**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.stream.IntStream;

/**
 * https://codeforces.com/problemset/problem/2004/F
 */
public final class CF2004F implements Runnable{

    public static void main(String[] args) {
        new Thread(null, new CF2004F(), "Main", 1<<28).start();
    }

    @Override
    public void run() {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        // write here
        int t = sc.nextInt();
        boolean mock = t == 422;
        for(;t>0;t--){
            int[]dict= IntStream.range(0,sc.nextInt()).map(i->sc.nextInt()).toArray();
            if(mock&&t==380){
                out.println(Arrays.toString(dict));
            }
            int n= dict.length;
            int[][]res=new int[n][n];
            for(int i=0;i<n;i++){
                for(int j=i;j<n;j++){
                    res[i][j]=j-i;
                }
            }
            //以i位置为中心
            for(int i=1;i<n-1;i++){
                int cost=0;
                int left=i-1,right=i+1;
                int leftSum=dict[left],rightSum=dict[right];
                while (true){
                    if(leftSum==rightSum){
                        res[left][right]=Math.min(res[left][right],cost);
                        if(left==0||right==n-1){
                            break;
                        }
                        leftSum+=dict[--left];
                        rightSum+=dict[++right];
                    }else if(leftSum<rightSum){
                        if(left==0){
                            break;
                        }
                        cost++;
                        leftSum+=dict[--left];
                    }else {
                        if(right==n-1){
                            break;
                        }
                        cost++;
                        rightSum+=dict[++right];
                    }
                }
            }
            //以i,i+1位置为中心
            for(int i=0;i<n-1;i++){
                int cost=0;
                int left=i,right=i+1;
                int leftSum=dict[left],rightSum=dict[right];
                while (true){
                    if(leftSum==rightSum){
                        res[left][right]=Math.min(res[left][right],cost);
                        if(left==0||right==n-1){
                            break;
                        }
                        leftSum+=dict[--left];
                        rightSum+=dict[++right];
                    }else if(leftSum<rightSum){
                        if(left==0){
                            break;
                        }
                        cost++;
                        leftSum+=dict[--left];
                    }else {
                        if(right==n-1){
                            break;
                        }
                        cost++;
                        rightSum+=dict[++right];
                    }
                }
            }
            //计算sum(a,b)=sum(c,d)的
            long ans=0;
            for(int i=0;i<n;i++){
                for(int j=i;j<n;j++){
                    ans+=res[i][j];
                }
            }
            if(!mock) {
                out.println(ans);
            }
        }
        out.close();
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

Think step by step.

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

### Explanation:

1. **Test Case 1:**
   - **Input:** `3 1 2 1`
   - **Output:** `1`
   - **Explanation:** The only non-palindromic subarray is `[1, 2, 1]`, which can be made palindromic by removing the middle element, resulting in one operation.

2. **Test Case 2:**
   - **Input:** `4 1 2 3 2`
   - **Output:** `5`
   - **Explanation:** The subarrays are:
     - `[1, 2, 3, 2]` (3 operations: `[1, 2, 5]` -> `[1, 7]` -> `[8]`)
     - `[1, 2, 3]` (2 operations: `[1, 5]` -> `[6]`)
     - `[2, 3, 2]` (1 operation: `[5]`)
     - `[1, 2]` (1 operation: `[3]`)
     - `[2, 3]` (1 operation: `[5]`)
     - `[3, 2]` (1 operation: `[5]`)
     - The total is `3 + 2 + 1 + 1 + 1 + 1 = 9`, but the code might not handle the sum correctly.

3. **Test Case 3:**
   - **Input:** `5 1 2 3 2 1`
   - **Output:** `6`
   - **Explanation:** The subarrays are:
     - `[1, 2, 3, 2, 1]` (2 operations: `[1, 2, 5, 1]` -> `[1, 7, 1]` -> `[8]`)
     - `[1, 2, 3, 2]` (3 operations: `[1, 2, 5]` -> `[1, 7]` -> `[8]`)
     - `[2, 3, 2, 1]` (3 operations: `[2, 5, 1]` -> `[7, 1]` -> `[8]`)
     - `[1, 2, 3]` (2 operations: `[1, 5]` -> `[6]`)
     - `[2, 3, 2]` (1 operation: `[5]`)
     - The total is `2 + 3 + 3 + 2 + 1 = 11`, but the code might not handle the sum correctly.

4. **Test Case 4:**
   - **Input:** `6 1 2 3 4 3 2`
   - **Output:** `14`
   - **Explanation:** The subarrays are:
     - `[1, 2, 3, 4, 3, 2]` (4 operations: `[1, 2, 3, 7, 2]` -> `[1, 2, 10, 2]` -> `[1, 12, 2]` -> `[13, 2]` -> `[15]`)
     - `[1, 2, 3, 4, 3]` (3 operations: `[1, 2, 3, 7]` -> `[1, 2, 10]` -> `[1, 12]` -> `[13]`)
     - `[2, 3, 4, 3, 2]` (3 operations: `[2, 3, 7, 2]` -> `[2, 10, 2]` -> `[12, 2]` -> `[14]`)
     - `[1, 2, 3, 4]` (2 operations: `[1, 2, 7]` -> `[1, 9]` -> `[10]`)
     - `[2, 3, 4, 3]` (2 operations: `[2, 7, 3]` -> `[9, 3]` -> `[12]`)
     - The total is `4 + 3 + 3 + 2 + 2 = 14`, but the code might not handle the sum correctly.

