**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
import java.io.*;
import java.util.*;
public class ques4 {
    private static final int mod = 1000000007;
    // Custom class to hold pairs
    static class Pair {
        int value;
        int index;
        Pair(int value, int index) {
            this.value = value;
            this.index = index;
        }
    }
    // Euclidean algorithm for finding GCD for int type
    public static int gcdInt(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
       return a;
    }
    // Euclidean algorithm for finding GCD for long type
    public static long gcdLong(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
       return a;
    }
    // Binary exponentiation for int type
    public static int power_Int(int base, int exponent) {
        int result = 1;
        while (exponent > 0) {
           if ((exponent & 1) == 1) {
               result *= base;
           }
            base *= base;
            exponent >>= 1;
        }
        return result;
    }
    // Binary exponentiation for long type
    public static long power_Long(long base, long exponent) {
        long result = 1;
        while (exponent > 0) {
           if ((exponent & 1) == 1) {
               result *= base;
           }
            base *= base;
            exponent >>= 1;
        }
        return result;
    }
    // Modular exponentiation for int type with MOD = 1e9 + 7
    public static int power_modInt(int base, int exponent) {
       int result = 1;
       base %= mod;
       while (exponent > 0) {
           if ((exponent & 1) == 1) {
           result = (int) ((result * (long) base) % mod);
           }
           base = (int) ((base * (long) base) % mod);
           exponent >>= 1;
           }
           return result;
       }
    // greater than or equal to key
    public static long long_lowerBound(long[] arr, long key) {
        long low = 0, high = arr.length - 1;
        long ans = arr.length;
        while (low <= high) {
            long mid = (low + high) / 2;
            if (arr[(int)mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than or equal to key
    public static long long_upperBound(long[] arr, long key) {
        long low = 0, high = arr.length - 1;
        long ans = arr.length;
        while (low <= high) {
            long mid = (low + high) / 2;
            if (arr[(int)mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than equal to key
    public static int int_lowerBound(int[] arr, int key) {
        int low = 0, high = arr.length - 1;
        int ans = arr.length;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // greater than key
    public static int int_upperBound(int[] arr, int key) {
        int low = 0, high = arr.length - 1;
        int ans = arr.length;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (arr[mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }
    // solve
    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        String test = br.readLine();
        int t = Integer.parseInt(test);
        while (t-- > 0) {
            String[] nk = br.readLine().split(" ");
            int n = Integer.parseInt(nk[0]);
            int m = Integer.parseInt(nk[1]);
            int[][] a = new int[n][m];
            int[][] cal = new int[n][m];
            for (int i = 0; i < n; i++) {
                for(int j = 0; j<m; j++){
                    cal[i][j]=0;
                }
            }
            for (int i = 0; i < n; i++) {
                String s = br.readLine();
                int val = Integer.parseInt(s);
                for(int j = 0; j<m; j++){
                    a[i][m-j-1]=val%10;
                    val/=10;
                }
            }
            long ans = 0;
            for (int i = 0; i < n; i++) {
                for(int j = 0; j<m; j++){
                    if(a[i][j]==1){
                        if(j+1<m && a[i][j+1]==5){
                            if(j+2<m && a[i][j+2]==4){
                                if(j+3<m && a[i][j+3]==3){
                                    ans++;
                                }
                                if(i+1<n && a[i+1][j+2]==3){
                                    ans++;
                                }
                            }
                            if(i+1<n && a[i+1][j+1]==4){
                                if(i+2<n && a[i+2][j+1]==3){
                                    ans++;
                                }
                                if(a[i+1][j]==3){
                                    ans++;
                                }
                            }
                        }
                        if(i+1<n && a[i+1][j]==5){
                            if(i+2<n && a[i+2][j]==4){
                                if(j-1>=0 && a[i+2][j-1]==3){
                                    ans++;
                                }
                            }
                            if(j-1>=0 && a[i+1][j-1]==4){
                                if(a[i][j-1]==3){
                                    ans++;
                                }
                                if(j-2>=0 && a[i+1][j-2]==3){
                                    ans++;
                                }
                            }
                        }
                        if(i-1>=0 && a[i-1][j]==5){
                            if(j+1<m && a[i-1][j+1]==4){
                                if(a[i][j+1]==3){
                                    ans++;
                                }
                                if(j+2<m && a[i-1][j+2]==3){
                                    ans++;
                                }
                            }
                            if(i-2>=0 && a[i-2][j]==4){
                                if(j+1<m && a[i-2][j+1]==3){
                                    ans++;
                                }
                                if(i-3>=0 && a[i-3][j]==3){
                                    ans++;
                                }
                            }
                        }
                        if(j-1>=0 && a[i][j-1]==5){
                            if(j-2>=0 && i-1>=0 && a[i][j-2]==4 && a[i-1][j-2]==3){
                                ans++;
                            }
                            if(i-1>=0 && a[i-1][j-1]==4){
                                if(i-2>=0 && a[i-2][j-1]==3){
                                    ans++;
                                }
                                if(a[i-1][j]==3){
                                    ans++;
                                }
                            }
                        }
                    }
                }
            }
            out.println(ans);
        }
        out.flush();
        out.close();
        br.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases and scenarios where the code might fail. The code is designed to count occurrences of the number "1543" in the layers of a carpet when traversed clockwise. Here are the steps to think through this:

1. **Edge Cases for Input Size**:
   - Ensure the input size is within the constraints but small enough to be manageable.
   - Use even values for `n` and `m`.

2. **Edge Cases for Carpet Content**:
   - Include patterns that might cause the code to fail, such as:
     - Patterns where "1543" appears in the corners or edges.
     - Patterns where "1543" appears in the middle of the carpet.
     - Patterns where "1543" appears in layers that are not the outermost layer.
     - Patterns where "1543" appears in multiple layers.

3. **Test Cases**:
   - Create a test case with a small carpet where "1543" appears in the outer layer.
   - Create a test case with a small carpet where "1543" appears in an inner layer.
   - Create a test case with a small carpet where "1543" appears in both the outer and inner layers.

Based on these considerations, here is a small-scale test input that should expose potential bugs in the code:

```plaintext
3
2 4
1543
7777
4 4
1543
7777
7777
1543
6 4
1543
7777
7777
1543
7777
1543
```

This test input includes:
1. A 2x4 carpet where "1543" appears in the outer layer.
2. A 4x4 carpet where "1543" appears in both the outer and inner layers.
3. A 6x4 carpet where "1543" appears in multiple layers, including the outer and inner layers.

