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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define ft first
#define sd second
#define pb push_back
#define nl "\n"
#define sz size()
#define int ll
typedef long long ll;
typedef long double ld;
using namespace std;
const ll mod = 1e9 + 7, N = 200200;

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        int a[n];
        vector<int> v;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        int k = -1;
        for (int i = 0; i < n - 1; i++) {
            if (a[i] > a[i + 1]) {
                k = i;
            }
        }
        for (int i = 0; i < n; i++) {
            if (i > k) {
                v.pb(a[i]);
                continue;
            }
            deque<int> u;
            while(a[i] > 0) {
                int c = a[i] % 10;
                u.push_front(c);
                a[i] /= 10;
            }
            for (auto x : u) {
                v.pb(x);
            }
        }
        bool f = 1;
        for (int i = 0; i < v.size() - 1; i++) {
            if (v[i] > v[i + 1]) {
                f = 0;
                break;
            }
        }
        if (f) {
            cout << "YES" << nl;
        } else {
            cout << "NO" << nl;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
4
12 3 45 67
3
10 1 0
2
9 9
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
n = int(input())
ans = []

for _ in range(n):
    m = int(input())
    list_temp = list(map(int, input().split()))
    count = 1

    if m == 2:
        if list_temp[0] <= list_temp[1]:
            ans.append("YES")
        else:
            x, y = str(list_temp[0])[0], str(list_temp[0])[1]
            if int(x) > int(y)  or int(x) < list_temp[0]:
                ans.append('NO')
                count -= 1
            else:
                ans.append("YES")
                count -= 1
    else:
        for i in range(m-1):
            if list_temp[i] > list_temp[i+1] and count == 1:
                x, y = str(list_temp[i])[0], str(list_temp[i])[1]
                if int(x) > int(y) or int(y) > list_temp[i+1]:
                    if i==0:
                        ans.append("NO")
                        count-=1
                    else:
                        if int(x) < list_temp[i-1]:
                            ans.append('NO')
                            count -= 1
                else:
                    pass
            if i == m-2 and count == 1:
                ans.append("YES")
                count -= 1

for item in ans:
    print(item)

**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
10 9 8
3
10 1 9
2
9 10
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
def sort_array(unfixed_arr):
    l = len(unfixed_arr)
    flag = 1
    for i in range(1,l):
        #print(unfixed_arr[l-1-i])

        # single single
        if unfixed_arr[l-i]//10==0 and unfixed_arr[l-i-1]//10==0:
            if unfixed_arr[l-i] < unfixed_arr[l-i-1]:
                # print(list2)
                # print(1)
                return "NO"
            else:
                # print(10)
                pass

        # single double
        elif unfixed_arr[l-i]//10!=0 and unfixed_arr[l-i-1]//10==0:
            # print(9)
            pass

        #double double
        elif unfixed_arr[l-i]//10!=0 and unfixed_arr[l-i-1]//10!=0:
            if unfixed_arr[l-i]>=unfixed_arr[l-i-1]:
                # print(8)
                pass
            else:
                #split
                b = int(str(unfixed_arr[l-1-i])[1])
                a = int(str(unfixed_arr[l-1-i])[0])
                c = unfixed_arr[l-i]
                if a<=b and b<=c:
                    list2 = []
                    for j in range(0,l-i):
                        for z in str(unfixed_arr[j]):
                            list2.append(int(z))
                    for j in unfixed_arr[l-i:]:
                        list2.append(j)
                    string1 = str(list2)
                    list2.sort()
                    string2 = str(list2)
                    if string1==string2:
                        # print(list2)
                        # print(2)
                        return "YES"
                    else:
                        # print(list2)
                        # print(3)
                        return "NO"

        # double sin1gle
        elif unfixed_arr[l-i]//10==0 and unfixed_arr[l-1-i]//10!=0:
            b = int(str(unfixed_arr[l-1-i])[1])
            a = int(str(unfixed_arr[l-1-i])[0])
            c = unfixed_arr[l-i]
            #print(c)
            list2 = []
            if a<=b and b<=c:
                for j in range(0,l-i):
                    for z in str(unfixed_arr[j]):
                        list2.append(int(z))
                for j in unfixed_arr[l-i:]:
                    list2.append(j)
                string1 = str(list2)
                list2.sort()
                string2 = str(list2)
                if string1==string2:
                    # print(list2)
                    # print(4)
                    return "YES"
                else:
                    # print(list2)
                    # print(4)
                    return "NO"
            #     unfixed_arr.delete(l-1-i)
            #     unfixed_arr.insert(l-1-i,b)
            #     unfixed_arr.insert(l-i,a)
            else:
                # print(list2)
                # print(5)
                return "NO"

        else:
            # print(7)
            pass
    if flag ==1:
        return "YES"

num = int(input())
for j in range(0, num):
  a=int(input())
  arrr = input()
  ar = list(map(int, arrr.split(' ')))
  print(sort_array(ar))

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

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

Think step by step.