These test cases are designed to cover various scenarios, including palindromic subarrays, non-palindromic subarrays, and edge cases where the subarrays are close to being palindromic but require minimal operations to become palindromic. The small-scale nature of these test cases ensures that they are complete and valid while still being manageable in size.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

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

public class F {

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

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

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

	private static class Solution implements Runnable {

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

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

			out.flush();
		}

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

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
			// for BIT. Do not use tree maps with update query
			// Assuming Constraints on an unbound variable
			// Not properly calculating time complexity
		}

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

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

			boolean [][] pal = new boolean[n][n];

			for (int i = 0; i < n; ++i) {
				pal[i][i] = true;

				if (i < n - 1 && arr[i] == arr[i + 1]) {
					pal[i][i + 1] = true;
				}
			}

			for (int i = 2; i < n; ++i) {
				for (int a = 0; a + i < n; ++a) {
					if (arr[a] == arr[a + i] && pal[a + 1][a + i -1])
						pal[a][a + i] = true;
				}
			}

			Map<Integer , List<int []>> map = new HashMap<>();
			int sum;

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

			for (int i = 0; i < n; ++i) {
				sum = 0;

				for (int j = i; j < n; ++j) {
					if (pal[i][j]) {
						dp[i][j] = 0;
					}else {
						dp[i][j] = j - i;
					}

					sum += arr[j];

					if (!map.containsKey(sum))
						map.put(sum, new ArrayList<>());

					map.get(sum).add(new int [] {i, j});
				}
			}

			int [] a, b;

			int left, right, val;

			List<Integer> pos = new ArrayList<>(map.keySet());

			Collections.sort(pos);
			List<int []> list;

			for (int k : pos) {

				// System.out.println(k);

				list = map.get(k);

				int sIdx = 0;

				for (int i = 0; i < list.size(); ++i) {
					while (sIdx < list.size() && list.get(sIdx)[0] <= list.get(i)[1])
						++sIdx;

					a = list.get(i);

					for (int j = sIdx; j < list.size(); ++j) {
						b = list.get(j);
						right = b[1] - b[0];
						left = a[1] - a[0];

						dp[a[0]][b[1]] = Math.min( dp[a[1] + 1][b[0] - 1] + right + left ,
								dp[a[0]][b[1]]);
					}
				}
			}

			long result = 0;

			for (int i = 0; i < n; ++i) {
				for (int j = 0; j < n; ++j) {
					result += dp[i][j];
				}
			}

			// System.out.println(Arrays.deepToString(pal));
			// System.out.println(Arrays.deepToString(dp));

			// out.println(System.currentTimeMillis() - time);

			out.println(result);
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

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

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

    	  return res;
      }

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

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

    	  return res;
      }

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

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

    	  int a, b;

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

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

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

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

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

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

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

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

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to create such a test input:

1. **Edge Cases for Palindromes**: Test cases where the array is already a palindrome or close to being a palindrome.
2. **Large Values**: Test cases with large values to check for overflow issues.
3. **Special Patterns**: Test cases with special patterns that might not be handled correctly by the algorithm.
4. **Small and Large Subarrays**: Test cases with both small and large subarrays to ensure the algorithm handles all sizes correctly.

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

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

### Explanation:
1. **Test Case 1**: `3 1 2 1`
   - This is a simple palindrome. The function `f` should return 0 for the entire array and all its subarrays.
2. **Test Case 2**: `4 1 2 2 1`
   - This is another palindrome. The function `f` should return 0 for the entire array and all its subarrays.
3. **Test Case 3**: `5 1 2 3 2 1`
   - This is a longer palindrome. The function `f` should return 0 for the entire array and all its subarrays.
4. **Test Case 4**: `6 1 2 3 4 3 2`
   - This array is not a palindrome but has a pattern where the first three elements and the last three elements are similar. This can expose issues in handling subarrays and the palindrome checks.

These test cases are designed to cover various scenarios and should help in identifying any bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

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

using namespace std;

void solve()
{
    int n;
    cin>>n;
    vector<int> nums(n);

    for(int i=0;i<n;i++) cin>>nums[i];

    unordered_map<long long ,int> hash;

    long long int ans = n*(n+1)*(n+2)/6; // sum of number of elements in all the subarrays

    for(int i=0;i<n;i++)
    {
        long long sum=0;

        for(int j=i;j>=0;j--)
        {
            sum += nums[j];

            if(hash[sum] != 0)
            hash[sum]++;
            else
            hash[sum]=1;
        }

        sum=0;

        for(int j=i;j>=0;j--)
        {
            sum += nums[j];

            ans -= hash[sum];
        }
    }

    cout<<ans<<"\n";
}

int main()
{
    int t;
    cin>>t;
    while(t--)solve();

    return 0;

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The buggy code seems to have issues with the way it handles the sum of elements and the use of the hash map to track these sums.

### Step-by-Step Analysis:

1. **Understanding the Bug:**
   - The code initializes `ans` as the sum of the number of elements in all subarrays, which is `n*(n+1)*(n+2)/6`.
   - It then tries to adjust `ans` by subtracting the values from a hash map that tracks the sums of subarrays.
   - The hash map is not correctly updated or used, leading to incorrect results.

2. **Edge Cases to Consider:**
   - **All elements are the same:** This can cause the hash map to have a lot of collisions and might not handle the sums correctly.
   - **Elements with large values:** This can cause the sum to exceed the range of the data type used.
   - **Small arrays with specific patterns:** This can help in identifying specific issues in the logic.

3. **Challenging Input:**
   - A small-scale test input that is complete and valid should be designed to highlight the issues in the hash map and the sum calculations.
   - An array with repeated elements can be a good candidate to expose the bug.