This should help in identifying any issues with the code's handling of different layers and patterns.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

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

public class challenge {
    static String target = "1543";
    static HashSet<String> hash = new HashSet<>();

    // 1= ul to ur;
    // 2= ur to lr
    // 3= lr to ll
    // 4= ll to ul
    static int solve(String[] k, int x, int y, int type, int j, int total, int minX, int minY, int maxX, int maxY,
            int c) {
        if (c == 2) {
            return total;
        }
        int r = total;
        if (k[y].charAt(x) == target.charAt(j)) {
            j++;
            j %= 4;
            if (!hash.contains(x + "," + y) && j == 0) {
                hash.add(x + "," + y);
                r += j == 0 ? 1 : 0;
            }
        } else {

            j = 0;
        }
        switch (type) {
            case 1:
                if (x + 1 >= maxX) {
                    r = solve(k, x, y + 1, 2, j, r, minX, minY, maxX, maxY, c);
                    break;
                }
                r = solve(k, x + 1, y, 1, j, r, minX, minY, maxX, maxY, c);
                break;
            case 2:
                if (y + 1 >= maxY) {
                    r = solve(k, x - 1, y, 3, j, r, minX, minY, maxX, maxY, c);
                    break;
                }
                r = solve(k, x, y + 1, 2, j, r, minX, minY, maxX, maxY, c);
                break;
            case 3:
                if (x - 1 < minX) {
                    r = solve(k, x, y - 1, 4, j, r, minX, minY, maxX, maxY, c);
                    break;
                }
                r = solve(k, x - 1, y, 3, j, r, minX, minY, maxX, maxY, c);
                break;
            case 4:
                if (y - 1 < minY) {
                    return r = solve(k, x + 1, y, 1, j, r, minX, minY, maxX, maxY, c + 1);
                }
                r = solve(k, x, y - 1, 4, j, r, minX, minY, maxX, maxY, c);
                break;
            default:
                break;
        }
        return r;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        outer: for (int q = 0; q < t; q++) {
            String p[] = br.readLine().split(" ");
            int n = Integer.parseInt(p[0]);
            int m = Integer.parseInt(p[1]);
            hash = new HashSet<>();

            String[] k = new String[n];
            for (int i = 0; i < n; i++) {
                k[i] = br.readLine();
            }

            int min = Math.min(n, m) / 2;
            int r = 0;
            for (int i = 0; i < min; i++) {
                r += solve(k, i, i, 1, 0, 0, i, i, m - i, n - i, 0);
            }
            System.out.println(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
4 4
1543
1543
1543
1543
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
t = int(input())

def check(n, m, matrix, layer):
    ans = 0
    sc = 0
    seq = '1543'
    for i in range(m-2*layer):
        if matrix[layer][layer+i] == seq[sc]:
            sc += 1
        else:
            sc = 0

        if sc == 4:
            ans += 1
            sc = 0

    if sc > 0:
        sc -= 1

    for i in range(n-2*layer):
        # print(layer+i, m-1-layer)
        if matrix[layer+i][m-1-layer] == seq[sc]:
            sc += 1
        else:
            sc = 0

        if sc == 4:
            ans += 1
            sc = 0

    if sc > 0:
        sc -= 1

    for i in range(m-2*layer):
        # print(n-1-layer, m-1-layer-i)
        if matrix[n-1-layer][m-1-layer-i] == seq[sc]:
            sc += 1
        else:
            sc = 0

        if sc == 4:
            ans += 1
            sc = 0

    if sc > 0:
        sc -= 1

    for i in range(n-2*layer):
        # print(n-1-layer-i, layer)
        if matrix[n-1-layer-i][layer] == seq[sc]:
            sc += 1
        else:
            sc = 0

        if sc == 4:
            ans += 1
            sc = 0

    if sc > 0:
        sc -= 1

    # print()

    if m - 2*layer > 2:
        for i in range(3):
            # print(layer, layer+i)
            if matrix[layer][layer+i] == seq[sc]:
                sc += 1
            else:
                sc = 0

            if sc == 4:
                ans += 1
                sc = 0
    else:
        # print("here", sc)
        if sc == 0:
            pass
        elif sc == 1:
            if matrix[layer][layer] == seq[1] and matrix[layer][layer+1] == seq[2] and matrix[layer+1][layer+1] == seq[3]:
                ans += 1

        elif sc == 2:
            if matrix[layer][layer] == seq[2] and matrix[layer][layer+1] == seq[3]:
                ans += 1
        elif sc == 3:
            if matrix[layer][layer] == seq[3]:
                ans += 1

    return ans

for _ in range(t):
    n, m = map(int,input().split())
    matrix = []

    for i in range(n):
        a = input()
        matrix.append(a)

    ans = 0

    for layer in range(min(n, m)//2 + 1):
        #print("layer", layer)
        x = check(n, m, matrix, layer)
        # print("layer", layer, x)
        ans += x

    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
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
def solve():
    n, m = list(map(int, input().split()))
    arr = []
    for _ in range(n):
        arr.append([int(e) for e in list(input())])
    w1 = m - 1
    w2 = n - 1
    w3 = 0
    w4 = 0
    dontdec = False
    if arr[0][0] == 1:
        i, j = 0, 0
        dire = 0
    elif arr[1][0] == 1:
        i, j = 1, 0
        dire = 3
        dontdec = True
    elif n > 2 and arr[2][0] == 1:
        i, j = 2, 0
        dire = 3
        dontdec = True
    elif arr[1][1] == 1:
        i, j = 1, 1
        dire = 2
        dontdec = True
    elif m > 2 and arr[1][2] == 1:
        i, j = 1, 2
        dire = 2
        dontdec = True
    else:
        i, j = 0, 0
        dire = 0
    num = []
    cnt = 0
    while w1 >= w3 and w2 >= w4:
        if dire == 0:
            num.append(arr[i][j])
            if len(num) > 4:
                num.pop(0)
            if j == w1:
                w1 -= 1
                dire = 1
                j -= 1
                i += 1
                if dontdec:
                    dontdec = False
                    w1 += 1
            if num == [1, 5, 4, 3]:
                cnt += 1
            j += 1
        elif dire == 1:
            num.append(arr[i][j])
            if len(num) > 4:
                num.pop(0)
            if i == w2:
                w2 -= 1
                dire = 2
                i -= 1
                j -= 1
                if dontdec:
                    dontdec = False
                    w2 += 1
            if num == [1, 5, 4, 3]:
                cnt += 1
            i += 1
        elif dire == 2:
            num.append(arr[i][j])
            if len(num) > 4:
                num.pop(0)
            if j == w3:
                w3 += 1
                dire = 3
                j += 1
                i -= 1
            if num == [1, 5, 4, 3]:
                cnt += 1
            j -= 1
        elif dire == 3:
            num.append(arr[i][j])
            if len(num) > 4:
                num.pop(0)
            if i == w4:
                w4 += 1
                dire = 0
                i += 1
                j += 1
                if dontdec:
                    w4 -= 1
                    dontdec = False
            if num == [1, 5, 4, 3]:
                cnt += 1
            i -= 1
        # print(num, w3, dire, i, j, w3, w1, w2, w4)
    print(cnt)
t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
2 4
1543
7777
2 4
7154
8903
4 4
1543
1435
5181
7634
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
import java.util.Scanner;
public class ilove1543 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for (int k = 0; k < t; k++) {
            int n = in.nextInt();
            int m = in.nextInt();
            int[][] mat = new int[n][m];
            for (int i = 0; i < n; i++) {

                String line = in.next();
                for (int j = 0; j < m; j++) {
                    mat[i][j] = line.charAt(j) - '0';
                }
                         }

        }
        System.out.println(1);
        System.out.println(1);
        System.out.println(0);
        System.out.println(1);
        System.out.println(0);
        System.out.println(2);
        System.out.println(2);
        System.out.println(2);

    }
}

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

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

Think step by step.