Response:
```plaintext
3
4
12 23 45 56
3
29 12 3
2
9 8
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

public class CodeForces {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());
        while (t-- > 0) {
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            st = new StringTokenizer(br.readLine());

            int prev = -1;
            boolean works = true;
            for (int i = 0; i < n; i++) {
                int num = Integer.parseInt(st.nextToken());
                if (num >= 10 && num / 10 >= prev && num % 10 >= num / 10) {
                    prev = num % 10;
                } else if (num < 10 && num >= prev) {
                    prev = num;
                } else {
                    works = false;
                    break;
                }
            }

            System.out.println(works ? "YES" : "NO");
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5
12 3 45 6 7
4
12 23 4 5
2
9 10
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

public class cp2 {
//11190400

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

        int t = Sc.nextInt();
        while (t-- > 0) {
            int n = Sc.nextInt();
            String Output = "YES";
            char prev = '0';
            int prevNumber = -1;
            for (int i  = 0; i < n ; i++){
                String a = Sc.next();
                int curra = Integer.parseInt(a);
//                System.out.println(prev+":"+prevNumber+"{"+a+"}"+curra);
                if(prevNumber <= curra){
                    prevNumber = curra;
                    prev = a.charAt(a.length()-1);
                    continue;
                }
                prevNumber = curra;
                if (a.length() == 2 && (  a.charAt(0) <= a.charAt(1) && a.charAt(0) >= prev)){
                    prev = a.charAt(1);
                    continue;
                }else if( a.length() == 1 && a.charAt(0) >= prev){
                    prev = a.charAt(0);
                    continue;
                }
                Output = "NO";
            }

            System.out.println(Output);

        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
4
12 3 45 67
3
12 28 5
2
0 0
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        int a[n];
        vector < int > v;
        for ( int i = 0; i < n; i++) cin >> a[i];
        for ( int i = 0; i < n - 1; i++){
            if ( a[i] > a[i+1]){
                if ( a[i] >= 10 ) {
                    v.push_back( a[i] / 10);
                    v.push_back( a[i] % 10);
                }
                else v.push_back( a[i] );
            }
            else v.push_back( a[i] );
        }
        v.push_back( a[n-1]);
        int oke = true;
        for ( int i = 0; i < v.size() - 1; i++){
            if ( v[i] > v[i + 1] ) {
                oke = false;
                break;
            }
        }
        if ( oke == true ) cout << "YES\n";
        else cout << "NO\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:
To expose the bug in the given code, we need to focus on scenarios where the code might incorrectly determine that the array can be sorted. The code has a flaw in how it processes the elements and checks the sorted condition. Specifically, it only processes elements that are out of order, and it does not correctly handle cases where an element is split into multiple digits that might still cause the array to be unsorted.

Let's consider the following edge cases:

1. **Array with a large number that, when split, creates a sequence that cannot be sorted:**
   - For example, an array like `[12, 3, 45, 56]` where `45` and `56` are split into `4, 5` and `5, 6` respectively, which would still result in an unsorted array.

2. **Array with a large number that, when split, creates a sequence that can be sorted:**
   - For example, an array like `[12, 3, 45, 67]` where `45` is split into `4, 5`, which can be sorted.

3. **Array with a small number that is out of order but can be sorted by splitting a larger number:**
   - For example, an array like `[12, 28, 5]` where `28` is split into `2, 8`, which can be sorted.

4. **Array with a small number that is out of order and cannot be sorted by splitting any number:**
   - For example, an array like `[12, 3, 5]` where `12` is split into `1, 2`, which still results in an unsorted array.

Given these considerations, let's create a test input that exposes the bug:

```plaintext
3
4
12 3 45 56
3
12 3 45
2
12 3
```

### Explanation:
1. **Test Case 1: `[12, 3, 45, 56]`**
   - Splitting `45` results in `[12, 3, 4, 5, 56]`, which is unsorted.
   - Splitting `56` results in `[12, 3, 4, 5, 5, 6]`, which is also unsorted.
   - The code should output `NO`.

2. **Test Case 2: `[12, 3, 45]`**
   - Splitting `45` results in `[12, 3, 4, 5]`, which is sorted.
   - The code should output `YES`.

3. **Test Case 3: `[12, 3]`**
   - Splitting `12` results in `[1, 2, 3]`, which is sorted.
   - The code should output `YES`.

This test input should help identify the bug in the code by providing cases where the array cannot be sorted even after splitting elements.


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

 int main(){
    int t; cin >> t;
    while (t--)
    {
        int n; cin >> n;
        int arr[n];
        vector<int> v;
        bool res = true , sorted = true;
        for (int i = 0; i < n; i++)
        {
            cin >> arr[i];
            if(i>0 && arr[i]<arr[i-1]){
                sorted = false;
            }
        }
        if(sorted){
            cout << "YES" << endl;
            continue;
        }
        for (int i = 0; i < n-1; i++)
        {
           if(arr[i]>arr[i+1]){

            v.push_back(arr[i]/10);
            v.push_back(arr[i]%10);
        }
        else{
            v.push_back(arr[i]);
        }
        }
        for(int i=0 ; i<v.size()-1 ; i++){
            if(v[i]>v[i+1]){
                res = false;
                break;
            }
        }

        cout << (res ? "YES" : "NO") << endl;

    }

    return 0;
 }

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

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

Think step by step.

Response:
```plaintext
3
4
10 9 8 7
3
99 1 2
2
10 10
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
from collections import deque

def solve():
    n = int(input())
    a = list(map(int, input().split()))
    mi = 0
    for i in range(n):
        # if (list(map(int, list(str(a[i])))) == sorted(map(int, list(str(a[i])))) and
        #         a[i - 1] <= int(str(a[i])[0]) and a[i + 1] >= int(str(a[i])[-1])):
        #     a[i] = int(str(a[i])[-1])
        # else:
        #     print("no")
        #     return
        if a[i] >= mi:
            if list(map(int, list(str(a[i])))) == sorted(map(int, list(str(a[i])))) and int(str(a[i])[-1]) > mi:
                mi = int(str(a[i])[-1])
            else:
                mi = a[i]
        else:
            print("no")
            return
    print("yes")

# ttt = 1
ttt = int(input())
for ___ in range(ttt):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
4
10 11 12 13
3
10 10 10
5
9 10 11 12 13
```


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
def solve():
    n=input()
    lis=[int(j) for i in input().split() for j in i]
    if lis ==sorted(lis):
        print("YES")
        return
    else :
        print("NO")
        return
for i in range(int(input())):
       solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the following aspects:

1. **Input Parsing**: The buggy code incorrectly parses the input. It splits the entire input string into individual digits, which is not the intended behavior. We need to provide an input where this incorrect parsing leads to a wrong result.

2. **Edge Cases**: We should consider edge cases where the array contains elements that are at least 10, and the array is close to being sorted but not quite.

3. **Small Scale**: The input should be small and valid to ensure it fits within the constraints and is easy to verify.

Let's construct a test input that will expose the bug:

- **Test Case 1**: An array where the elements are close to being sorted, but the buggy code will incorrectly split the digits.
- **Test Case 2**: An array where the elements are already sorted, but the buggy code might still fail due to incorrect parsing.

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

```plaintext
2
3
12 3 4
2
1 10
```

### Explanation:

1. **Test Case 1**:
   - Input: `3 12 3 4`
   - Correct Output: `YES`
   - Explanation: The array can be transformed to `[1, 2, 3, 4]` by splitting the element `12`. The buggy code will incorrectly split `12` into `1` and `2` and then parse `3` and `4` individually, leading to an incorrect result.

2. **Test Case 2**:
   - Input: `2 1 10`
   - Correct Output: `YES`
   - Explanation: The array can be transformed to `[1, 1, 0]` by splitting the element `10`. The buggy code will incorrectly split `10` into `1` and `0`, but the array is already sorted, so the correct output should be `YES`.

This test input should help expose the bug in the provided buggy code.


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

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

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

public class Main {
    static int p=61;static long mod=3037000493l;
     static TreeSet<Integer> ss=new TreeSet<>();
    public static void main(String args[]) throws IOException {
        Scanner sc=new Scanner(System.in);
       int t=sc.nextInt();
        while(t-->0)
        {
            int n=sc.nextInt();
            ArrayList<Integer> al=new ArrayList<>();
            for(int i=0;i<n;i++)
            al.add(sc.nextInt());
            int flag=0;
            for(int i=0;i<n-1;i++)
            {
                if(al.get(i)>al.get(i+1))
                {
                    if(al.get(i)<10)
                    {
                        flag=1;break;
                    }
                    else{
                        int d1=al.get(i)%10;
                        int d2=al.get(i)/10;
                        if(d1<d2 || al.get(i+1)<d1)
                        {
                            flag=1;break;
                        }
                        else{
                            al.remove(i);
                            al.add(i,d1);al.add(i,d2);
                        }
                    }
                }
            }
            //System.out.println(al);
            for(int i=0;i<al.size()-1;i++)
            {
                if(al.get(i)>al.get(i+1)){ flag=1;break;}
            }
            if(flag==1)
            System.out.println("NO");
            else System.out.println("YES");
        }
    }

}

**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 where the algorithm might fail. Specifically, we need to look at scenarios where:

1. The array contains elements that are exactly 10, as these can be split into 1 and 0, which might cause issues with the sorting logic.
2. The array contains elements that, when split, might still result in a non-sorted array.
3. The array contains elements that are just below the threshold of 10, which can cause the code to incorrectly flag the array as unsortable.

Let's construct a test case that fits these criteria:

- The array contains elements that are exactly 10.
- The array contains elements that, when split, might still result in a non-sorted array.
- The array contains elements that are just below the threshold of 10.

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

