**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
t=int(input())
for _ in range(t):
  n=int(input())
  arr=list(map(int,input().split()))
  c=0
  d={}
  for num in arr:
    if num not in d:
      d[num]=1
    else:
      d[num]+=1
  for i in d:
    if d[i]>2:
      c+=1
  print(c)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to create a test input that highlights the incorrect logic. The buggy code counts the number of stick lengths that appear more than twice, but it does not consider the requirement to form regular polygons. Specifically, it does not check if the count of a particular stick length can form a polygon (i.e., the count must be a multiple of the number of sides of the polygon).

Here are the steps to create a challenging test input:

1. **Identify the Bug**: The code incorrectly counts the number of stick lengths that appear more than twice without ensuring they can form a polygon.
2. **Create a Test Case**: We need a test case where the count of a particular stick length is more than 2 but cannot form a polygon.

Let's construct a test case:

- **Test Case 1**: A single stick length that appears more than twice but cannot form a polygon (e.g., 3 sticks of length 1).
- **Test Case 2**: A combination of stick lengths where some can form a polygon and others cannot, to ensure the code's logic is thoroughly tested.

Here is the test input:

```plaintext
2
3
1 1 1
6
2 2 2 3 3 3
```

### Explanation:
1. **Test Case 1**:
   - Input: `3 1 1 1`
   - Expected Output: `0`
   - Explanation: We have 3 sticks of length 1, but 3 sticks cannot form a regular polygon (minimum is 3 sides, and all sides must be equal).

2. **Test Case 2**:
   - Input: `6 2 2 2 3 3 3`
   - Expected Output: `0`
   - Explanation: We have 3 sticks of length 2 and 3 sticks of length 3. Neither set can form a regular polygon (3 sticks cannot form a polygon with more than 3 sides).

This test input should effectively expose the bug in the given code, as it counts the number of stick lengths that appear more than twice without considering the polygon formation requirement.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
# Function to count the frequency of stick lengths
def count_sticks(sticks):
    freq = {}
    for stick in sticks:
        if stick in freq:
            freq[stick] += 1
        else:
            freq[stick] = 1
    return freq

# Function to calculate the maximum number of polygons
def max_polygons(freq):
    max_poly = 0
    for length in freq:
        max_poly += freq[length] // 4
    return max_poly

# Input the number of test cases
t = int(input())

# Iterate through each test case
for _ in range(t):
    # Input the number of sticks and their lengths
    n = int(input())
    sticks = list(map(int, input().split()))

    # Count the frequency of stick lengths
    freq = count_sticks(sticks)

    # Calculate the maximum number of polygons
    max_poly = max_polygons(freq)

    # Output the result
    print(max_poly)

**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 1 1 1
5
1 1 1 1 1
12
2 2 2 2 2 2 2 2 3 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
def solution():
    n:int = int(input())
    nums = list(map(int,input().split()))
    freq = {}
    ans:int =0
    for i in nums:
        if i in freq:
            freq[i]+=1
        else:
            freq[i]=1
    for key in freq:
        if freq[key]>2:
            no:int=3
            while freq[key]%no!=0:
                no+=1
            if freq[key]!=no:
               ans+=freq[key]//no
            else:
                ans+=freq[key]//3
    return ans
n:int = int(input())
for i in range(n):
    print(solution())

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases where the code might fail to correctly count the number of regular polygons that can be formed. The key issues in the buggy code are:

1. The code incorrectly handles the case where the number of sticks of a particular length is exactly 3.
2. The code uses a while loop to find the smallest number of sides that can be formed, which is inefficient and might not always find the correct number of polygons.

Given these issues, let's construct a test input that will likely expose the bug:

- We need a test case where the number of sticks of a particular length is exactly 3.
- We also need a test case where the number of sticks of a particular length is a multiple of 3 but not exactly 3, to ensure the code handles it correctly.

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

```plaintext
2
3
3 3 3
6
3 3 3 3 3 3
```

Explanation:
- The first test case has exactly 3 sticks of length 3. The correct output should be 1 (one triangle).
- The second test case has 6 sticks of length 3. The correct output should be 2 (two triangles).

This test input should highlight the issue in the buggy code where it incorrectly handles the case with exactly 3 sticks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
/*

    ॐ भूर् भुवः स्वः।
    तत् सवितुर्वरेण्यं।
    भर्गो देवस्य धीमहि।
    धियो यो नः प्रचोदयात् ॥

  @Author  : ANKUSH
  @Country : INDIA

*/
import java.io.*;
import java.util.*;

public class Main {// Codechef {
    static class pair {
        long val1;
        long val2;

        public pair(long val1, long val2) {
            this.val1 = val1;
            this.val2 = val2;
        }

        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;
            pair pair = (pair) o;
            return val1 == pair.val1 && val2 == pair.val2;
        }

        @Override
        public int hashCode() {
            return 31 * Long.hashCode(val1) + Long.hashCode(val2);
        }

        @Override
        public String toString() {
            return "pair{" + "x=" + val1 + ", y=" + val2 + '}';
        }
    }

    static long MOD = (long) (1e9 + 7);
    static long dp[][];

    public static void main(String[] args) {
        int TTT = 0;
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
           int testCases = in.nextInt();
            while (TTT++ != testCases) {
                // write code here
              int n=in.nextInt();
              int a[]=new int[n];
              int f[]=new int[n+1];
              for(int i=0;i<n;i++){
                  a[i]=in.nextInt();
                  f[a[i]]++;
              }int res=0;
              for(int i=0;i<n;i++){
                  res+=f[i]/3;
              }

              out.println(res);
            }
            out.close();
        } catch (Exception e) {
            System.out.println(e);
            return;
        }
    }
    /*-------------------------------------------  MATHS     MATHS     MATHS   ------------------------------------ */
    public static long sqrt(long x) {
        long low = 0, high = 3037000499L;
        while (low <= high) {
            long mid = low + (high - low) / 2;
            long square = mid * mid;

            if (square == x) {
                return mid;
            } else if (square < x) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return high;
    }
    public static boolean isPrime(long number) {
        if (number <= 1) {
            return false;
        }
        if (number <= 3) {
            return true;
        }
        if (number % 2 == 0 || number % 3 == 0) {
            return false;
        }

        long sqrt = (long) Math.sqrt(number);
        for (long i = 5; i <= sqrt; i += 6) {
            if (number % i == 0 || number % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }

    public static long power(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    public static long power_mod(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans *= a;
                ans %= MOD;
            }
            a *= a;
            a %= MOD;
            b >>= 1;
        }
        return ans;
    }

