**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

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

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

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

using namespace std;

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

#define int long long
#define endl "\n"
#define inf LLONG_MAX
#define _inf LLONG_MIN
#define fors(i, a, b) for (long long i = a; i <= b; i++)
#define fors_(i, a, b) for (long long i = a; i >= b; i--)

void iLLuSiOnZ()
{
    int n,q;
    cin>>n>>q;
    vector<int> a(n);
    fors(i,0,n-1){
        cin>>a[i];
    }
    vector<int> x(n+1,0);
    fors(i,1,n){
        x[i]=x[i-1]^a[i-1];
    }
    while(q--){
        int l,r;
        cin>>l>>r;
        int val=x[r]^x[l-1];
        if(val==0){
            cout<<"YES"<<endl;
            continue;
        }
        int curr=0;
        int ct=0;
        fors(i,l,r){
            curr^=a[i-1];
            if(curr==val){
                ct++;
                curr=0;
            }
        }
        if(ct>1){
            cout<<"YES"<<endl;
        }
        else{
            cout<<"NO"<<endl;
        }
    }
    cout<<endl;
    return;
}

signed main()
{
    auto begin = std::chrono::high_resolution_clock::now();
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int _t = 1;
    cin >> _t;
    for (int _ = 1; _ <= _t; _++)
    {
        iLLuSiOnZ();
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds." << endl;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code has a specific issue in handling the XOR of subarrays and determining if they can be split into parts with equal XOR values.

### Key Points to Consider:
1. **Zero XOR Subarrays**: The code might incorrectly handle subarrays where the XOR of the entire subarray is zero.
2. **Multiple Segments**: The code might fail to correctly identify if a subarray can be split into multiple segments with the same XOR value.
3. **Edge Cases**: Small arrays, arrays with repeated elements, and arrays with all elements being zero.

### Test Input:
Let's create a small-scale test input that includes these edge cases:

```plaintext
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
6 4
0 0 0 0 0 0
1 6
2 5
3 4
1 4
7 4
1 2 3 4 5 6 7
1 7
2 6
3 5
1 4
8 4
1 2 2 3 3 4 4 5
1 8
2 7
3 6
1 6
```

### Explanation:
1. **First Test Case**:
   - Array: `[1, 1, 2, 3, 0]`
   - Queries:
     - `1 5`: The subarray `[1, 1, 2, 3, 0]` can be split as `[1], [1], [2, 3, 0]` with XOR values `1, 1, 1`. (Expected: YES)
     - `2 4`: The subarray `[1, 2, 3]` can be split as `[1, 2], [3]` with XOR values `3, 3`. (Expected: YES)
     - `3 5`: The subarray `[2, 3, 0]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `1 3`: The subarray `[1, 1, 2]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `3 4`: The subarray `[2, 3]` cannot be split into segments with equal XOR values. (Expected: NO)

2. **Second Test Case**:
   - Array: `[0, 0, 0, 0, 0, 0]`
   - Queries:
     - `1 6`: The subarray `[0, 0, 0, 0, 0, 0]` can be split into any number of segments with XOR value `0`. (Expected: YES)
     - `2 5`: The subarray `[0, 0, 0, 0]` can be split into any number of segments with XOR value `0`. (Expected: YES)
     - `3 4`: The subarray `[0, 0]` can be split into any number of segments with XOR value `0`. (Expected: YES)
     - `1 4`: The subarray `[0, 0, 0, 0]` can be split into any number of segments with XOR value `0`. (Expected: YES)

3. **Third Test Case**:
   - Array: `[1, 2, 3, 4, 5, 6, 7]`
   - Queries:
     - `1 7`: The subarray `[1, 2, 3, 4, 5, 6, 7]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `2 6`: The subarray `[2, 3, 4, 5, 6]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `3 5`: The subarray `[3, 4, 5]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `1 4`: The subarray `[1, 2, 3, 4]` cannot be split into segments with equal XOR values. (Expected: NO)

4. **Fourth Test Case**:
   - Array: `[1, 2, 2, 3, 3, 4, 4, 5]`
   - Queries:
     - `1 8`: The subarray `[1, 2, 2, 3, 3, 4, 4, 5]` cannot be split into segments with equal XOR values. (Expected: NO)
     - `2 7`: The subarray `[2, 2, 3, 3, 4, 4]` can be split as `[2, 2], [3, 3], [4, 4]` with XOR values `0, 0, 0`. (Expected: YES)
     - `3 6`: The subarray `[2, 3, 3, 4]` can be split as `[2, 3, 3, 4]` with XOR value `0`. (Expected: YES)
     - `1 6`: The subarray `[1, 2, 2, 3, 3, 4]` cannot be split into segments with equal XOR values. (Expected: NO)

This test input should help expose the bugs in the provided code by covering a variety of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

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

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

Buggy Code:
from sys import stdin
input=lambda :stdin.readline()[:-1]
def solve():
	n,q = map(int,input().split())

	a = [0] + list(map(int,input().split()))
	id_map = {0: [0]}

	for i in range(1, n + 1):
		a[i] ^= a[i - 1]
		if a[i] not in id_map:
			id_map[a[i]] = []
		id_map[a[i]].append(i)

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

		if a[r] == a[l - 1]:
			print("YES")
			continue
		pleft = []
		for i in id_map[a[l-1]]:
			if i < r:
				pleft.append(i)
		if pleft:
			pl = pleft[-1]
		else:
			pl = -1
		pright = []
		for i in id_map[a[r]]:
			if i >= l:
				pright.append(i)
		if pright:
			pr = pright[0]
		else:
			pr = n+1

		if pl > pr :
			print("YES")
		else:
			print("NO")

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

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

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

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

Buggy Code:
import sys
input = sys.stdin.readline

from random import randint
RANDOM = randint(1, 10 ** 9)
class myHash(int):
    def __init__(self, x):
        int.__init__(x)

    def __hash__(self):
        return super(myHash, self).__hash__() ^ RANDOM

def readList():
    return list(map(int, input().split()))
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()

# BF-OB5, EC, DB, CC, CL
def solve():
    n, q = readInts()
    a = readList()
    xv = [0]
    for v in a:
        xv.append(xv[-1] ^ v)
    a = [0] + a

    se_r, se_l = [[] for _ in range(n+1)], [[] for _ in range(n+1)]
    ans = ["NO"] * q
    for i in range(q):
        l, r = readInts()
        if xv[r] ^ xv[l-1] == 0:
            ans[i] = "YES"
            continue
        se_r[r].append([l, i, -1])
        se_l[l].append([r, i, -1])

    for i in range(1, n+1):
        se_l[i].sort()
        se_r[i].sort()

    val = {}
    v = 0
    for i in range(1, n+1):
        v ^= a[i]
        hv = myHash(v)
        if hv in val:
            b = val[hv]
            idx, m = 0, len(b)
            for j in range(len(se_r[i])):
                while idx < m and b[idx] < se_r[i][j][0]:
                    idx += 1
                if idx < m and b[idx] + 1 < i:
                    se_r[i][j][2] = b[idx] + 1
        else:
            val[hv] = []
        val[hv].append(i)

    val = {}
    v = 0
    for i in range(n, 0, -1):
        v ^= a[i]
        hv = myHash(v)
        if hv in val:
            b = val[hv]
            idx, m = 0, len(b)
            for j in range(len(se_l[i])-1, -1, -1):
                while idx < m and b[idx] > se_l[i][j][0]:
                    idx += 1
                if idx < m and b[idx] - 1 > i:
                    se_l[i][j][2] = b[idx] - 1
        else:
            val[hv] = []
        val[hv].append(i)

    cnt = [[-1, -1] for _ in range(q)]
    for i in range(1, n+1):
        for j in range(len(se_l[i])):
            if se_l[i][j][-1] == -1:
                continue
            r, idx, rp = se_l[i][j]
            cnt[idx][1] = rp

    for i in range(1, n+1):
        for j in range(len(se_r[i])):
            if se_r[i][j][-1] == -1:
                continue
            l, idx, lp = se_r[i][j]
            cnt[idx][0] = lp
            if l <= lp <= cnt[idx][1] <= i:
                if xv[lp-1] ^ xv[l-1] == xv[i] ^ xv[cnt[idx][1]] == xv[i] ^ xv[l-1] > 0:
                    ans[idx] = "YES"
    print("\n".join(ans))
    return

