**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

def init(tree, N):
    for i in range(N - 1, 0, -1):
        tree[i] = tree[i * 2] + tree[i * 2 + 1]

def query(tree, N, l, r):
    l += N
    r += N
    ret = 0
    while l < r:
        if l % 2 == 1:
            ret += tree[l]
            l += 1
        l //= 2

        if r % 2 == 1:
            ret += tree[r - 1]
            r -= 1
        r //= 2
    return ret

for _ in range(int(input())):
    n = int(input())
    a = list(input().rstrip())
    b = list(input().rstrip())

    cnt1 = []
    for i in range(1, n - 2):
        if a[i] == '0' and a[i + 2] == '0' and b[i - 1] == '1' and b[i + 1] == '0':
            cnt1.append(1)
        else:
            cnt1.append(0)
    l1 = len(cnt1)
    tree1 = [0 for _ in range(l1 * 2)]
    for i in range(l1):
        tree1[l1 + i] = cnt1[i]
    init(tree1, l1)

    cnt2 = []
    for i in range(1, n - 1):
        if a[i] == '0' and b[i - 1] == '1' and b[i + 1] == '1':
            cnt2.append(1)
        else:
            cnt2.append(0)
    l2 = len(cnt2)
    tree2 = [0 for _ in range(len(cnt2) * 2)]
    for i in range(l2):
        tree2[l2 + i] = cnt2[i]
    init(tree2, l2)

    cnt = []
    for i in range(n):
        if a[i] == '1':
            cnt.append(1)
        else:
            cnt.append(0)
    tree = [0 for _ in range(n * 2)]
    for i in range(n):
        tree[n + i] = cnt[i]
    init(tree, n)

    for x in range(int(input())):
        l, r = map(int, input().split())
        t = r - l + 1

        res = 0
        if t >= 3:
            res += query(tree2, l2, l - 1, r - 2)
        if t >= 4:
            res += query(tree1, l1, l - 1, r - 3)
        res += query(tree, n, l - 1, r)
        print(res)

**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
00000
11111
3
1 3
2 4
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
from collections import defaultdict
from functools import lru_cache
from itertools import count
from math import floor, ceil, log2, log10
from copy import deepcopy

def _i():
    return input()

def _ii():
    return int(input())

def _lii():
    return list(map(int, input().split()))

def _si():
    return input().split()

def _li():
    return list(input())

def _lm(t, v):
    return list(map(t, v))

INF = float("inf")
NINF = float("-inf")

ans = []

def solve():
    n = _ii()
    a = _lm(int, _li())
    b = _lm(int, _li())
    out = []
    a_changes = set()
    b_changes = set()

    for i, v in enumerate(a[:-2]):
        if v == a[i + 2] == 0 and b[i + 1] != 1:
            b[i + 1] = 1
            b_changes.add(i + 1)
    for i, v in enumerate(b[:-2]):
        if v == b[i + 2] == 1 and a[i + 1] != 1:
            a[i + 1] = 1
            a_changes.add(i + 1)
    for _ in range(_ii()):
        l, r = _lii()
        c = sum(a[l - 1:r])
        if l == r:
            c = 0 if l - 1 in a_changes else a[l - 1]
        else:
            adjusted = set()
            if l - 1 in a_changes:
                adjusted.add(l - 1)
                c -= 1
            if r - 1 not in adjusted and r - 1 in a_changes:
                adjusted.add(l - 1)
                c -= 1
            if l not in adjusted and l in a_changes and l - 1 in b_changes:
                adjusted.add(l - 1)
                c -= 1
            if r - 2 not in adjusted and r - 1 in b_changes and r - 2 in a_changes:
                c -= 1
        out.append(max(c, 0))

    return '\n'.join(map(str, out))

for _ in range(_ii()):
    ans.append(solve())

print("\n".join(map(str, ans)))

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

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

Think step by step.

Response:
```plaintext
1
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

public class Computing_Machine {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n=sc.nextInt();
            String a=sc.next();
            String b=sc.next();
            char []arr5=a.toCharArray();
            char []arr6=b.toCharArray();
            int []arr1=new int[n+1];
            int []arr2=new int[n+1];
            int []arr3=new int[n+1];
            int []arr4=new int[n+1];
            for (int j = 0; j < n; j++) {
                arr1[j+1]=arr5[j]-'0';
                arr3[j+1]=arr5[j]-'0';
                arr2[j+1]=arr6[j]-'0';
                arr4[j+1]=arr6[j]-'0';
            }
            int []pre=new int[n+1];
            for (int j = 1; j < n-1; j++) {
                if (arr1[j]==0&&arr1[j+2]==0) {
                    arr4[j+1]=1;
                }
            }
            for (int j = 1; j < n-1; j++) {
                if (arr4[j]==1&&arr4[j+2]==1) {
                    arr3[j+1]=1;
                }
            }
            for (int j = 0; j < n; j++) {
                pre[j+1]=pre[j]+arr3[j+1];
            }
            int q=sc.nextInt();
            for (int j = 0; j < q; j++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                int ans = pre[r] - pre[l - 1];
                if (l == r) {
                    System.out.println(arr1[l]);
                } else if (l + 1 == r) {
                    System.out.println(arr1[l] + arr1[r]);
                } else if (l + 2 == r) {
                    int v = 0;
                    if (arr1[l + 1] == 0 && arr2[l] == 0 && arr2[r] == 0) {
                        v++;
                    }
                    System.out.println(arr1[l + 1] + arr1[l + 2] + arr1[l] + v);
                } else {
                    ans -= arr3[l] - arr1[l];
                    ans -= arr3[r] - arr1[r];
                    if (arr1[l + 1] != arr3[l + 1] && arr2[l] != arr4[l]) {
                        ans--;
                    }
                    if (arr1[r - 1] != arr3[r - 1] && arr2[r] != arr4[r]) {
                        ans--;
                    }
                    System.out.println(ans);
                }
            }

        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The main issues in the buggy code seem to be related to how it handles the operations and the preprocessing of the arrays.

### Steps to Create the Test Input:

1. **Edge Case: Single Character Strings**
   - Test with single character strings to see if the code handles them correctly.
2. **Edge Case: Two Character Strings**
   - Test with two character strings to see if the code handles them correctly.
3. **Edge Case: Three Character Strings**
   - Test with three character strings to see if the code handles the operations correctly.
4. **General Case: Longer Strings**
   - Test with longer strings to see if the code handles multiple operations correctly.
5. **Query Edge Cases**
   - Test with queries that cover the entire string, single character, and sub-strings of various lengths.

### Test Input:
```plaintext
4
1
1
0
1
1 1
2
10
01
1
1 2
3
101
110
2
1 3
1 2
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```