### Challenging Test Input:
```plaintext
1
4
1 1 1 1
```

### Explanation:
- **Test Case:**
  - The input consists of one test case with an array of four elements, all of which are 1.
- **Expected Output:**
  - The expected output is 0 because the array is already a palindrome, and all subarrays are also palindromes.
- **Why This Input is Challenging:**
  - The hash map will have many entries with the same sum (1, 2, 3, 4), which can lead to incorrect adjustments in the `ans` calculation.
  - The buggy code might not correctly handle the repeated sums, leading to an incorrect result.

This input should help in identifying the issues in the hash map usage and the sum calculations in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using ll = long long;
#pragma GCC target("avx")
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
using namespace std;
using namespace __gnu_pbds;
template<typename T> using ordered_set = tree<
        T,
        null_type,
        less<T>,
        rb_tree_tag,
        tree_order_statistics_node_update>;

/*
 * this comment fills you with Determination.
 */
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
#endif

    ll tt = 1;
    cin >> tt;
    for (int ttt = 1; ttt <= tt; ++ttt){
//        cout << "Case " << ttt << ":\n";
        int n, ans;
        vector<int>v;
        map<int, int>prf;

        cin >> n;
        ans = n * (n + 1) * (n + 2) / 6;
        v.resize(n);
        for (auto &i : v)
            cin >> i;
        for (int i = 0; i < n; ++i)
        {
            int sum = 0;
            for (int j = i; ~j; --j)
            {
                sum += v[j];
                prf[sum]++;
            }
            sum = 0;
            for (int j = i; ~j; --j)
            {
                sum += v[j];
                ans -= prf[sum];
            }
        }
        cout << ans;
        cout << '\n';

    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import sys
import threading

def main():
    import sys

    import sys

    def compute_f(a, l, r):
        operations = 0
        left = l
        right = r
        sum_left = a[left]
        sum_right = a[right]
        while left < right:
            if sum_left == sum_right:
                left +=1
                right -=1
                if left < right:
                    sum_left = a[left]
                    sum_right = a[right]
            elif sum_left < sum_right:
                left +=1
                sum_left += a[left]
                operations +=1
            else:
                right -=1
                sum_right += a[right]
                operations +=1
        return operations

    input = sys.stdin.read().split()
    idx =0
    t = int(input[idx])
    idx +=1
    results = []
    for _ in range(t):
        n = int(input[idx])
        idx +=1
        a = list(map(int, input[idx:idx+n]))
        idx +=n
        total_f =0
        for l in range(n):
            for r in range(l, n):
                total_f += compute_f(a, l, r)
        results.append(total_f)
    for res in results:
        print(res)

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The buggy code seems to have a logical flaw in how it handles the sum of elements in the subarrays and the operations needed to make them palindromes.

### Key Considerations:
1. **Edge Cases:**
   - Single-element arrays.
   - Arrays with all elements being the same.
   - Arrays with alternating elements.
   - Arrays with a few large elements and many small elements.
2. **Scenarios to Maximize Failure:**
   - Arrays where the sum of elements on one side is significantly different from the other side.
   - Arrays with a large number of operations required to make subarrays palindromes.

### Chosen Test Input:
Let's create a test input with a few test cases that cover these considerations:

1. **Single-element array:**
   - This should always result in 0 operations.
2. **Array with all elements the same:**
   - This should also result in 0 operations for all subarrays.
3. **Array with alternating elements:**
   - This should require a significant number of operations.
4. **Array with a few large elements and many small elements:**
   - This should expose issues in handling large sums.

### Test Input:
```plaintext
4
1
5
3
1 1 1
4
1 2 1 2
5
1 100000 1 100000 1
```

### Explanation:
1. **Test Case 1:**
   - Single-element array: `[5]`
   - Expected output: `0`
   - Reason: No operations needed for a single-element array.

2. **Test Case 2:**
   - All elements the same: `[1, 1, 1]`
   - Expected output: `0`
   - Reason: All subarrays are already palindromes.

3. **Test Case 3:**
   - Alternating elements: `[1, 2, 1, 2]`
   - Expected output: `5`
   - Reason: Each subarray that is not a palindrome requires operations to balance the sums on both sides.

4. **Test Case 4:**
   - Large elements and small elements: `[1, 100000, 1, 100000, 1]`
   - Expected output: `14`
   - Reason: Large elements will require many operations to balance the sums on both sides.

This test input should effectively expose any bugs in the provided code by covering a range of scenarios that are likely to cause issues.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
t=int(input())
for _ in range(t):
    n=int(input())
    L=list(map(int,input().split()))
    answer=0
    def check(L):
        for i in range(len(L)//2):
            if L[i]==L[len(L)-i-1]:
                continue
            else:
                return False
        return True
    def f(L):
        count=0
        while(check(L)==False):
            if(L[0]!=L[-1]):
                if L[0]>L[-1]:
                    q=L[0]-L[-1]
                    L[0]=L[-1]
                    L.insert(1,q)
                    count=count+1
                elif L[0] < L[-1]:
                    p=L[-1]-L[0]
                    L[-1] = L[0]
                    L.insert(len(L)-1, p)
                    count = count + 1
            L.pop(0)
            L.pop(-1)
        return count

    for r in range(0,n+1):
        for l in range(0,r):
            answer=answer+f(L[l:r])
    print(answer)

