**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
import sys,math,random
from heapq import heappush,heappop
from bisect import bisect_right,bisect_left
from collections import Counter,deque,defaultdict

# functions #
MOD = 998244353
MOD = 10**9 + 7
RANDOM = random.randrange(2**62)
def gcd(a,b):
    if a%b==0:
        return b
    else:
        return gcd(b,a%b)
def lcm(a,b):
    return a//gcd(a,b)*b
def w(x):
    return x ^ RANDOM
##

#String hashing : sh, fenwick sortedlist : fsortl, sieve : sieve, SparseTable : SparseTable
#bucket sorted list : bsortl, segment tree(lazy propogation) : SegmentTree, bootstrap : bootstrap
#binary indexed tree : BIT, segment tree(point updates) : SegmentPoint, Convex Hull : hull
#Combinatorics : pnc
#Template : https://github.com/OmAmar106/Template-for-Competetive-Programming

def solve():
    n = int(sys.stdin.readline().strip())
    L = list(map(int, sys.stdin.readline().split()))
    ans = 0
    flag = False
    for i in range(len(L)):
        if L[i]==0:
            flag = False
            continue
        elif L[i]<=2:
            if flag:
                flag = False
                continue
            flag = True
            ans += 1
        else:
            if flag:
                if L[i]<=4:
                    ans += 1
                    if i+1<len(L) and L[i+1]<=4:
                        L[i+1] = (L[i]-2)
                    flag = False
                else:
                    ans += 1
                    flag = False
            else:
                ans += 1
    print(ans)
    #L1 = list(map(int, sys.stdin.readline().split()))
    #st = sys.stdin.readline().strip()
for _ in range(int(sys.stdin.readline().strip())):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
def solve(arr):
    ops =0
    for i in range(len(arr)):
        if arr[i]>4:
            ops+=1
        else:
            if i==0:
                if arr[i]>2: ops+=1
                elif arr[i]<3 and arr[i]!=0:
                    ops+=1
                    if n!=1: arr[i+1]-=2
            else:
                if arr[i]>2: ops+=1
                elif arr[i]<3 and arr[i]!=0:
                    ops+=1
                    if n!=1 and i!=n-1: arr[i+1] = max(arr[i+1]-2,0)
        # print(arr)
    return(ops)

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    print(solve(a))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. The main issues in the buggy code are:

1. The code incorrectly handles cases where `arr[i] > 4` and does not account for the correct number of operations needed.
2. The code does not properly handle the transition between rows when `arr[i]` is less than 3.
3. The code does not correctly propagate the effect of operations to the next row.

Let's consider a small-scale test input that can expose these issues:

1. A case where `arr[i] > 4` but the number of operations is not correctly calculated.
2. A case where `arr[i] < 3` and the effect on the next row is not correctly handled.
3. A case where the transition between rows is not handled correctly.

Here is a test input that can expose these issues:

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

Let's break down each test case:

1. **Test Case 1: `4 5 4 3 2`**
   - The first row has 5 black cells, which requires at least 2 operations (one for the 2x2 subgrid and one for the whole row).
   - The second row has 4 black cells, which requires 1 operation (one for the whole row).
   - The third row has 3 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - The fourth row has 2 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - Total operations: 5

2. **Test Case 2: `3 2 1 3`**
   - The first row has 2 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - The second row has 1 black cell, which requires 1 operation (one for the 2x2 subgrid).
   - The third row has 3 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - Total operations: 3

3. **Test Case 3: `4 2 2 2 2`**
   - Each row has 2 black cells, which requires 1 operation per row (one for the 2x2 subgrid).
   - Total operations: 4

4. **Test Case 4: `5 3 2 1 0 4`**
   - The first row has 3 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - The second row has 2 black cells, which requires 1 operation (one for the 2x2 subgrid).
   - The third row has 1 black cell, which requires 1 operation (one for the 2x2 subgrid).
   - The fourth row has 0 black cells, which requires 0 operations.
   - The fifth row has 4 black cells, which requires 1 operation (one for the whole row).
   - Total operations: 4

This test input should help expose the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

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

#define int long long
#define need_for_speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

using namespace std;

int32_t main() {
    need_for_speed

    int t; cin >> t;
    while (t--)
    {
        int n; cin >> n;
        vector<int> v(n + 1);
        for (int i = 0; i < n; i++) cin >> v[i];

        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            if (v[i] > 4) {ans++; v[i] = 0;}
            if ((v[i] == 1 || v[i] == 2) && (v[i + 1] == 1 || v[i + 1] == 2)) {ans += 1; v[i] = 0; v[i + 1] = 0;}
        }
        for (int i = 0; i < n; i++)
        {
            if (v[i] == 1 || v[i] == 2)
            {
                int c = i;
                i++;
                while (v[i] > 2) {i++;}
                if ((i - c) % 2 == 1 && v[i] > 0) {ans += i - c;}
                else {ans += i - c; i--;}
            }
            else if (v[i] > 2) ans++;
        }
        cout << ans << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;

template <typename T1, typename T2>
using indexed_map = tree<T1, T2, less<T1>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using indexed_set = indexed_map<T, null_type>;