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

1. **Input Validation**: Ensure the input is valid and follows the constraints.
2. **Edge Cases**: Consider edge cases such as the smallest and largest possible inputs.
3. **Complex Patterns**: Create a pattern that requires the algorithm to correctly handle multiple layers and rotations.

Given the constraints:
- \(2 \leq n, m \leq 10^3\)
- \(n\) and \(m\) are even integers
- The sum of \(n \cdot m\) across all test cases does not exceed \(10^6\)

We will create a test input that includes:
- A single test case with a small \(n\) and \(m\) to ensure it fits within the constraints.
- A pattern that includes the number \(1543\) in different layers to test the algorithm's ability to handle multiple layers and rotations correctly.

### Test Input
```plaintext
1
4 4
1543
5431
4354
3154
```

### Explanation
- **Test Case 1**: A \(4 \times 4\) carpet.
- **Pattern**: The number \(1543\) appears in the first layer and the second layer when traversed clockwise.

This test input should expose the bug in the buggy code because:
- The buggy code does not actually process the carpet to find occurrences of \(1543\).
- The buggy code simply prints predefined outputs, which will not match the correct output for this test case.

### Expected Output
For the provided test input, the correct output should be:
```plaintext
2
```

This is because the number \(1543\) appears twice in the carpet:
- Once in the first layer.
- Once in the second layer.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

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

public class Template {
    public static void main(String[] args) throws FileNotFoundException {
        Locale.setDefault(Locale.ROOT);
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
//        InputReader in = new InputReader(new FileInputStream("sum.in"));
//        PrintWriter out = new PrintWriter(new FileOutputStream("sum.out"));
        solve(in, out);
        out.close();
    }

    public static void solve(InputReader in, PrintWriter out) {
        int t = in.nextInt();
        char[] kk = new char[]{'1', '5', '4', '3'};
        for (int i = 0; i < t; i++) {
            int n = in.nextInt();
            int m = in.nextInt();
            char[][] matr = new char[n][m];
            int ans = 0;

            for (int j = 0; j < n; j++) {
                matr[j] = in.next().toCharArray();
            }

            for (int layer = 0; layer < Math.min(n, m) / 2; layer++) {
                int lookup = 0;
                for (int j = layer; j < m - layer; j++) {
                    if (matr[layer][j] == kk[lookup]) {
                        lookup++;
                        if (lookup == 4) {
                            ans++;
                            lookup = 0;
                        }
                    } else {
                        lookup = 0;
                    }
                }

                for (int j = layer + 1; j < n - layer; j++) {
                    if (matr[j][m - layer - 1] == kk[lookup]) {
                        lookup++;
                        if (lookup == 4) {
                            ans++;
                            lookup = 0;
                        }
                    } else {
                        lookup = 0;
                    }
                }

                for (int j = m - layer - 2; j >= layer; j--) {
                    if (matr[n - layer - 1][j] == kk[lookup]) {
                        lookup++;
                        if (lookup == 4) {
                            ans++;
                            lookup = 0;
                        }
                    } else {
                        lookup = 0;
                    }
                }

                for (int j = n - layer - 2; j > layer; j--) {
                    if (matr[j][layer] == kk[lookup]) {
                        lookup++;
                        if (lookup == 4) {
                            ans++;
                            lookup = 0;
                        }
                    } else {
                        lookup = 0;
                    }
                }

                //++
                if (lookup > 0) {
                    int lookup1 = lookup;
                    for (int j = layer; j < Math.min(5 - lookup1, m - layer); j++) {
                        if (matr[layer][j] == kk[lookup]) {
                            lookup++;
                            if (lookup == 4) {
                                ans++;
                                lookup = 0;
                            }
                        } else {
                            lookup = 0;
                        }
                    }

                    if ((m - layer - layer == 2 || n - layer - layer == 2) && lookup > 0) {
                        lookup1 = lookup;
                        for (int j = layer + 1; j < Math.min(5 - lookup1, n - layer); j++) {
                            if (matr[j][m - layer - 1] == kk[lookup]) {
                                lookup++;
                                if (lookup == 4) {
                                    ans++;
                                    lookup = 0;
                                }
                            } else {
                                lookup = 0;
                            }
                        }

                        if (lookup > 0) {
                            lookup1 = lookup;
                            for (int j = m - layer - 2; j >= Math.min(5 - lookup1, layer); j--) {
                                if (matr[n - layer - 1][j] == kk[lookup]) {
                                    lookup++;
                                    if (lookup == 4) {
                                        ans++;
                                        lookup = 0;
                                    }
                                } else {
                                    lookup = 0;
                                }
                            }

                        }

                    }
                }

            }
            out.println(ans);
        }
    }

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

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

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

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

        String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }

}

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest and largest possible even values for \( n \) and \( m \).
2. **Special Patterns**: Create patterns where the number 1543 appears in tricky ways, such as overlapping layers or very close to the boundaries.
3. **Single Layer**: Test with a carpet that has only one layer to ensure the code handles this correctly.
4. **Multiple Layers**: Test with a carpet that has multiple layers to ensure the code correctly traverses all layers.

Let's create a test input that includes these scenarios:

1. A small carpet with \( n = 2 \) and \( m = 2 \).
2. A small carpet with \( n = 2 \) and \( m = 4 \).
3. A larger carpet with \( n = 4 \) and \( m = 4 \) to test multiple layers.
4. A carpet with a single layer that has the number 1543.
5. A carpet with overlapping 1543 sequences in different layers.

Here is the test input:

```plaintext
5
2 2
15
43
2 4
1543
7890
4 4
1543
5431
1543
5431
6 4
1543
1543
1543
1543
1543
1543
4 4
1543
1543
1543
1543
```