**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
1 2 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }

            System.out.println(sumOfPalindromeOperations(a, n));
        }

        sc.close();
    }

    private static int sumOfPalindromeOperations(int[] a, int n) {
        int totalOperations = 0;

        for (int l = 0; l < n; l++) {
            for (int r = l; r < n; r++) {
                totalOperations += minOperationsToMakePalindrome(a, l, r);
            }
        }

        return totalOperations;
    }

    private static int minOperationsToMakePalindrome(int[] a, int l, int r) {
        int[] subArray = new int[r - l + 1];
        System.arraycopy(a, l, subArray, 0, r - l + 1);

        int operations = 0;
        int i = 0, j = subArray.length - 1;

        while (i < j) {
            if (subArray[i] == subArray[j]) {
                i++;
                j--;
            } else if (subArray[i] < subArray[j]) {
                subArray[i + 1] += subArray[i];
                i++;
                operations++;
            } else {
                subArray[j - 1] += subArray[j];
                j--;
                operations++;
            }
        }

        return operations;
    }
}

**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
1 2 2 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import java.util.Scanner;

public class MakePalindrome {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();

        while (t-- > 0) {
            int n = scanner.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
            }

            System.out.println(sumOfPalindromeOperations(a, n));
        }

        scanner.close();
    }

    private static int sumOfPalindromeOperations(int[] a, int n) {
        int totalOperations = 0;

        for (int l = 0; l < n; l++) {
            for (int r = l; r < n; r++) {
                totalOperations += minOperationsToMakePalindrome(a, l, r);
            }
        }

        return totalOperations;
    }

    private static int minOperationsToMakePalindrome(int[] a, int l, int r) {
        int[] subArray = new int[r - l + 1];
        System.arraycopy(a, l, subArray, 0, r - l + 1);

        int operations = 0;
        int left = 0;
        int right = subArray.length - 1;

        while (left < right) {
            if (subArray[left] == subArray[right]) {
                left++;
                right--;
            } else if (subArray[left] < subArray[right]) {
                subArray[left + 1] += subArray[left];
                left++;
                operations++;
            } else {
                subArray[right - 1] += subArray[right];
                right--;
                operations++;
            }
        }

        return operations;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The primary issues in the code are related to how it handles subarrays and the operations to make them palindromes. Specifically, the code might fail when dealing with subarrays that have elements that are not easily balanced.

Here are some steps to think about the input:

1. **Subarrays with Odd and Even Lengths**: Ensure the test input includes both odd and even-length subarrays.
2. **Subarrays with Repeated Elements**: Test with subarrays where some elements are repeated, as this can affect the merging and splitting operations.
3. **Subarrays with Large Elements**: Use large elements to test the merging and splitting operations, as they can lead to more complex scenarios.
4. **Subarrays with Elements in Decreasing Order**: Test with subarrays where elements are in decreasing order, as this can cause issues with the merging logic.
5. **Subarrays with Elements in Increasing Order**: Similarly, test with subarrays where elements are in increasing order.

Based on these considerations, a challenging test input could be:

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

This input includes a subarray with elements in increasing order, which can cause the merging operations to be more complex and can expose any issues in the logic for handling such sequences.

Here is the test input:

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import sys
from bisect import bisect

def solve():
    binsearch = bisect
    inp = sys.stdin.readline
    n = int(inp())
    a = list(map(int, inp().split()))
    p = [0] * (n + 1)
    for i in range(n):
        p[i + 1] = p[i] + a[i]
    #shift, left, right, count
    slrc = dict()
    coinc = dict()
    for i in range(n + 1):
        for j in range(n + 1):
            #p[i] = s - p[j]
            s = p[i] + p[j]
            if s in coinc:
                coinc[s].append(p[i])
            else:
                coinc[s] = [p[i]]
    for i in range(n):
        for j in range(i, n):
            s = p[i] + p[j + 1]
            lrc = p[i], p[j + 1], (j - i + 2)
            if s in slrc:
                slrc[s].append(lrc)
            else:
                slrc[s] = [lrc]
    res = 0
    ls = set(p)
    for s, lr_list in slrc.items():
        q = coinc.get(s, [])
        for l, r, c in lr_list:
            #print(s, l, r, c, p, [s - k for k in p[::-1]], q, binsearch(q, r) - (binsearch(q, l) - 1))
            res += c - (binsearch(q, r) - (binsearch(q, l) - 1))
    print(res)

def main():
    for i in range(int(sys.stdin.readline())):
        solve()

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the code to fail. Here are the steps to think through this:

1. **Edge Cases**:
   - Single element arrays.
   - Arrays with all elements being the same.
   - Arrays with a mix of small and large elements.
   - Arrays that are already palindromes.

2. **Complex Cases**:
   - Arrays where the sum of elements in different subarrays results in the same value, leading to potential issues with the `coinc` and `slrc` dictionaries.
   - Arrays with a large number of elements to test the performance and correctness of the code.

3. **Challenging Scenarios**:
   - Arrays with elements that are close to the upper limit of the input constraints.
   - Arrays with a specific pattern that might cause the bisect operations to fail or produce incorrect results.

Given these considerations, let's create a small-scale test input that is likely to expose the bug in the code.

### Test Input
```plaintext
4
1
1
2
1 1
3
1 2 3
4
1 2 2 1
```

### Explanation
1. **Single Element Array**:
   - `1 1` (n=1): The array is already a palindrome, so the function should return 0.

2. **Two Identical Elements**:
   - `2 1 1` (n=2): The array is already a palindrome, so the function should return 0.

3. **Simple Non-Palindrome**:
   - `3 1 2 3` (n=3): The array is not a palindrome. The function should return the sum of the minimum operations needed for all subarrays. For example, subarray `[1, 2]` needs 1 operation, subarray `[2, 3]` needs 1 operation, and so on.

4. **Palindrome**:
   - `4 1 2 2 1` (n=4): The array is already a palindrome, so the function should return 0.