#define loop(x, i) for (int i = 0; i < (x); i++)
#define loop1(x, i) for (int i = 1; i <= (x); i++)
#define rev(x, i) for (int i = (int)(x) - 1; i >= 0; i--)
#define int long long
#define INF (4e18 + 1)
#define ALL(x) begin(x), end(x)
#define RALL(x) rbegin(x), rend(x)
#define existsIn(x, l) (count(ALL(l), x) > 0)
#define removeIn(x, l) l.erase(find(ALL(l), x))
#define pb push_back
#define sz(x) (int)(x).size()
#define F first
#define S second
#define var const auto &
#define foreach(l) for (var e : l)

typedef pair<int, int> ii;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<vii> vvii;
typedef vector<viii> vviii;
typedef set<int> si;
typedef set<ii> sii;
typedef set<iii> siii;
typedef vector<si> vsi;
typedef vector<sii> vsii;
typedef vector<vsi> vvsi;
typedef vector<string> vstr;
typedef vector<vector<string>> vvstr;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vi::iterator viit;
typedef si::iterator siit;
typedef int8_t in8;
typedef int16_t in16;
typedef int32_t in32;
typedef int64_t in64;

void solve()
{
    int n, res = 0;
    cin >> n;
    vi a;
    loop(n, i)
    {
        int v;
        cin >> v;
        if (v > 4)
        {
            res++;
            a.pb(0);
        }
        else if (v == 3)
            a.pb(4);
        else if (v == 1)
            a.pb(2);
        else if (v != 0 || a.empty() || a.back() != 0)
            a.pb(v);
    }
    n = a.size();
    // vi dp(n);
    loop(n, i)
    {
        if (a[i] == 0)
            continue;
        if (i + 3 < n && a[i] == 2 && a[i + 1] == 4 && a[i + 2] == 4 && a[i + 3] == 2)
        {
            i += 3;
            res += 3;
        }
        else if (i + 1 < n && a[i] == 2 && a[i + 1] == 2)
        {
            res++;
            i++;
        }
        else
            res++;
    }
    cout << res << endl;
}

signed main()
{
    cin.tie(0)->sync_with_stdio(0);
    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:
```plaintext
1
5
3 2 1 0 4
```

### Explanation:
This test case is designed to expose potential issues in the buggy code by including a mix of values that are greater than 4, equal to 3, and less than 3. Specifically:

1. **Value 3**: This should be handled by the condition `if (v == 3) a.pb(4);`, but the code might incorrectly handle it.
2. **Value 2**: This should be handled by the condition `if (v == 1) a.pb(2);`, but the code might incorrectly handle it.
3. **Value 1**: This should be handled by the condition `if (v == 1) a.pb(2);`, but the code might incorrectly handle it.
4. **Value 0**: This should be skipped, but the code might incorrectly handle it.
5. **Value 4**: This should be handled by the condition `if (v > 4) { res++; a.pb(0); }`, but the code might incorrectly handle it.

The sequence `3 2 1 0 4` is designed to test the correctness of the conditions and the logic for handling different values. The expected output for this test case should be `3`, but the buggy code might produce a different result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

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

public class CF_1990_D{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int N = ni();
        int[] A = new int[1+N];
        for(int i = 1; i<= N; i++)A[i] = ni();
        int[] ans = new int[1+N];
        Arrays.fill(ans, 2*N);
        ans[0] = 0;
        for(int i = 1; i<= N; i++){
            ans[i] = ans[i-1] + Math.min(A[i], 1);
            for(int j = Math.max(0, i-5); j< i; j++){
                ans[i] = Math.min(ans[i], ans[j] + calc(A, j+1, i));
            }
        }
        pn(ans[N]);
    }
    int calc(int[] A, int l, int r){
        int sum = 0, max = 0;
        for(int i = l; i<= r; i++){
            sum += A[i];
            max = Math.max(max, A[i]);
        }
        if(sum > 4*(r-l+1))return A.length*2;
        if(max > 2*(r-l+1))return A.length*2;

        int[][] g = new int[r-l+1][max];
        for(int i = l, row = 0; i<= r; i++, row++){
            for(int j = 0; j< A[i]; j++)g[row][j] = 1;
        }
        int ans = 0;
        for(int i = 0; i< g.length; i++){
            for(int j = 0; j< g[i].length; j++){
                if(g[i][j] == 1){
                    ans++;
                    for(int ii = 0; ii< 2; ii++)
                        for(int jj = 0; jj< 2; jj++)
                            if(i+ii < g.length && j+jj < g[i].length)
                                g[i+ii][j+jj] = 0;
                }
            }
        }
        return ans;
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399;
    static boolean multipleTC = true, memory = true, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = multipleTC? ni():1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println("Runtime: "+(System.currentTimeMillis() - ct));
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1990_D().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
        else new CF_1990_D().run();
    }
    int[][] make(int n, int e, int[] from, int[] to, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object... o){for(Object oo:o)out.print(oo+" ");}
    void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
    void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

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

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Grid Puzzle
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array $a$ of size $n$.

There is an $n \times n$ grid. In the $i$-th row, the first $a_i$ cells are black and the other cells are white. In other words, note $(i,j)$ as the cell in the $i$-th row and $j$-th column, cells $(i,1), (i,2), \ldots, (i,a_i)$ are black, and cells $(i,a_i+1), \ldots, (i,n)$ are white.

You can do the following operations any number of times in any order:

  * Dye a $2 \times 2$ subgrid white;    * Dye a whole row white. Note you can not dye a whole column white.

Find the minimum number of operations to dye all cells white.

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

For each test case:

  * The first line contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the size of the array $a$.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq n$).

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

Output: For each test case, output a single integer — the minimum number of operations to dye all cells white.

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

Note: In the first test case, you don't need to do any operation.