This input includes:
- A 2x2 carpet with a single layer.
- A 2x4 carpet with a single layer.
- A 4x4 carpet with two layers.
- A 6x4 carpet with multiple layers and overlapping 1543 sequences.
- A 4x4 carpet with a single layer and repeating 1543 sequences.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
#define _USE_MATH_DEFINES
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <cstring>
#include <fstream>
#include <utility>
#include <cstdlib>
#include <vector>
#include <string>
#include <limits>
#include <cmath>
#include <stack>
#include <queue>
#include <regex>
#include <ctime>
#include <set>
#include <map>
int main()
{
    int t;
    std::cin >> t;
    while (t) {
        --t;
        int n, m;
        std::string s;
        std::cin >> n >> m;
        getline(std::cin, s);
        int** a = new int* [n];
        for (int i = 0; i < n; ++i) {
            a[i] = new int[m];
        }
        for (int i = 0; i < n; ++i) {
            getline(std::cin, s);
            for (int j = 0; j < m; ++j) {
                a[i][j] = s[j]-'0';
            }
        }
        int q = 0,d=0,r=0;
        while (q < n * m) {
            int* b = new int[2 * n + 2 * m + 10];
            for (int i = d; i < m-d; ++i) {
                if (q >= n * m)break;
                b[q] = a[d][i];
                //std::cout << b[q] << " ";
                ++q;
            }
            for (int j = d+1; j < n-d; ++j) {
                if (q >= n * m)break;
                b[q] = a[j][m-d-1];
                //std::cout << b[q] << " ";
                ++q;
            }
            for (int i = d+1; i < m-d; ++i) {
                if (q >= n * m)break;
                b[q] = a[n-d-1][m-1-i];
                 //std::cout << b[q] << " ";
                ++q;
            }
            for (int j = d+1; j<n-d-1; ++j) {
                if (q >= n * m)break;
                b[q] = a[n-j-1][d];
                //std::cout << b[q] << " ";
                ++q;
            }
            int y = q;
            for (int i = d; i <std::min(3, m - d); ++i) {
                b[y] = a[d][i];
                //std::cout << b[y] << " ";
                ++y;
            }
            if (m - d == 2) {
                for (int j = d + 1; j < n - d; ++j) {
                    b[y] = a[j][m - d - 1];
                    //std::cout << b[q] << " ";
                    ++y;
                    break;
                }
            }
            for (int i = 3; i < y; ++i) {
                if (b[i - 3] == 1 && b[i - 2] == 5 && b[i - 1] == 4 && b[i] == 3)++r;
            }
            ++d;
        }
        std::cout << r << "\n";
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The primary issues in the code seem to be related to how the layers are processed and how the sequence "1543" is checked.

### Step-by-Step Analysis

1. **Layer Processing**:
   - The code processes each layer by traversing the outermost layer first and then moving inward.
   - It uses a dynamic array `b` to store the values of the current layer, but it does not correctly handle the traversal of the inner layers.

2. **Sequence Checking**:
   - The code checks for the sequence "1543" in the array `b` but does not handle the case where the sequence spans multiple layers or is partially in one layer and partially in another.

3. **Edge Cases**:
   - Small carpets (e.g., 2x2, 2x4, 4x4) where the sequence "1543" might appear in the first layer.
   - Carpets with exactly the sequence "1543" in the middle or near the edges of the carpet.
   - Carpets where the sequence "1543" is split across layers.

### Challenging Test Input

Given the above analysis, a small-scale test input that can expose the bug is:

```plaintext
1
4 4
1543
3451
5143
3154
```

### Explanation

- **Carpet Description**:
  - The carpet is a 4x4 grid.
  - The sequence "1543" appears multiple times in different layers.

- **Layer Traversal**:
  - The first layer (outermost) contains: `1543345151433154`
  - The second layer (inner) contains: `51433154`

- **Sequence Checking**:
  - The sequence "1543" appears in both the first and second layers.
  - The code should correctly count these occurrences, but if it has issues with layer processing or sequence checking, it might fail to count all occurrences.

This test input is small enough to be manageable and should effectively expose any issues with layer processing and sequence checking in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

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

int main(){
	int t;
	cin>>t;
	while(t--){
		int n,m ;
		cin >> n >> m ;

		vector<vector<int>> arr(n,vector<int>(m,0));
		for(int i=0 ; i<n ; i++){
			int num ; cin >> num ;
			for(int j=m-1 ; j>=0 ; j--){
				arr[i][j]= num%10 ;
				num=num/10;
				// cout << arr[i][j] << " " ;
			}
			// cout << endl;
		}

		int top=0, bottom = n-1 , left=0, right = m-1 ;
		int ans = 0 ;

		while(top < bottom && left < right) {
			vector<int> temp ;

			for(int i=left ; i<=right ; i++){
				temp.push_back(arr[top][i]) ;
			}

			top++ ;
			for(int i=top ; i<=bottom ; i++){
				temp.push_back(arr[i][right]);
			}
			right-- ;

			if(top<=bottom)
			{for(int i=right ; i>=left ; i--){
							temp.push_back(arr[bottom][i]);
						}
						bottom--;}

			if (left <= right)
			{for(int i=bottom; i>=top ; i--){
							temp.push_back(arr[i][left]);
						}
						left++;}

			// cout << "TEMP\n" ;
			// for(auto i : temp)
			// 	cout << i<< "\t" ;
			// cout << endl ;

			int n_ = temp.size();
			for(int i=0 ; i< n_ - 3 ; i++){
				if(temp[i]==1 && temp[i+1]==5 && temp[i+2]==4 && temp[i+3]==3)ans++;
			}

			if(n_ > 3)
			{
				if(temp[n_-3]==1 && temp[n_-2]==5 && temp[n_-1]==4 && temp[0]==3) ans++;
				else if(temp[1]==3 && temp[n_-2]==1 && temp[n_-1]==5 && temp[0]==4) ans++;
				else if(temp[1]==4 && temp[2]==3 && temp[n_-1]==1 && temp[0]==5) ans++;
			}
		}
		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
1
4 4
1543
3451
1543
3451
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
t = int(input())

for ii in range(t):
    n,m = list(map(int,input().split()))
    t = [[0]*m for i in range(n)]
    for jj in range(n):
        s = int(input())
        for kk in range(m):
            t[jj][kk] = s // 10**(m-kk-1) % 10
    bound = min(n//2,m//2)
    count = 0
    rotate = 0
    for jj in range(bound):
        upperBound = jj
        rightBound = m - jj - 1
        lowerBound = n - jj - 1
        leftBound = jj
        hash = set()
        numList = []
        currentX = jj
        currentY = jj
        nextX = currentX
        nextY = currentY + 1
        direction = 0
        hash.add((currentX,currentY))
        numList.append(t[currentX][currentY])
        while (nextX,nextY) not in hash:
            currentX = nextX
            currentY = nextY
            hash.add((currentX,currentY))
            numList.append(t[currentX][currentY])
            if direction == 0 and currentY != rightBound:
                nextX = currentX
                nextY = currentY + 1
            elif direction == 0 and currentY == rightBound:
                nextX = currentX + 1
                nextY = currentY
                direction = 1
            elif direction == 1 and currentX != lowerBound:
                nextX = currentX + 1
                nextY = currentY
            elif direction == 1 and currentX == lowerBound:
                nextX = currentX
                nextY = currentY - 1
                direction = 2
            elif direction == 2 and currentY != leftBound:
                nextX = currentX
                nextY = currentY - 1
            elif direction == 2 and currentY == leftBound:
                nextX = currentX - 1
                nextY = currentY
            elif direction == 3 and currentX != upperBound:
                nextX = currentX - 1
                nextY = currentY
                direction = 0
        lengthList = len(numList)

        for kk in range(lengthList):
            rotate = rotate + 1
            if numList[kk % lengthList] == 1 and numList[(kk + 1) % lengthList] == 5 and numList[(kk + 2) % lengthList] == 4 and numList[(kk + 3) % lengthList] == 3:
                count = count + 1
    print(count)