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

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

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

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

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));;
    static StringTokenizer st;
    static int n, q, l, r, a[], xor[];
    public static void main(String[] args) throws IOException {
        int T = readInt();
        while (T-- > 0) {
            solve();
        }
    }
    static void solve() throws IOException {
        n = readInt(); q = readInt(); a = new int[n + 1]; xor = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = readInt();
            xor[i] = xor[i - 1] ^ a[i];
        }
        while (q-- > 0) {
            l = readInt(); r = readInt();
            if ((xor[r] ^ xor[l - 1]) == 0) { // two parts, where we have a prefix xor x, and a suffix xor x between a[l...r]
                System.out.println("YES"); continue;
            }
            boolean flg = false;
            int x = xor[r] ^ xor[l - 1];
            for (int i = l; i < r - 1; i++) {
                if ((xor[i] ^ xor[l - 1]) == x) {
                    for (int j = i; j < r; j++) {
                        if ((xor[j] ^ xor[i]) == x) {
                            flg = true; break;
                        }
                    }
                    break;
                }
            }
            System.out.println(flg ? "YES" : "NO");
        }
        System.out.println();
    }

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }

    static long readLong() throws IOException {
        return Long.parseLong(next());
    }

    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }

    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }

    static char readCharacter() throws IOException {
        return next().charAt(0);
    }

    static String readLine() throws IOException {
        return br.readLine().trim();
    }
}

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

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

Think step by step.

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

1. **Zero XOR Subarrays**: The code might not handle subarrays where the XOR of the entire subarray is zero correctly.
2. **Single Element Subarrays**: The code should handle queries where the subarray is a single element.
3. **Multiple Segments**: The code should correctly identify subarrays that can be split into multiple segments with the same XOR.
4. **Boundary Conditions**: The code should handle boundary conditions, such as subarrays starting from the first element or ending at the last element.

Given these points, let's create a small-scale test input that includes these edge cases:

```plaintext
2
6 4
0 1 2 3 4 5
1 6
2 4
3 6
1 3
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
```

### Explanation:
1. **Test Case 1**:
   - Array: `[0, 1, 2, 3, 4, 5]`
   - Queries:
     - `1 6`: The subarray is `[0, 1, 2, 3, 4, 5]`. The XOR of the entire subarray is 1. This subarray cannot be split into multiple segments with the same XOR.
     - `2 4`: The subarray is `[1, 2, 3]`. The XOR of the entire subarray is 0. This subarray can be split into `[1, 2]` and `[3]`.
     - `3 6`: The subarray is `[2, 3, 4, 5]`. The XOR of the entire subarray is 4. This subarray cannot be split into multiple segments with the same XOR.
     - `1 3`: The subarray is `[0, 1, 2]`. The XOR of the entire subarray is 3. This subarray cannot be split into multiple segments with the same XOR.

2. **Test Case 2**:
   - Array: `[1, 1, 2, 3, 0]`
   - Queries:
     - `1 5`: The subarray is `[1, 1, 2, 3, 0]`. This subarray can be split into `[1], [1], [2, 3, 0]`.
     - `2 4`: The subarray is `[1, 2, 3]`. The XOR of the entire subarray is 0. This subarray can be split into `[1, 2]` and `[3]`.
     - `3 5`: The subarray is `[2, 3, 0]`. The XOR of the entire subarray is 1. This subarray cannot be split into multiple segments with the same XOR.
     - `1 3`: The subarray is `[1, 1, 2]`. The XOR of the entire subarray is 2. This subarray cannot be split into multiple segments with the same XOR.
     - `3 4`: The subarray is `[2, 3]`. The XOR of the entire subarray is 1. This subarray cannot be split into multiple segments with the same XOR.

This test input covers a variety of cases that should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

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

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

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

public class codeforce1 {

  // static class Reader {
  // final private int BUFFER_SIZE = 1 << 16;
  // private DataInputStream din;
  // private byte[] buffer;
  // private int bufferPointer, bytesRead;

  // public Reader() {
  // din = new DataInputStream(System.in);
  // buffer = new byte[BUFFER_SIZE];
  // bufferPointer = bytesRead = 0;
  // }

  // public Reader(String file_name) throws IOException {
  // din = new DataInputStream(
  // new FileInputStream(file_name));
  // buffer = new byte[BUFFER_SIZE];
  // bufferPointer = bytesRead = 0;
  // }

  // public String readLine() throws IOException {
  // byte[] buf = new byte[64]; // line length
  // int cnt = 0, c;
  // while ((c = read()) != -1) {
  // if (c == '\n') {
  // if (cnt != 0) {
  // break;
  // } else {
  // continue;
  // }
  // }
  // buf[cnt++] = (byte) c;
  // }
  // return new String(buf, 0, cnt);
  // }

  // public int nextInt() throws IOException {
  // int ret = 0;
  // byte c = read();
  // while (c <= ' ') {
  // c = read();
  // }
  // boolean neg = (c == '-');
  // if (neg)
  // c = read();
  // do {
  // ret = ret * 10 + c - '0';
  // } while ((c = read()) >= '0' && c <= '9');

  // if (neg)
  // return -ret;
  // return ret;
  // }

  // public long nextLong() throws IOException {
  // long ret = 0;
  // byte c = read();
  // while (c <= ' ')
  // c = read();
  // boolean neg = (c == '-');
  // if (neg)
  // c = read();
  // do {
  // ret = ret * 10 + c - '0';
  // } while ((c = read()) >= '0' && c <= '9');
  // if (neg)
  // return -ret;
  // return ret;
  // }

  // public double nextDouble() throws IOException {
  // double ret = 0, div = 1;
  // byte c = read();
  // while (c <= ' ')
  // c = read();
  // boolean neg = (c == '-');
  // if (neg)
  // c = read();

  // do {
  // ret = ret * 10 + c - '0';
  // } while ((c = read()) >= '0' && c <= '9');

  // if (c == '.') {
  // while ((c = read()) >= '0' && c <= '9') {
  // ret += (c - '0') / (div *= 10);
  // }
  // }

  // if (neg)
  // return -ret;
  // return ret;
  // }

  // private void fillBuffer() throws IOException {
  // bytesRead = din.read(buffer, bufferPointer = 0,
  // BUFFER_SIZE);
  // if (bytesRead == -1)
  // buffer[0] = -1;
  // }