In the second test case, you can do:

  * Dye $(1,1), (1,2), (2,1)$, and $(2,2)$ white;    * Dye $(2,3), (2,4), (3,3)$, and $(3,4)$ white;    * Dye $(3,1), (3,2), (4,1)$, and $(4,2)$ white.

It can be proven $3$ is the minimum number of operations.

In the third test case, you can do:

  * Dye the first row white;    * Dye $(2,1), (2,2), (3,1)$, and $(3,2)$ white.

It can be proven $2$ is the minimum number of operations.

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

public class cp {
    static class Datamax implements Comparable<Datamax>{
        int idx;
        int val;
        public Datamax(int val,int idx){
            this.idx = idx;
            this.val = val;
        }
        @Override
        public int compareTo(Datamax p1){
            if(this.val<p1.val) return 1;
            return -1;
        }
    }
    static class Datamin implements Comparable<Datamin>{
        int idx;
        int val;
        public Datamin(int val,int idx){
            this.idx = idx;
            this.val = val;
        }
        @Override
        public int compareTo(Datamin p1){
            if(this.val>p1.val) return 1;
            else if(this.val<p1.val) return -1;
            if(this.idx>p1.idx) return 1;
            return -1;
        }
    }
    // static class Edge {
    //     int dst;
    //     int dir;
    //     public Edge(int dst,int dir){
    //         this.dir = dir;
    //         this.dst = dst;
    //     }
    // }
    public static PP.Reader cs = new PP.Reader(System.in);
    public static int[] wts;
    public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException{
        int tests = cs.nextInt();
        while(tests>0){
            tests--;
            solve();
        }
        // writer.flush();
        writer.close();
    }
    public static long mod = (long)Math.pow(10,9) + 7;
    public static long MOD = 998244353;
    // public static int ptr = 0;
    public static void solve() throws IOException{
        // long n = cs.nextLong();
        // int it =(int)n;
        int n = cs.nextInt();
        //int m = cs.nextInt();
        //long x = cs.nextLong();
        // int k = cs.nextInt();
        // String str = cs.next();
        // String strb = cs.next();
        long[] arr = new long[n];
        for(int i = 0; i < n; i++){
            arr[i] = cs.nextLong();
        }
        int ans = 0;
        for(int i = 0; i < n; i++){
            if(arr[i] == 0) continue;
            if(arr[i]>2 || i==n-1 || arr[i+1]>4) ans++;
            else{
                if(arr[i+1]<3){
                    i++;
                    ans++;
                }
                else if(i < n-3 && arr[i+1]<5 && arr[i+2] < 5 && arr[i+3]<3 && arr[i+1] != 0 && arr[i+2] != 0 && arr[i+3] != 0){
                    ans += 3;
                    i += 3;
                }
                else ans++;
            }
        }
        System.out.println(ans);

       // writer.flush();
    }
    //Write fxn here

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = (int)start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (int)(start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }
    public static long invmod(long A, long M)
    {

        for (int X = 1; X < (int)M; X++)
            if (((A % M) * (X % M)) % M == 1)
                return X;
        return 1;
    }
    public static long gcd(long a,long b){
        long c = Math.min(a,b);
        long d = Math.max(a,b);
        if(d%c==0) return c;
        return gcd(c,d%c);
    }
    public static long divmod(long a,long b,long m){   //claculate (a/b)%m
        // a = a%m;b = b%m  // Include if not working
        long inv =  powermod(b, m-2, m);
        return (inv*a)%m;
    }
    static long powermod(long x, long y, long p) //calculate (x^y)%p
    {
        long res = 1;
        x = x % p;
        while (y > 0) {
            if ((y & 1) > 0)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
}

class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

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

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
#include <bits/stdc++.h>
#define F(i,x,y) for(int i=(x);i<=(y);i++)
#define DF(i,x,y) for(int i=(x);i>=(y);i--)
#define ms(x,y) memset(x,y,sizeof(x))
#define SZ(x) (int)x.size()-1
#define all(x) x.begin(),x.end()
#define pb push_back
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> pii;
template<typename T> void chkmax(T &x,T y){ x=max(x,y);}
template<typename T> void chkmin(T &x,T y){ x=min(x,y);}
template<typename T> void read(T &FF){
    FF=0;int RR=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') RR=-1;
    for(;isdigit(ch);ch=getchar()) FF=(FF<<1)+(FF<<3)+ch-48;
    FF*=RR;
}
const int N=110;
int n,a[N];
bool f[N][N][N];
void work(){
    read(n);
    F(i,1,n) read(a[i]);
    f[0][0][0]=1;
    F(i,1,n) F(j,0,(i-1)/2) F(k,0,i-1-j*2) if(f[i-1][j][k]){
        f[i][j][k]=1;
        if(k) f[i][j+1][k-1]=1;
        if(a[i]>=i&&j>=a[i]-i) f[i][j][k+1]=1;
    }
    DF(i,n/2,0) if(f[n][i][0]){ printf("%d\n",i);break;}
    F(i,0,n) F(j,0,i/2) F(k,0,i-j*2) f[i][j][k]=0;
}
int main(){
    int T;read(T);
    while(T--) work();
    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 given code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The code seems to be implementing a dynamic programming solution, but there might be issues with how it handles certain conditions, such as when the array has multiple valid indices for removal or when the array is small.

Let's consider the following edge cases:
1. An array where no valid index exists for removal.
2. An array where multiple valid indices exist for removal.
3. An array with a very small size.
4. An array where the elements are in a specific order that might cause the dynamic programming table to be filled incorrectly.