```plaintext
3
5
10 10 11 12 13
3
10 9 10
4
10 10 10 10
```

Explanation:
1. The first test case: `[10, 10, 11, 12, 13]` should be sortable because splitting 10 results in `[1, 0, 1, 0, 1, 1, 1, 2, 1, 3]`, which is sorted.
2. The second test case: `[10, 9, 10]` should be unsortable because splitting 10 results in `[1, 0, 9, 1, 0]`, which is not sorted.
3. The third test case: `[10, 10, 10, 10]` should be sortable because splitting 10 results in `[1, 0, 1, 0, 1, 0, 1, 0]`, which is sorted.

This test input should help 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: Array Fix
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation any number of times (possibly zero): take any element of the array $$$a$$$, which is at least $$$10$$$, delete it, and instead insert the digits that element consisted of in the same position, in order they appear in that element.

For example:

- if we apply this operation to the $$$3$$$-rd element of the array $$$[12, 3, 45, 67]$$$, then the array becomes $$$[12, 3, 4, 5, 67]$$$.
- if we apply this operation to the $$$2$$$-nd element of the array $$$[2, 10]$$$, then the array becomes $$$[2, 1, 0]$$$.

Your task is to determine whether it is possible to make $$$a$$$ sorted in non-descending order using the aforementioned operation any number of times (possibly zero). In other words, you have to determine if it is possible to transform the array $$$a$$$ in such a way that $$$a_1 \le a_2 \le \dots \le a_k$$$, where $$$k$$$ is the current length of the array $$$a$$$.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

Each test case consists of two lines:

- the first line contains a single integer $$$n$$$ ($$$2 \le n \le 50$$$).
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 99$$$).

Output: For each test case, print YES if it is possible to make $$$a$$$ sorted in non-decreasing order using the aforementioned operation; otherwise, print NO.

You can print each letter in any case. For example, yes, Yes, YeS will all be recognized as a positive answer.

Examples:
input:
3
4
12 3 45 67
3
12 28 5
2
0 0
output:
YES
NO
YES

Note: In the first example, you can split the first element, then the array becomes $$$[1, 2, 3, 45, 67]$$$.

In the second example, there is no way to get a sorted array.

In the third example, the array is already sorted.

Buggy Code:
import java.util.*;

public class Test {
    static Scanner sr = new Scanner(System.in);
    static Integer[]a;
    public static void solve(){
        int n = sr.nextInt();
        a = new Integer[n];
        for(int i=0;i<n;++i)a[i] = sr.nextInt();
        List<Integer> list = new ArrayList<>(Arrays.asList(a));
        for(int i=0;i<n-1;++i){
            if(list.get(i) - list.get(i+1) > 0){
                int tmp = list.get(i);
                int c = tmp % 10;
                tmp /= 10;
                int d = tmp;
                list.remove(i);
                list.add(i,d);
                list.add(i+1,c);
                i++;
                n++;
            }
        }
        for(int i=0;i<list.size()-1;++i){
            if(list.get(i) - list.get(i+1) > 0){
                System.out.println("NO");
                return;
            }
        }
        System.out.println("YES");
    }
    public static void main(String[] args){
        int t = sr.nextInt();
        while(t-- > 0){
            solve();
        }
    }
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Arrays with elements exactly equal to 10.
   - Arrays with elements that are just below 10.
   - Arrays with elements that are just above 10.
   - Arrays where the first element is 10 and the second element is a two-digit number.
   - Arrays where the last element is 10 and the second last element is a two-digit number.

2. **Scenarios to Maximize Failure Probability:**
   - Arrays where the operation on a specific element does not immediately resolve the sorting issue.
   - Arrays where multiple operations are needed to achieve a sorted array.
   - Arrays where the operation on one element creates a new sorting issue.

3. **Small-Scale Test Input:**
   - We need a small-scale test input to ensure it fits within the output length limit.

Given these considerations, let's construct a test input that is likely to expose the bug in the code:

```plaintext
1
5
10 20 1 2 3
```

### Explanation:
- **Array: [10, 20, 1, 2, 3]**
  - The first element is 10, which when split becomes [1, 0].
  - The second element is 20, which when split becomes [2, 0].
  - After splitting, the array becomes [1, 0, 2, 0, 1, 2, 3].
  - The array is not sorted, and the code might not handle the splitting and repositioning correctly, leading to an incorrect output.

This test input is small and complete, and it should expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

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

#ifdef LOCAL
#include "debug.h"
#else
#define dbg(...)
#endif

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

void solve() {
    int n, q;
    cin >> n >> q;
    vector<int64_t> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] ^= a[i - 1] ^ rng();
    }
    dbg(a);
    while (q--) {
        int l, r;
        cin >> l >> r;
        dbg(l, r);
        cout << ((r - l) % 2 == 1 && a[l - 1] ^ a[r] == 0 ? "YES" : "NO") << endl;
    }
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cout << fixed << setprecision(20);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define debug(a) cout << #a << " = " << (a) << nl;
#define ll long long
#define int long long
#define nl '\n'

const int MX = 1e6 + 5;
int L, R, odd;
int block_size;
int mp[MX], a[MX], ans[MX];

bool cmp(array<int, 3> &x, array<int, 3> &y)
{
    if (x[0] / block_size == y[0] / block_size)
        return ((x[0] / block_size) & 1) ? x[1] > y[1] : x[1] < y[1];
    return x[0] / block_size < y[0] / block_size;
}
void add(int x)
{
    mp[a[x]]++;
    odd += (mp[a[x]] & 1 ? 1 : -1);
}
void del(int x)
{
    mp[a[x]]--;
    odd += (mp[a[x]] & 1 ? 1 : -1);
}

void jAVA()
{
    L = 1, R = 1;
    odd = 0;
    int n, q;
    cin >> n >> q;
    block_size = sqrtl(n);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    vector<array<int, 3>> Q(q);
    for(int i =0;i<q;++i) cin >> Q[i][0] >> Q[i][1], Q[i][2] = i;
    sort(Q.begin(), Q.end(), cmp);
    for(auto [l,r, id] : Q){
        while (L < l) del(L++);
        while (L > l) add(--L);
        while (R <= r) add(R++);
        while (R > r + 1) del(--R);

        ans[id] = odd == 0;
    }
    // for(int i = 0; i < q; ++i) cout << (ans[i]) << nl;
    for(int i = 0; i < q; ++i) cout << (ans[i] ? "YES" : "NO") << nl;
    for(int i=0;i<=n;++i) mp[i] = 0;
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int t = 1, cs = 0;
    cin >> t;
    while (t--){
        // cout << "Case " << ++cs << ": ";
        jAVA();
    }
    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
2
3 3
1 2 2
1 2
1 3
2 3
5 3
2 1 2 1 1
1 2
1 3
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

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

using namespace std;

#define ll long long
#define ii array<int, 2>

constexpr int mxN = 1e6+1;
constexpr int block = 1e3;

struct MoQuery {
	int l, r, idx;

	bool operator<(MoQuery& other) {
		return make_pair(l / block, r) < make_pair(other.l / block, other.r);
	}
};

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

	vector<int> freq(mxN);
	int count = 0;

	auto add = [&](int idx) -> void {
		freq[a[idx]]++;
		if (freq[a[idx]] & 1)
			count++;
		else
			count--;
	};

	auto remove = [&](int idx) -> void {
		freq[a[idx]]--;
		if (freq[a[idx]] & 1)
			count++;
		else
			count--;
	};

	vector<MoQuery> queries(q);
	vector<int> ans(q);
	for (int i = 0; i < q; i++) {
		int l, r; cin >> l >> r; l--; r--;
		queries[i] = {l, r, i};
	}
	sort(queries.begin(), queries.end());

	int l = 0, r = -1;
	for (auto qq : queries) {
		while (l > qq.l) add(--l);
		while (l < qq.l) remove(l++);
		while (r < qq.r) add(++r);
		while (r > qq.r) remove(r--);
		ans[qq.idx] = !count;
	}

	for (int i = 0; i < q; i++)
		cout << (ans[i] ? "YES" : "NO") << '\n';
}

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

	int t; cin >> t;
	while (t--)
		solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. The code uses a Mo's algorithm (or square root decomposition) to handle range queries efficiently. The key points to consider are:

1. **Edge Cases**: Small values of `n` and `q`.
2. **Large Ranges**: Queries that span the entire range or very large ranges.
3. **Small Ranges**: Queries that span very small ranges.
4. **Repetitive Values**: Target values that are repeated multiple times.
5. **Unique Values**: Target values that are all unique.
6. **Sorted and Unsorted Values**: Target values that are sorted and unsorted.

Given the constraints and the nature of the problem, let's create a test input that includes a mix of these scenarios to ensure the code is thoroughly tested.

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

### Explanation:
1. **Test Case Count**: 1
2. **Number of Targets and Queries**: 5 targets and 3 queries
3. **Target Values**: [1, 2, 3, 2, 1]
   - This sequence includes repeated values and is symmetric.
4. **Queries**:
   - Query 1: [1, 5] (entire range)
   - Query 2: [1, 3] (small range at the beginning)
   - Query 3: [3, 5] (small range at the end)

This test input is designed to:
- Test the entire range, which should be a challenging scenario for the algorithm.
- Test small ranges at the beginning and end, which can help expose issues with the add and remove operations.
- Use repeated values to ensure the frequency count logic is correct.

This should help in identifying any potential bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
for _ in range(int(input())):
    n, q = map(int, input().split())
    a = [0] + list(map(int, input().split()))
    ckd = [0] * (10**6 + 5)
    for i in a: ckd[i] += 1
    for _ in range(q):
        l, r = map(int, input().split())
        if (l - r + 1) & 1: print('no'); continue
        for i in range(l, r + 1):
            if ckd[i] & 1: print('no'); break
            else: print('yes'); break

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
import sys
from bisect import bisect_right

class SegmentTree:
    def __init__(self, data):
        self.n = len(data)
        self.tree = [None] * (4 * self.n)
        self._build(data, 0, 0, self.n - 1)

    def _build(self, data, node, tl, tr):
        if tl == tr:
            self.tree[node] = [data[tl]]
        else:
            tm = (tl + tr) // 2
            self._build(data, 2 * node + 1, tl, tm)
            self._build(data, 2 * node + 2, tm + 1, tr)
            self.tree[node] = sorted(self.tree[2 * node + 1] + self.tree[2 * node + 2])

    def query(self, node, tl, tr, l, r):
        if l > r:
            return []
        if l == tl and r == tr:
            return self.tree[node]
        tm = (tl + tr) // 2
        left_part = self.query(2 * node + 1, tl, tm, l, min(r, tm))
        right_part = self.query(2 * node + 2, tm + 1, tr, max(l, tm + 1), r)
        return sorted(left_part + right_part)

def solve():
    input = sys.stdin.readline
    t = int(input())
    results = []

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

        seg_tree = SegmentTree(a)

        for _ in range(q):
            l, r = map(int, input().split())
            l -= 1
            r -= 1

            targets = seg_tree.query(0, 0, n - 1, l, r)
            robin_score = 0
            sheriff_score = 0
            turn = 0

            while targets:
                if turn % 2 == 0:
                    robin_score += targets.pop()
                else:
                    sheriff_score += targets.pop()
                turn += 1

            if sheriff_score >= robin_score:
                results.append("YES")
            else:
                results.append("NO")

    print("\n".join(results))

if __name__ == "__main__":
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
for _ in range(int(input())):
    n, q = map(int, input().split())
    a = [int(i) for i in input().split()]
    b = [[int(i) for i in input().split()] for j in range(q)]
    ans = []
    m = sorted(a)[::-1]
    for i, j in b:
        c1, c2 = 0, 0
        for i in range(i, j):
            if i % 2 == 1:
                c1 += m[i]
            else:
                c2 += m[i]
        if c1 >= c2:
            print('NO')
        else:
            print('YES')

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
import java.util.*;

public class SheriffArchery {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int q = sc.nextInt();
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                a[i] = sc.nextInt();
            }

            long[] prefixSum = new long[n + 1];
            for (int i = 1; i <= n; i++) {
                prefixSum[i] = prefixSum[i - 1] + a[i];
            }

            for (int i = 0; i < q; i++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                long totalSum = prefixSum[r] - prefixSum[l - 1];
                long robinScore = (totalSum + 1) / 2;
                long sheriffScore = totalSum - robinScore;
                if (sheriffScore >= robinScore) {
                    System.out.println("YES");
                } else {
                    System.out.println("NO");
                }
            }
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
import static java.lang.Math.*;
import java.io.*;
import java.util.*;

public class Main {

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

    static void solve() {
        int n = sc.nextInt();
        int q = sc.nextInt();

        int[] a = sc.nextIntArray(n);

        List<List<Integer>> ll = new ArrayList<>();

        for(int i = 0;i < q;i++) {
            List<Integer> l = new ArrayList<>();
            l.add(sc.nextInt() - 1);
            l.add(sc.nextInt() - 1);
            l.add(i);
            ll.add(l);
        }

        int block = (int) sqrt(n);

        ll.sort((x, y) -> {
            if (x.get(0) / block != y.get(0) / block)
                return x.get(0) - y.get(0);
            return x.get(1) - y.get(1);
        });

        boolean[] res = new boolean[q];

        int[] cnt = new int[1000001];

        int curL = 0, curR = 0, odd = 1;
        cnt[a[0]]++;

        for(List<Integer> qu : ll) {

            int L = qu.get(0), R = qu.get(1);

            while(curR < R) {
                cnt[a[++curR]]++;
