++) {
				ops.add(ops.get(j));
			}
		}

		ops.add(new int[] {1, N});

		io.println(sum + " " + ops.size());
		for (int i = 0; i < ops.size(); i++) {
			io.println(ops.get(i)[0] + " " + ops.get(i)[1]);
		}
		io.close();
	}
}

class FastIO extends PrintWriter {

	private InputStream stream;

	private byte[] buf = new byte[1 << 16];

	private int curChar;

	private int numChars;

	// standard input

	public FastIO() { this(System.in, System.out); }

	public FastIO(InputStream i, OutputStream o) {

		super(o);

		stream = i;

	}

	// file input

	public FastIO(String i, String o) throws IOException {

		super(new FileWriter(o));

		stream = new FileInputStream(i);

	}

	// throws InputMismatchException() if previously detected end of file

	private int nextByte() {

		if (numChars == -1) { throw new InputMismatchException(); }

		if (curChar >= numChars) {

			curChar = 0;

			try {

				numChars = stream.read(buf);

			} catch (IOException e) { throw new InputMismatchException(); }

			if (numChars == -1) {

				return -1;  // end of file

			}

		}

		return buf[curChar++];

	}

	// to read in entire lines, replace c <= ' '

	// with a function that checks whether c is a line break

	public String next() {

		int c;

		do { c = nextByte(); } while (c <= ' ');

		StringBuilder res = new StringBuilder();

		do {

			res.appendCodePoint(c);

			c = nextByte();

		} while (c > ' ');

		return res.toString();

	}

	public int nextInt() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		int res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

	public long nextLong() {  // nextLong() would be implemented similarly

		int c;

		do { c = nextByte(); } while (c <= ' ');

		int sgn = 1;

		if (c == '-') {

			sgn = -1;

			c = nextByte();

		}

		long res = 0;

		do {

			if (c < '0' || c > '9') { throw new InputMismatchException(); }

			res = 10 * res + c - '0';

			c = nextByte();

		} while (c > ' ');

		return res * sgn;

	}

	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 create 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. Let's analyze the code and identify potential issues:

1. **Handling of Zeros**: The code seems to have special handling for zeros, but it might not correctly handle all cases where zeros are present.
2. **MEX Calculation**: The code does not correctly calculate the MEX of subarrays, which is crucial for the problem.
3. **Operations**: The code adds operations in a way that might not always maximize the sum.

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

- **Input with Zeros**: Ensure the input contains zeros to test the special handling.
- **Input with Non-Zeros**: Ensure the input contains non-zeros to test the MEX calculation and operation handling.
- **Input with Mixed Values**: Ensure the input contains a mix of zeros and non-zeros to test the overall logic.

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

```plaintext
5
0 1 2 0 3
```

### Explanation:
- **Array Length**: 5, which is within the limit and small enough to test manually.
- **Array Content**: `0 1 2 0 3`
  - Contains zeros to test the special handling.
  - Contains non-zeros to test the MEX calculation.
  - Contains a mix of values to test the overall logic.

### Expected Behavior:
- The correct solution should maximize the sum of the array by performing the MEX operation optimally.
- The buggy code might fail to correctly handle the zeros or the MEX calculation, leading to an incorrect sum or sequence of operations.

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