Given these considerations, let's generate a test input that might expose the bug:

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - Valid indices: 1, 2, 3, 4
   - Expected output: 2 (e.g., `[1, 2, 3, 4, 5] -> [1, 3, 5] -> [5]`)

2. **Test Case 2:**
   - Array: `[2, 1, 3, 4, 5, 6]`
   - Valid indices: 1, 2, 3, 4, 5
   - Expected output: 3 (e.g., `[2, 1, 3, 4, 5, 6] -> [2, 4, 6] -> [2, 6] -> [6]`)

3. **Test Case 3:**
   - Array: `[1, 2, 3]`
   - Valid indices: 1, 2
   - Expected output: 1 (e.g., `[1, 2, 3] -> [1]`)

4. **Test Case 4:**
   - Array: `[1, 1]`
   - Valid indices: None
   - Expected output: 0 (no valid indices for removal)

This test input covers a variety of scenarios, including arrays with multiple valid indices, arrays with no valid indices, and small arrays, which should help in identifying any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define ld long double
using namespace std;
using namespace __gnu_pbds;
typedef pair<int, int> node;
typedef tree<node, null_type, less<node>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
const int N = 5e5 + 5;
const ll MOD = 1e9 + 7, MAX = 1e18;
const double ep = 1e-6, pi = atan(1.0) * 4;
long long inv(long long a, long long b = MOD)
{
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
int dx[8] = {1, 0, -1, 0, -1, 1, -1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};
ll n, m = 0, k;
void fastio()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int comp(ld x, ld y)
{
  // return 1 if x<=y
  if (abs(x - y) <= 1e-9)
    return 1;
  if (x < y)
    return 1;
  return 0;
}
ll arr[N];
ll dp[105][105]; // cur opnumber
ll solve(int i, int op)
{
  if (i > n + 1)
    return -1500;
  if (i == n + 1)
    return op;
  ll &ret = dp[i][op];
  if (~ret)
    return ret;
  ret = solve(i + 1, op);
  ll x = arr[i] - i;
  if (arr[i] % 2 != i % 2 || i > arr[i] || x / 2 > op)
    return ret;
  ret = max(solve(i + 2, op + 1), ret);
  for (int j = i + 1; j <= n; j++)
  {
    if (arr[j] % 2 != j % 2 || j > arr[j] || arr[j] - j > x)
      break;
    x = min(x, arr[j] - j);
    ret = max(ret, solve(i + (j - i + 1) * 2, op + (j - i + 1)));
  }
  return ret;
}
void fn()
{
  cin >> n;
  for (int i = 1; i <= n; i++)
  {
    cin >> arr[i];
  }
  memset(dp, -1, sizeof dp);
  ll ans = solve(1, 0);
  cout << ans << '\n';
}
int main()
{
  fastio();
  int t = 1;
  cin >> t;
  while (t--)
    fn();
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's break down the problem and identify potential issues:

1. **Invalid Operations**: The code should only perform operations where `a_i = i` and `1 <= i < n`.
2. **Optimal Sequence**: The code should find the maximum number of operations, which might involve complex sequences of removals.
3. **Edge Cases**: Small arrays, arrays with no valid operations, and arrays with multiple valid operations.

Given these points, let's construct a test case that can expose potential bugs:

### Test Case Analysis
- **Test Case 1**: An array where no valid operations can be performed.
- **Test Case 2**: An array where a single valid operation can be performed.
- **Test Case 3**: An array where multiple valid operations can be performed in a specific sequence.
- **Test Case 4**: An array with a sequence that might cause the code to fail due to incorrect conditions.

### Chosen Test Case
Let's focus on a test case where multiple valid operations can be performed, but the sequence is tricky to handle. This will help in identifying if the code correctly manages the sequence of operations.

### Test Input
```plaintext
4
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
```

### Explanation
1. **Test Case 1**: `5 1 5 3 2 4`
   - Optimal sequence: `[1, 5, 3, 2, 4] -> [1, 5, 4] -> [4]`
   - Expected output: `2`

2. **Test Case 2**: `8 2 1 3 4 5 6 7 8`
   - Optimal sequence: `[2, 1, 3, 4, 5, 6, 7, 8] -> [2, 3, 4, 5, 6, 7] -> [2, 4, 5, 6] -> [2, 5]`
   - Expected output: `3`

3. **Test Case 3**: `3 1 2 3`
   - Optimal sequence: `[1, 2, 3] -> [1]`
   - Expected output: `1`

4. **Test Case 4**: `4 1 2 4 4`
   - Optimal sequence: `[1, 2, 4, 4] -> [4, 4]`
   - Expected output: `2`

This test input covers a range of scenarios, including no valid operations, single valid operations, and multiple valid operations with a specific sequence. It should help in exposing any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
// Har Har Mahadev
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define int long long
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define repin rep(i,0,n)
#define precise(i) cout<<fixed<<setprecision(i)
#define vi vector<int>
#define si set<int>
#define mii map<int,int>
#define take(a,n) for(int j=0;j<n;j++) cin>>a[j];
#define give(a,n) for(int j=0;j<n;j++) cout<<a[j]<<' ';
#define vpii vector<pair<int,int>>
#define db double
#define be(x) x.begin(),x.end()
#define pii pair<int,int>
#define pb push_back
#define pob pop_back
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define bpc(x) __builtin_popcountll(x)
#define btz(x) __builtin_ctz(x)
using namespace std;

using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<pair<int, int>, null_type,less<pair<int, int> >, rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;

const long long INF=1e18;
const long long M=1e9+7;
const long long MM=998244353;

int power( int N, int M){
    int power = N, sum = 1;
    if(N == 0) sum = 0;
    while(M > 0){if((M & 1) == 1){sum *= power;}
    power = power * power;M = M >> 1;}
    return sum;
}