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
5431
4315
3154
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
from collections import defaultdict
from sys import stdin

def I(): return int(stdin.readline().strip())

def II(): return map(int, stdin.readline().strip().split())

def IL(): return list(map(int, stdin.readline().strip().split()))

def SIL(): return sorted(map(int, stdin.readline().strip().split()),)

def S() : return stdin.readline().strip()

def SL() : return list(stdin.readline().strip().split())

def count(part):
    ans = 0
    for i in range(len(part) - 3):
        if part[i:i + 4] == "1543":
            ans += 1
    return ans

def solve():
    n, m = II()
    mat = []
    for _ in range(n):
        mat.append(list(S()))
    result = []
    while mat:
        result += mat.pop(0)
        mat = (list(zip(*mat)))[::-1]

    broken = []
    i = 0
    j = i + (2 * n + 2 * m - 4)
    while i < len(result):
        broken.append(result[i:j])
        n -= 2
        m -= 2
        i = j
        j = i + (2 * n + 2 * m - 4)

    for i in range(len(broken)):
        broken[i] = "".join(broken[i] + broken[i][0:(3 if len(broken[i]) >= 4 else 0)])
    ans = 0
    for part in broken:
        ans += count(part)

    print(ans)

T = I()
for _ in range(T):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
4315
5431
3154
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define op()                      \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define srt(a) sort(a.begin(), a.end())
#define rsrt(a) sort(a.begin(), a.end(), greater<>())
#define fo(n) for (int i = 0; i < n; i++)
#define fo1(n) for (int i = 1; i <= n; i++)
#define all(a) a.begin(), a.end()
#define pb push_back
#define ll long long
#define Inf 10000000000000000
#define pi 3.1415926536
#define gcd __gcd
#define vi vector<int>
#define vl vector<ll>
#define nl '\n'
#define min_priority priority_queue<ll, vector<ll>, greater<ll>>
#define ordered_set tree<pair<ll, ll>, null_type, greater<pair<ll, ll>>, rb_tree_tag, tree_order_statistics_node_update>
#define show(a)           \
    for (auto e : a)      \
        cout << e << " "; \
    cout << nl;
#define rY               \
    cout << "YES" << nl; \
    return;
#define rN              \
    cout << "NO" << nl; \
    return;
#define rr return
const ll mod = 1e9 + 7;

void Solve()
{
    ll n, m, x, y, z, i, j, k, l, r;
    cin >> n >> m;
    char a[n + 3][m + 2];
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= m; j++)
            cin >> a[i][j];
    }

    ll n1 = n, m1 = m;
    l = 1, r = 1;
    ll ans = 0;
    while (l < n1)
    {
        string s;
        for (i = r; i <= m1; i++)
            s.push_back(a[l][i]);
        for (i = l + 1; i <= n1; i++)
            s.push_back(a[i][m1]);
        for (i = m1 - 1; i >= r; i--)
            s.push_back(a[n1][i]);
        for (i = n1 - 1; i > l; i--)
            s.push_back(a[i][r]);
        l++, r++, m1--, n1--;
        for (i = 0; i < min((int)s.size(), 3); i++)
            s.push_back(s[i]);
        for (i = 0; i < s.size() - 3; i++)
        {
            if (s[i] == '1' && s[i + 1] == '5' && s[i + 2] == '4' && s[i + 3] == '3')
                ans++;
        }
    }
    cout << ans << nl;
}
int main()
{
    op();
    int test = 1, c = 1;
    cin >> test;
    while (test--)
    {
        // cin.ignore();
        // cout << "Case " << c++ << ":\n";
        Solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The bug in the code seems to be related to how the layers are processed and how the string is constructed. Specifically, the code constructs the string for each layer and then checks for the sequence "1543" in that string. However, it might not correctly handle the transition between layers or the edge cases where the sequence spans different layers.

2. **Identify Edge Cases**:
   - **Single Layer**: Test a carpet that has only one layer.
   - **Multiple Layers**: Test a carpet with multiple layers where the sequence "1543" spans different layers.
   - **Corner Cases**: Test a carpet where the sequence "1543" is present at the corners or edges of the layers.

3. **Construct Test Inputs**:
   - **Single Layer**: A 2x2 carpet.
   - **Multiple Layers**: A 4x4 carpet.
   - **Corner Cases**: A 4x4 carpet with "1543" at the corners.

Given these considerations, let's construct a small-scale test input that is complete and valid:

```plaintext
3
2 2
15
43
4 4
1543
7890
3215
4321
4 4
1543
5432
2154
3215
```

This test input includes:
1. A 2x2 carpet with "1543" in the first layer.
2. A 4x4 carpet with "1543" in the first layer.
3. A 4x4 carpet with "1543" in the first layer and another "1543" in the second layer.

These cases should help expose any issues in the code's handling of layer transitions and sequence detection.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
#include<iostream>
#include<vector>
using namespace std;
 int n,m;
 void print(vector<int>&layer)
 {
     for(auto&i:layer)
     {
         cout<<i<<" ";
     }
     cout<<endl;

 }
void strips(int w,vector<int>&layer,vector<vector<int>>&arr)
{
    layer.clear();
    for(int i=w,j=w;j<=m-w;j++)
    {
        layer.push_back(arr[i][j]);
    }
    for(int i=w+1,j=m-w;i<=n-w;i++)
    {
         layer.push_back(arr[i][j]);
    }
    for(int i=n-w,j=m-w-1;j>=w;j--)
    {
         layer.push_back(arr[i][j]);
    }
    for(int i=n-w-1, j=w;i>w;i--)
    {
         layer.push_back(arr[i][j]);
    }
}

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

        cin>>n>>m;

        vector<vector<int>>arr(n,vector<int>(m));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                char c;
                cin>>c;
                arr[i][j]=c-'0';
            }
        }
        n--;m--;
        vector<int>layer;
        int w=0;
        long long ans=0;
        vector<int>tar={1,5,4,3};
        while(w<=n&&w<=m)
        {
            strips(w,layer,arr);
            int yoyo=0;
            for(int i=0;i<(int)layer.size();i++)
            {
                if(layer[i]==tar[yoyo])
                {
                    yoyo++;
                    if(yoyo==4)
                    {
                        yoyo=0;ans++;
                    }
                }
                else {yoyo=0;if(layer[i]==tar[yoyo])yoyo++;}
            }
            if(yoyo!=0)
            {
                for(int i=0;i<(int)layer.size();i++)
                {
                    if(layer[i]==tar[yoyo])
                    {
                    yoyo++;
                    if(yoyo==4) {yoyo=0;ans++; break; }
                    }
                    else {yoyo=0;break;}
                }
            }
            w++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
4 4
1543
3154
4315
5431
6 6
154315
431543
315431
543154
431543
315431
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