    static final int SIZE = (int) (2000);
    private static long[] fact = new long[SIZE + 1];
    private static boolean[] isPrime = new boolean[SIZE + 1];

    public static long add_mod(long a, long b) {
        return ((a % MOD) + (b % MOD)) % MOD;
    }

    public static long sub_mod(long a, long b) {
        return (((a % MOD) - (b % MOD)) + MOD) % MOD;
    }

    public static long multiply_mod(long a, long b) {
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    public static long inv(long a) {
        return power_mod(a, MOD - 2);
    }

    public static long divide_mod(long a, long b) {
        return multiply_mod(a, inv(b));
    }

    public static long nCr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], multiply_mod(fact[(int) r], fact[(int) (n - r)]));
    }

    public static long nPr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], fact[(int) (n - r)]);
    }

    public static void preFactorial() {
        fact[0] = 1;
        for (int i = 1; i <= SIZE; i++) {
            fact[i] = multiply_mod(i, fact[i - 1]);
        }
    }

    public static void Sieve_Of_Eratosthenes() {
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;

        for (int i = 2; i * i <= SIZE; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= SIZE; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    public static long gcd(long x, long y) {
        if (x > y)
            return gcd(y, x);
        if (y == 0) {
            return x;
        }
        if (x == 0) {
            return y;
        }
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    /*----------------------------------------  ADVANCED DATA STRUCTURES  ADVANCED DATA STRUCTURES  ------------------------------------ */
    public static class DisjointSet {
        private long[] parent;
        private long[] size;

        public DisjointSet(int n) {
            parent = new long[n];
            size = new long[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        public long findParent(long node) {
            if (node == parent[(int) node]) {
                return node;
            }
            return parent[(int) node] = findParent(parent[(int) node]);
        }

        public void unionSize(long node1, long node2) {
            long p1 = findParent(node1);
            long p2 = findParent(node2);
            if (p1 == p2)
                return;
            if (size[(int) p1] > size[(int) p2]) {
                size[(int) p1] += size[(int) p2];
                parent[(int) p2] = p1;
            } else {
                size[(int) p2] += size[(int) p1];
                parent[(int) p1] = p2;
            }
        }

        public long findSize(long node) {
            long p1 = findParent(node);
            return size[(int) p1];
        }
    }

    static class Segment_Tree {
        long arr[];
        int n;
        long seg[];

        public Segment_Tree(long arr[], int n) {
            this.arr = arr;
            this.n = n;
            this.seg = new long[(4 * n) + 1];
            build(0, 0, n - 1);
        }

        public void build(int ind, int low, int high) {
            if (low == high) {
                seg[ind] = arr[low];
                return;
            }
            int mid = (low + high) / 2;
            build((2 * ind) + 1, low, mid);
            build((2 * ind) + 2, mid + 1, high);
            seg[ind] = Math.max(seg[2 * ind + 1], seg[2 * ind + 2]);
        }

        public long query2(int ind, int low, int high, int left, int right) {
            if (low >= left && high <= right) {
                return seg[ind];
            }
            if (low > right || high < left) {
                return 0;
            }
            int mid = (low + high) / 2;
            long l = query2(2 * ind + 1, low, mid, left, right);
            long r = query2(2 * ind + 2, mid + 1, high, left, right);
            return Math.max(l, r);
        }

        public long query(int l, int r) {
            return query2(0, 0, n - 1, l, r);
        }

        public void pointupdate2(int ind, int low, int high, int node, int val) {
            if (low == high) {
                seg[ind] = val;
            } else {
                int mid = (low + high) / 2;
                if (node <= mid && node >= low)
                    pointupdate2(2 * ind + 1, low, mid, node, val);
                else
                    pointupdate2(2 * ind + 2, mid + 1, high, node, val);
                seg[ind] = Math.max(seg[2 * ind + 1], seg[2 * ind + 2]);
            }
        }

        public void point_update(int node, int val) {
            pointupdate2(0, 0, n - 1, node, val);
        }
    }

    /*-------------------------------------  FAST READER AND WRITER      FAST READER AND WRITER  ------------------------------------ */
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {

        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

}

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

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

Think step by step.

Response:
```plaintext
1
12
3 3 3 3 3 3 3 3 3 3 3 3
```

### Explanation:
This test case is designed to expose the bug in the buggy code. Here’s how it works:

1. **Input Description:**
   - The first line contains `1`, indicating there is one test case.
   - The second line contains `12`, indicating there are 12 sticks.
   - The third line contains the lengths of the sticks: `3 3 3 3 3 3 3 3 3 3 3 3`.

2. **Expected Output:**
   - The expected output is `4`, because you can form 4 regular polygons (each a triangle) with the 12 sticks of length 3.

3. **Buggy Code Behavior:**
   - The buggy code incorrectly calculates the number of polygons by using `f[i]/3` for each stick length `i`. This means it only considers triangles (3-sided polygons) and does not account for other possible polygons (like squares, pentagons, etc.).
   - In this test case, the buggy code will output `0` because it does not correctly handle the case where multiple sticks of the same length can form polygons with more than 3 sides.

4. **Correct Approach:**
   - The correct approach should consider all possible regular polygons (triangles, squares, pentagons, etc.) and use the maximum number of sticks to form the largest possible polygons first, then move to smaller polygons.

This test case is designed to highlight the failure of the buggy code in handling multiple sticks of the same length to form polygons with more than 3 sides.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

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

public class Main {
  public static void main(String[] args) {

    var sc = new Scanner(System.in);

    var t = sc.nextInt();

    while (t-- > 0) {
      var n = sc.nextInt();
      int[] nums = new int[n];

      for (var i=0; i < n; i++)
        nums[i] = sc.nextInt();

      var map = addToMap(nums);
      var cnt = 0;

      for (var v : map.values())
        if (v == 3 || v == 4 || v == 6 || v == 8 || v == 5)
          cnt++;

      System.out.println(cnt);
    }

  }

  static HashMap<Integer, Integer> addToMap(int[] nums) {
    var map = new HashMap<Integer, Integer>();

    for (var num : nums) {
      if (!map.containsKey(num))
        map.put(num, 1);
      else {
        var index = map.get(num);
        map.put(num, ++index);
      }
    }

    return map;
  }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the logic that counts the number of valid polygons. The code incorrectly assumes that any count of sticks that is 3, 4, 5, 6, or 8 can form a polygon. This is incorrect because a polygon can be formed by any number of sticks that is a multiple of 3 (for triangles), 4 (for squares), 5 (for pentagons), 6 (for hexagons), etc.

To create a challenging test input, we should include stick counts that are multiples of these numbers but are not directly 3, 4, 5, 6, or 8. For example, 9 sticks of the same length can form 3 triangles, but the buggy code will not count this correctly.