### Explanation:
1. **First Test Case:**
   - String length: 1
   - `s = "1"`, `t = "0"`
   - Query: `1 1`
   - Expected output: `1` (since the string is just one character and it is '1')

2. **Second Test Case:**
   - String length: 2
   - `s = "10"`, `t = "01"`
   - Query: `1 2`
   - Expected output: `1` (since no operations can be performed on such a short string)

3. **Third Test Case:**
   - String length: 3
   - `s = "101"`, `t = "110"`
   - Queries: `1 3`, `1 2`
   - Expected output: `3` (for `1 3`), `2` (for `1 2`)

4. **Fourth Test Case:**
   - String length: 6
   - `s = "010101"`, `t = "011010"`
   - Queries: `2 3`, `1 6`, `2 5`, `4 4`, `3 6`
   - Expected output: `1`, `4`, `3`, `1`, `2`

This test input covers a variety of edge cases and general cases to ensure that the buggy code's issues are exposed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();

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

            String a = sc.next();
            String b = sc.next();

            int[] leftRead = new int[n];
            int[] rightRead = new int[n];

            int[] lArr = new int[n];

            for(int i = 1; i < n-1; i++){
                if(a.charAt(i) == '0'){
                    int l = -1;
                    int r = -1;

                    if(b.charAt(i-1) == '1'){
                        l = i-1;
                    }
                    else{
                        if(i >= 2 && a.charAt(i-2) == '0')
                            l = i-2;
                    }

                    if(b.charAt(i+1) == '1'){
                        r = i+1;
                    }
                    else{
                        if(i < n-2 && a.charAt(i+2) == '0')
                            r = i+2;
                    }

                    if(l != -1 && r != -1){
                        leftRead[l]++;
                        rightRead[r]++;
                        for(int k = l+1; k < r; k++){
                            lArr[k] = r;
                        }
                    }
                }
            }

            int[] leftReadPre = new int[n];
            int[] rightReadPre = new int[n];
            leftReadPre[0] = leftRead[0];

            for(int i = 1; i < n; i++){
                leftReadPre[i] = leftReadPre[i-1] + leftRead[i];
                rightReadPre[i] = rightReadPre[i-1] + rightRead[i];
            }

            int[] leftPrefix = new int[n];
            int[] rightPrefix =new int[n];

            rightPrefix[0] = a.charAt(0)-'0';

            for(int i = 1; i < n; i++){
                leftPrefix[i] = leftPrefix[i-1] + (a.charAt(i-1)-'0');
                rightPrefix[i] = rightPrefix[i-1] + (a.charAt(i) - '0');
            }

            for(int i = 1; i < n; i++){
                leftPrefix[i] += leftReadPre[i-1];
            }

            for(int i = 1; i < n; i++){
                rightPrefix[i] += rightReadPre[i];
            }

            int q = sc.nextInt();

            for(int i = 0; i < q; i++){
                int l = sc.nextInt()-1;
                int r = sc.nextInt()-1;
                if(l == r) System.out.println(a.charAt(l) - '0');
                else{
                if(lArr[l] > r) System.out.println(rightPrefix[r] - leftPrefix[l] + 1);
                else System.out.println(rightPrefix[r] - leftPrefix[l]);
                }
            }

        }
    }
}

**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
00000
11111
2
1 3
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define V vector
#define all0(x) (x).begin(),(x).end()
#define all1(x) (x).begin()+1,(x).end()
#define pb push_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define cin std::cin
#define cout std::cout
typedef long long LL;
typedef pair<int, int> pi;
typedef pair<LL, LL> pl;

//const int MN = 2e5 + 20;//MaxN 记得改一下
const int INF = 2e9 + 1000;//INF
const LL INFLL = 8e18 + 1000;//INF long long
mt19937 mrand(random_device{}());
//模板区域~~~~~~~

//模板结束~~~~~~~

void solve() {
    int n; cin >> n;
    string s, t; cin >> s >> t;
    s = " " + s, t = " " + t;
    int q; cin >> q;
    V<int> a(n + 1), b(n + 1);
    for(int i = 1; i <= n; i++) a[i] = s[i] - '0';
    for(int i = 1; i <= n; i++) b[i] = t[i] - '0';
    for(int i = 2; i < n; i++) b[i] += a[i - 1] == a[i + 1] && a[i - 1] == 0;
    V<int> c(n + 1);
    for(int i = 1; i <= n; i++) {
        c[i] = a[i];
        if(i != 1 && i != n) c[i] |= b[i - 1] && b[i + 1];
    }
    V<int> sum(n + 1);
    for(int i = 1; i <= n; i++) sum[i] = c[i] + sum[i - 1];
    V<int> vis(n + 1);
    V<int> cur_b = b;
    for(int i = 1; i <= q; i++) {
        int l, r; cin >> l >> r;
        if(l == r) {
            cout << a[i] << "\n";
            continue;
        }
        int ans = sum[r] - sum[l - 1];
        if(vis[l] == 0) {
            vis[l] = 1;
            ans += a[l] - c[l];
        }
        if(vis[r] == 0) {
            vis[r] = 1;
            ans += a[r] - c[r];
        }
        cur_b[l] -= l != 1 && l != n && a[l - 1] == 0 && a[l + 1] == 0;
        cur_b[r] -= r != 1 && r != n && a[r - 1] == 0 && a[r + 1] == 0;
        if(vis[l + 1] == 0) {
            vis[l + 1] = 1;
            ans += (a[l + 1] || l + 2 <= n && cur_b[l] && cur_b[l + 2]) - c[l + 1];
        }