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

public class D_I_Love_1543{
public static int mod = (int)(1e9 + 7);
    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int t = sc.nextInt();
        while (t-- > 0)
        solve(sc, out);
        out.close();
    }

   public static void solve(FastScanner sc, PrintWriter out) {
        int n = sc.nextInt();
        int m = sc.nextInt();
        char[][] carpet = new char[n][m];

        for (int i = 0; i < n; i++) {
            carpet[i] = sc.next().toCharArray();
        }

        int ans = 0;
        String target = "1543";
        int r = 0, c = 1;
        while (n >= c && m >= r) {
            StringBuilder layer = new StringBuilder();

            int count = 0;
            String str = "";
            for (int j = r; j < m; j++) {
                layer.append(carpet[c-1][j]);
                if(count < 3){
                    str += "" + carpet[c-1][j];
                    count++;
                }
            }
            for (int i = c; i < n; i++) {
                layer.append(carpet[i][m - 1]);
                if(count < 3){
                    str += "" + carpet[i][m - 1];
                    count++;
                }
            }

            for (int j = m - 2; j >= r; j--) {
                layer.append(carpet[n - 1][j]);
            }
            for (int i = n - 2; i >= c; i--) {
                layer.append(carpet[i][r]);
            }
            layer.append(str);
            ans += countSub(layer.toString(), target);
            n -= 1; r++;
            m -= 1; c++;
        }
        out.println(ans);
   }
   private static int countSub(String str, String target) {
        int count = 0;
        for (int i = 0; i <= str.length() - target.length(); i++) {
            if (str.startsWith(target, i)) {
                count++;
            }
        }
        return count;
    }
    public static class Pair implements Comparable<Pair>{
    int first;
    int second;

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

    @Override
    public boolean equals(Object obj) {
        if (this == obj) return true;
        if (obj == null || getClass() != obj.getClass()) return false;

        Pair pair = (Pair) obj;

        if (first != pair.first) return false;
        return second == pair.second;
    }

    @Override
    public int hashCode() {
        int result = first;
        result = 31 * result + second;
        return result;
    }
    @Override
public int compareTo(Pair other) {
   if (this.first != other.first) {
      return Integer.compare(this.first, other.first); // Compare by 'first'
 } else {
    return Integer.compare(this.second, other.second); // If 'first' values are equal, compare by 'second'
}
}
}
private static long gcd(long a, long b){
    while(a != 0 && b != 0){
        if(a > b) a = a % b;
        else b =  b % a;
    }
    if(a == 0) return b;
    return a;
}

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The key issues to consider are:

1. **Incorrect layer traversal**: The code might not correctly traverse the layers of the carpet.
2. **Incorrect counting of "1543"**: The code might miss occurrences of "1543" or count them incorrectly.

### Edge Cases to Consider:
1. **Single Layer Carpet**: A carpet with only one layer.
2. **Multiple Layers**: A carpet with multiple layers where "1543" appears in different layers.
3. **Different Orientations**: "1543" appearing in different orientations (e.g., starting from different positions in the layer).

### Small-Scale Test Input:
We will create a test input with a small carpet that has multiple layers and includes "1543" in different positions to ensure the code handles all layers correctly.

```plaintext
4
4 4
1543
1543
1543
1543
4 4
1234
5154
3154
3154
6 4
5432
1152
4542
2432
2302
5942
6 4
1543
1543
1543
1543
1543
1543
```

### Explanation:
1. **First Test Case (4x4 Carpet)**:
   - Contains "1543" in all cells of the outer layer.
   - Should count 4 occurrences of "1543".

2. **Second Test Case (4x4 Carpet)**:
   - Contains "1543" in different positions within the outer layer.
   - Should count 2 occurrences of "1543".

3. **Third Test Case (6x4 Carpet)**:
   - Contains "1543" in different layers.
   - Should count 2 occurrences of "1543".

4. **Fourth Test Case (6x4 Carpet)**:
   - Contains "1543" in all cells of the outer layer and inner layer.
   - Should count 4 occurrences of "1543".

This test input should help expose any issues in the layer traversal and counting of "1543" in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: I Love 1543
time_limit_ms: 2000
memory_limit_mb: 256
Description: One morning, Polycarp woke up and realized that $1543$ is the most favorite number in his life.

The first thing that Polycarp saw that day as soon as he opened his eyes was a large wall carpet of size $n$ by $m$ cells; $n$ and $m$ are even integers. Each cell contains one of the digits from $0$ to $9$.

Polycarp became curious about how many times the number $1543$ would appear in all layers$^{\text{∗}}$ of the carpet when traversed clockwise.

$^{\text{∗}}$The first layer of a carpet of size $n \times m$ is defined as a closed strip of length $2 \cdot (n+m-2)$ and thickness of $1$ element, surrounding its outer part. Each subsequent layer is defined as the first layer of the carpet obtained by removing all previous layers from the original carpet.

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

The first line of each test case contains a pair of numbers $n$ and $m$ ($2 \leq n, m \leq 10^3$, $n, m$ — even integers).

This is followed by $n$ lines of length $m$, consisting of digits from $0$ to $9$ — the description of the carpet.

It is guaranteed that the sum of $n \cdot m$ across all test cases does not exceed $10^6$.

Output: For each test case, output a single number — the total number of times $1543$ appears in all layers of the carpet in the order of traversal clockwise.

Examples:
input:
8
2 4
1543
7777
2 4
7154
8903
2 4
3451
8888
2 2
54
13
2 2
51
43
2 6
432015
512034
4 4
5431
1435
5518
7634
6 4
5432
1152
4542
2432
2302
5942
output:
1
1
0
1
0
2
2
2

Note: ![](CDN_BASE_URL/9c851ea8888390da76b9eeeb247a1d68) Occurrences of $1543$ in the seventh example. Different layers are colored in different colors.

Buggy Code:
import java.util.*;
public class Epic{
	public static void main(String [] args){
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
			int n = sc.nextInt();
			int m = sc.nextInt();
			char [][] arr = new char[n][m];
			for(int i=0;i<n;i++){
				arr[i] = sc.next().toCharArray();
			}
			int times = n/2;
			int count = 0;
			int x = 0;
			while(times-- > 0){
				StringBuilder st = new StringBuilder();
				for(int j=x;j<m-x;j++){
					st.append(arr[x][j]);
				}
				for(int row=x+1;row<n-x;row++){
					st.append(arr[row][m-1-x]);
				}
				for(int col=m-2-x;col>=x;col--){
					st.append(arr[n-1-x][col]);
				}
				for(int row = n-2-x;row >x; row--){
					st.append(arr[row][x]);
				}
				int size = st.length();
				st.append(st.toString());
				for(int i=0;i<size;i++){
					if(st.charAt(i) == '1' && st.charAt(i+1) == '5' && st.charAt(i+2) == '4' && st.charAt(i+3) == '3'){
						count++;
						i+=3;
					}
				}
				x++;
			}
			System.out.println(count);
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
4 4
1543
4315
5431
3154
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Go Learn! (Easy Version)
time_limit_ms: 5000
memory_limit_mb: 256
Description: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.

Well, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.

Bessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\ldots,a_n]$ with $n$ elements.

              let l = 1     let h = n          while l < h:       let m = floor((l + h) / 2)            if a[m] < k:         l = m + 1       else:         h = m          return l

Bessie submitted her code to Farmer John's problem with $m$ ($1 \leq m \leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \leq x, k \leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.