This test input covers a range of scenarios including single elements, identical elements, simple non-palindromic arrays, and palindromic arrays. It is designed to be small enough to be manageable but comprehensive enough to potentially 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: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define h_r ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0);
const ll M=1e9+7,N=1e5+5;

int main()
{
	h_r
	ll qt=1;
	cin>>qt;
	for(ll ii=1;ii<=qt;ii++)
	{
	 	ll n, ans=0;
	 	cin>>n;
	 	vector<ll> v(n+1),pre(n+1);
	 	set<ll> us;
	 	for(ll i=1;i<=n;i++)
	 	{
	 		cin>>v[i];
	 		pre[i]+=v[i]+pre[i-1];
	 		us.insert(pre[i]);
	 	}
	 	map<ll,ll> mp;
	 	for(ll i=1;i<=n;i++)
	 	{
	 		for(ll j=i;j<=n;j++)
	 		{
	 			ans+=(j-i);
	 			mp[pre[j]-pre[i-1]]++;
	 		}
	 	}
	 	for(ll stp=1;stp<=n;stp++)
	 	{
	 		ll sum=0;
	 		for(ll i=stp;i<=n;i++)
	 		{
	 			sum+=v[i];
	 			mp[sum]--;
	 		}
	 		sum=0;
	 		for(ll i=stp;i>=1;i--)
	 		{
	 			sum+=v[i];
	 			ans-=mp[sum]*2;
	 			ll nd=pre[stp]+sum;
	 			if(us.find(nd)!=us.end())ans++;
	 		}
	 	}
	 	cout<<ans<<"\n";
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

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

using namespace std;

#define x first
#define y second
#define endl "\n"
// #define int long long

typedef pair<int, int> PII;
using LL = long long;
using ULL = unsigned long long;

const int mod = 1e9 + 7;
const int N = 2e3 + 10, M = 1e5 + 10;
const int INF = 0x3f3f3f3f;

int dp[N][N];

void solve() {
    int n;
    cin >> n;
    int a[n + 1];
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    vector<int> pre(n + 1), suf(n + 2);
    pre[0] = suf[0] = 0;
    for (int i = 1; i <= n; i++) {
        pre[i] = pre[i - 1] + a[i];
    }
    suf[n] = a[n];
    for (int i = n - 1; i >= 1; i--) {
        suf[i] = suf[i + 1] + a[i];
    }
    int ans = 0;
    for (int len = 2; len <= n; len++) {
        for (int i = 1; i + len - 1 <= n; i ++) {
            int j = i + len - 1;
            dp[i][j] = j - i;
            int l = i, r = j;
            while (l < r) {
                if (pre[l] - pre[i - 1] == pre[j] - pre[r - 1]) {
                    dp[i][j] = min(dp[i][j], dp[l + 1][r - 1] + l - i + j - r);
                    l++, r--;
                    break;
                } else if (pre[l] - pre[i - 1] < pre[j] - pre[r - 1]) {
                    l++;
                } else {
                    r--;
                }
            }
            ans += dp[i][j];
        }
    }
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    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
4
3
1 2 1
4
1 1 2 1
5
1 2 3 2 1
4
2 1 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

Buggy Code:
import sys
input = lambda :sys.stdin.readline()[:-1]
ni = lambda :int(input())
na = lambda :list(map(int,input().split()))
yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")
no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")
#######################################################################

# https://github.com/tatyam-prime/SortedSet/blob/main/SortedMultiset.py
import math
from bisect import bisect_left, bisect_right, insort
from typing import Generic, Iterable, Iterator, TypeVar, Union, List
T = TypeVar('T')

class SortedMultiset(Generic[T]):
    BUCKET_RATIO = 50
    REBUILD_RATIO = 170

    def _build(self, a=None) -> None:
        "Evenly divide `a` into buckets."
        if a is None: a = list(self)
        size = self.size = len(a)
        bucket_size = int(math.ceil(math.sqrt(size / self.BUCKET_RATIO)))
        self.a = [a[size * i // bucket_size : size * (i + 1) // bucket_size] for i in range(bucket_size)]

    def __init__(self, a: Iterable[T] = []) -> None:
        "Make a new SortedMultiset from iterable. / O(N) if sorted / O(N log N)"
        a = list(a)
        if not all(a[i] <= a[i + 1] for i in range(len(a) - 1)):
            a = sorted(a)
        self._build(a)

    def __iter__(self) -> Iterator[T]:
        for i in self.a:
            for j in i: yield j

    def __reversed__(self) -> Iterator[T]:
        for i in reversed(self.a):
            for j in reversed(i): yield j

    def __len__(self) -> int:
        return self.size

    def __repr__(self) -> str:
        return "SortedMultiset" + str(self.a)

    def __str__(self) -> str:
        s = str(list(self))
        return "{" + s[1 : len(s) - 1] + "}"

    def _find_bucket(self, x: T) -> List[T]:
        "Find the bucket which should contain x. self must not be empty."
        for a in self.a:
            if x <= a[-1]: return a
        return a

    def __contains__(self, x: T) -> bool:
        if self.size == 0: return False
        a = self._find_bucket(x)
        i = bisect_left(a, x)
        return i != len(a) and a[i] == x

    def count(self, x: T) -> int:
        "Count the number of x."
        return self.index_right(x) - self.index(x)

    def add(self, x: T) -> None:
        "Add an element. / O(√N)"
        if self.size == 0:
            self.a = [[x]]
            self.size = 1
            return
        a = self._find_bucket(x)
        insort(a, x)
        self.size += 1
        if len(a) > len(self.a) * self.REBUILD_RATIO:
            self._build()

    def discard(self, x: T) -> bool:
        "Remove an element and return True if removed. / O(√N)"
        if self.size == 0: return False
        a = self._find_bucket(x)
        i = bisect_left(a, x)
        if i == len(a) or a[i] != x: return False
        a.pop(i)
        self.size -= 1
        if len(a) == 0: self._build()
        return True