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

```plaintext
1
9
1 1 1 1 1 1 1 1 1
```

Explanation:
- The input consists of 9 sticks, all of length 1.
- According to the correct logic, these 9 sticks can form 3 triangles (since 9 is a multiple of 3).
- However, the buggy code will only check if the count is exactly 3, 4, 5, 6, or 8, and it will miss the fact that 9 can form 3 triangles.
- The correct output should be 3, but the buggy code will output 0.

This test input should effectively expose the bug 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: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include <bits/stdc++.h>
#define all(x) x.begin(),x.end()
#define pp push_back
#define ones(x) __builtin_popcountll(x)
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
// 32bit(1)=2147483647
using namespace std;
#define ll  long long
bool  const sorted (const pair<long long,long long>&p1,const pair<long long,long long>&p2){
    if(p1.second==p2.second){return p1.first>p2.first;}
   // else{return p1.first<p2.first;}
    else{return p1.second>p2.second;}
}
int fastpower(int x,int n){
    int ans=1;
    while(n>0){
        if(n%2==1){ans*=x;}
        x*=x;
        n/=2;
    }
    return ans;
}
vector<ll>v; map<ll,ll>mp;
vector<ll> piles(ll n,vector<ll> &v,map<ll,ll>&mp){
    if(n%3==0){
       v.pp(n/3);
       v.pp(((n/3)*2));
       if(mp[n/3]<1){ piles((n/3),v,mp);mp[n/3]++;}
       if(mp[((n/3)*2)]<1){ piles(((n/3)*2),v,mp);mp[((n/3)*2)]++;}

    }
    return v;

}

void solve() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
   ll t;cin>>t;
   while(t--) {
      ll n;cin>>n;
      map<ll,ll>mp;
       for (long long i = 0; i < n; ++i) {
           ll x;cin>>x;mp[x]++;
       }
       ll ans=0;
       for(auto it:mp){
           ll v=it.second;
           if(v<4){ continue;}
           else if(v>=4&&v<7){ans++;}
           else if (v==7){ans+=2;}
           else{ans+=2;v-=7;ans+=(v/2);}
       }
       cout<<ans<<'\n';
   }
}
int main(){

    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 given code, we need to focus on edge cases that the code might not handle correctly. Here are some potential edge cases to consider:

1. **Single Stick:** The code should correctly handle the case where there is only one stick.
2. **Two Sticks:** The code should correctly handle the case where there are only two sticks.
3. **Sticks with Lengths that Form Only One Polygon:** The code should correctly handle the case where the sticks can form exactly one polygon.
4. **Sticks with Lengths that Form Multiple Polygons:** The code should correctly handle the case where the sticks can form multiple polygons.
5. **Sticks with Lengths that Form a Combination of Different Polygons:** The code should correctly handle the case where the sticks can form a combination of different polygons.

Given the code's logic, it seems to have issues with correctly handling the distribution of sticks to form multiple polygons, especially when the number of sticks of a particular length is large.

Let's create a test case that includes a combination of different lengths and quantities that should form multiple polygons but might be mishandled by the code.

```plaintext
3
1
1
2
1 1
9
4 2 2 2 2 4 2 4 4
```

This test case includes:
1. A single stick (should output 0).
2. Two sticks (should output 0).
3. A set of sticks that can form multiple polygons (should output 2, but might be mishandled by the code).

However, to make it more challenging, let's add a test case with a larger number of sticks that can form multiple polygons but in a more complex distribution:

```plaintext
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
12
5 5 5 5 5 5 5 5 5 5 5 5
```

This test case includes:
1. A single stick (should output 0).
2. Two sticks (should output 0).
3. A set of sticks that can form one polygon (should output 1).
4. A set of sticks that can form two polygons (should output 2).
5. A set of sticks that can form multiple polygons (should output 2, but might be mishandled by the code).

The last test case with 12 sticks of length 5 should form 2 hexagons (6 sides each), but the code might mishandle the distribution of these sticks.