Test $i$ is correct if the following hold:

  1. The $x_i$-th element in the array is $k_i$.    2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$.

It might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\ldots,a_n]$ with $1 \leq a_i \leq n$ so that all remaining tests are correct.

In addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\ldots,a_n]$ with $1 \leq a_i \leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\,244\,353$.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq m \leq n \leq 3000$) denoting the number of the array and the number of tests.

The following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \leq x_i, k_i \leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.

It is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.

Output: For each test case, output two integers, $r$ — the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\,244\,353$.

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

input:
2
6 6
1 3
2 5
3 1
4 2
5 4
6 6
30 8
19 22
6 12
12 1
28 27
3 4
14 25
29 14
11 15
output:
3 78
3 839271911

Note: Consider the first example.

In the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.

In the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int p=998244353;
int t,n,m,f[3011],dp[3011],a[3011],rt;
int fa[6011],id[6011],sz,lc[6011],rc[6011],siz[6011],dep[6011],cl[3011][21],cr[3011][21],lca[3011][3011],pw[3011];
void build(int L,int R,int &x,int d)
{
	if(L==R)
	{
		x=L;//printf("build([%d,%d],%d) dep:%d\n",L,R,x,d);
		dep[x]=d;siz[x]=0;
		return;
	}
	if(!x)x=++sz;
	dep[x]=d;//printf("build([%d,%d],%d) dep:%d\n",L,R,x,d);
	id[x]=L+R>>1;
	for(int i=L;i<=(L+R>>1);++i)for(int j=(L+R>>1)+1;j<=R;++j)lca[i][j]=x;
	build(L,L+R>>1,lc[x]=0,d+1);build((L+R>>1)+1,R,rc[x]=0,d+1);
	fa[lc[x]]=fa[rc[x]]=x;siz[x]=siz[lc[x]]+siz[rc[x]]+1;
}
void proccl(int x)
{
	// printf(">>>>>>>>proccl(%d)\n",x);
	int ans=1,u=x;
	while(u!=rt)
	{
		cl[x][dep[u]-1]=ans;
		// printf("u:%d\n",u);
		if(u==lc[fa[u]])
		{
			if(id[fa[u]]!=x)ans=1ll*ans*(n-a[x]+1)%p;
			ans=1ll*ans*pw[siz[rc[fa[u]]]]%p;
			// printf("node (%d) subtree pw^%d\n",n-a[x]+1,siz[rc[fa[u]]]);
		}
		u=fa[u];
	}
	cl[x][0]=ans;
}
void proccr(int x)
{
	int ans=1,u=x;
	while(u!=rt)
	{
		cr[x][dep[u]-1]=ans;
		if(u==rc[fa[u]])
		{
			if(id[fa[u]]!=x)ans=1ll*ans*(a[x]-1)%p;
			ans=1ll*ans*pw[siz[lc[fa[u]]]]%p;
		}
		u=fa[u];
	}
	cr[x][0]=ans;
}
int main()
{
	scanf("%d",&t);while(t--)
	{
		scanf("%d%d",&n,&m);
		pw[0]=1;for(int i=1;i<=n;++i)pw[i]=1ll*pw[i-1]*n%p;
		for(int i=1;i<=n;++i)a[i]=-1;
		for(int i=1;i<=m;++i)
		{
			int x,k;scanf("%d%d",&x,&k);
			if(k==1&&x!=1)k=-1;
			a[x]=k;
		}
		int cnt=0;
		for(int i=1;i<=n;++i)
		{
			cnt+=(a[i]!=-1);
		}
		if(cnt==0)
		{
			printf("%d\n",pw[n]);
			continue;
		}
		for(int i=1;i<=n;++i)
		{
			if(a[i]==-1)dp[i]=-1e9;
			else
			{
				dp[i]=1;
				for(int j=1;j<i;++j)if(a[j]<a[i])dp[i]=max(dp[i],dp[j]+1);
			}
		}
		sz=n;
		rt=0;
		build(1,n,rt,1);
		for(int i=1;i<=n;++i)if(~a[i])proccl(i),proccr(i);
		for(int i=1;i<=n;++i)f[i]=0;
		for(int i=1;i<=n;++i)if(~a[i])
		{
			if(dp[i]==1)f[i]=cr[i][0];
			else
			{
				for(int j=1;j<i;++j)if(dp[j]+1==dp[i]&&a[j]<a[i])
				{
					int L=lca[j][i],coe=1;
					if(id[L]!=j)
					{
						coe=a[i]-a[j];
					}
					// printf("j:%d i:%d L:%d cl:%d cr:%d coe:%d\n",j,i,L,cl[j][dep[L]],cr[i][dep[L]],coe);
					f[i]=(f[i]+1ll*f[j]*cl[j][dep[L]]%p*cr[i][dep[L]]%p*coe)%p;
				}
			}
		}
		// printf("a:");for(int i=1;i<=n;++i)printf("%d ",a[i]);putchar(10);
		// printf("dp:");for(int i=1;i<=n;++i)printf("%d ",dp[i]);putchar(10);
		// printf("f:");for(int i=1;i<=n;++i)printf("%d ",f[i]);putchar(10);
		int ans=0,mx=0;
		for(int i=1;i<=n;++i)mx=max(mx,dp[i]);
		for(int i=1;i<=n;++i)if(dp[i]==mx)
		{
			ans=(ans+1ll*f[i]*cl[i][0]%p*(i!=n?n:1))%p;
		}
		if(dp[n]!=mx)ans=1ll*ans*n%p;
		printf("%d %d\n",m-mx,(ans%p+p)%p);
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Go Learn! (Easy Version)
time_limit_ms: 5000
memory_limit_mb: 256
Description: The differences between the easy and hard versions are the constraints on $n$ and the sum of $n$. In this version, $n \leq 3000$ and the sum of $n$ does not exceed $10^4$. You can only make hacks if both versions are solved.

Well, well, well, let's see how Bessie is managing her finances. She seems to be in the trenches! Fortunately, she is applying for a job at Moogle to resolve this issue. Moogle interviews require intensive knowledge of obscure algorithms and complex data structures, but Bessie received a tip-off from an LGM on exactly what she has to go learn.

Bessie wrote the following code to binary search for a certain element $k$ in a possibly unsorted array $[a_1, a_2,\ldots,a_n]$ with $n$ elements.

              let l = 1     let h = n          while l < h:       let m = floor((l + h) / 2)            if a[m] < k:         l = m + 1       else:         h = m          return l

Bessie submitted her code to Farmer John's problem with $m$ ($1 \leq m \leq n$) tests. The $i$-th test is of the form $(x_i, k_i)$ ($1 \leq x, k \leq n$). It is guaranteed all the $x_i$ are distinct and all the $k_i$ are distinct.

Test $i$ is correct if the following hold:

  1. The $x_i$-th element in the array is $k_i$.    2. If Bessie calls the binary search as shown in the above code for $k_i$, it will return $x_i$.

It might not be possible for all $m$ tests to be correct on the same array, so Farmer John will remove some of them so Bessie can AC. Let $r$ be the minimum of tests removed so that there exists an array $[a_1, a_2,\ldots,a_n]$ with $1 \leq a_i \leq n$ so that all remaining tests are correct.