  // private byte read() throws IOException {
  // if (bufferPointer == bytesRead)
  // fillBuffer();
  // return buffer[bufferPointer++];
  // }

  // public void close() throws IOException {
  // if (din == null)
  // return;
  // din.close();
  // }
  // }

  static class FastReader {
    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

    String nextLine() {
      String str = "";
      try {
        if (st.hasMoreTokens()) {
          str = st.nextToken("\n");
        } else {
          str = br.readLine();
        }
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }
  }

  static void merge(int arr[], int l, int m, int r, int brr[]) {
    // Find sizes of two subarrays to be merged
    int n1 = m - l + 1;
    int n2 = r - m;

    // Create temp arrays
    int L[] = new int[n1];
    int R[] = new int[n2];
    int L1[] = new int[n1];
    int R1[] = new int[n2];

    // Copy data to temp arrays
    for (int i = 0; i < n1; ++i) {
      L[i] = arr[l + i];
      L1[i] = brr[l + i];
    }
    for (int j = 0; j < n2; ++j) {
      R[j] = arr[m + 1 + j];
      R1[j] = brr[m + 1 + j];
    }

    // Merge the temp arrays
    // Initial indexes of first and second subarrays
    int i = 0, j = 0;

    // Initial index of merged subarray array
    int k = l;
    while (i < n1 && j < n2) {
      if (L[i] <= R[j]) {
        arr[k] = L[i];
        brr[k] = L1[i];
        i++;
      } else {
        arr[k] = R[j];
        brr[k] = R1[j];
        j++;
      }
      k++;
    }

    // Copy remaining elements of L[] if any
    while (i < n1) {
      arr[k] = L[i];
      brr[k] = L1[i];
      i++;
      k++;
    }

    // Copy remaining elements of R[] if any
    while (j < n2) {
      arr[k] = R[j];
      brr[k] = R1[j];
      j++;
      k++;
    }
  }

  // Main function that sorts arr[l..r] using
  // merge()
  static void sort(int arr[], int l, int r, int brr[]) {
    if (l < r) {
      // Find the middle point
      int m = (l + r) / 2;

      // Sort first and second halves
      sort(arr, l, m, brr);
      sort(arr, m + 1, r, brr);

      // Merge the sorted halves
      merge(arr, l, m, r, brr);
    }
  }

  public static int gcdExtended(int a, int b, int x,
      int y) {
    // Base Case
    if (a == 0) {
      x = 0;
      y = 1;
      return b;
    }

    int x1 = 1,
        y1 = 1; // To store results of recursive call
    int gcd = gcdExtended(b % a, a, x1, y1);

    // Update x and y using results of recursive
    // call
    x = y1 - (b / a) * x1;
    y = x1;

    return gcd;
  }

  static boolean isPossible(char a[], char b[], int mid) {
    if (mid == 0)
      return true;
    int i = 0;
    int j = 0;
    int cnt = 0;
    while (i < a.length && j < b.length) {
      if (a[i] == b[j]) {
        i++;
        j++;
        cnt++;
        if (cnt == mid)
          break;
      } else {
        j++;
      }
    }
    if (cnt == mid)
      return true;
    else
      return false;
  }

  public static void main(String[] args) throws java.lang.Exception {

    FastReader sc = new FastReader();
    int test = 1;
    test = sc.nextInt();
    while (test-- > 0) {
      int n = sc.nextInt();
      int q = sc.nextInt();
      int a[] = new int[n];
      for (int i = 0; i < a.length; i++) {
        a[i] = sc.nextInt();
      }

      long x[] = new long[n];
      x[0] = a[0];
      for (int i = 1; i < n; i++) {
        x[i] = x[i-1] ^ a[i];
      }
      int l, r;
      for (int i = 0; i < q; i++) {
        l = sc.nextInt() - 1;
        r = sc.nextInt() - 1;
        long p = 0;
        if (l != 0)
          p = x[l - 1];
        long temp = x[r] ^ p;
        if (temp == 0) {
          System.out.println("YES");
          continue;
        } else {
          int k = 0;
          int j;
          for (j = l; j < r; j++) {
            if ((p ^ x[j]) == temp) {
              k = 1;
              break;
            }
          }
          for (int m = r; m > j; m--) {
            if ((x[r] ^ x[m - 1]) == temp) {
              k++;
              break;
            }
          }
          if (k == 2) {
            System.out.println("YES");
          } else {
            System.out.println("NO");
          }
        }
      }
    }
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;

int n;

int s_toi(string s){
//	cout<<"STOI: S="<<s<<"   ";
	int ret = 0;
	for(int i=s.size()-1; i>=0;i--){
		ret=ret*10+(s[i]-'0');
	}
//	cout<<"RET "<<ret<<endl;
	return ret;
}

int joinn(string s, int times){
//	cout<<"JOIN: S,TIMES = "<<s<<" "<<times<<"   ";
	//���� sƴ��������Ϊtimes �õ����ַ��� תΪ��ֵ�Ľ��
	int ret = 0;
	int cnt = 0;
	string tmp = "";
	while(1){
		if(cnt>=times)break;
		cnt+=s.size();
		tmp+=s;
	}
	//�ȸ��s�ĺܳ���һ��ƴ���ַ��� �ٽ�ȡǰtimesλ
	tmp = tmp.substr(0,times);//substr(��ʼλ��,����)
	ret = s_toi(tmp);
//	cout<<"= "<<ret<<endl;
	return ret;
}

int main(){
	int T; cin>>T;
	while(T--){
		int cnt=0;
		vector<pair<int,int> > v;
		cin>>n;
		int lenn = to_string(n).size();
		for(int a=1;a<=10000;a++){
			for(int lenans1 = 1; lenans1<=6; lenans1++){
				int b = a*lenn-lenans1;//�� b = a*len(n) - len(ans1) ����b
				//����Ҫ���� b = n*a - ans1, �� ans1����nƴ��������Ϊ(����)a*len(n)-bҲ����lenans1�ó���
				if(1<=b && b<=min(10000,a*n)){//��Ŀ�е�b��Χ
					int anstmp = joinn(to_string(n),lenans1);
					if(b==n*a-anstmp){
						cnt++;
						v.push_back({a,b});
					}
				}
			}
		}
		cout<<cnt<<endl;
		for(int i=0;i<v.size();i++){
			cout<<v[i].first<<" "<<v[i].second<<endl;
		}
	}
	/*
	ans1 = n*a-b
	ans2 = nƴ��a��-ĩβbλ
	ans1=ans2
	==>
	b = n*a - ans1
	b = a*len(n) - len(ans1)
	n*a-ans1 = a*len(n)-len(ans1)
	����: n:1e2,a:1e4 -> ans1:1e6, len(ans1)<=6. ö��a��ans1�����ʣ��
	*/
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a specific issue with how it handles the string operations and conversions. Here are the steps to create such a test input:

1. **Edge Cases for `n`**:
   - Use values of `n` that are single digits, double digits, and triple digits.
   - Use values of `n` that are close to the upper limit (100).

2. **Edge Cases for `a` and `b`**:
   - Use values of `a` that are close to the upper limit (10000).
   - Use values of `b` that are close to the upper limit (10000 or `a * n`).
   - Use values of `b` that are just below or equal to the length of the string `s` formed by concatenating `n` `a` times.

3. **Combining Edge Cases**:
   - Combine the above edge cases to create test cases that are likely to expose the bug in the string operations and conversions.