// int funo(vi &a){
//     if(a.size() == 1)return 0;
//     // vi v;
//     int n = a.size();
//     int ans = 0;
//     vi b(2,1);
//     rrep(i,n-2,0){
//         if(a[i] == i && b[i%2]){
//             vi v;
//             rep(j,0,n){
//                 if(j == i)continue;
//                 if(j == i+1)continue;
//                 v.pb(a[j]);
//             }
//             b[i%2] = 0;
//             ans = max(ans,1+funo(v));
//         }
//     }
//     return ans;
// }

int fin(vi &a){
    int n = a.size();
    // return 0;
    vector<vector<int>> dp(n,vector<int>(n,0));
    // return 0;
    rep(i,1,n){
        rep(j,0,n-i+1){
            rep(k,j,i+j){
                // continue;
                if(k >= n)continue;
                if(a[k] == k-j){
                    if(i+j >= n)continue;
                    // continue;
                    dp[j][i+j] = max(dp[j][i+j],(k-1>=j?dp[j][k-1]:0) + 1 + (k+2<=i+j?dp[k+2][i+j]:0));
                    int temp = k-1;
                    int cnt = 0;
                    while(temp >= j){
                        cnt++;
                        if(a[temp] == temp-j && k+2+cnt <= i+j+1){
                            dp[j][i+j] = max(dp[j][i+j],(k-1-cnt>=j?dp[j][k-1-cnt]:0) + cnt + 1 + (k+2+cnt<=i+j?dp[k+2+cnt][i+j]:0));
                            temp--;
                        }
                        else break;
                    }
                }
            }
        }
    }
    // return dp[0][n-1];
    // repin{
    //     rep(j,0,n){
    //         cout << dp[i][j] << " ";
    //     }cout << "\n";
    // }
    return dp[0][n-1];
}

void solve()
{

    int n;
    cin >> n;
    vi a(n);
    take(a,n);
    for(auto &x : a)x--;
    cout << fin(a) << "\n";
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    #ifdef NCR
        init();
    #endif
    #ifdef SIEVE
        sieve();
    #endif
        int t;
        cin >> t;
        while(t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The code is designed to find the maximum number of times the specified operation can be performed on an array. The operation involves removing elements at specific indices and concatenating the remaining parts.

### Steps to Identify a Challenging Test Input

1. **Edge Cases**:
   - **Single Element Array**: The operation cannot be performed if the array has only one element.
   - **No Valid Operations**: An array where no valid operations can be performed, e.g., an array without any element equal to its index.
   - **All Valid Operations**: An array where every element can be removed in a valid operation, e.g., `[1, 2, 3, 4]`.

2. **Complex Scenarios**:
   - **Mixed Valid and Invalid Operations**: An array with a mix of elements that can and cannot be removed, e.g., `[1, 3, 2, 4]`.
   - **Consecutive Valid Operations**: An array where multiple consecutive elements can be removed, e.g., `[1, 2, 3, 4, 5]`.

3. **Large Input**:
   - Although the problem constraints are small, we can still create a small-scale test input that is complex enough to challenge the algorithm.

### Challenging Test Input

Let's create a test input that includes a mix of valid and invalid operations, and ensures that the algorithm must handle multiple steps to find the correct solution.

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

### Explanation

1. **Test Case 1: `[1, 3, 2, 4, 5]`**
   - The array has elements at indices that can be removed, but not all are consecutive. This tests the ability to handle multiple valid operations and ensure the correct maximum count.

2. **Test Case 2: `[2, 1, 3, 4, 5, 6, 7, 8]`**
   - This array has a sequence of valid operations but starts with an element that cannot be removed immediately. This tests the ability to handle initial invalid operations and find the correct sequence.

3. **Test Case 3: `[1, 2, 3]`**
   - A simple case where all elements can be removed in valid operations. This tests the basic functionality.

4. **Test Case 4: `[1, 2, 4, 4]`**
   - An array with a mix of valid and invalid operations. This tests the ability to handle elements that cannot be removed and still find the correct maximum count.

5. **Test Case 5: `[4, 4, 1, 3, 5]`**
   - An array with no valid operations at the start, but valid operations can be performed after some steps. This tests the ability to handle initial invalid operations and find the correct sequence.

6. **Test Case 6: `[1]`**
   - A single element array. This tests the edge case where no operations can be performed.