        if(vis[r - 1] == 0) {
            vis[r - 1] = 1;
            ans += (a[r - 1] || r - 2 >= 1 && cur_b[r] && cur_b[r - 2]) - c[r - 1];
        }

        vis[l] = vis[r] = vis[l + 1] = vis[r - 1] = 0;
        cur_b[l] = b[l], cur_b[r] = b[r];
        cout << ans << "\n";
    }
}

int32_t main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int tt=1;
    cin >> tt;
    while (tt--)
    solve();
}

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

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

Think step by step.

Response:
```plaintext
3
4
0000
1111
2
1 2
2 4
4
1101
1010
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'
#define x first
#define y second
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int N = 1e7 + 100;
const int mod = 1e9 + 7;
void solve() {
    int n, q;
    cin >> n;
    string a, b;
    cin >> a >> b;
    a = '#' + a, b = '#' + b;
    vector<pii> A(n + 1, {-1, 1e9}), B(n + 1, {-1, 1e9});
    for (int i = 1; i <= n; i++) {
        if (a[i] == '1')
            A[i] = {i, i};
        if (b[i] == '1')
            B[i] = {i, i};
    }
    for (int i = 1; i + 2 <= n; i++)
        if (a[i] == '0' && a[i + 2] == '0')
            B[i + 1].x = max(B[i + 1].x, i), B[i + 1].y = min(B[i + 1].y, i + 2), b[i + 1] = '1';

    for (int i = 1; i + 2 <= n; i++)
        if (b[i] == '1' && b[i + 2] == '1')
            A[i + 1].x = max(A[i + 1].x, B[i].x), A[i + 1].y = min(A[i + 1].y, B[i + 2].y), a[i] = '1';
    vector<int> s(n + 1);
    for (int i = 1; i <= n; i++)
        s[i] = s[i - 1] + (a[i] == '1');
    // for (int i = 1; i <= n; i++)
    //     cout << b[i];
    // cout << endl;
    // for (int i = 1; i <= n; i++)
    //     cout << A[i].x << " " << A[i].y << endl;
    cin >> q;
    for (int i = 1; i <= q; i++) {
        int l, r;
        cin >> l >> r;
        int ans = s[r] - s[l - 1];
        auto chk = [&](int pos) -> bool {
            if (a[pos] == '0')
                return 1;
            if (A[pos].x >= l && A[pos].y <= r)
                return 1;
            return 0;
        };
        for (int j = 1, x = l, y = r; j <= 3 && x <= y; j++, x++, y--) {
            if (!chk(x))
                ans--;
            if (x != y && !chk(y))
                ans--;
        }
        cout << ans << endl;
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

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 = readInt()
    s, t = [int(c) for c in readStr()], [int(c) for c in readStr()]
    ps = [0] * (n+1)
    v = [0]
    for i in range(1, n+1):
        if i == 1 or i == n or s[i-1] == 1:
            val = s[i-1]
        else:
            if i == 2:
                val = 1 if t[i-2] == 1 and ((i+1 < n and s[i+1] == 0) or t[i] == 1) else 0
            elif i == n-1:
                val = 1 if t[i] == 1 and ((i-3 >= 0 and s[i-3] == 0) or t[i-2] == 1) else 0
            else:
                if s[i-3] == s[i+1] == 0:
                    val = 1
                elif s[i-3] == 0:
                    val = 1 if t[i-2] == 1 else 0
                elif s[i+1] == 0:
                    val = 1 if t[i] == 1 else 0
                else:
                    val = 1 if t[i-2] == t[i] == 1 else 0
        ps[i] = ps[i-1] + val
        v.append(val)

    q = readInt()
    for i in range(q):
        l, r = readInts()
        curr = ps[r] - ps[l-1]
        if s[l-1] == 0 and v[l] == 1:
            curr -= 1
        if r > l and s[r-1] == 0 and v[r] == 1:
            curr -= 1
        if l+1 < r and s[l] == 0 and v[l+1] == 1:
            if t[l-1] == 1 and ((l+2 <= r and t[l+1] == 1) or ((l+3) <= r and s[l+2] == 0)):
                pass
            else:
                curr -= 1
        if r-1 > l+1 and s[r-2] == 0 and v[r-1] == 1:
            if t[r-1] == 1 and (t[r-3] == 1 or s[r-4] == 0):
                pass
            else:
                curr -= 1
        print(curr)
    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
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    s = input()
    t = input()
    q = int(input())

    tt = []
    ts = []
    for i in range(n-2):
        if s[i] == s[i+2] == '0': tt.append('1')
        else: tt.append(t[i])
    tt.append(t[n-2])
    tt.append(t[n-1])
    for i in range(n-2):
        if tt[i] == tt[i+2] == '1': ts.append('1')
        else: ts.append(s[i])
    ts.append(s[n-2])
    ts.append(s[n-1])
    trans = ''.join(ts)

    uns = [0]
    for c in trans:
        uns.append(uns[-1])
        if c == '1': uns[-1] += 1

    for _ in range(q):
        l, r = list(map(int, input().split()))
        if r-l == 0:
            ans = int(s[l-1])
        elif r-l == 1:
            ans = int(s[l-1]) + int(s[r-1])
        elif r-l == 2:
            ans = int(s[l-1]) + int(s[r-1]) + int(s[l] == '1' or (t[l-1] == '1' and t[r-1] == '1'))
        else:
            ans = uns[r-2] - uns[l+1]
            if s[l-1] == '1': ans += 1
            if s[r-1] == '1': ans += 1
            if s[l] == '1' or (t[l-1] == '1' and (t[l+1] == '1' or s[l+2] == '0')): ans += 1
            if s[r-2] == '1' or (t[r-1] == '1' and (t[r-3] == '1' or t[r-4] == '0')): ans += 1

        print(ans)

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

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

Think step by step.

Response:
```plaintext
2
6
000000
111111
4
1 3
2 4
3 5
4 6
8
01010101
10101010
5
1 8
2 7
3 6
4 5
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();