Here is a small-scale test input that includes these edge cases:

```plaintext
4
1
9
99
100
```

This test input includes:
- The smallest possible value for `n` (1).
- A single-digit value for `n` (9).
- A double-digit value for `n` (99).
- The largest possible value for `n` (100).

These values will help to test the edge cases and ensure that the buggy code's string operations and conversions are thoroughly checked.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
using namespace std;

int main()
{
	int m = 0;
	cin >> m;
	while (m--)
	{
		int n = 0;
		cin >> n;
		vector<pair<int, int>> arr;
		if (n < 10)
		{
			int count = 0;
			for (int a = 1; a < 2300; a++)
			{
				for (int b = n; b < 2300; b += n)
				{
					if (a <= b)
						continue;
					if (a * n - b > pow(10, a - b) || a * n - b < pow(10, a - b - 1))
						continue;
					else
					{
						string c = to_string(n);
						string temp(a - b, c[0]);
						int string = stoi(temp);
						if (string == a * n - b)
						{
							count++;
							pair<int, int> ans(a, b);
							arr.push_back(ans);
						}
					}

				}
			}
			cout << count<<endl;
			for (int i = 0; i < arr.size(); i++)
			{
				cout << arr[i].first << " " << arr[i].second << endl;
			}
		}
		else if (n >= 10 && n < 100)
		{
			int count = 0;
			for (int a = 1; a < 2000; a++)
			{
				for (int b = n; b < 2600; b++)
				{
					if (a * 2 <= b)
						continue;
					if (a * n - b < pow(10, 2 * a - b - 1))
						continue;
					if (a * n - b > pow(10, 2 * a - b))
						break;
					else
					{
						string c = to_string(n);
						string string1(a * 2, ' ');
						for (int i = 0; i < a * 2; i += 2)
						{
							string1[i] = c[0];
						}
						for (int i = 1; i < a * 2; i += 2)
						{
							string1[i] = c[1];
						}
						string string2(string1.begin(), string1.begin() + a * 2 - b);
						int string3 = stoi(string2);
						if (string3 == a * n - b)
						{
							count++;
							pair<int, int> ans(a, b);
							arr.push_back(ans);
						}
					}

				}
			}
			cout << count<<endl;
			for (int i = 0; i < arr.size(); i++)
			{
				cout << arr[i].first << " " << arr[i].second << endl;
			}
		}
		else if (n == 100)
		{
			cout << 0 << 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
1
9
99
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
/*
author tryandtryagain（codeforces，atcoder）
nowcowid 觅者
luoguid zkystudent2027
*/
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
#include<iomanip>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<utility>
#include<cstdio>
#include<cstdlib>
#include<climits>
#include<cstdint>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
using namespace std;
#define int long long
#define INF 0x3f3f3f
#define f(i,s,e) for(int i=s;i<e;i++)
#define cf(i,s,e) for(int i=s;i<=e;i++)
#define rf(i,e,s) for(int i=e;i>s;i--)
#define pb push_back
#define db double
#define pii pair<int,int>
#define fstdio ios_base::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define endl '\n'
template <typename T1>istream& operator>>(istream& cin, vector<T1>& a) { for (auto& x : a)cin >> x; return cin; }
const double Pi = acos(-1.0);
constexpr int N = 2e5 + 10;
int T;
int a[10], b[10];
signed main()
{
	fstdio;
	cin >> T;
	while (T--) {
		int n;
		cin >> n;
		int cnt = 0;
		if (n <= 9) {
			if (n == 1) {
				cout << 0 << endl;
				continue;
			}
			if ((11 * n - 2) % (n - 1) == 0) {
				if ((11 * n - 2) / (n - 1) <= 10000) {
					a[++cnt] = (11 * n - 2) / (n - 1);
					b[cnt] = a[cnt] - 2;
				}
			}
			if ((111 * n - 3) % (n - 1) == 0) {
				if ((111 * n - 3) / (n - 1) <= 10000) {
					a[++cnt] = (111 * n - 3) / (n - 1);
					b[cnt] = a[cnt] - 3;
				}
			}
			if ((1111 * n - 4) % (n - 1) == 0) {
				if ((1111 * n - 4) / (n - 1) <= 10000) {
					a[++cnt] = (1111 * n - 4) / (n - 1);
					b[cnt] = a[cnt] - 4;
				}
			}
			if ((11111 * n - 5) % (n - 1) == 0) {
				if ((11111 * n - 5) / (n - 1) <= 10000) {
					a[++cnt] = (11111 * n - 5) / (n - 1);
					b[cnt] = a[cnt] - 5;
				}
			}
			if ((111111 * n - 6) % (n - 1) == 0) {
				if ((111111 * n - 6) / (n - 1) <= 10000) {
					a[++cnt] = (111111 * n - 6) / (n - 1);
					b[cnt] = a[cnt] - 6;
				}
			}
			cout << cnt << endl;
			cf(i, 1, cnt) {
				cout << a[i] << " " << b[i] << endl;
			}
		}
		else if (n >= 10 && n < 100) {
			if ((10 * n - 2) % (n - 2) == 0) {
				//cout << "YES" << endl;
				if ((10 * n - 2) / (n - 2) <= 10000 && (2 * (10 * n - 2) / (n - 2)) - 3 <= 10000) {
					a[++cnt] = (10 * n - 2) / (n - 2);
					b[cnt] = a[cnt] * 2 - 3;
				}
			}
			if ((101 * n - 4) % (n - 2) == 0) {
				//cout << "YES" << endl;
				if ((101 * n - 4) / (n - 2) <= 10000 && (2 * (101 * n - 4) / (n - 2)) - 4 <= 10000) {
					a[++cnt] = (101 * n - 4) / (n - 2);
					b[cnt] = a[cnt] * 2 - 4;
				}
			}
			if ((1010 * n - 4) % (n - 2) == 0) {
				//cout << "YES" << endl;
				if ((1010 * n - 4) / (n - 2) <= 10000 && (2 * (1010 * n - 4) / (n - 2)) - 5 <= 10000) {
					a[++cnt] = (1010 * n - 4) / (n - 2);
					b[cnt] = a[cnt] * 2 - 5;
				}
			}
			if ((10101 * n - 6) % (n - 2) == 0){
				//cout << "NO" << endl;
				if ((10101 * n - 6) / (n - 2) <= 10000 && (2 * (10101 * n - 6) / (n - 2)) - 6 <= 10000) {
					a[++cnt] = (10101 * n - 6) / (n - 2);
					b[cnt] = a[cnt] * 2 - 6;
				}
			}
			if ((101010 * n - 6) % (n - 2) == 0) {
				//cout << "NO" << endl;
				if ((101010 * n - 6) / (n - 2) <= 10000 && (2 * (101010 * n - 6) / (n - 2)) - 7 <= 10000) {
					a[++cnt] = (101010 * n - 6) / (n - 2);
					b[cnt] = a[cnt] * 2 - 7;
				}
			}
			cout << cnt << endl;
			cf(i, 1, cnt) {
				cout << a[i] << " " << b[i] << endl;
			}
		}
		else {
			cout << 0 << endl;
		}
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key issues in the buggy code are related to the string manipulations and the conditions under which the solution gives the correct answer.

### Steps to Generate the Test Input:
1. **Edge Case for Single-Digit \( n \):**
   - For \( n = 1 \), the solution should output 0 because there are no valid pairs \((a, b)\) that satisfy the conditions.
   - For \( n = 9 \), we need to check if the conditions for \( a \) and \( b \) are met, especially for the cases where \( n \) is close to 10.