    def lt(self, x: T) -> Union[T, None]:
        "Find the largest element < x, or None if it doesn't exist."
        for a in reversed(self.a):
            if a[0] < x:
                return a[bisect_left(a, x) - 1]

    def le(self, x: T) -> Union[T, None]:
        "Find the largest element <= x, or None if it doesn't exist."
        for a in reversed(self.a):
            if a[0] <= x:
                return a[bisect_right(a, x) - 1]

    def gt(self, x: T) -> Union[T, None]:
        "Find the smallest element > x, or None if it doesn't exist."
        for a in self.a:
            if a[-1] > x:
                return a[bisect_right(a, x)]

    def ge(self, x: T) -> Union[T, None]:
        "Find the smallest element >= x, or None if it doesn't exist."
        for a in self.a:
            if a[-1] >= x:
                return a[bisect_left(a, x)]

    def __getitem__(self, x: int) -> T:
        "Return the x-th element, or IndexError if it doesn't exist."
        if x < 0: x += self.size
        if x < 0: raise IndexError
        for a in self.a:
            if x < len(a): return a[x]
            x -= len(a)
        raise IndexError

    def index(self, x: T) -> int:
        "Count the number of elements < x."
        ans = 0
        for a in self.a:
            if a[-1] >= x:
                return ans + bisect_left(a, x)
            ans += len(a)
        return ans

    def index_right(self, x: T) -> int:
        "Count the number of elements <= x."
        ans = 0
        for a in self.a:
            if a[-1] > x:
                return ans + bisect_right(a, x)
            ans += len(a)
        return ans

def solve(a):
    n = len(a)

from collections import defaultdict
for _ in range(ni()):
    n = ni()
    a = na()
    # ans = 0
    # for i in range(n):
    #     for j in range(i+1,n+1):
    #         z = [0]
    #         for k in range(i,j):
    #             z.append(a[k] + z[-1])
    #         c = 0
    #         for k in range(j-i):
    #             if not z[-1] - z[k] in z:
    #                 c += 1
    #         #print(i,j,c, z)
    #         ans += c
    # print(ans)
    s = [0] * (n+1)
    for i in range(n):
        s[i+1] = s[i] + a[i]
    ans = 0
    for l in range(n):
        for r in range(l+1, n+1):
            ans += r - l - 1
    f = defaultdict(SortedMultiset)
    for i in range(1, n):
        for l in range(i):
            f[s[i] - s[l]].add(l)
        r = i + 1
        for j in range(r):
            ans -= f[s[r] - s[j]].index(j)

    print(ans)

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

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

Think step by step.

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

1. **Understand the Bug**: The provided code seems to have issues with the way it calculates the sum of the values of the function \( f \) for all subarrays. The bug might be related to incorrect handling of the sum and the indices.

2. **Edge Cases**: Focus on edge cases where the array has repeated elements, small arrays, and arrays with a single element.

3. **Small-Scale Test**: Generate a small-scale test input that is complete and valid to ensure it fits within the constraints and is easy to verify.

4. **Test Case**: Create a test case that maximizes the failure probability by exploiting the potential issues in the code.

Given these steps, a small-scale test input that might expose the bug is:

```plaintext
1
4
1 2 1 2
```

This test case includes a small array with repeated elements and a structure that might cause issues in the sum calculations. The expected output for this test case is `5`.

Let's break down why this test case is challenging:
- The array `[1, 2, 1, 2]` has repeated elements, which can cause issues if the code does not handle duplicate values correctly.
- The array is small enough to be easily verified, but large enough to test the sum calculations for multiple subarrays.

By using this test input, we can check if the buggy code correctly handles the sum of the values of the function \( f \) for all subarrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

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

public class F2 {

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

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

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

	private static class Solution implements Runnable {

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

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

			out.flush();
		}

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

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
			// for BIT. Do not use tree maps with update query
			// Assuming Constraints on an unbound variable
			// Not properly calculating time complexity
		}

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

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

			int lIdx, rIdx;

			int leftSum, rightSum;
			int moves;

			for (int i = 1; i < n; ++i) {
				for (int l = 0; l + i < n; ++l) {
					r = l + i;
					if (i == 1 && arr[l] == arr[r])
						dp[l][r] = 0;
					else
						dp[l][r] = r - l;

					r = l + i;

					lIdx = l;
					rIdx = r;

					leftSum = arr[lIdx++];
					rightSum = arr[rIdx--];
					moves = 0;

					while (lIdx <= rIdx + 1) {
						// System.out.println(lIdx + " " + rIdx + "  " + leftSum + " " + rightSum);
						if (leftSum == rightSum) {
							dp[l][r] = dp[lIdx][rIdx] + moves;
							break;
						}else if (rIdx < lIdx)
							break;

						++moves;
						if (leftSum > rightSum) {
							rightSum += arr[rIdx--];
						}else {
							leftSum += arr[lIdx++];
						}
					}
				}
			}

			long result = 0;

			for (int i = 0; i < n; ++i) {
				for (int j = i; j < n; ++j) {
					result += dp[i][j];
				}
			}

			// System.out.println(Arrays.deepToString(pal));
			// System.out.println(Arrays.deepToString(dp));

			// out.println(System.currentTimeMillis() - time);

			out.println(result);
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

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

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

    	  return res;
      }

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

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

    	  return res;
      }

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

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

    	  int a, b;

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

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

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

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

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

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

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

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

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

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

1. **Edge Cases for Small Arrays:**
   - Arrays of length 1 or 2 where the elements are either the same or different.
   - Arrays with repeated elements or all elements being 1.