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

            String a = sc.next();
            String b = sc.next();

            int[] leftRead = new int[n];
            int[] rightRead = new int[n];

            int[] lArr = new int[n];

            for(int i = 1; i < n-1; i++){
                if(a.charAt(i) == '0'){
                    int l = -1;
                    int r = -1;

                    if(b.charAt(i-1) == '1'){
                        l = i-1;
                    }
                    else{
                        if(i >= 2 && a.charAt(i-2) == '0')
                            l = i-2;
                    }

                    if(b.charAt(i+1) == '1'){
                        r = i+1;
                    }
                    else{
                        if(i < n-2 && a.charAt(i+2) == '0')
                            r = i+2;
                    }

                    if(l != -1 && r != -1){
                        leftRead[l]++;
                        rightRead[r]++;
                        for(int k = l+1; k < r; k++){
                            lArr[k] = r;
                        }
                    }
                }
            }

            int[] leftReadPre = new int[n];
            int[] rightReadPre = new int[n];
            leftReadPre[0] = leftRead[0];

            for(int i = 1; i < n; i++){
                leftReadPre[i] = leftReadPre[i-1] + leftRead[i];
                rightReadPre[i] = rightReadPre[i-1] + rightRead[i];
            }

            int[] leftPrefix = new int[n];
            int[] rightPrefix =new int[n];

            rightPrefix[0] = a.charAt(0)-'0';

            for(int i = 1; i < n; i++){
                leftPrefix[i] = leftPrefix[i-1] + (a.charAt(i-1)-'0');
                rightPrefix[i] = rightPrefix[i-1] + (a.charAt(i) - '0');
            }

            for(int i = 1; i < n; i++){
                leftPrefix[i] += leftReadPre[i-1];
            }

            for(int i = 1; i < n; i++){
                rightPrefix[i] += rightReadPre[i];
            }

            int q = sc.nextInt();

            for(int i = 0; i < q; i++){
                int l = sc.nextInt()-1;
                int r = sc.nextInt()-1;

                if(rightPrefix[r] < leftPrefix[l]) System.out.println(0);
                else{
                if(lArr[l] > r) System.out.println(rightPrefix[r] - leftPrefix[l] + 1);
                else System.out.println(rightPrefix[r] - leftPrefix[l]);
                }
            }

        }
    }
}

**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
00000
11111
2
1 3
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

public class Main {
    private FastScanner in;
    private PrintWriter out;

    private void solve() throws IOException {
        int n = in.nextInt();
        int[] s = new int[n];
        char[] strS = in.next().toCharArray();
        int[] t = new int[n];
        char[] strT = in.next().toCharArray();

        for (int i = 0; i < n; i++) {
            s[i] = strS[i] - '0';
            t[i] = strT[i] - '0';
        }

        for (int i = 0; i + 2 < n; i++) {
            if (s[i] == 0 && s[i + 2] == 0 && t[i + 1] == 0) {
                t[i + 1] = 2;
            }
        }

        for (int i = 0; i + 2 < n; i++) {
            if (s[i + 1] == 1)
                continue;
            if (t[i] != 0 && t[i + 2] != 0) {
                // 1 1 -> 2
                // 1 2 -> 3
                // 2 1 -> 4
                // 2 2 -> 5
                s[i + 1] = (t[i] - 1) * 2 + t[i + 2] - 1 + 2;
            }
        }

        int[][] pref = new int[6][n + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 5; j++) {
                pref[j][i + 1] = pref[j][i];
            }
            pref[s[i]][i + 1]++;
        }

        int q = in.nextInt();
        while (q-- > 0) {
            int l = in.nextInt(), r = in.nextInt();
            out.println(pref[1][r] - pref[1][l - 1] + (r - l + 1 < 3 ? 0 :
                    pref[2][r - 1] - pref[2][l] +
                            pref[3][r - 2] - pref[3][l] +
                            pref[4][r - 1] - pref[4][l + 1] +
                            (r - l + 1 < 4 ? 0 : pref[5][r - 2] - pref[5][l + 1])));
            // out.flush();
        }
    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

        boolean hasNext() throws IOException {
            return br.ready() || (st != null && st.hasMoreTokens());
        }