This test input should effectively challenge the algorithm and expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
dp = {}
def f(i,a,k):
    if i<0 or len(a)==0:
        return k
    if i>=len(a)-1:
        i = len(a)-2
    if (i,a) in dp:
        return dp[(i,a)]
    # print(i,a,a[i])
    ans = 0
    if i>=0 and a[i]==i+1:
        # x1,x2 = a[i],a[i+1]
        x = f(i,a[:i]+a[i+2:],k+1)
        # print(a[i],x)
        ans = max(ans,x)
    # print(a)
    ans = max(ans,f(i-1,a,k))
    dp[(i,a)]=ans
    return ans
for _ in range(int(input())):
    n = int(input())
    a = tuple(map(int,input().split()))
    print(f(n-2,a,0))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
from sys import stdin,stdout
# from os import _exit
# from bisect import bisect_left,bisect
# from heapq import heapify,heappop,heappush
# from sys import setrecursionlimit
# from collections import defaultdict,Counter,deque
# from itertools import permutations
# from math import gcd,ceil,sqrt,factorial
# setrecursionlimit(int(1e5))
input,print = stdin.readline,stdout.write

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    b = [i for i in range(1,n+1)]
    ans = 0
    for i in range(len(a)-2,-1,-1):
        if a[i]==b[i] and i<len(a)-1:
            #print(a,i)
            del a[i]
            del a[i]
            ans+=1

    print(f"{ans}\n")
    #print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
import sys
input = lambda :sys.stdin.readline()[:-1]
ni = lambda :int(input())
na = lambda :list(map(int,input().split()))
yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")
no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")
#######################################################################
inf = 10 ** 18
def f(a):
    n = len(a)
    z = []
    for i in range(n-1, -1, -1):
        z = []
        for j in range(len(a)-1, -1, -1):
            if a[j] == i:
                if z:
                    z.pop()
            else:
                z.append(a[j])
        a = z[::-1]
    return (n-len(a)) // 2
for _ in range(ni()):
    n = ni()
    a = [x-1 for x in na()]
    dp = [-inf] * (n+1)
    dp[0] = 0
    for i in range(n):
        sk = []
        c = 0
        for j in range(n):
            if (a[j] - j) % 2 or (j - a[j]) // 2 < 0 or (j - a[j]) // 2 > dp[i]:
                x = -1
            else:
                x = (j - a[j]) // 2
            sk.append(x)
            #print(i, j, c, sk)
            #print(sk, f(sk))
            dp[j+1] = max(dp[j+1], f(sk))
    #print(dp)
    print(dp[-1])

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases**: Test with arrays where the elements are very close to their indices or far from their indices.
2. **Special Patterns**: Test with arrays that have repeating patterns or specific sequences that might cause the buggy code to fail.
3. **Boundary Conditions**: Test with the minimum and maximum allowed values for `n`.

Given the constraints and the nature of the problem, a good test case would be one where the array elements are arranged in a way that the buggy code's logic fails to correctly identify and remove the elements as per the rules.

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

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

### Explanation:
1. **Test Case 1**:
   - Array: `[1, 2, 3, 4, 5]`
   - Expected Output: `2`
   - Explanation: The optimal sequence of operations is `[1, 2, 3, 4, 5] → [1, 2, 4] → [4]`.

2. **Test Case 2**:
   - Array: `[1, 2, 4, 4]`
   - Expected Output: `2`
   - Explanation: The optimal sequence of operations is `[1, 2, 4, 4] → [4, 4] → [4]`.

3. **Test Case 3**:
   - Array: `[1, 2, 3, 4, 5, 6]`
   - Expected Output: `3`
   - Explanation: The optimal sequence of operations is `[1, 2, 3, 4, 5, 6] → [1, 2, 4, 6] → [1, 2, 6] → [6]`.

These test cases are designed to challenge the buggy code by presenting arrays where the elements are close to their indices and where the removal operations can be performed multiple times. The buggy code's logic of using a dynamic programming approach with a fixed offset might fail to correctly handle these cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

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

public class F {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	static Integer [][][] memo;

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Sources of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
		}

		private void solve(int t) {
			int n = sc.nextInt();

			int [] arr = sc.readArray2(n);

			memo = new Integer [n + 1][n + 1][n + 1];
			int result = solve(arr, 1, 1, 0);

			out.println(result);
		}

		private int solve(int [] arr, int idx, int pos, int rem) {
			if (idx >= arr.length - 1) {
				return rem;
			}else if (memo[idx][pos][rem] != null)
				return memo[idx][pos][rem];

			int target;

			/*
			List<Integer> list = new ArrayList<>();

			for (int i = 0; i <= rem; ++i) {
				target = pos - i * 2;
				list.add(target);
			}*/

			int result = rem;

			result = Math.max(result, solve(arr, idx + 1, pos + 1, rem));

			int added, nextIdx;

			int allowed = rem;

			int diff;

			boolean qualifies = false;

			if (arr[idx] > pos) {
				qualifies = false;
			}else if ((pos % 2) != (arr[idx] % 2)) {
				qualifies = false;
			}else {
				diff = pos - arr[idx];

				diff /= 2;
				qualifies = diff <= rem;
			}

			int needed;

			if (qualifies) {
				result = Math.max(result, solve(arr, idx + 2, pos, rem + 1));

				for (int i = idx, j = pos, count = 1; idx + 2*count <= arr.length;
						++count, ++i, ++j) {

					if (i > idx) {
						if ((arr[i] % 2) == (arr[i - 1] % 2) || arr[i] <= arr[i - 1]) {
							break;
						}
					}

					needed = arr[i] - pos;
					needed -= (i - idx);

					if (needed > rem)
						break;

					added = count;
					nextIdx = idx + 2 * count;

					int test = solve(arr, nextIdx, pos, added + rem);
					result = Math.max(result, test);
				}
			}

			memo[idx][pos][rem]  = result;
			// System.out.println(idx + " " + pos + " " + rem + " " + result);
			return result;
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - **Array:** [1, 2, 3, 4, 5]
   - **Expected Output:** 2
   - **Reason:** The optimal sequence of operations is:
     - [1, 2, 3, 4, 5] → [3, 4, 5] (remove 1 and 2)
     - [3, 4, 5] → [5] (remove 3 and 4)
   - This test case checks the basic functionality of the algorithm, ensuring it can handle a simple sequence where the operation can be performed multiple times.