2. **Edge Case for Two-Digit \( n \):**
   - For \( n = 10 \), the solution should handle the transition from single-digit to two-digit values correctly.
   - For \( n = 99 \), we need to check if the conditions for \( a \) and \( b \) are met, especially for the cases where \( n \) is close to 100.

3. **Intermediate Values:**
   - For \( n = 50 \), we need to check if the conditions for \( a \) and \( b \) are met, especially for values that are neither close to the edge cases nor trivial.

### Test Input:
```plaintext
3
1
9
50
```

### Explanation:
- **For \( n = 1 \):**
  - The solution should output 0 because there are no valid pairs \((a, b)\) that satisfy the conditions.
- **For \( n = 9 \):**
  - The solution should handle the transition from single-digit to two-digit values correctly. The conditions for \( a \) and \( b \) need to be checked for values close to 10.
- **For \( n = 50 \):**
  - The solution should handle intermediate values correctly. The conditions for \( a \) and \( b \) need to be checked for values that are neither close to the edge cases nor trivial.

This test input is designed to expose potential bugs in the buggy code by covering a range of \( n \) values, including edge cases and intermediate values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
from collections import *
from math import *
from heapq import *
import sys
from bisect import *
from random import randint
rrr = randint(8888,88888)

def sol():
    N = int(input())
    if N == 1:return print(0)
    my = N >= 10

    s = str(N) * 5
    ans = []
    for i in range(1,min(len(s),7)):
        BT = T = int(s[:i])
        T -= i
        deno = N -1 - my
        if T % deno:continue
        T //= deno
        # print(i,BT)
        # print((N-1 - my),N,T)
        x = T
        y = N * x - BT
        if ( 1 <= x <= 10000) and (1 <= y <= 10000) and (y <= x * N):
            # print(T , N , T , i, BT)
            # print(T * N - T + i)
            if (x * N - y == BT):
                ans.append((x,y))

    print(len(ans))
    for i,j in ans:
        print(i,j)

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

**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
1
9
99
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
for _ in range(int(input())):

    n = input()

    ans = []
    l = len(n)

    for i in range(1 , 10001):
        mnb = max(1 , l*i - 5)
        mxb = l*i
        for b in range(mnb , mxb):
            x = int(n)*i   - b
            y = 0
            for j in range(l*i - b):
                y = y*10  + int(n[j%l])

            if x==y:
                ans.append((i , b))

    print(len(ans))

    for p in ans:
        print(*p)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases that highlight the incorrect string manipulation and subtraction logic. Specifically, we need to ensure that the string representation of `n` and the operations on it produce unexpected results that differ from the correct integer arithmetic.

### Steps to Generate the Test Input:
1. **Choose a value of `n` that has a unique and distinct string representation**: This will help in identifying issues with string manipulation.
2. **Select values of `a` and `b` that will maximize the difference between the correct integer result and the string result**:
   - For large values of `a`, the string `n * a` will be very long.
   - For values of `b` close to the length of the string `n * a`, the subtraction will remove a significant portion of the string, potentially leaving a different result than the integer subtraction.
3. **Ensure the values of `a` and `b` are within the constraints**:
   - `1 <= a <= 10000`
   - `1 <= b <= min(10000, a * n)`

### Chosen Test Case:
- **`n = 9`**: This is a single-digit number, which will help in clearly seeing the string manipulation issues.
- **`a = 11`**: This will create a string `n * a` of length 11.
- **`b = 10`**: This will remove the last 10 characters from the string, leaving a single character.

### Expected Behavior:
- **Correct Integer Calculation**: `9 * 11 - 10 = 99 - 10 = 89`
- **Buggy String Calculation**: `99999999999 - 10 = "99999999999" - 10 = "9999999999"` (which is 9999999999, not 89)

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

### Explanation:
- The input `9` will be treated as a string in the buggy code.
- For `a = 11`, the string `n * a` will be `"99999999999"`.
- For `b = 10`, the string `"99999999999"` will be shortened by 10 characters, resulting in `"9999999999"`.
- The integer value of `"9999999999"` is 9999999999, which is not equal to the correct result of 89.

This test case should expose the bug in the string manipulation and subtraction logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
import java.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import java.util.List;

public class Main{
    public static void ArrayInput(String [] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            Seq[i] = fr.next();
        }
    }
    public static void ArrayInput(long [][] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            for(int j=0; j<Seq[0].length; j++){
                Seq[i][j] = fr.nextLong();
            }
        }
    }
    public static void ArrayInput(long [] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            Seq[i] = fr.nextLong();
        }
    }
    public static void ArrayInput(float [] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            Seq[i] = Float.parseFloat(fr.next());
        }
    }
    public static void ArrayInput(double [] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            Seq[i] = Double.parseDouble(fr.next());
        }
    }
    public static void ArrayInput(int [] Seq, FastReader fr){
        for(int i=0; i<Seq.length; i++){
            Seq[i] = fr.nextInt();
        }
    }
    public static void Prefix2D(int [][] grid){
        int n = grid.length;
        int m = grid[0].length;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                long case1 = 0, case2 = 0, case3 = 0;
                if(i-1 >= 0 && j-1 >= 0){
                    case1 = grid[i-1][j-1];
                }
                if(i-1 >= 0){
                    case2 = grid[i-1][j];
                }
                if(j-1 >= 0){
                    case3 = grid[i][j-1];
                }
                grid[i][j] += case2 + case3 - case1;
//                System.out.println(i + " " + j + " " + grid[i][j]);
            }
        }
    }
    static int [] nr = new int [] {-1, 0, 1, 0};
    static int [] nc = new int [] {0, 1, 0, -1};

    //    ---------------------------------------------MAIN CODE-------------------------------------------------
    public static void Solve(FastReader fr, PrintWriter out) {
        int n = fr.nextInt();
        int lenN = len(n);
        String str = String.valueOf(n);
        int rem = 0;
        int count = 0;
        int ans = 0;
        List<int []> list = new ArrayList<>();

        if(n == 1){
            out.println("0");
            return;
        }
        while (count < 6){
            rem *= 10;
            rem += str.charAt(count % lenN)-'0';

            int gap = len(rem);
            if((rem - gap) % (n-lenN) == 0 && (rem - gap != 0)){
                int a = (rem - gap) / (n-lenN);
                int b = (a*lenN) - gap;
                if(a > 0 && b > 0 && a <= 10000 && b <= min(a*n, 10000)) {
                    ans++;
                    list.add(new int[]{a, b});
                }
            }
            count++;
        }

        out.println(ans);
        for(int [] a : list){
            out.println(a[0] + " " + a[1]);
        }
    }

    public static int len(int n){
        int len = 0;
        while (n != 0){
            len++;
            n/=10;
        }
        return len;
    }

    public static long[] get_subset_sums(long[] a, int l, int r) {
        int len = r - l + 1;
        int index = 0;
        long[] sums = new long[1 << len];
        for (int i = 0; i < (1 << len); i++) {
            for (int j = 0; j < len; j++) {
                if ((i & (1 << j)) != 0) { sums[index] += a[j + l]; }
            }
            index++;
        }
        return sums;
    }

    public static int CustomCompare(long a, long b, long k){
        if(a % k < b % k){
            return -1;
        } else if (a % k > b % k) return 1;
        return Long.compare(a, b);
    }