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

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

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

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

        boolean hasNextLine() throws IOException {
            return br.ready();
        }
    }

    private void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        // in = new FastScanner(new FileInputStream(".in"));
        // out = new PrintWriter(new FileOutputStream(".out"));

        int t = in.nextInt();
        while (t-- > 0)
            solve();

        out.flush();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }
}

**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
10101
01010
2
1 5
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int mod=1e9+7;
int dp[5005][5005],s[5005],n,q,a[500005],sum[500005],b[500005];
string s1,s2;
int ck(string s3,string s4)
{
	int r=0;
	for (int i=0;i+2<s3.size();i++){
		if (s3[i]==s3[i+2] && s3[i]=='0'){
			s4[i+1]='1';
		}
	}
	for (int i=0;i+2<s3.size();i++){
		if (s4[i]==s4[i+2] && s4[i]=='1'){
			s3[i+1]='1';
		}
	}
	for (int i=0;i<s3.size();i++){
		r+=s3[i]-'0';
	}
	return r;
}
signed main()
{
	int t;
	cin>>t;
	while (t--){
		cin>>n>>s1>>s2>>q;
		string s3=s1,s4=s2;
		for (int i=0;i+2<s3.size();i++){
			if (s3[i]==s3[i+2] && s3[i]=='0'){
				s4[i+1]='1';
			}
		}
		for (int i=0;i+2<s3.size();i++){
			if (s4[i]==s4[i+2] && s4[i]=='1'){
				s3[i+1]='1';
			}
		}
		for (int i=1;i<=s3.size();i++){
			sum[i]=sum[i-1]+s3[i-1]-'0';
		}
		while (q--){
			int l,r,ans=0;
			string s5="",s6="";
			cin>>l>>r;
			if (r-l>=25){
				ans+=sum[r-9]-sum[l+10];
				for (int i=l;i<=l+10;i++){
					s5+=s1[i-1];
					s6+=s2[i-1];
				}
				ans+=ck(s5,s6);
				s5=s6="";
				for (int i=r-10;i<=r;i++){
					s5+=s1[i-1];
					s6+=s2[i-1];
				}
				ans+=ck(s5,s6);
			}else{
				for (int i=l-1;i<r;i++){
					s5+=s1[i];
					s6+=s2[i];
				}
				ans+=ck(s5,s6);
			}
			cout<<ans<<endl;
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
3
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
4
101010
110110
4
1 3
1 4
2 5
2 4
10
1010101010
1101101101
10
1 10
2 9
3 8
4 7
5 6
1 5
6 10
1 3
8 10
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

using namespace std;

#define int long long
#define inf 0x3F3F3F3F3F3F3F3F

const int MXN = 1e5 + 5;

int solve(string a, string b, int t)
{
	int res = 0;
	string c = a, d = b;
	for (int i = 0; i + 2 < a.length(); i++)
	{
		if (a[i] == '0' && a[i + 2] == '0') d[i + 1] = '1';
	}
	for (int i = 0; i + 2 < a.length(); i++)
	{
		if (d[i] == '1' && d[i + 2] == '1') c[i + 1] = '1';
	}
	if (t == -100)
	{
		for (int i = 0; i < a.length(); i++) res += c[i] - '0';
	}
	else if (t < 0)
	{
		for (int i = (int)a.length() + t; i < a.length(); i++) res += c[i] - '0';
	}
	else
	{
		for (int i = 0; i < t; i++) res += c[i] - '0';
	}
	return res;
}

signed main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		string a, b;
		cin >> n >> a >> b;
		string c = a, d = b;
		for (int i = 0; i + 2 < n; i++)
		{
			if (a[i] == '0' && a[i + 2] == '0') d[i + 1] = '1';
		}
		for (int i = 0; i + 2 < n; i++)
		{
			if (d[i] == '1' && d[i + 2] == '1') c[i + 1] = '1';
		}
		int p[n];
		for (int i = 0; i < n; i++) p[i] = (i ? p[i - 1] : 0) + c[i] - '0';
		int q;
		cin >> q;
		while (q--)
		{
			int l, r, res = 0;
			cin >> l >> r;
			l--, r--;
			if (r - l + 1 <= 12)
			{
				string x, y;
				for (int i = l; i <= r; i++) x.push_back(a[i]), y.push_back(b[i]);
				cout << solve(x, y, -100) << '\n';
				continue;
			}
			string x, y;
			for (int i = l; i < l + 5; i++) x.push_back(a[i]), y.push_back(b[i]);
			res += solve(x, y, 2);
			x.clear(), y.clear();
			for (int i = r - 4; i <= r; i++) x.push_back(a[i]), y.push_back(b[i]);
			res += solve(x, y, 2);
			res += p[r - 2] - p[l + 1];
			cout << res << '\n';
		}
	}
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the operations might not be applied correctly. The code seems to handle small segments of the strings and then combines the results, but it might miss some important details when the operations can be applied multiple times in a sequence.

Here's a step-by-step approach to generating a challenging test input:

1. **Small String Length**: Use a small string length to ensure the test case is manageable and can be easily validated.
2. **Multiple Operations**: Create strings where multiple operations can be applied to ensure the code handles these cases correctly.
3. **Edge Cases**: Test with strings that have patterns where the operations can be applied in a specific order, and the results can be different based on the order of operations.

Let's create a test case with a string length of 6 and a few queries to test the operations:

```plaintext
1
6
001000
110111
3
1 6
2 5
4 6
```

Explanation:
- **String Length**: 6
- **String s**: `001000`
- **String t**: `110111`
- **Queries**:
  - **1 6**: Query the entire string.
  - **2 5**: Query a substring that can be modified.
  - **4 6**: Query a small substring.

This test case will help expose any issues in the code where the operations are not applied correctly or where the results are not combined properly. The expected output for this test case should be:

```plaintext
4
3
2
```

This test case ensures that the operations are applied multiple times and that the results are correctly combined for different substrings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
def naive(a, b):
    a = list(map(int, list(a)))
    b = list(map(int, list(b)))
    for i in range(len(a) - 2):
        if not a[i] and not a[i + 2]:
            b[i + 1] = 1

    for i in range(len(a) - 2):
        if b[i] and b[i + 2]:
            a[i + 1] = 1

    return a

t = int(input())
for _ in range(t):
    n = int(input())
    s = input()
    t = input()
    a = naive(s, t)
    ac = [0]
    for i in a:
        ac.append(ac[-1] + i)

    q = int(input())
    for i in range(q):
        l, r = map(int, input().split())
        l -= 1
        r -= 1
        if r - l <= 8:
            print(sum(naive(s[l : r + 1], t[l : r + 1])))
            continue

        ans = sum(a[l + 2 : r - 1])
        ans = ac[r - 1] - ac[l + 2]
        ans += sum(naive(s[l : l + 3], t[l : l + 3])[:-1])
        ans += sum(naive(s[r - 2 : r + 1], t[r - 2 : r + 1])[1:])
        print(ans)

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

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

Think step by step.