Here is the final test input:

```plaintext
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
12
5 5 5 5 5 5 5 5 5 5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

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

constexpr auto sp = " ";
constexpr auto en = "\n";

using ull = unsigned long long;
using ll = long long;
using ld = long double;

#define isz(x) int((x).size())
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define fin(x) for (auto& it : x) cin >> it;
#define foutsp(x) for (auto& it : x) cout << it << " "; cout << en;
#define fouten(x) for (auto& it : x) cout << it << en;
#define fyesno(x) cout << (x ? "YES" : "NO") << en;

using namespace std;

void solve();
void global();

ull gcd(ull a, ull b) {
    return b == 0 ? a : gcd(b, a % b);
}

ull lcm(ull a, ull b) {
    return a * b / gcd(a, b);
}

vector<int> to_z(string& s, bool flag = false) {
    vector<int> z(isz(s));
    if (flag) z[0] = isz(z);
    int l = 0, r = 1;
    for (int i = 1; i < isz(z); i++) {
        if (r > i) z[i] = min(z[i - l], r - i);
        while (z[i] + i < isz(z) && s[z[i]] == s[z[i] + i]) z[i]++;
        if (r < z[i] + i) { l = i; r = z[i] + i; }
    }
    return z;
}

ull get_c(ull n, ull k) {
    ull r = 1, dif = n - k;
    for (int i = 1; i <= k; i++)
        r += r * dif / i;
    return r;
}

string pi_str = "31415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679";

double get_pi() {
    return acos(-1);
}

int my_rand(int left, int right) {
    return rand() % (right - left + 1) + left;
}

void test() {
    solve();
}

void tests() {
    int t; cin >> t;
    while (t--) solve();
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    cout.setf(ios::fixed); cout.precision(12);

    srand(time(0));
    global();
    tests();

    return 0;
}

void global() {

}

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

    map<int, int> d;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;

        d[x]++;
    }

    int answer = 0;
    for (auto& p : d) {
        int x = 3;

        while (p.second - x >= 0) {
            p.second -= x++;
            answer++;
        }
    }

    cout << answer << en;
}

**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 code, we need to focus on edge cases where the code might fail. The key issue in the provided code is that it tries to form polygons by decrementing the count of sticks by increasing side lengths (3, 4, 5, etc.), which can lead to suboptimal solutions.

### Step-by-Step Analysis:
1. **Single Stick Case**: This is already handled correctly.
2. **Two Sticks Case**: This is also handled correctly.
3. **Multiple Sticks of the Same Length**: The code might fail if there are multiple sticks of the same length but the number of sticks is not a multiple of 3, 4, 5, etc. For example, if there are 8 sticks of length 1, the code might not form the optimal number of polygons.
4. **Mixed Lengths**: The code might fail if there are sticks of different lengths, and the optimal solution involves forming polygons with different side lengths.

### Challenging Test Input:
Let's create a test case where there are multiple sticks of the same length, but the number of sticks is not a multiple of 3, 4, 5, etc. This will expose the bug in the code.

### Test Input:
```plaintext
1
8
1 1 1 1 1 1 1 1
```

### Explanation:
- **Input**: The test case consists of 8 sticks, all of length 1.
- **Expected Output**: The optimal solution is to form 2 squares (4 sticks each), resulting in 2 polygons.
- **Code's Output**: The provided code will form 1 triangle (3 sticks), 1 square (4 sticks), and then fail to form any more polygons, resulting in 2 polygons.

This test case will expose the bug because the code does not form the optimal number of polygons by trying to form polygons with increasing side lengths.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
t=int(input())
while t:
    n=int(input())
    side_length=[]
    count={}
    x=input().split()
    side_length=list(x)
    for i in side_length:
        if i in count:
            count[i]=count[i]+1
        else:
            count[i]=1
    sum=0
    for i in count:
        if count[i]>=3:
            sum=sum+1

    print(sum)

    t=t-1

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

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

Think step by step.

Response:
```plaintext
1
12
1 1 1 1 1 1 2 2 2 2 2 2
```

### Explanation:
This test case is designed to expose the bug in the buggy code. The buggy code incorrectly increments the sum by 1 for each stick length that appears 3 or more times, without considering the actual number of polygons that can be formed.

### Analysis:
- The input has 12 sticks with lengths: 6 sticks of length 1 and 6 sticks of length 2.
- The correct output should be 2, because:
  - You can form two triangles (each using 3 sticks of the same length).
  - One triangle can be formed with the 6 sticks of length 1.
  - Another triangle can be formed with the 6 sticks of length 2.
- The buggy code, however, will output 2 (once for each length), but it will not correctly account for the fact that each length can form more than one polygon.

This test case should help identify the issue with the buggy code's logic for counting the number of polygons.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
from sys import stdin
def input(): return stdin.readline()

def solve():
    l=int(input())
    a = list(map(int, input().split()))
    c=[0]*100
    i=0
    while(i<l):
        c[a[i]]=c[a[i]]+1
        i=i+1
    i=0
    count=0
    while(i<l):
        if(c[i]>=4):
            count=count+1
        i=i+1
    print(count)

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

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that the code might handle incorrectly. The primary issue in the code is that it only checks if the count of any stick length is greater than or equal to 4, which is not sufficient to determine the maximum number of regular polygons that can be formed.

Here are some key points to consider:
1. **Multiple Stick Lengths**: The code should handle cases where multiple stick lengths can form different polygons.
2. **Polygon Sides**: The code should correctly account for the number of sides needed for different polygons (e.g., 3 for a triangle, 4 for a square, 5 for a pentagon, etc.).
3. **Optimal Usage**: The code should maximize the number of polygons formed, not just count the sticks that can form a square.

Let's create a test case that exposes these issues:

1. **Multiple Stick Lengths**: Include sticks of different lengths that can form different polygons.
2. **Optimal Usage**: Ensure that the test case can form multiple polygons, and the optimal solution is not just forming squares.

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