2. **Test Case 2:**
   - **Array:** [1, 2, 3, 4, 5, 6]
   - **Expected Output:** 3
   - **Reason:** The optimal sequence of operations is:
     - [1, 2, 3, 4, 5, 6] → [3, 4, 5, 6] (remove 1 and 2)
     - [3, 4, 5, 6] → [5, 6] (remove 3 and 4)
     - [5, 6] → [] (remove 5 and 6)
   - This test case checks if the algorithm can handle a longer sequence and perform the operation the maximum number of times.

3. **Test Case 3:**
   - **Array:** [1, 2, 3, 4, 5, 6, 7]
   - **Expected Output:** 3
   - **Reason:** The optimal sequence of operations is:
     - [1, 2, 3, 4, 5, 6, 7] → [3, 4, 5, 6, 7] (remove 1 and 2)
     - [3, 4, 5, 6, 7] → [5, 6, 7] (remove 3 and 4)
     - [5, 6, 7] → [7] (remove 5 and 6)
   - This test case checks if the algorithm can handle a sequence that is one element longer than the previous test case, ensuring it can still perform the operation the maximum number of times.

These test cases are designed to expose any issues in the algorithm's logic, particularly in handling sequences where the operation can be performed multiple times.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

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

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

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

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

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

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

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

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

    public static void main(String[] args) throws java.lang.Exception {
        try {
            FastReader scan = new FastReader();
            int t = scan.nextInt();

            while (t-- > 0) {
                int n = scan.nextInt();

                int[] a = new int[n];

                for (int i = 0; i < n; i++) {
                    a[i] = scan.nextInt();
                }
                if(n==4 && a[0] == 1&& a[1] == 2 && a[2] == 4 && a[3]==4) {
                	System.out.println("2");
                	continue;
                }
                int[] dp = new int[n];
                Arrays.fill(dp, -1);

                int maxOperations = 0;
                for (int i = 0; i < n; i++) {
                    maxOperations = Math.max(maxOperations, calculateMaxOperations(a, dp, i));
                }

                System.out.println(maxOperations);
            }

        } catch (Exception e) {
            return;
        }
    }

    private static int calculateMaxOperations(int[] a, int[] dp, int index) {
        if (index >= a.length - 1) {
            return 0;
        }

        if (dp[index] != -1) {
            return dp[index];
        }

        int maxOperations = 0;
        for (int i = index; i < a.length - 1; i++) {
            if (a[i] == i + 1) {
                maxOperations++;
                i++; // Skip the next element since we are removing a[i] and a[i+1]
            }
        }

        return dp[index] = maxOperations;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the logic in the `calculateMaxOperations` method might fail. The method is supposed to calculate the maximum number of operations that can be performed, but it has a flaw in how it handles the removal of elements and the dynamic programming (DP) array.

Here are some key points to consider:
1. **Edge Cases**: Test cases where the array has very few elements or where the elements do not match the condition `a[i] == i + 1`.
2. **Specific Patterns**: Test cases where the array has specific patterns that might not be handled correctly by the logic.
3. **Dynamic Programming Array**: Test cases that might cause the DP array to be filled incorrectly.

Given these considerations, a small-scale test input that could expose the bug is:

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

### Explanation:
1. **Test Case 1: `4 1 2 4 4`**
   - This is a specific case that the code has a special handling for, but we want to see if the general logic works correctly.

2. **Test Case 2: `3 1 2 3`**
   - This is a straightforward case where the operations can be performed once.

3. **Test Case 3: `5 1 2 3 4 5`**
   - This case should allow for multiple operations, but the logic in the `calculateMaxOperations` method might not handle the removal of elements correctly.

4. **Test Case 4: `6 1 2 3 4 5 6`**
   - This is a longer array where multiple operations should be possible, but the dynamic programming logic might fail to correctly calculate the maximum number of operations.

These test cases are designed to cover a range of scenarios, including edge cases and specific patterns, to ensure that the bug in the code is exposed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
from copy import deepcopy as dc
def perform(l):
    ma=0
    for i in range(1,len(l)-1):
        ai=l[i]
        if i==ai:
            ldash = dc(l)
            ldash=ldash[:i]+ldash[i+2:]
            ma=max(ma,1+perform(ldash))
    return ma

t=int(input())
for _ in range(t):
    input()
    print(perform(list(map(int, ('69 '+input()).strip().split(' ')))), flush=False)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
import sys
import math
# Fast input
input = sys.stdin.readline
# Fast output
def print(*args, **kwargs):
    sep = kwargs.get('sep', ' ')
    end = kwargs.get('end', '\n')
    file = kwargs.get('file', sys.stdout)
    flush = kwargs.get('flush', False)
    output = sep.join(map(str, args)) + end
    file.write(output)
    if flush:
        file.flush()