    //    -------------------------------------------------------------------------------------------------------
    public static int max(int a, int b) {
        return Math.max(a, b);
    }
    public static int min(int a, int b) {
        return Math.min(a, b);
    }
    public static long max(long a, long b) {
        return Math.max(a, b);
    }
    public static long min(long a, long b) {
        return Math.min(a, b);
    }
    public static long XorTillN(long n){
        if(n % 4L == 0L) return n;

        if(n % 4L == 1L) return 1;

        if(n % 4L == 2L) return n+1L;

        return 0L;
    }

    public static long RangedXOR(long a, long b){
        long X1 = XorTillN(a);
        long X2 = 0L;
        if(b-1L >= 0){
            X2 = XorTillN(b-1L);
        }

        X1 ^= X2;
        return X1;
    }
    static long [] fact;
    static long [] inv;

    public static void Factorials(int len){
        fact = new long[len+1];
        inv = new long[len+1];
        fact[0] = 1;
        fact[1] = 1;
        inv[0] = 1;
        inv[1] = BinaryExpo(1, MOD - 2);
        for(int i=2; i<=len; i++){
            fact[i] = ModMul(i, fact[i-1]);
            inv[i] = ModMul(inv[i-1], BinaryExpo(i, MOD-2));
        }
        return;
    }

    public static long nCr(int a, int b){
        if(a < b || b < 0){
            return 0L;
        }
        if(b == 0){
            return 1L;
        }
        long ans = fact[a];
        long denominator = ModMul(inv[b], inv[a-b]) % MOD;
        ans = ModMul(ans, denominator) % MOD;
        return ans;
    }
    public static void main(String args[]) {
        FastReader fr = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = fr.nextInt();
//        int t = 1;
        while (t-- > 0) {
            Solve(fr, out);
        }
        out.close();
    }

    //    -------------------------------------------------------------------------------------------------------

    //    ---------------------------------------GCD OF TWO NUMBERS-----------------------------------------------
    public static long gcd(long a, long b){
        if(a == 0){
            return b;
        }

        return gcd(b % a, a);
    }

    //    ----------------------------------LCM OF TWO NUMBERS------------------------------------------------
    public static long LCM(long a, long b){
        return  (a * b) / gcd(a, b);
    }

    //    ----------------------------LOWER BOUND-------------------------------------------------
    static int LowerBound(List<Integer> arr, long target){
        int s = 0, e = arr.size()-1;
        int index = -1;
        while (s <= e){
            int mid = s + (e - s) / 2;
            if(arr.get(mid) > target){
                e = mid - 1;
            } else {
                index = mid;
                s = mid +  1;
            }
        }
        return index;
    }

    //    ----------------------------UPPER BOUND-------------------------------------------------
    static int UpperBound(List<Integer> arr, long target){
        int s = 0, e = arr.size()-1;
        int index = -1;
        while (s <= e){
            int mid = s + (e - s) / 2;
            if(arr.get(mid) >= target){
                index = mid;
                e = mid - 1;
            } else {
                s = mid +  1;
            }
        }
        return index;
    }

    //    -----------------------------x ^ y binary exponentiation-----------------------------------------
    static long BinaryExpo(long x, long y) {
        if (y == 0) return 1;
        long temp = BinaryExpo(x, y / 2);
        temp = ModMul(temp, temp);
        if (y % 2 != 0) return ModMul(temp, x);
        return temp;
    }

    static long BinaryExpo2(long x, long y) {
        long res = 1l;
        x = x % MOD;
        while (y > 0){
            if((y & 1) != 0){
                res = (res * x) % MOD;
            }
            y >>= 1;
            x = (x * x) % MOD;
        }
        return res;
    }

    //    ------------------------------- Multiplicative Inverse ----------------------------------------

    static long modInverse(long a, long MOD){
        return BinaryExpo(a, MOD - 2);
    }

    //    -------------------------------(X * Y) % MOD-------------------------------------------------------

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

    static int[] Give_Primes(int n){
        int prime[] = new int[n+1];
        for(int j=2;(j*1*j)<=n;j++){
            if(prime[j] == 0){
                for(int k=j*j;k<=n;k+=j) prime[k]=j;
            }
        }
//        int count=0;
//        for(int j=2;j<=n;j++){
//            if(!prime[j]) count++;
//        }
//        int arr[]  = new int[count],t=0;
//        for(int j=2;j<=n;j++){
//            if(!prime[j]) arr[t++]=j;
//        }
        return prime;
    }
    //    --------------------------------FOR TAKING FASTER INPUTS----------------------------------------

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

            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }
        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
    static int MOD = (int)(1e9)+7;
    static int maxI = Integer.MAX_VALUE;
    static int minI = Integer.MIN_VALUE;
    static long maxL = Long.MAX_VALUE;

    static long minL = Long.MIN_VALUE;
}

class Pair{
    int x;
    int y;
    int steps;
    public Pair(int x, int y, int z){
        this.x = x;
        this.y = y;
        this.steps = z;
    }
}

class DisjointIntervals {
    int [] parent;
    public DisjointIntervals(int n) {
        parent = new int[n+1];
        for(int i=1; i<=n; i++){
            parent[i] = i;
        }
    }

    public int Setparent(int x){
        if(parent[x] == x) return x;
        int p = Setparent(parent[x]);
        parent[x] = p;
        return p;
    }

    public void Join(int a, int b){
        int p = Setparent(a);
        int p2 = Setparent(b);
        parent[p2] = p;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases where the string manipulation and integer operations can lead to incorrect results. Here are the key points to consider:

1. **Small `n` values**: Test with small values of `n` (e.g., 1, 2, 3) to see if the string manipulation and integer operations are handled correctly.
2. **Large `a` values**: Test with large values of `a` (e.g., 10000) to see if the string concatenation and subtraction operations are handled correctly.
3. **Special `b` values**: Test with `b` values that are close to the length of the concatenated string to see if the subtraction operation is handled correctly.
4. **Boundary conditions**: Test with boundary conditions for `a` and `b` (e.g., `a = 10000` and `b = 10000`).

Given these considerations, let's generate a small-scale test input:

```plaintext
3
1
2
100
```

### Explanation:
1. **Test Case 1: `n = 1`**
   - This is a boundary condition where `n` is the smallest possible value.
   - The correct answer should be 0, as there are no pairs `(a, b)` that satisfy the conditions for `n = 1`.

2. **Test Case 2: `n = 2`**
   - This is a small value of `n` to check if the string manipulation and integer operations are handled correctly.
   - Example pairs could be `(20, 18)` and `(219, 216)` as given in the example.

3. **Test Case 3: `n = 100`**
   - This is a larger value of `n` to check if the string manipulation and integer operations are handled correctly.
   - Example pairs could be `(10000, 9999)` and `(10000, 10000)` to test boundary conditions.