Response:
```plaintext
3
4
0000
1111
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
import sys

# sys.setrecursionlimit(10000)

def input_general():
    return sys.stdin.readline().rstrip('\r\n')

def input_num():
    return int(sys.stdin.readline().rstrip("\r\n"))

def input_multi(x=int):
    return map(x, sys.stdin.readline().rstrip("\r\n").split())

def input_list(x=int):
    return list(input_multi(x))

def main():
    n = input_num()
    s = input_general()
    t = input_general()
    q = input_num()
    queries = []
    for _ in range(q):
        left, right = input_multi()
        queries.append((left, right))

    s = list(int(c) for c in s)
    t = list(int(c) for c in t)

    ss = s.copy()
    tt = t.copy()

    for i, (c, d) in enumerate(zip(ss, ss[2:]), 1):
        if c == d == 0:
            tt[i] = 1

    for i, (c, d) in enumerate(zip(tt, tt[2:]), 1):
        if c == d == 1:
            ss[i] = 1

    def direct(a, b, left=None, right=None):
        for i, (c, d) in enumerate(zip(a, a[2:]), 1):
            if c == d == 0:
                b[i] = 1

        for i, (c, d) in enumerate(zip(b, b[2:]), 1):
            if c == d == 1:
                a[i] = 1

        return sum(a[left:right]) if left is None else sum(a)

    psum = [0]
    for c in ss:
        psum.append(psum[-1] + c)

    for left, right in queries:
        left -= 1

        if right - left <= 8:
            print(direct(s[left:right], t[left:right]))
        else:
            answer = (
                psum[right - 4] - psum[left + 4]
                + direct(s[left:left + 6], t[left:left + 6], 0, 4)
                + direct(s[right - 6:right], t[right - 6:right], 2, 6)
            )
            print(answer)

if __name__ == "__main__":
    cases = input_num()

    for _ in range(cases):
        main()

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

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

Think step by step.

Response:
To generate a 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 code has a few potential issues:

1. **Handling of small segments**: The code uses a special case for segments of length 8 or less, which might not be handled correctly.
2. **Prefix sum logic**: The prefix sum logic might not correctly account for the changes made during the operations.
3. **Boundary conditions**: The code might not handle boundary conditions correctly, especially when the segment length is close to the special case threshold.

Given these points, let's create a test case that maximizes the chances of exposing these issues.

### Test Case
We will create a test case with a small number of test cases and a string length that is just above the threshold (8) to ensure the special case is not triggered. We will also include a segment that requires multiple operations to maximize the number of 1s.

```plaintext
1
10
0000000000
1111111111
5
1 10
2 9
3 8
4 7
5 6
```

### Explanation
- **Test Case**: We have one test case with a string length of 10.
- **Strings**:
  - `s = 0000000000`
  - `t = 1111111111`
- **Queries**:
  - `1 10`: The entire string.
  - `2 9`: A segment of length 8.
  - `3 8`: A segment of length 6.
  - `4 7`: A segment of length 4.
  - `5 6`: A segment of length 2.

### Expected Output
- For the query `1 10`, the maximum number of 1s that can be obtained is 10.
- For the query `2 9`, the maximum number of 1s that can be obtained is 8.
- For the query `3 8`, the maximum number of 1s that can be obtained is 6.
- For the query `4 7`, the maximum number of 1s that can be obtained is 4.
- For the query `5 6`, the maximum number of 1s that can be obtained is 2.

### Test Input
```plaintext
1
10
0000000000
1111111111
5
1 10
2 9
3 8
4 7
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

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

public class CFTemp {

    static int MOD = 998244353;
    static int INF = 0x3f3f3f3f;

    public static void main(String[] args) throws IOException {
        int t = Reader.nextInt();
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
            int n = Reader.nextInt();
            String a = Reader.next();
            String b = Reader.next();
            int c = Reader.nextInt();
            int[][] q = new int[c][2];
            for (int i = 0; i < c; i++) {
                q[i][0] = Reader.nextInt();
                q[i][1] = Reader.nextInt();
            }
            sb.append(solve(a, b, q));
        }

        System.out.print(sb.toString());
    }

    static int mod1 = 998244353;
    static int mod2 = (int) 1e9 + 7;

    static String solve(String a, String b, int[][] q) {
        StringBuilder sb = new StringBuilder();
        int n = a.length();
        int m = q.length;

        int[] blow = new int[n];
        Arrays.fill(blow, -1);
        int[] bhigh = new int[n];
        Arrays.fill(bhigh, -1);
        for (int i = 0; i < n; i++) {
            if (b.charAt(i) == '1') {
                blow[i] = i;
                bhigh[i] = i;
            } else if (i > 0 && i < n - 1 && a.charAt(i - 1) == '0' && a.charAt(i + 1) == '0') {
                blow[i] = i - 1;
                bhigh[i] = i + 1;
            }
        }

        int[] alow = new int[n];
        Arrays.fill(alow, -1);
        int[] ahigh = new int[n];
        Arrays.fill(ahigh, -1);
        int[] presum = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (a.charAt(i) == '1') {
                alow[i] = i;
                ahigh[i] = i;
            } else if (i > 0 && i < n - 1 && blow[i - 1] >= 0 && blow[i + 1] >= 0) {
                alow[i] = Math.min(i - 1, Math.min(blow[i - 1], blow[i + 1]));
                ahigh[i] = Math.max(i + 1, Math.max(bhigh[i - 1], bhigh[i + 1]));
            }
            presum[i + 1] = presum[i] + (alow[i] >= 0 ? 1 : 0);
        }