                if(cnt[a[curR]] % 2 == 1)
                    odd++;
                else
                    odd--;
            }

            while(curL > L) {
                cnt[a[--curL]]++;
                if(cnt[a[curL]] % 2 == 1)
                    odd++;
                else
                    odd--;
            }

            while(curL < L) {
                cnt[a[curL]]--;
                if(cnt[a[curL++]] % 2 == 1)
                    odd++;
                else
                    odd--;
            }

            while(curR > R) {
                cnt[a[curR]]--;
                if(cnt[a[curR--]] % 2 == 1)
                    odd++;
                else
                    odd--;
            }

            res[qu.get(2)] = odd == 0;
        }

        for(boolean b : res)
            out.println(b ? "YES" : "NO");
    }

    public static void main(String[] args) {

        int t = sc.nextInt();

        while(t-- > 0) {
            solve();
            out.flush();
        }
    }

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

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

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextLong();
            }
            return a;
        }

        double[] nextDoubleArray(int n) {
            double[] a = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextDouble();
            }
            return a;
        }

        List<Integer>[] nextGraph(int n) {
            List<Integer>[] graph = new List[n + 1];
            for (int i = 0; i <= n; i++)
                graph[i] = new ArrayList<>();
            for (int i = 0; i < n - 1; i++) {
                int u = sc.nextInt();
                int v = sc.nextInt();
                graph[u].add(v);
                graph[v].add(u);
            }
            return graph;
        }

        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 {
                if (st.hasMoreTokens())
                    str = st.nextToken("\n");
                else
                    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 test input that exposes the bug in the given buggy code, let's focus on the following aspects:

1. **Edge Cases**: Consider cases where the number of targets is very small or the range of queries is very specific.
2. **Odd and Even Counts**: The bug in the code seems to be related to the handling of the `odd` count, which is used to determine if the sum of the target values is even. We need to create scenarios where the `odd` count might be miscalculated.
3. **Queries with Overlapping Ranges**: Create queries that overlap in such a way that the code's handling of the `curL` and `curR` pointers might lead to incorrect results.

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

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

### Explanation:
- **Test Case 1**:
  - **Number of targets (n)**: 5
  - **Number of queries (q)**: 4
  - **Target values (a)**: 1, 2, 3, 4, 5
  - **Queries**:
    - **Query 1**: (1, 3) -> Targets: 1, 2, 3
    - **Query 2**: (2, 4) -> Targets: 2, 3, 4
    - **Query 3**: (1, 4) -> Targets: 1, 2, 3, 4
    - **Query 4**: (3, 5) -> Targets: 3, 4, 5

### Expected Output:
- **Query 1**: (1, 3) -> Sum = 1 + 2 + 3 = 6 (Even) -> **YES**
- **Query 2**: (2, 4) -> Sum = 2 + 3 + 4 = 9 (Odd) -> **NO**
- **Query 3**: (1, 4) -> Sum = 1 + 2 + 3 + 4 = 10 (Even) -> **YES**
- **Query 4**: (3, 5) -> Sum = 3 + 4 + 5 = 12 (Even) -> **YES**

This test input should help expose the bug in the code by testing various scenarios with overlapping ranges and different sums of target values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
for _ in range(int(input())):
    n,q=list(map(int,input().split()))
    a=list(map(int,input().split()))
    for i in range(q):
        l,r=list(map(int,input().split()))
        if (r-l)%2==0:print('NO')
        else:
            for j in range(l-1,r,2):
                if a[j]!=a[j+1]:print('NO');break
            else:
                print('YES')

# for _ in range(int(input())):
#     n,d,k=list(map(int,input().split()))
#     mp=[0]*100010
#     for i in range(k):
#         l,r=list(map(int,input().split()))
#         if l-d+1>1:mp[l-d+1]+=1
#         else:mp[1]+=1
#         mp[r+1]-=1
#     pre=0
#     l = [0] * 100010
#     for i in range(1,n+1):
#         pre+=mp[i]
#         l[i]=pre
#     mn=99999999999999;mx=-1;np=0
#     mni=-1;mxi=-1
#     for i in range(1-d,n-d+2):
#         if i>=1:
#             if mn>l[i]:
#                 mn=l[i]
#                 mni=i
#             if mx<l[i]:
#                 mx=l[i]
#                 mxi=i
#     print(mxi,mni)
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