```plaintext
3
6
2 2 2 2 2 2
9
2 2 2 2 2 2 4 4 4
12
2 2 2 2 2 2 3 3 3 3 3 3
```

### Explanation:
1. **Test Case 1**:
   - Sticks: 2, 2, 2, 2, 2, 2
   - Expected Output: 2
   - Reason: We can form two hexagons (6 sides each).

2. **Test Case 2**:
   - Sticks: 2, 2, 2, 2, 2, 2, 4, 4, 4
   - Expected Output: 2
   - Reason: We can form one hexagon (6 sides) with the 2s and one triangle (3 sides) with the 4s.

3. **Test Case 3**:
   - Sticks: 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3
   - Expected Output: 3
   - Reason: We can form two hexagons (6 sides each) with the 2s and one hexagon (6 sides) with the 3s.

This test input should help expose the bug in the given code by requiring it to handle multiple stick lengths and form different polygons optimally.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
import java.util.*;

public class A940 {
    static class Pair implements Comparable<Pair>{
        int first;
        int second;
        public Pair(int first, int second){
            this.first=first;
            this.second=second;
        }
        @Override
        public int compareTo(Pair other){
            return this.first-other.first;
        }
    }
    public static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
    public static int gcdOfArray(int[] arr) {
        int result = arr[0];
        for (int i = 1; i < arr.length; i++) {
            result = gcd(result, arr[i]);
        }
        return result;
    }
    public static int findMax(int[] arr) {
        int max = arr[0];
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] > max) {
                max = arr[i];
            }
        }
        return max;
    }
    public static int findMin(int[] arr) {
        int min = arr[0];
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] < min) {
                min = arr[i];
            }
        }
        return min;
    }
    public static int[] update(int[] arr, int tickets, int day){
        for (int i = day; i < arr.length; i++) {
            arr[i]+=tickets;
        }
        return arr;
    }
    public static int Solve(int n, int[] arr){
        if (n==1 || n==2) {
            return 0;
        }
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            map.put(arr[i], map.getOrDefault(arr[i], 0)+1);
        }
        int count = 0;
        for (int value : map.values()) {
            if (value > 2) {
                count++;
            }
        }
        return count;
    }
    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n= sc.nextInt();
            int[] arr= new int[n];
            for (int j = 0; j < n; j++) {
                arr[j]= sc.nextInt();
            }
            System.out.println(Solve(n ,arr));
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3
3 3 3
4
1 1 1 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: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
import java.lang.*;
import java.util.*;
public class A
{
      public static void main(String args[])
      {
            Scanner s=new Scanner(System.in);
            int t=s.nextInt();
            while(t-->0)
            {
                  int n=s.nextInt();int a[]=new int[n];int b[]=new int[n];
                  // if(n<3)
                  // System.out.println(0);
                  // else{
                  for(int i=0;i<n;i++)
                  {
                        a[i]=s.nextInt();
                  }
                  if(n<3)
                  System.out.println(0);
                  else{
                  Arrays.sort(a);int c=1,k=0;
                  for(int i=0;i<n-1;i++)
                  {
                        if(a[i]==a[i+1])
                        c++;
                        else{

                              b[k]=c;
                              k=k+1;
                              c=1;
                        }
                  }
                  b[k]=c;
                  int no=0;
                  for(int i=0;i<n;i++)
                  {
                        if(b[i]>=3)
                        no++;
                  }
                  System.out.println(no);
                  }

            }
      }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include <bits/stdc++.h>
#define	endl '\n'
#define	int long long
using namespace std;
const int N=1e5+5;
const int mod=998244353;
const int INF=0x3f3f3f3f;
int arr[N];
signed main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		int n,cnt=0;
		cin>>n;
		priority_queue<int,vector<int>,greater<int>> q;
		for(int i=0;i<n;i++)
		{
			int a;
			cin>>a;
			q.push(a);
		}
		if(n<=2)
		{
			cout<<0<<endl;
			continue;
		}
		while(q.size()>=3)
		{
			int arr[5];
			for(int i=1;i<=3;i++)
			{
				arr[i]=q.top();
				q.pop();
			}
			if(arr[1]==arr[2]&&arr[2]==arr[3])
				cnt++;
		}
		cout<<cnt<<endl;
	}

	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the code might fail. The code currently only considers the case where three sticks of the same length are used to form a polygon. However, it does not account for the possibility of forming polygons with more than three sides (e.g., squares, pentagons, etc.).

### Steps to Create a Challenging Test Input:

1. **Test with a Single Stick:**
   - This should output `0` because a single stick cannot form a polygon.
   - Input: `1\n1`

2. **Test with Two Sticks:**
   - This should also output `0` because two sticks cannot form a polygon.
   - Input: `1\n2\n1 1`

3. **Test with Three Sticks of Different Lengths:**
   - This should output `0` because three sticks of different lengths cannot form a regular polygon.
   - Input: `1\n3\n1 2 3`

4. **Test with Three Sticks of the Same Length:**
   - This should output `1` because three sticks of the same length can form a triangle.
   - Input: `1\n3\n1 1 1`

5. **Test with Multiple Sticks of the Same Length:**
   - This should output the number of triangles that can be formed.
   - Input: `1\n9\n1 1 1 1 1 1 1 1 1`

6. **Test with Sticks that Can Form a Square:**
   - This should output `1` because four sticks of the same length can form a square.
   - Input: `1\n4\n2 2 2 2`

7. **Test with Sticks that Can Form a Pentagon:**
   - This should output `1` because five sticks of the same length can form a pentagon.
   - Input: `1\n5\n3 3 3 3 3`

8. **Test with Mixed Lengths:**
   - This should output the maximum number of polygons that can be formed.
   - Input: `1\n8\n2 2 2 2 3 3 3 3`

### Challenging Test Input:
```plaintext
1
8
2 2 2 2 3 3 3 3
```

This input will challenge the code to handle the case where there are enough sticks to form multiple polygons, but the code only considers forming triangles. The expected output should be `2` (one square and one triangle), but the code will likely output `1` because it only checks for triangles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define vi vector<long long >
#define vpi vector<pair<long long, long long >>
#define vvi vector<vector<int>>
#define sl set<ll>;
#define f(a,i) a[i].second
#define f(i, a, b) for (ll i = a; i < b; i++)
#define fors(n) for (ll  i = 0; i < n; i++)
#define show(c) cout << c << "\n";
#define spcshow(c) cout << c << ' ';
//const ll m1 = 1e12 + 7;
//const int m2 = 998244353;
const int big = 1e6;
#define min3(a, b, c) min(c, min(a, b)),
#define min4(a, b, c, d) min(d, min(c, min(a, b))),
#define newline cout<<endl;
long long  factorial(int n)
{
      if(n==0)
           return 1;
      else
           return(n*factorial(n-1));
}

long long  ncr(long long  n,long long  r)
{
      long long  value;
      value=factorial(n)/(factorial(r)*factorial(n-r));
      return value;
}
long long len(long long x){
    ll p = int(log10(x) + 1);
    return p;
}
ll power(ll a, ll b, ll mod = 0){
    ll result = 1;
    if(mod == 0){
        while(b){
            if(b & 1) result *= a;
            a *= a;
            b = b >> 1;
        }
    }
    else {
        while(b){
            if(b & 1){
                result *= a;
                result  = result % mod;
            }
            a *= a;
            a = a % mod;
            b = b >> 1;
        }
    }
    return result;
}
bool checkprime(ll n)
{
    for(ll i=2;i*i<=n;i++)
    {
            if(n==4)
            {return false;}
        if(n%i==0)
        {
            return false;
        }
    }
    return true;
}
//Code starts Here
//Dont forget while(t--) for questions which donot involve it.
//made by kushal.
void solve()
{
  ll n;
  cin>>n;
  vi a(n);
  map<ll,ll>m;
  fors(n){
    cin>>a[i];
    m[a[i]]++;
  }
  ll ans=0;
  auto it=m.begin();
  while(it!=m.end()){
    if(it->second>2)ans++;
    it++;
  }
  show(ans)
}

int main()
{
 ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll t = 1;
    cin >> t;

    while(t--){
        solve();
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The current code has a significant flaw: it only checks if the frequency of each stick length is greater than 2 and increments the answer accordingly. This approach is incorrect because it does not consider the number of sides required to form a regular polygon.

### Steps to Generate the Test Input:
1. **Edge Case: Single Stick**
   - A single stick cannot form a polygon.
   - Input: `1 1`

2. **Edge Case: Two Sticks**
   - Two sticks cannot form a polygon.
   - Input: `2 1 1`

3. **Edge Case: Three Sticks of Different Lengths**
   - Three sticks of different lengths cannot form a polygon.
   - Input: `3 1 2 3`

4. **Edge Case: Multiple Sticks of the Same Length**
   - A large number of sticks of the same length should form multiple polygons.
   - Input: `12 3 3 3 3 3 3 3 3 3 3 3 3` (Should form 3 squares)

5. **Edge Case: Mixed Sticks with Possible Polygons**
   - A combination of sticks that can form multiple polygons.
   - Input: `9 4 2 2 2 2 4 2 4 4` (Should form 2 polygons: one pentagon and one square)

6. **Edge Case: Sticks with Frequencies Just Enough for Polygons**
   - Sticks with frequencies just enough to form a single polygon.
   - Input: `4 3 3 3 3` (Should form 1 square)

### Chosen Test Input:
```plaintext
6
1 1
2 1 1
3 1 2 3
12 3 3 3 3 3 3 3 3 3 3 3 3
9 4 2 2 2 2 4 2 4 4
4 3 3 3 3
```

This test input covers a range of edge cases and should effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
x = int(input())
for _ in range(x):
    y = int(input())
    z = list(map(int, input().split()))
    z.sort()
    z2 = z[:]
    sum = 0
    for i in range(len(z)-1):
        if z[i] == z[i+1]:
            z2.remove(z[i])
    for k in z2:
        if z.count(k)>2:
            sum += 1
    print(sum)

**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 buggy code, let's analyze the issues in the code:

1. **Sorting and Removing Elements**: The code sorts the list `z` and then creates a copy `z2`. It then iterates through `z` and removes elements from `z2` if they are equal to the next element. This means that if there are consecutive equal elements, they will be removed from `z2`.

2. **Counting and Summing**: The code then iterates through `z2` and counts how many times each element appears in `z`. If the count is greater than 2, it increments `sum`.

The main issue with this approach is that it does not correctly handle the formation of regular polygons. It only counts elements that appear more than twice, but it does not consider the actual formation of polygons (e.g., triangles, squares, etc.).

To expose this bug, we need a test case where the code incorrectly counts the number of polygons. A good candidate is a test case where there are multiple sticks of the same length, but the number of sticks is not sufficient to form the expected number of polygons.

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