In addition to finding $r$, Farmer John wants you to count the number of arrays $[a_1, a_2,\ldots,a_n]$ with $1 \leq a_i \leq n$ such that there exists a way to remove exactly $r$ tests so that all the remaining tests are correct. Since this number may be very large, please find it modulo $998\,244\,353$.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq m \leq n \leq 3000$) denoting the number of the array and the number of tests.

The following $m$ lines each contain two integers, describing the tests. The $i$-th line contains two integers $x_i$ and $k_i$ ($1 \leq x_i, k_i \leq n$) denoting the index and value of the test. It is guaranteed all $x_i$ are distinct and all $k_i$ are distinct.

It is guaranteed the sum of $n$ across all test cases does not exceed $10^4$.

Output: For each test case, output two integers, $r$ — the minimum of tests removed so that there exists an array so that all remaining tests are correct, and the number of arrays such that it is possible to remove $r$ tests to make all remaining tests correct modulo $998\,244\,353$.

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

input:
2
6 6
1 3
2 5
3 1
4 2
5 4
6 6
30 8
19 22
6 12
12 1
28 27
3 4
14 25
29 14
11 15
output:
3 78
3 839271911

Note: Consider the first example.

In the first test case, the array $[1,2,2,3,4]$ satisfies all $m$ tests, so the minimum number of tests Bessie has to remove is $0$. Note that this is also the only array that satisfies all $m$ tests.

In the second test case, the minimum number of tests Bessie has to remove is $1$. The only test Bessie can remove is $(2,5)$. If Bessie removes test $(2,5)$, then the arrays satisfying the remaining $m-1$ tests are $[2,2,3,1,4]$, $[2,2,3,2,4]$, $[2,2,3,3,4]$.

Buggy Code:
#include <bits/stdc++.h>
#define Loop(x,l,r) for (ll x = (l); x < (ll)(r); ++x)
#define LoopR(x,l,r) for (ll x = (r)-1; x >= (ll)(l); --x)
typedef long long ll;
typedef std::pair<int, int> pii;
typedef std::pair<ll , ll > pll;
using namespace std;

const int N = 3*1024;
const int mod = 998244353;
int inv[N];

ll pw(ll x, ll y)
{
	ll a = 1;
	while (y) {
		if (y&1)
			a = a*x % mod;
		x = x*x % mod;
		y /= 2;
	}
	return a;
}

int par[N], height[N];

void make_tree(int l, int r, int p, int h)
{
	if (l == r)
		return;
	int m = (l + r)/2;
	par[m] = p;
	height[m] = h;
	make_tree(l, m, m, h+1);
	make_tree(m+1, r, m, h+1);
}

int get(int l, int r, int i)
{
	if (l == r)
		return -1;
	int m = (l + r)/2;
	int x;
	if (i <= m) {
		x = get(l, m, i);
	} else {
		x = get(m+1, r, i);
	}
	if (x == -1)
		return m;
	return x;
}

int n;

ll cnt1(int p, int i, int k)
{
	ll ans = inv[n];
	for (int v = i; v != -1; v = par[v]) {
		ans %= mod;
		if (v < p) {
			ans *= inv[n];
			ans %= mod;
			ans *= k;
		} else if (v > p) {
			ans *= inv[n];
			ans %= mod;
			ans *= n-k;
		}
		ans %= mod;
	}
	return ans;
}

ll cnt1r(int x, int k)
{
	ll ans = cnt1(x, get(0, n-1, x), k);
	//cerr << "cnt1 " << x << ' ' << k << " = " << ans << '\n';
	return ans;
}

bool anc(int v, int a)
{
	while (v != -1) {
		if (v == a)
			return 1;
		v = par[v];
	}
	return 0;
}

ll cnt2(int p1, int i1, int k1, int p2, int i2, int k2)
{
	ll ans = anc(i1, p2)? inv[n-k1]: inv[n];
	while (height[i1] > height[i2])
		i1 = par[i1];
	//while (height[i1] < height[i2])
	//	i1 = par[i1];
	for (int v = i2; v != -1; v = par[v]) {
		while (height[i1] > height[v])
			i1 = par[i1];
		if (i1 == v) {
			if (v > p2) {
				ans *= inv[n];
				ans %= mod;
				ans *= n-k2;
			} else if (p1 < v && v < p2) {
				ans *= inv[n-k1];
				ans %= mod;
				//ans *= k1..k2;
				ans *= (k2-k1);
			}
		} else {
			if (v < p2) {
				ans *= inv[n];
				ans %= mod;
				ans *= k2;
			} else if (v > p2) {
				ans *= inv[n];
				ans %= mod;
				ans *= n-k2;
			}
		}
		ans %= mod;
	}
	return ans;
}

ll cnt2r(int x1, int k1, int x2, int k2)
{
	ll ans = cnt2(x1, get(0, n-1, x1), k1, x2, get(0, n-1, x2), k2);
	//cerr << "cnt2 " << x1 << ' ' << k1 << ' ' << x2 << ' ' << k2 << " = " << ans << '\n';
	return ans;
}

void solve()
{
	int m;
	cin >> n >> m;
	make_tree(0, n-1, -1, 0);
	vector<pii> vec;
	int bad = 0;
	Loop (i,0,m) {
		int x, k;
		cin >> x >> k;
		x--; k--;
		if (x != 0 && k == 0) {
			bad++;
			continue;
		}
		vec.push_back({x, k});
	}
	if (n == 1) {
		cout << "0 1\n";
		return;
	}
	m = vec.size();
	sort(vec.begin(), vec.end());
	vector<ll> sz(m), cnt(m);
	int mx = 0;
	Loop (i,0,m) {
		sz[i] = 1;
		Loop (j,0,i) {
			if (vec[j].second < vec[i].second)
				sz[i] = max(sz[i], 1 + sz[j]);
		}
		mx = max<int>(mx, sz[i]);
	}
	Loop (i,0,m) {
		if (sz[i] == 1) {
			cnt[i] = cnt1r(vec[i].first, vec[i].second);
			continue;
		}
		cnt[i] = 0;
		Loop (j,0,i) {
			if (vec[j].second >= vec[i].second)
				continue;
			if (sz[j] + 1 != sz[i])
				continue;
			ll x = cnt2r(vec[j].first, vec[j].second, vec[i].first, vec[i].second);
			cnt[i] += x * cnt[j];
			cnt[i] %= mod;
		}
	}
	ll ans = 0;
	Loop (i,0,m) {
		if (sz[i] == mx) {
			ans += cnt[i];
			ans %= mod;
		}
	}
	ans *= pw(n, n);
	ans %= mod;
	cout << (m-mx) + bad << ' ' << ans << '\n';
}

signed main()
{
	Loop (i,0,N)
		inv[i] = pw(i, mod-2);
	cin.tie(0) -> sync_with_stdio(false);
	int t;
	cin >> t;
	while (t--)
		solve();
}

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

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

Think step by step.

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