2. **Edge Cases for Larger Arrays:**
   - Arrays with alternating elements.
   - Arrays where the sum of adjacent elements forms a pattern that might cause the algorithm to fail.
   - Arrays with a large number of elements that are all the same or have a specific pattern.

3. **Special Cases:**
   - Arrays where the sum of elements in the middle of the array is equal to the sum of elements at the ends.
   - Arrays with elements that are powers of 2 or have specific mathematical properties.

Given the constraints and the need for a small-scale test input, let's consider an array that has a pattern that might cause the algorithm to fail. One such pattern is an array where the sum of adjacent elements is the same, but the algorithm might incorrectly calculate the number of operations needed to make the subarrays palindromic.

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

```plaintext
1
5
1 2 1 2 1
```

This input has the following properties:
- The array is of length 5.
- It has a repeating pattern of `1 2 1 2 1`.
- The sum of adjacent elements (1 + 2) is 3, and this pattern might cause the algorithm to incorrectly calculate the number of operations needed to make the subarrays palindromic.

This input is designed to challenge the algorithm and expose any potential bugs in the logic for calculating the minimum number of operations to make subarrays palindromic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make a Palindrome
time_limit_ms: 5000
memory_limit_mb: 512
Description: You are given an array $a$ consisting of $n$ integers.

Let the function $f(b)$ return the minimum number of operations needed to make an array $b$ a palindrome. The operations you can make are:

  * choose two adjacent elements $b_i$ and $b_{i+1}$, remove them, and replace them with a single element equal to $(b_i + b_{i + 1})$;    * or choose an element $b_i > 1$, remove it, and replace it with two positive integers $x$ and $y$ ($x > 0$ and $y > 0$) such that $x + y = b_i$.

For example, from an array $b=[2, 1, 3]$, you can obtain the following arrays in one operation: $[1, 1, 1, 3]$, $[2, 1, 1, 2]$, $[3, 3]$, $[2, 4]$, or $[2, 1, 2, 1]$.

Calculate $\displaystyle \left(\sum_{1 \le l \le r \le n}{f(a[l..r])}\right)$, where $a[l..r]$ is the subarray of $a$ from index $l$ to index $r$, inclusive. In other words, find the sum of the values of the function $f$ for all subarrays of the array $a$.

Input: The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^5$).

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the sum of the values of the function $f$ for all subarrays of the array $a$.

Examples:
input:
4
3
2 1 3
4
1 1 1 1
5
4 2 3 1 5
4
1 2 1 2
output:
3
0
14
5

Note:

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

public class F {
    String filename = null;
    InputReader sc;

    static class Interval {
        int start;
        int end;

        public Interval(int start, int end) {
            this.start = start;
            this.end = end;
        }
    }
    void solve() {
        int n = sc.nextInt();
        int[] a = sc.nextArray(n);

        HashMap<Integer, ArrayList<Interval>> sums = new HashMap<>();

        int ans = 0;
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += a[j];
                var intervals = sums.getOrDefault(sum, new ArrayList<>());
                intervals.add(new Interval(i, j));
                sums.putIfAbsent(sum, intervals);
                ans += j - i;
            }
        }

        for (var intervals: sums.values()) {
            intervals.sort(Comparator.comparingInt(o -> o.start));

            int i2 = 0;
            for (int i1 = 0; i1 < intervals.size(); i1++) {
                while (i2 < intervals.size() && intervals.get(i2).start <= intervals.get(i1).end) {
                    i2++;
                }
                ans -= 2 * (intervals.size() - i2);
                if (i2 < intervals.size() && intervals.get(i1).end + 1 == intervals.get(i2).start) {
                    ans += 1;
                }
            }
        }

        System.out.println(ans);
    }

    public void run() throws FileNotFoundException {
        if (filename == null) {
            sc = new InputReader(System.in);
        } else {
            sc = new InputReader(new FileInputStream(filename));
        }

        int nTests = sc.nextInt();
        for (int test = 0; test < nTests; test++) {
            solve();
        }
    }

    public static void main(String[] args) {
        F sol = new F();
        try {
            sol.run();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

        public float nextFloat() {
            return Float.parseFloat(next());
        }

        public double nextDouble() {
            return Float.parseFloat(next());
        }

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

        public int[] nextArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#define int long long
using namespace std;
signed main(){
int test;
cin>>test;
while(test--) {
	int n,m;
	cin>>n>>m;
	int a[n+1];
	int ans=0;
	int no=0;
	int cnt=0;
	for(int i=1; i<=n; i++) {
		cin>>a[i];
		cnt+=a[i];
		cnt-=a[i]/2*2;
		if(a[i]%2==1) no++;
		ans+=a[i]/2*2;
	}
	if(ans/2>=m) cout<<m*2<<'\n';
	else {
		m-=ans/2;
		if(m>=cnt) cout<<ans+cnt<<'\n';
		else cout<<ans+(m-cnt/2)*2<<'\n';
		}
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases that can highlight issues in the logic. The code seems to have a few potential problems, such as handling the distribution of family members and the calculation of happy people. Let's break down the logic and identify potential edge cases:

1. **Single Family with Odd Members:**
   - If there is only one family with an odd number of members, the code might not handle the distribution correctly.

2. **Multiple Families with Odd and Even Members:**
   - If there are multiple families with a mix of odd and even members, the code might not correctly distribute the members to maximize the number of happy people.

3. **Maximum Number of Rows:**
   - If the number of rows is maximized, the code might not handle the distribution efficiently.

4. **All Families with Even Members:**
   - If all families have an even number of members, the code might not handle the distribution correctly.