```plaintext
1
6
2 2 2 2 2 2
```

### Explanation:
- **Test Case**: The test case has 6 sticks, all of length 2.
- **Expected Output**: The expected output should be 1, because we can form one hexagon (6-sided polygon) with the 6 sticks.
- **Buggy Code Output**: The buggy code will likely output 0 or an incorrect number because it does not correctly handle the formation of polygons.

This test case should clearly show that the code is not correctly identifying the formation of regular polygons.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
T = int(input())
for t in range(T):
    n = int(input())
    l = [int(i) for i in input().split()]
    # print(l)
    freq = {}
    for i in l:
        freq[i] = freq.get(i, 0)+1
    count  = 0
    for k, v in freq.items():
        if v > 2:
            count += 1
    print(count)

**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
3
1 1 1
4
2 2 2 2
5
3 3 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
import java.util.Scanner;

public class Stickogon {
    private static int solve(int n, int[] sticks) {
        int[] freq = new int[101];

        int res = 0;

        for (int stick : sticks) {
            freq[stick]++;

            if (freq[stick] == 3) {
                res++;
            }
        }

        return res;
    }

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

        int T = scan.nextInt();

        for (int t = 0; t < T; t++) {
            int n = scan.nextInt();

            int[] sticks = new int[n];

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

            System.out.println(solve(n, sticks));
        }

        scan.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
1
1
2
1 1
6
2 2 3 3 3 3
9
4 2 2 2 2 4 2 4 4
```

This test input includes a variety of cases that cover different scenarios:

1. **Single Stick Case**: A single stick cannot form a polygon.
2. **Two Sticks Case**: Two sticks cannot form a polygon.
3. **Multiple Sticks Case with a Square**: Six sticks with lengths that can form a square (using the four sticks of length 3).
4. **Multiple Sticks Case with Multiple Polygons**: Nine sticks with lengths that can form a pentagon (using the five sticks of length 2) and a square (using the four sticks of length 4).

The buggy code only counts polygons when a stick length appears exactly three times, which is incorrect. This test input will expose the bug by showing that the code fails to count polygons correctly in the third and fourth test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Stickogon
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given $$$n$$$ sticks of lengths $$$a_1, a_2, \ldots, a_n$$$. Find the maximum number of regular (equal-sided) polygons you can construct simultaneously, such that:

- Each side of a polygon is formed by exactly one stick.
- No stick is used in more than $$$1$$$ polygon.

Note: Sticks cannot be broken.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the number of sticks available.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 100$$$) — the stick lengths.

Output: For each test case, output a single integer on a new line — the maximum number of regular (equal-sided) polygons you can make simultaneously from the sticks available.

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

Note: In the first test case, we only have one stick, hence we can't form any polygon.

In the second test case, the two sticks aren't enough to form a polygon either.

In the third test case, we can use the $$$4$$$ sticks of length $$$3$$$ to create a square.

In the fourth test case, we can make a pentagon with side length $$$2$$$, and a square of side length $$$4$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class lol{
     static Scanner input = new Scanner(System.in);
    static PrintWriter output = new PrintWriter(System.out);

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

      int n=input.nextInt();
      while (n-->0) {
        int length=input.nextInt();
        int[] nums=new int[length];
        HashMap<Integer,Integer> hmap=new HashMap<>();
        int[] count=new int[1];
        for(int i=0;i<length;i++) {
            nums[i]=input.nextInt();
            hmap.put(nums[i],hmap.getOrDefault(nums[i],0)+1);
        }
        hmap.forEach((key,value)->{
            count[0]+=value>2? 1:0;
        });
        output.println(count[0]);

      }

          output.flush();
        }

     //**********************Scanner&helpers**************\\
        static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }
        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }
        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }
        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
       public String nextLine() throws IOException {
            return br.readLine();
        }
        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

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

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

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        public boolean ready() throws IOException {
            return br.ready();
        }
    }
    static class Pair implements Comparable<Pair> {
        long first;
        long second;

        public Pair(long first, long second) {
            this.first = first;
            this.second = second;
        }

        public int compareTo(Pair p) {
            if (first != p.first)
                return Long.compare(first, p.first);
            else if (second != p.second)
                return Long.compare(second, p.second);
            else
                return 0;
        }
        public String toString() {
            return first + " " + second;

        }
    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to consider the logic and potential issues in the code. The main issue in the provided code is that it simply counts the number of stick lengths that occur more than twice, without considering the actual formation of regular polygons. This means the code will incorrectly count any stick length that appears more than twice as a valid polygon, regardless of the number of sides required for a regular polygon.

### Steps to Generate the Test Input:
1. **Understand the Bug**: The bug is in the logic that counts the number of stick lengths that occur more than twice. This does not correctly account for the number of sides required to form a regular polygon.
2. **Edge Cases**:
   - **Single Stick**: This should return 0.
   - **Two Sticks**: This should return 0.
   - **Three Sticks of the Same Length**: This should return 1 (a triangle).
   - **Four Sticks of the Same Length**: This should return 1 (a square).
   - **More Than Four Sticks of the Same Length**: This should return the maximum number of polygons that can be formed.
   - **Mixed Lengths**: This should correctly count only the valid polygons.
3. **Construct a Test Case**:
   - Use a test case with a mix of stick lengths, including some that can form multiple polygons and some that cannot.

### Test Input:
```plaintext
4
1
1
2
1 1
4
1 1 1 1
9
4 2 2 2 2 4 2 4 4
```

### Explanation:
1. **Test Case 1**:
   - 1 stick of length 1. Expected output: 0 (cannot form any polygon).
2. **Test Case 2**:
   - 2 sticks of length 1. Expected output: 0 (cannot form any polygon).
3. **Test Case 3**:
   - 4 sticks of length 1. Expected output: 1 (can form a square).
4. **Test Case 4**:
   - 9 sticks with lengths 4, 2, 2, 2, 2, 4, 2, 4, 4. Expected output: 2 (can form a square with side length 4 and a pentagon with side length 2).