        for (int i = 0; i < m; i++) {
            int l = q[i][0] - 1;
            int r = q[i][1] - 1;
            int result = 0;

            if (r - l <= 5) {
                for (int j = l; j <= r; j++) {
                    result += alow[j] >= l && ahigh[j] <= r ? 1 : 0;
                }
            } else {
                result = presum[r - 2] - presum[l + 1];
                int[] cal = new int[] { l, l + 1, r, r - 1 };
                for (int j : cal) {
                    if (alow[j] >= l && ahigh[j] <= r) {
                        result++;
                    }
                }
            }

            sb.append(result).append("\n");
        }

        return sb.toString();
    }

    static class Reader {
        static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        static StringTokenizer tokenizer = new StringTokenizer("");

        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }

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

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

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

}

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

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

Think step by step.

Response:
```plaintext
3
6
000000
111111
3
1 2
3 4
5 6
4
1010
0101
2
1 3
2 4
5
10101
01010
4
1 3
2 4
3 5
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

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

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
import java.util.Scanner;

public class problemE {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int tests = sc.nextInt();
    for (int test = 0; test < tests; test++) {
      int l = sc.nextInt();
      String s = sc.next();
      String t = sc.next();

      int[] prefix = new int[l + 1];

      for (int k = 0; k < l; k++) {
        prefix[k + 1] = prefix[k] + (getRange(s, k, t)[0] != -1 ? 1 : 0);
      }
      int q = sc.nextInt();
      for (int query = 0; query < q; query++) {
        int a = sc.nextInt();
        int b = sc.nextInt();
        if (b - a <= 5) {
          int res = 0;
          for (int i = a - 1; i < b; i++) {
            int[] indices = getRange(s, i, t);
            int lo = indices[0], hi = indices[1];
            if (lo >= a - 1 && hi < b) {
              res++;
            }
          }
          System.out.println(res);
        } else {
          int res = prefix[b] - prefix[a - 1];
          for (int m = a - 1; m < b; m++) {
            int[] indices = getRange(s, m, t);
            int lo = indices[0], hi = indices[1];
            if (lo != -1 && lo < a || hi >= b) {
              res--;
            }
          }
          System.out.println(res);
        }
      }
    }
  }

  public static int[] getRange(String s, int i, String t) {
    if (s.charAt(i) == '1') {
      return new int[] {i, i};
    }
    int l = -1, r = -1;
    if (i > 0 && t.charAt(i - 1) == '1') {
      l = i - 1;
    } else if (i > 1 && t.charAt(i - 1) == '0' && s.charAt(i - 2) == '0') {
      l = i - 2;
    }
    if (i < s.length() - 1 && t.charAt(i + 1) == '1') {
      r = i + 1;
    } else if (i < s.length() - 2 && t.charAt(i + 1) == '0' && s.charAt(i + 2) == '0') {
      r = i + 2;
    }
    if (l == -1) {
      r = -1;
    }
    if (r == -1) {
      l = -1;
    }
    return new int[] {l, r};
  }
}

**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
8
00110011
11001100
4
1 8
2 7
3 6
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Simultaneous Coloring
time_limit_ms: 6000
memory_limit_mb: 512
Description: You are given a matrix, consisting of $n$ rows and $m$ columns.

You can perform two types of actions on it:

  * paint the entire column in blue;    * paint the entire row in red.

Note that you cannot choose which color to paint the row or column.

In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform $k > 1$ actions at the same time, it will cost $k^2$ coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.

You are asked to process $q$ queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the $i$-th query, a restriction of the following form is added:

  * $x_i~y_i~c_i$ — the cell in row $x_i$ in column $y_i$ should be painted in color $c_i$.

Thus, after $i$ queries, there are $i$ restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions.

Input: The first line contains three integers $n, m$ and $q$ ($1 \le n, m, q \le 2 \cdot 10^5$) — the size of the matrix and the number of queries.

In the $i$-th of the next $q$ lines, two integers $x_i, y_i$ and a character $c_i$ ($1 \le x_i \le n$; $1 \le y_i \le m$; $c_i \in$ {'R', 'B'}, where 'R' means red, and 'B' means blue) — description of the $i$-th restriction. The cells in all queries are pairwise distinct.

Output: Print $q$ integers — after each query, output the minimum cost of painting the matrix according to the restrictions.

Examples:
input:
2 2 4
1 1 R
2 2 R
1 2 B
2 1 B
output:
0
0
0
16

input:
3 5 10
1 1 B
2 5 B
2 2 B
2 3 R
2 1 B
3 2 R
3 3 B
1 2 R
1 3 B
3 1 B
output:
0
0
0
0
0
0
16
16
25
25

Note:

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

#define ll long long

inline int read() {
    int x = 0; char ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return x;
}

const int N = 400005;

int f[N], sz[N];
int row[N], col[N], tot;
int n, m, q;

ll ans;

int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}
ll calc(int x) {
    if (x == 1) return 0;
    else return 1ll * x * x;
}
void merge(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return ;
    ans -= calc(sz[x]) + calc(sz[y]);
    f[x] = y, sz[y] += sz[x];
    ans += calc(sz[y]);
}

vector<int> adj[N], event[N];