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
# @bootstrap
# def dfs(node, parent, x):
#     sx = 1
#     for i in graph[node]:
#         if i==parent:
#             continue
#         sx += yield dfs(i, node, x)
#     if sx >= x:
#         ans[0] += 1
#         sx = 0
#     yield sx

from collections import Counter, deque, defaultdict
from itertools import accumulate, product, groupby, combinations
import operator
import math
import bisect
import heapq

def readIntLine():
    return [int(char) for char in input().split()]

def readCharLine():
    return [char for char in input().split()]

def modmul(a, b, mod):
    a %= mod
    b %= mod
    return (a * b) % mod

def modadd(a, b, mod):
    a %= mod
    b %= mod
    return (a+b)%mod

def modsub(a, b, mod):
    a %= mod
    b %= mod
    return (a - b + mod)%mod

def modInverse(A, M):
    for X in range(1, M):
        if (((A % M) * (X % M)) % M == 1):
            return X
    return -1

def gcd(a,b):
    if b== 0:
        return a
    return gcd(b, a%b)

def is_coprime(a,b):
    return gcd(a,b)==1

def expo(a, n, mod):
    res = 1
    while n:
        if n&1:
            res = modmul(res,a,mod)
            n -= 1
        else:
            a = modmul(a,a,mod)
            n >>= 1
    return res

def get_mex(arr, N):
    arr = sorted(arr)
    mex = 0
    for idx in range(N):
        if arr[idx] == mex:
            mex += 1
    return mex

def primeFactors(n):
    res = Counter()
    while n % 2 == 0:
        res[2] += 1
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2):
        while n % i== 0:
            res[i] += 1
            n //= i
    if n > 2:
        res[n] += 1
    return res

def printDivisors(n) :
    i = 1
    divs = []
    while i <= math.sqrt(n):
        if (n % i == 0) :
            if (n / i == i) :
                divs.append(i)
            else :
                divs.append(i)
                divs.append(n//i)
        i = i + 1
    return sorted(divs)

def LcmOfArray(arr, idx):
    # lcm(a,b) = (a*b/gcd(a,b))
    if (idx == len(arr)-1):
        return arr[idx]
    a = arr[idx]
    b = LcmOfArray(arr, idx+1)
    return int(a*b/math.gcd(a,b)) # __gcd(a,b) is inbuilt library function

UF = {}

def find(x):
    if x not in UF:
        UF[x] = x
    if UF[x] != x:
        UF[x] = find(UF[x])
    return UF[x]

def union(x, y):
    rX, rY = find(x), find(y)
    if rX < rY:
        UF[rY] = rX
    else:
        UF[rX] = rY

def solve(idx):
    n = int(input())
    # n, k = list(map(int,input().split()))
    # s = input().rstrip('\n')
    A = readIntLine()
    # a = input().rstrip('\n')

    q = deque()
    q.append([A[:], 0])
    ans = 0

    while q:
        arr, res = q.popleft()
        ans = max(ans, res)

        if len(arr) <= 1:
            continue

        for i, a in enumerate(arr[:-1]):
            if i+1 == a:
                new_arr = arr[:i] + arr[i+2:]
                q.append([new_arr, res+1])

    print(ans)

# solve(0)
for i in range(int(input())):
    solve(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

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

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

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

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

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

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

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

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

	public static void main(String[] args) throws java.lang.Exception {
		try {
			FastReader scan = new FastReader();
			int t = scan.nextInt();

			while (t-- > 0) {
				int n = scan.nextInt();

				int[] a = new int[n];

	            for (int i = 0; i < n; i++) {
	                a[i] = scan.nextInt();
	            }

	            int maxOperations = calculateMaxOperations(a, 0);
	            System.out.println(maxOperations);
			}

		} catch (Exception e) {
			return;
		}
	}
	private static int calculateMaxOperations(int[] a, int operations) {
        boolean found = false;
        int maxOperations = operations;

        for (int i = 0; i < a.length - 1; i++) {
            if (a[i] == i + 1) {
                found = true;
                int[] newArray = removeElements(a, i);
                int result = calculateMaxOperations(newArray, operations + 1);
                maxOperations = Math.max(maxOperations, result);
            }
        }

        if (!found) {
            return operations;
        }

        return maxOperations;
    }

    private static int[] removeElements(int[] a, int index) {
        int[] newArray = new int[a.length - 2];
        int j = 0;

        for (int i = 0; i < a.length; i++) {
            if (i != index && i != index + 1) {
                newArray[j++] = a[i];
            }
        }

        return newArray;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 100$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $100$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

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

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=105;
int n,T;
int a[N],c[N],nx[N];
int ans=0;
int cnt=0;
void solve(int len){
	cnt++;
	if(cnt>=1e5)return;
	ans=max(ans,n-len>>1);
	int f=0;
//	cout<<len<<endl;
	for(int i=1;i<=n;i++){
//		cout<<c[i]<<" "<<nx[i]<<endl;;
		if(a[i]>=i)f++;
	}
//	cout<<endl;
	if(f<=ans)return;
	int r=0,ls=0;
	for(int i=1;i<=n;i++){
		if(!c[i])continue;
		r++;
		if(a[i]==r){
//			cout<<"FFFF"<<i<<endl;
			int p=nx[i];
			if(p==0)continue;
			c[i]=0;c[p]=0;
			if(ls)nx[ls]=nx[p];
			solve(len-2);
			c[i]=1,c[p]=1;
			nx[ls]=i;
		}ls=i;
	}
}
int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]),c[i]=1,nx[i]=i+1;
		cnt=0;
		nx[n]=0;
		ans=0;
		solve(n);
		int ans2=0;
		int len=n;
		while(1){int p=0;
			for(int i=len-1;i>=1;i--){
				if(a[i]==i){
					p=i;
					break;
				}
			}
			if(p==0)break;
			len-=2;
			for(int i=p;i<=len;i++){
				swap(a[i],a[i+2]);
			}ans2++;
		}
		printf("%d\n",max(ans2,ans));
	}
	return 0;
}

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

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

Think step by step.

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