This test input should effectively expose the bug in the provided code, as the code will incorrectly count the number of polygons based on the frequency of stick lengths rather than the actual number of sides required.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Orangutan Approved Subarrays
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:

  * In one operation, select indices $l$ and $r$ ($1 \leq l \leq r \leq |b|$) such that $v = b_l = b_{l+1} = \ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \leq i \leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \ldots$ as $b_l, b_{l+1}, \ldots$ accordingly.

You are given an array $a$ of length $n$ and $q$ queries.

Each query consists of two indices $l$ and $r$ ($1 \le l \le r \le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \ldots, a_r$ is orangutan-approved.

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

The first line of each test case contains integers $n$ and $q$ ($1 \leq n,q \leq 2 \cdot 10^5$) — the size of $a$ and the number of queries, respectively.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array $a$.

The following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \leq l \leq r \leq n$).

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

Output: For each query, output "YES" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3
4 2
1 2 2 1
1 4
1 3
5 3
1 2 1 2 1
2 5
3 5
1 3
8 4
1 2 3 2 1 3 2 3
1 5
2 8
3 5
6 8
output:
YES
YES
NO
YES
YES
YES
NO
YES
YES

Note: In the first query of the first testcase, the answer is YES.

  * Initially, $S=\\{1,2\\}$ and $b=[1,2,2,1]$    * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\{1\\}$ and the array becomes $[1,1]$.    * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\{\\}$ and the array becomes $[]$.    * Since the array is now empty, we can say the original array is orangutan-approved.

In the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define int long long
typedef long long ll;
int INF = 1e18;
const int N = 2e5+10;

//int mod =  1e9+7;
//int ways[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};

int n,q;
int oc[N],a[N],nex[N],ac[N];
vector<int> have[N];

#define se second
#define fi first

typedef struct{
    #define ls(x) (x<<1)
    #define rs(x) (x<<1|1)
    const int MAXN = N;

    typedef struct info{
        int val;
    }info;

    typedef struct{
        info val;
    }Node;

    friend info operator + (const info &a,const info &b){
        info c;
        c.val = max(a.val , b.val);
        return c;
    };

    Node tree[N*4];
    int input[N];

    void update(int p) {
        tree[p].val = tree[ls(p)].val + tree[rs(p)].val;
    }

    void build(int p,int l,int r){
        if(l==r) tree[p] = {input[l]};
        else{
            int mid = (l+r)>>1;
            build(ls(p),l,mid);
            build(rs(p),mid+1,r);
            update(p);
        }
    }

    info ask(int p, int l,int r,int tl, int tr) {
        if (tl <= l and r <= tr) return tree[p].val;
        int mid = l + r >> 1;
        if (tr <= mid) return ask(ls(p),l,mid,tl, tr);
        else if (tl > mid) return ask(rs(p),mid+1,r,tl, tr);
        else return ask(ls(p),l,mid,tl, tr) + ask(rs(p),mid+1,r,tl, tr);
    }

    void change(int p, int val,int l,int r,int tl,int tr) {
        if (l == r) {
            input[tl] = val;
            tree[p].val = {val};
            return;
        }
        int mid = (l+r) >> 1;
        if (tl <= mid) change(ls(p), val,l,mid, tl,tr);
        else change(rs(p), val,mid+1,r, tl,tr);
        update(p);
    }
}SGT;

SGT tree;

void solve() {
    cin>>n>>q;

    for(int i = 0;i<=n+1;i++){
        have[i].clear();
        nex[i] = ac[i] = oc[i] = 0;
    }

    for(int i = 1;i<=n;i++) cin>>a[i];

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

    for(int i = n;i>=1;i--){
        if(a[i] == a[i+1]) nex[i] = nex[i+1];
        else nex[i] = oc[a[i]];

        have[nex[i]].push_back(i);
        oc[a[i]] = i;
    }

    for(int i = 1;i<=n;i++) tree.input[i] = nex[i];
    tree.build(1,1,n);

    int l = n;
    for(auto i:have[n+1]) tree.change(1,0,1,n,i,i);

    for(int r = n;r>=1;r--){
        while(l >= 1 && tree.ask(1,1,n,l,nex[l]).val <= nex[l]){
            l--;
        }

        for(auto i:have[r]) tree.change(1,0,1,n,i,i);
        ac[r] = l+1;
    }

    int r;
    while(q--){
        cin>>l>>r;

        if(ac[r] <= l) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }

}
signed main() {
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //cout<<fixed<<setprecision(10);

    int caset = 1;
    cin >> caset;
    while (caset--) { solve(); }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Orangutan Approved Subarrays
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:

  * In one operation, select indices $l$ and $r$ ($1 \leq l \leq r \leq |b|$) such that $v = b_l = b_{l+1} = \ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \leq i \leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \ldots$ as $b_l, b_{l+1}, \ldots$ accordingly.

You are given an array $a$ of length $n$ and $q$ queries.

Each query consists of two indices $l$ and $r$ ($1 \le l \le r \le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \ldots, a_r$ is orangutan-approved.

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

The first line of each test case contains integers $n$ and $q$ ($1 \leq n,q \leq 2 \cdot 10^5$) — the size of $a$ and the number of queries, respectively.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array $a$.

The following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \leq l \leq r \leq n$).

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

Output: For each query, output "YES" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3
4 2
1 2 2 1
1 4
1 3
5 3
1 2 1 2 1
2 5
3 5
1 3
8 4
1 2 3 2 1 3 2 3
1 5
2 8
3 5
6 8
output:
YES
YES
NO
YES
YES
YES
NO
YES
YES

Note: In the first query of the first testcase, the answer is YES.

  * Initially, $S=\\{1,2\\}$ and $b=[1,2,2,1]$    * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\{1\\}$ and the array becomes $[1,1]$.    * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\{\\}$ and the array becomes $[]$.    * Since the array is now empty, we can say the original array is orangutan-approved.

In the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.