stack<int> stk;
bool instk[N];
int dfn[N], low[N], dtot, scc[N], stot;
void tarjan(int u) {
    dfn[u] = low[u] = ++dtot;
    stk.push(u), instk[u] = 1;
    for (auto v: adj[u]) {
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (dfn[u] == low[u]) {
        int y;
        ++stot;
        do {
            y = stk.top(); stk.pop();
            instk[y] = 0, scc[y] = stot;
        } while (y != u);
    }
}

bool vis[N];
void solve(int L, int R, vector<array<int, 3>> queries) {
    int mid = L + R >> 1;
    vector<int> nodes;
    for (auto [u, v, id]: queries) {
        if (!vis[u]) vis[u] = 1, dfn[u] = 0, adj[u].clear(), nodes.push_back(u);
        if (!vis[v]) vis[v] = 1, dfn[v] = 0, adj[v].clear(), nodes.push_back(v);
        if (id <= mid) {
            adj[u].push_back(v);
        }
    }
    dtot = stot = 0;
    for (auto u: nodes) {
        if (!dfn[u]) tarjan(u);
        vis[u] = 0;
    }

    // printf("solve [%d, %d], ", L, R);
    // for (auto [u, v, id]: queries) printf("(%d, %d) ", u, v);
    // puts("");
    // for (auto u: nodes) {
    //     printf("dfn[%d] = %d, ", u, dfn[u]);
    // }

    vector<array<int, 3>> qL, qR;
    for (auto [u, v, id]: queries) {
        if (scc[u] == scc[v]) {
            if (id <= mid)
                event[mid].push_back(id), qL.push_back({u, v, id});
        } else {
            qR.push_back({scc[u], scc[v], id});
        }
    }
    if (L == R) return ;
    solve(L, mid, qL), solve(mid + 1, R, qR);
}

int main() {
    n = read(), m = read(), q = read();
    for (int i = 1; i <= n; i++) row[i] = ++tot;
    for (int i = 1; i <= m; i++) col[i] = ++tot;
    vector<array<int, 3>> queries;
    for (int i = 1; i <= q; i++) {
        int x = read(), y = read();
        char c;
        c = getchar();
        while (c != 'R' && c != 'B') c = getchar();
        if (c == 'B') queries.push_back({row[x], col[y], i});
        else queries.push_back({col[y], row[x], i});
    }
    solve(1, q, queries);

    for (int i = 1; i <= tot; i++) f[i] = i, sz[i] = 1;
    for (int i = 1; i <= q; i++) {
        for (auto t: event[i])
            merge(queries[t - 1][0], queries[t - 1][1]);
        printf("%lld\n", ans);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 3 6
1 1 R
1 2 B
2 1 B
2 2 R
3 1 B
3 2 R
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Simultaneous Coloring
time_limit_ms: 6000
memory_limit_mb: 512
Description: You are given a matrix, consisting of $n$ rows and $m$ columns.

You can perform two types of actions on it:

  * paint the entire column in blue;    * paint the entire row in red.

Note that you cannot choose which color to paint the row or column.

In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform $k > 1$ actions at the same time, it will cost $k^2$ coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.

You are asked to process $q$ queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the $i$-th query, a restriction of the following form is added:

  * $x_i~y_i~c_i$ — the cell in row $x_i$ in column $y_i$ should be painted in color $c_i$.

Thus, after $i$ queries, there are $i$ restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions.

Input: The first line contains three integers $n, m$ and $q$ ($1 \le n, m, q \le 2 \cdot 10^5$) — the size of the matrix and the number of queries.

In the $i$-th of the next $q$ lines, two integers $x_i, y_i$ and a character $c_i$ ($1 \le x_i \le n$; $1 \le y_i \le m$; $c_i \in$ {'R', 'B'}, where 'R' means red, and 'B' means blue) — description of the $i$-th restriction. The cells in all queries are pairwise distinct.

Output: Print $q$ integers — after each query, output the minimum cost of painting the matrix according to the restrictions.

Examples:
input:
2 2 4
1 1 R
2 2 R
1 2 B
2 1 B
output:
0
0
0
16

input:
3 5 10
1 1 B
2 5 B
2 2 B
2 3 R
2 1 B
3 2 R
3 3 B
1 2 R
1 3 B
3 1 B
output:
0
0
0
0
0
0
16
16
25
25

Note:

Buggy Code:
// https://codeforces.com/blog/entry/91608
#include <bits/stdc++.h>

using namespace std;

signed main() {
    cout<<"0\n0\n0\n16";
}

**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 create a test input that highlights the incorrect output. The buggy code always outputs "0\n0\n0\n16" regardless of the input, which is clearly incorrect for most inputs.

To challenge this, we should create an input that requires a different output pattern. A good approach is to use a small matrix and a sequence of queries that lead to different costs after each query.

Let's consider a 2x2 matrix and a sequence of queries that result in different costs:

1. The first query should not require any cost.
2. The second query should also not require any cost.
3. The third query should require a cost of 16.
4. The fourth query should require a cost of 0.

Here is a suitable test input:

```plaintext
2 2 4
1 1 R
1 2 B
2 1 B
2 2 R
```

This input will produce the following output:
- After the first query, the matrix is:
  - (1,1) = R
  - (1,2) = colorless
  - (2,1) = colorless
  - (2,2) = colorless
  - Cost: 0

- After the second query, the matrix is:
  - (1,1) = R
  - (1,2) = B
  - (2,1) = colorless
  - (2,2) = colorless
  - Cost: 0

- After the third query, the matrix is:
  - (1,1) = R
  - (1,2) = B
  - (2,1) = B
  - (2,2) = colorless
  - Cost: 16 (since we need to paint both rows and columns to satisfy the constraints)

- After the fourth query, the matrix is:
  - (1,1) = R
  - (1,2) = B
  - (2,1) = B
  - (2,2) = R
  - Cost: 0 (since we can paint the entire matrix in one action)

This input should clearly show that the buggy code is incorrect. Here is the test input:

```plaintext
2 2 4
1 1 R
1 2 B
2 1 B
2 2 R
```