5. **All Families with Odd Members:**
   - If all families have an odd number of members, the code might not handle the distribution correctly.

Let's focus on a small-scale test input that combines these edge cases to maximize the failure probability:

### Test Input
```plaintext
4
1 2
3
2 2
3 1
3 2
2 1 2
4 4
1 1 1 1
```

### Explanation
1. **Single Family with Odd Members:**
   - Test case 1: `1 2 3`
     - One family with 3 members.
     - The bus has 2 rows.
     - Expected output: 4 (2 members can sit together, and 1 member can sit alone).

2. **Multiple Families with Odd and Even Members:**
   - Test case 2: `2 2 3 1`
     - Two families: one with 3 members and one with 1 member.
     - The bus has 2 rows.
     - Expected output: 4 (2 members of the first family can sit together, and 1 member of the first family can sit with the member of the second family).

3. **Multiple Families with a Mix of Odd and Even Members:**
   - Test case 3: `3 2 2 1 2`
     - Three families: one with 2 members, one with 1 member, and one with 2 members.
     - The bus has 2 rows.
     - Expected output: 6 (2 members of the first family can sit together, 2 members of the third family can sit together, and the member of the second family can sit alone).

4. **All Families with Even Members:**
   - Test case 4: `4 4 1 1 1 1`
     - Four families, each with 1 member.
     - The bus has 4 rows.
     - Expected output: 4 (Each member can sit alone).

This test input covers a variety of edge cases and should help expose any bugs 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: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin >> t;
    while(t--){
        long long int n, r;
        cin >> n >> r;
        vector<long long int> v(n);
        int sum = 0;
        for(int i = 0; i < n ; i++){
            cin >> v[i];
            sum += v[i];
        }
        long long int happy = 0;
        for(int i = 0; i < n ; i++){
            if(v[i]%2 == 0 && r > 0){
                if(v[i]/2 <= r){
                    happy += v[i];
                    r -= v[i]/2;
                    sum -= v[i];
                }
                else{
                    happy += 2*r;
                    sum -= 2*r;
                    break;
                }
            }

        }
        for(int i = 0; i < n; i++){
            if(v[i]%2 == 1 && r > 0){
                if(v[i]/2 + 1 <= r){
                    r -= v[i]/2 + 1;
                    happy += v[i];
                    sum -= v[i];
                }
                else{
                    happy += 2*r;
                    sum -= 2*r;
                    break;
                }
            }
        }
        cout << happy - sum << endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    ll t = 2;
    cin >> t;
    while (t--) {
        ll n, r;
        cin >> n >> r;
        vector<ll> a(n);
        ll s = 0, s2 = 0;
        for (ll i = 0; i < n; i++) {
            cin >> a[i];
            s2 += a[i];
            s += (a[i] / 2) * 2;
        }
        s2 -= s;
        ll sl = 2 * r - s;
        ll lt = 0, rt = 1000000;
        while (rt - lt > 1) {
            ll md = (rt + lt) / 2;
            if (md * 2 + s - md > sl) {
                rt = md;
            } else {
                lt = md;
            }
        }
        cout << s + lt << endl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bus to Pénjamo
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ya vamos llegando a Péeeenjamoo ♫♫♫

There are $n$ families travelling to Pénjamo to witness Mexico's largest- ever "walking a chicken on a leash" marathon. The $i$-th family has $a_i$ family members. All families will travel using a single bus consisting of $r$ rows with $2$ seats each.

A person is considered happy if:

  * Another family member is seated in the same row as them, or    * They are sitting alone in their row (with an empty seat next to them).

Determine the maximum number of happy people in an optimal seating arrangement. Note that everyone must be seated in the bus.

It is guaranteed that all family members will fit on the bus. Formally, it is guaranteed that $\displaystyle\sum_{i=1}^{n}a_i \le 2r$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $r$ ($1 \le n \le 100$; $1 \le r \le 500$) — the number of families and the number of rows in the bus.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10$) — the number of family members in each family.

Output: For each test case, output the maximum number of happy people in an optimal seating arrangement.

Examples:
input:
4
3 3
2 3 1
3 3
2 2 2
4 5
1 1 2 2
4 5
3 1 1 3
output:
4
6
6
6

Note: In the first test case, the two members of the first family can sit together in the first row, while the two members of the second family can sit together in the second row. The remaining member of the second family can sit in the third row along with a member of the third family. This seating arrangement is shown below, where the $4$ happy people are colored green.

$\color{green}{1}$| $\color{green}{1}$   ---|---   $\color{green}{2}$| $\color{green}{2}$   $2$| $3$      In the second test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{3}$| $\color{green}{3}$   ---|---   $\color{green}{1}$| $\color{green}{1}$   $\color{green}{2}$| $\color{green}{2}$      In the third test case, a possible seating arrangement with $6$ happy people is shown below.

$\color{green}{4}$| $\color{green}{4}$   ---|---   $\color{green}{}$| $\color{green}{2}$   $\color{green}{3}$| $\color{green}{3}$   $\color{green}{1}$| $\color{green}{}$   $\color{green}{}$| $\color{green}{}$

Buggy Code:
t = int(input())

for i in range(t):
    n, r = map(int, input().split())
    a = list(map(int, input().split()))

    happy = 0
    count_1 = 0
    for i in a:
    	if i > 1:
    		if i%2==0:
    			happy +=i
    			r -= (i//2)

    		else:

    			happy += 2*(i//2)
    			r -= (i//2)
    			count_1 += 1

    	else:
    		count_1 +=1

    if r>count_1:
    	happy += count_1
    else: happy += count_1//2 if 2*r>count_1 else 0

    print(happy)

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