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

using pii = pair<int,int>;

#define int long long

const int N = 2e5 + 10;

struct Info {
    int mx;

    Info operator+(const Info &o) const {
        Info ret;
        ret.mx = max(mx, o.mx);
        return ret;
    }
} tr[N << 2];

int pre[N], pre1[N];

void push_up(int p) {
    tr[p] = tr[p << 1] + tr[p << 1|1];
}

void build(int s, int t, int p) {
    if (s == t) {
        tr[p].mx = pre1[s];
        return;
    }
    int mid = s + t >> 1;
    build(s,mid,p << 1);
    build(mid + 1,t,p << 1|1);
    push_up(p);
}

void modify(int k, int c, int s, int t, int p) {
    if (s == t) {
        tr[p].mx = c;
        return;
    }
    int mid = s + t >> 1;
    if (k <= mid) modify(k,c,s,mid,p << 1);
    else modify(k,c,mid + 1,t,p << 1|1);
    push_up(p);
}

Info query(int l, int r, int s, int t, int p) {
    if (l <= s && t <= r) {
        return tr[p];
    }
    int mid = s + t >> 1;
    if (l > mid) return query(l,r,mid + 1,t,p << 1|1);
    if (r <= mid) return query(l,r,s,mid,p << 1);
    return query(l,r,s,mid,p << 1) + query(l,r,mid + 1,t,p << 1|1);
}

void solve() {
    int n, q; cin >> n >> q;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    for (int i = 1; i <= n; i++) {
        modify(1,0,1,n,1);
        pre1[i] = pre[i] = 0;
    }

    vector<int> f(n + 1);
    vector<pii> b;
    for (int i = 1; i <= n; i++) {
        pre[i] = f[a[i]];
        f[a[i]] = i;
        // cout << pre[i] << " \n"[i == n];
        b.push_back({pre[i], i});
    }

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

    for (auto [pr, i] : b) {
        if (pr + 1 > i - 1) {
            pre1[i] = 0;
            modify(i,pr,1,n,1);
            continue;
        }
        Info ret = query(pr + 1, i - 1,1,n,1);
        pre1[i] = ret.mx;
        modify(i,pr,1,n,1);
    }

    build(1,n,1);
    while (q--) {
        int l,r; cin >> l >> r;
        if (query(l,r,1,n,1).mx >= l) cout << "NO\n";
        else cout << "YES\n";
    }

}

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Orangutan Approved Subarrays
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you have an array $b$. Initially, you also have a set $S$ that contains all distinct elements of $b$. The array $b$ is called orangutan-approved if it can be emptied by repeatedly performing the following operation:

  * In one operation, select indices $l$ and $r$ ($1 \leq l \leq r \leq |b|$) such that $v = b_l = b_{l+1} = \ldots = b_r$ and $v$ is present in $S$. Remove $v$ from $S$, and simultaneously remove all $b_i$ such that $l \leq i \leq r$. Then, reindex the elements $b_{r+1}, b_{r+2}, \ldots$ as $b_l, b_{l+1}, \ldots$ accordingly.

You are given an array $a$ of length $n$ and $q$ queries.

Each query consists of two indices $l$ and $r$ ($1 \le l \le r \le n$), and you need to determine whether or not the subarray $a_{l}, a_{l+1}, \ldots, a_r$ is orangutan-approved.

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

The first line of each test case contains integers $n$ and $q$ ($1 \leq n,q \leq 2 \cdot 10^5$) — the size of $a$ and the number of queries, respectively.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array $a$.

The following $q$ lines contain two integers $l$ and $r$ — the endpoints of the subarray for each query ($1 \leq l \leq r \leq n$).

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

Output: For each query, output "YES" (without quotes) if the subarray from $l$ to $r$ is orangutan-approved, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3
4 2
1 2 2 1
1 4
1 3
5 3
1 2 1 2 1
2 5
3 5
1 3
8 4
1 2 3 2 1 3 2 3
1 5
2 8
3 5
6 8
output:
YES
YES
NO
YES
YES
YES
NO
YES
YES

Note: In the first query of the first testcase, the answer is YES.

  * Initially, $S=\\{1,2\\}$ and $b=[1,2,2,1]$    * Select $l=2$ and $r=3$. Since $b_2=b_3=2$ is in $S$, we may erase $b_2$ and $b_3$ from the array, as well as erasing $2$ from $S$. The set $S$ becomes $\\{1\\}$ and the array becomes $[1,1]$.    * Select $l=1$ and $r=2$. Since $b_1=b_2=1$ is in $S$, we may erase $b_1$ and $b_2$ from the array, as well as erasing $1$ from $S$. The set $S$ becomes $\\{\\}$ and the array becomes $[]$.    * Since the array is now empty, we can say the original array is orangutan-approved.

In the first query of the second testcase, the answer is NO, because it can be shown that the subarray $[2,1,2,1]$ cannot become empty through any sequence of valid operations.

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

using namespace std;

typedef long long ll;

const int MAXN = 1e6 + 10;
const int mod = 998244353;

inline int add(int x, int y) { return x += y, x < mod || (x -= mod), x; }
inline void cadd(int &x, int y) { x += y, x < mod || (x -= mod); }

int T, n, m, a[MAXN], p[MAXN], pre[MAXN], nxt[MAXN], val[MAXN << 2], res[MAXN];

inline void pushup(int p) { val[p] = max(val[p << 1], val[p << 1 | 1]); }

void build(int l = 1, int r = n, int p = 1) {
	if (l == r) return val[p] = nxt[l], void(); int mid = l + r >> 1;
	build(l, mid, p << 1), build(mid + 1, r, p << 1 | 1), pushup(p);
}

void modify(int k, int l = 1, int r = n, int p = 1) {
	if (l == r) return val[p] = 0, void(); int mid = l + r >> 1;
	if (k <= mid) modify(k, l, mid, p << 1);
	else modify(k, mid + 1, r, p << 1 | 1); pushup(p);
}

int ask(int ql, int qr, int l = 1, int r = n, int p = 1) {
	if (l > r || r < ql || l > qr) return 0;
	if (ql <= l && r <= qr) return val[p]; int mid = l + r >> 1;
	return max(ask(ql, qr, l, mid, p << 1), ask(ql, qr, mid + 1, r, p << 1 | 1));
}

int __cnt = 0;

int main() {
	for (scanf("%d", &T); T--; ) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[i] = 0;
		for (int i = n; i; i--) nxt[i] = p[a[i]], pre[nxt[i]] = i, p[a[i]] = i; build();
		for (int i = n, j = n; i; i--) {
			for (; j > 1 && ask(j, nxt[j - 1] - 1) <= nxt[j - 1]; j--);
			res[i] = j; if (pre[i]) modify(pre[i]);
		}
		for (int l, r; m--; ) {
			scanf("%d%d", &l, &r);
			if (++__cnt == 8201) return printf("%d %d %d %d %d %d %d\n", a[1], a[2], a[3], a[4], a[5], l, r), 0;
			puts(res[r] <= l ? "YES" : "NO");
		}
	}
}

**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
1 2 1 2 1
2 5
3 5
1 3
```