This test input should help expose the bug in the buggy code by focusing on edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

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

public class Main {
    static int mod = 1000000007;
    static PrintWriter out;

    static class TrieNode {
        TrieNode[] childArray = new TrieNode[26];
        ArrayList<int[]> infoList = new ArrayList<>();
    }

    static class TreeNode {
        int value;
        TreeNode left;
        TreeNode right;

        TreeNode(int value) {
            this.value = value;
            right = null;
            left = null;
        }
    }

    private TrieNode buildTrie(String[] words, int[] costs) {
        TrieNode root = new TrieNode();
        for (int i = 0; i < words.length; i++) {
            TrieNode curr = root;
            for (char c : words[i].toCharArray()) {
                int charIndex = c - 'a';
                if (curr.childArray[charIndex] == null) {
                    curr.childArray[charIndex] = new TrieNode();
                }
                curr = curr.childArray[charIndex];
            }
            curr.infoList.add(new int[]{words[i].length(), costs[i]});
        }
        return root;
    }

    static class DSU {
        int[] par, size;

        DSU(int n) {
            par = new int[n + 1];
            size = new int[n + 1];
            Arrays.fill(size, 1);
            for (int i = 0; i <= n; i++)
                par[i] = i;
        }

        int findLeader(int node) {
            if (node == par[node])
                return node;
            return par[node] = findLeader(par[node]); // path comp
        }

        void unionBySize(int u, int v) {
            int uparU = findLeader(u);
            int uparV = findLeader(v);
            if (uparU == uparV)
                return;
            if (size[uparU] < size[uparV]) {
                par[uparU] = uparV;
                size[uparV] += size[uparU];
            } else {
                par[uparV] = uparU;
                size[uparU] += size[uparV];
            }
        }
    }

    public static class FenwickTree {
        long[] fen;
        long size;

        FenwickTree(int n) {
            fen = new long[n];
            size = n;
        }

        void update(long i, long add) {
            for (i++; i < size; i += i & -i) fen[(int) i] += add;
        }

        long sum(long i) {
            long sum = 0;
            for (; i > 0; i -= i & -i) sum += fen[(int) i];
            return sum;
        }
    }

    public static class Pair<L, R> {

        private final L left;
        private final R right;

        public Pair(L left, R right) {
            assert left != null;
            assert right != null;

            this.left = left;
            this.right = right;
        }

        public L getLeft() {
            return left;
        }

        public R getRight() {
            return right;
        }

        @Override
        public int hashCode() {
            return left.hashCode() ^ right.hashCode();
        }

        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Pair)) return false;
            Pair pairo = (Pair) o;
            return this.left.equals(pairo.getLeft()) &&
                    this.right.equals(pairo.getRight());
        }
    }

    public static long mod_sub(long a, long b, long m) {
        a = a % m;
        b = b % m;
        return (((a - b) % m) + m) % m;
    }

    public static long powerMod(long x, long y, long p) {
        long res = 1;
        x = x % p;
        if (x == 0)
            return 0;
        while (y > 0) {
            if ((y & 1) > 0)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

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

    static int sumOfSubArrayProd(int[] arr, int l, int r, int m) {
        int ans = 0;
        int res = 0;
        int sum = 0;
        for (int i = r - 1; i >= l; i--) {
            int incr = ((arr[i] % m) * ((1 + res) % m)) % m;
            ans = (ans % m + incr % m) % m;
            res = (incr % m);
            sum += arr[i];
        }
        return (ans % m - sum % m) % m;
    }

    public static void println(int[] a) {
        for (int i : a) out.print(i + " ");
        out.println();
    }

    static int lower_bound(int array[], int key) {

        int index = Arrays.binarySearch(array, key);
        if (index < 0) {
            return Math.abs(index) - 1;
        } else {
            while (index > 0) {
                if (array[index - 1] == key)
                    index--;
                else
                    return index;
            }

            return index;
        }
    }

    public static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n == 2 || n == 3)
            return true;

        if (n % 2 == 0 || n % 3 == 0)
            return false;

        for (int i = 5; i <= Math.sqrt(n); i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    static long getLargestDivisor(long number) {
        long i;
        for (i = 2; i <= Math.sqrt(number); i++) {
            if (number % i == 0) {
                return number / i;
            }
        }
        return 1L;
    }

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

    public static int lengthOfLIS(ArrayList<Integer> nums) {
        if (nums == null || nums.isEmpty()) {
            return 0;
        }

        int n = nums.size();
        int[] dp = new int[n];
        Arrays.fill(dp, 1);

        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums.get(i) > nums.get(j)) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }

        int maxLength = Arrays.stream(dp).max().orElse(0);
        return maxLength;
    }

    static ArrayList<ArrayList<Integer>> allPaths = new ArrayList<>();

    public static void getAllPaths(int s, int d, int n, ArrayList<ArrayList<Integer>> graph) {
        boolean[] isVisited = new boolean[n];
        ArrayList<Integer> pathList = new ArrayList<>();
        pathList.add(s);
        getAll(s, d, isVisited, pathList, graph);
    }

    private static void getAll(Integer u, Integer d, boolean[] isVisited, ArrayList<Integer> localPathList, ArrayList<ArrayList<Integer>> graph) {

        if (u.equals(d)) {
            allPaths.add(new ArrayList<>(localPathList));
            return;
        }

        isVisited[u] = true;

        for (Integer i : graph.get(u)) {
            if (!isVisited[i]) {
                localPathList.add(i);
                getAll(i, d, isVisited, localPathList, graph);
                localPathList.remove(i);
            }
        }

        isVisited[u] = false;
    }

    static ArrayList<Integer> getAllDivisors(int n) {
        ArrayList<Integer> arr = new ArrayList<>();
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i) {
                    arr.add(i);
                } else {
                    arr.add(i);
                    arr.add(n / i);
                }
            }
        }
        return arr;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        out = new PrintWriter(System.out);
        int tes = sc.nextInt();
        while (tes-- != 0) {
            int n = sc.nextInt();
            int len = Integer.toString(n).length();
            ArrayList<Pair<Integer, Integer>> arr = new ArrayList<>();
            for (int a = 1; a <= 10000; a++) {
                int posLen = a * len;
                int start = posLen - 8;
                int end = posLen;
                if (start < 0 || end < 0 || start > Math.min(10000, a * n)) continue;
                for (int b = start; b <= end; b++) {
                    String f = getNAsString(n, posLen - b);
                    int val = n * a - b;
                    if (Integer.toString(val).equals(f) && b >= 1 && b <= Math.min(10000, a * n)) {
                        arr.add(new Pair<>(a, b));
                    }

                }
            }
            out.println(arr.size());
            for (Pair<Integer, Integer> a : arr) {
                out.println(a.getLeft() + " " + a.getRight());
            }
        }
        sc.close();
        out.close();
    }

    private static String getNAsString(int n, int rep) {
        String g = Integer.toString(n);
        String h = "";
        while (rep >= g.length()) {
            h += g;
            rep -= g.length();
        }
        if (rep > 0) {
            for (char i : g.toCharArray()) {
                h += i;
                rep--;
                if (rep == 0) break;
            }
        }
        return h;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the string manipulation and integer operations differ significantly. The key issue in the buggy code is that it treats the number of friends \( n \) as a string, and then performs operations that are different from the expected integer operations.

### Key Considerations:
1. **String Multiplication**: When \( n \) is treated as a string, multiplying it by \( a \) results in a string where \( n \) is repeated \( a \) times.
2. **String Subtraction**: Subtracting \( b \) from this string removes the last \( b \) characters.
3. **Correct Answer**: The correct answer should be an integer \( n \cdot a - b \).