# # n=int(input())
# # ans=0
# # a=[]
# # s=0
# # for i in range(n):
# #     x,y,z=list(map(int,input().split()))
# #     a.append([x/y,x,y,z])
# #     s+=x
# # a.sort()
# # print(a)
# # for i in range(n):
# #     s-=a[i][1]
# #     ans+=a[i][2]-a[i][3]*s
# # 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 expose the bug in the given code, we need to focus on the conditions where the code might fail. The primary issue in the code is that it only checks if the number of targets is even or odd and then checks if all pairs of targets have the same value. This is incorrect because it doesn't consider the optimal play strategy of both players.

To create a test input that exposes this bug, we need to consider scenarios where the number of targets is odd, but the optimal play does not necessarily result in a tie or a win for the Sheriff. Here are the steps to create such a test input:

1. **Odd Number of Targets**: Ensure the number of targets is odd.
2. **Non-Equal Values in Pairs**: Ensure that the values in pairs of targets are not always equal to force the code to print "NO" even when the Sheriff can win.

Let's create a small-scale test input:

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

### Explanation:
- **Test Case 1**:
  - **Number of Targets (n)**: 5
  - **Number of Queries (q)**: 3
  - **Target Values (a)**: 1, 2, 3, 4, 5

### Queries:
1. **Query 1 (1, 3)**:
   - Targets: 1, 2, 3
   - Optimal Play: Robin Hood can always win by choosing the higher value targets first.
   - Expected Output: "NO"

2. **Query 2 (1, 5)**:
   - Targets: 1, 2, 3, 4, 5
   - Optimal Play: Robin Hood can always win by choosing the higher value targets first.
   - Expected Output: "NO"

3. **Query 3 (3, 5)**:
   - Targets: 3, 4, 5
   - Optimal Play: Robin Hood can always win by choosing the higher value targets first.
   - Expected Output: "NO"

### Why This Test Input Exposes the Bug:
- The code incorrectly assumes that if the number of targets is odd and all pairs of targets have the same value, the game will be a tie. However, in this test input, the values are not equal in pairs, and the optimal play strategy will always result in Robin Hood winning, which the code will fail to detect.

This test input should effectively expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Hood Archery
time_limit_ms: 3000
memory_limit_mb: 256
Description: At such times archery was always the main sport of the day, for the Nottinghamshire yeomen were the best hand at the longbow in all merry England, but this year the Sheriff hesitated...

Sheriff of Nottingham has organized a tournament in archery. It's the final round and Robin Hood is playing against Sheriff!

There are $n$ targets in a row numbered from $1$ to $n$. When a player shoots target $i$, their score increases by $a_i$ and the target $i$ is destroyed. The game consists of turns and players alternate between whose turn it is. Robin Hood always starts the game, then Sheriff and so on. The game continues until all targets are destroyed. Both players start with score $0$.

At the end of the game, the player with most score wins and the other player loses. If both players have the same score, it's a tie and no one wins or loses. In each turn, the player can shoot any target that wasn't shot before. Both play optimally to get the most score possible.

Sheriff of Nottingham has a suspicion that he might lose the game! This cannot happen, you must help Sheriff. Sheriff will pose $q$ queries, each specifying $l$ and $r$. This means that the game would be played only with targets $l, l+1, \dots, r$, as others would be removed by Sheriff before the game starts.

For each query $l$, $r$, determine whether the Sheriff can not lose the game when only considering the targets $l, l+1, \dots, r$.

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

The first line of each test case contains two integers $n$, $q$ ($1 \le n,q \le 2\cdot10^5$) — the number of targets and the queries Sheriff will pose.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the points for hitting each target.

Then follow $q$ lines, each with two integers $l$ and $r$ ($1 \le l \le r \le n$) — the range of the targets that is considered for each query.

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

Output: For each query, output "YES", if the Sheriff does not lose the game when only considering the targets $l, l+1, \dots, r$, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note:

Buggy Code:
def can_sheriff_win(t, test_cases):
    results = []
    for case in test_cases:
        n, q, a, queries = case
        for l, r in queries:
            l -= 1  # Convert to 0-based index
            r -= 1
            total_score = sum(a[l:r+1])
            max_score = max(a[l:r+1])

            if total_score % 2 == 1:  # Odd total score
                results.append("NO")
            else:
                # Even total score check
                if max_score > total_score // 2:
                    results.append("NO")
                else:
                    results.append("YES")
    return results

# Input handling
t = int(input())
test_cases = []

for _ in range(t):
    n, q = map(int, input().split())
    a = list(map(int, input().split()))
    queries = [tuple(map(int, input().split())) for _ in range(q)]
    test_cases.append((n, q, a, queries))

# Getting results and printing
results = can_sheriff_win(t, test_cases)
print('\n'.join(results))

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