### Edge Cases:
1. **Small \( n \)**: When \( n \) is a single-digit number, the string manipulation is straightforward.
2. **Large \( n \)**: When \( n \) is a multi-digit number, the string manipulation can lead to more complex and unexpected results.
3. **Edge Values for \( a \) and \( b \)**: Values of \( a \) and \( b \) that are close to the limits (1 and 10000 for \( a \), 1 and \( \min(10000, a \cdot n) \) for \( b \)) can expose issues.

### Test Input:
Let's create a small-scale test input that includes a mix of single-digit and multi-digit \( n \) values, and edge values for \( a \) and \( b \).

```plaintext
3
1
99
10
```

### Explanation:
1. **Test Case 1 (n = 1)**:
   - For \( n = 1 \), the string representation is simply "1".
   - The string "1" repeated \( a \) times will be "1111...1" (with \( a \) ones).
   - Subtracting \( b \) characters from this string should leave a string of length \( a - b \), which should equal \( 1 \cdot a - b \).

2. **Test Case 2 (n = 99)**:
   - For \( n = 99 \), the string representation is "99".
   - The string "99" repeated \( a \) times will be "9999...99" (with \( 2a \) characters).
   - Subtracting \( b \) characters from this string should leave a string of length \( 2a - b \), which should equal \( 99 \cdot a - b \).

3. **Test Case 3 (n = 10)**:
   - For \( n = 10 \), the string representation is "10".
   - The string "10" repeated \( a \) times will be "1010...10" (with \( 2a \) characters).
   - Subtracting \( b \) characters from this string should leave a string of length \( 2a - b \), which should equal \( 10 \cdot a - b \).

This test input should be sufficient to expose the bug in the given code, as it covers a range of \( n \) values and the edge values for \( a \) and \( b \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Novice's Mistake
time_limit_ms: 3000
memory_limit_mb: 256
Description: One of the first programming problems by K1o0n looked like this: "Noobish_Monk has $$$n$$$ $$$(1 \le n \le 100)$$$ friends. Each of them gave him $$$a$$$ $$$(1 \le a \le 10000)$$$ apples for his birthday. Delighted with such a gift, Noobish_Monk returned $$$b$$$ $$$(1 \le b \le \min(10000, a \cdot n))$$$ apples to his friends. How many apples are left with Noobish_Monk?"

K1o0n wrote a solution, but accidentally considered the value of $$$n$$$ as a string, so the value of $$$n \cdot a - b$$$ was calculated differently. Specifically:

- when multiplying the string $$$n$$$ by the integer $$$a$$$, he will get the string $$$s=\underbrace{n + n + \dots + n + n}_{a\ \text{times}}$$$
- when subtracting the integer $$$b$$$ from the string $$$s$$$, the last $$$b$$$ characters will be removed from it. If $$$b$$$ is greater than or equal to the length of the string $$$s$$$, it will become empty.

Learning about this, ErnKor became interested in how many pairs $$$(a, b)$$$ exist for a given $$$n$$$, satisfying the constraints of the problem, on which K1o0n's solution gives the correct answer.

"The solution gives the correct answer" means that it outputs a non-empty string, and this string, when converted to an integer, equals the correct answer, i.e., the value of $$$n \cdot a - b$$$.

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

For each test case, a single line of input contains an integer $$$n$$$ ($$$1 \le n \le 100$$$).

It is guaranteed that in all test cases, $$$n$$$ is distinct.

Output: For each test case, output the answer in the following format:

In the first line, output the integer $$$x$$$ — the number of bad tests for the given $$$n$$$.

In the next $$$x$$$ lines, output two integers $$$a_i$$$ and $$$b_i$$$ — such integers that K1o0n's solution on the test "$$$n$$$ $$$a_i$$$ $$$b_i$$$" gives the correct answer.

Examples:
input:
3
2
3
10
output:
3
20 18
219 216
2218 2214
1
165 162
1
1262 2519

Note: In the first example, $$$a = 20$$$, $$$b = 18$$$ are suitable, as "$$$\text{2}$$$" $$$\cdot 20 - 18 =$$$ "$$$\text{22222222222222222222}$$$"$$$- 18 = 22 = 2 \cdot 20 - 18$$$

Buggy Code:
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;

int n;

int s_toi(string s){
//	cout<<"STOI: S="<<s<<"   ";
	int ret = 0;
	for(int i=s.size()-1; i>=0;i--){
		ret=ret*10+(s[i]-'0');
	}
//	cout<<"RET "<<ret<<endl;
	return ret;
}

int joinn(string s, int times){
//	cout<<"JOIN: S,TIMES = "<<s<<" "<<times<<"   ";
	int ret = 0;
	int cnt = 0;
	string tmp = "";
	while(1){
		if(cnt>=times)break;
		cnt+=s.size();
		tmp+=s;
	}
	//????s????????????????? ?????times��
	tmp = tmp.substr(0,times);//substr(???��??,????)
	ret = s_toi(tmp);
//	cout<<"= "<<ret<<endl;
	return ret;
}

int main(){
	int T; cin>>T;
	while(T--){
		int cnt=0;
		vector<pair<int,int> > v;
		cin>>n;
		int lenn = to_string(n).size();
		for(int a=1;a<=10000;a++){
			for(int lenans1 = 1; lenans1<=6; lenans1++){
				int b = a*lenn-lenans1;//?? b = a*len(n) - len(ans1) ????b
				//????????? b = n*a - ans1, ?? ans1????n??????????(????)a*len(n)-b?????lenans1?��???
				if(1<=b && b<=min(10000,a*n)){//????��?b??��
					int anstmp = joinn(to_string(n),lenans1);
					if(b==n*a-anstmp){
						cnt++;
						v.push_back({a,b});
					}
				}
			}
		}
		cout<<cnt<<endl;
		for(int i=0;i<v.size();i++){
			cout<<v[i].first<<" "<<v[i].second<<endl;
		}
	}
	/*
	ans1 = n*a-b
	ans2 = n???a??-?��b��
	ans1=ans2
	==>
	b = n*a - ans1
	b = a*len(n) - len(ans1)
	n*a-ans1 = a*len(n)-len(ans1)
	????: n:1e2,a:1e4 -> ans1:1e6, len(ans1)<=6. ???a??ans1????????
	*/
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios where the incorrect handling of the string operations in the buggy code will lead to incorrect results. The key points to consider are:

1. **Edge Cases for `n`**: Values of `n` that have a small number of digits, such as single-digit and two-digit numbers.
2. **Values of `a` and `b`**: Values that will cause the string operations to produce different results compared to the correct integer operations.
3. **Special Cases**: Values of `a` and `b` that result in the string `s` being very long or very short, which can expose the bug in the string manipulation.

Given these considerations, let's construct a small-scale test input:

```plaintext
3
1
9
10
```

### Explanation:
1. **Test Case 1: `n = 1`**
   - For `n = 1`, the string `s` will be "1" repeated `a` times.
   - The bug in the code will be exposed when `a` is large enough to make the string very long, and `b` is just a few less than `a`.

2. **Test Case 2: `n = 9`**
   - For `n = 9`, the string `s` will be "9" repeated `a` times.
   - The bug will be exposed when `a` is large and `b` is just a few less than `a * 9`.

3. **Test Case 3: `n = 10`**
   - For `n = 10`, the string `s` will be "10" repeated `a` times.
   - The bug will be exposed when `a` is large and `b` is just a few less than `a * 10`.

These test cases are designed to highlight the incorrect handling of string concatenation and subtraction in the buggy code, especially when the length of the string `s` becomes significant.