**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

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

using namespace std;
typedef long long ll;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

// #define ordered_set tree<pair<ll,ll> , null_type,less<pair<ll,ll>>, rb_tree_tag,tree_order_statistics_node_update>
#define AQUA                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0);

// const ll N = 10000000 + 5;
const ll mod = 998244353;
const ll inf = 1e15 + 7;
#define el '\n'
// #define f first
// #define s second
// const ll inf = 1e15 + 1, N = 2000005;
// vector<ll> v;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#pragma GCC optimize("Ofast")

// #define F first
// #define S second

long long n, m, k;

ll fun(ll a, ll b, ll c, ll d, vector<vector<vector<vector<ll>>>> &dp, vector<vector<long long>> &v)
{
    ll ans = inf;
    if (dp[a][b][c][d] != -1)
        return dp[a][b][c][d];
    // if(a > n || b > m)return inf;

    if (b + 1 <= m)
    {
        ans = min(ans, fun(a, b + 1, c, 0, dp, v));
    }

    for (ll i = 0; i <= n; i++)
    {
        if (a + 1 <= n)
        {
            ans = min(ans, fun(a + 1, b, i, 1, dp, v));
        }
    }
    // }
    if (a == n && b == m)
        ans = 0;

    ll cst = 0;
    if (d)
    {
        cst = c * k;
    }
    ll i2 = b + c;
    if (i2 > m)
    {
        i2 = (i2 % (m + 1)) + 1;
    }
     if (i2 <= 0 || i2 > m) return dp[a][b][c][d] = inf;

    return dp[a][b][c][d] = ans + cst + v[a][i2];
}

int main()
{
    AQUA
        long long tt = 1;
    cin >> tt;

    while (tt--)
    {
        cin >> n >> m >> k;
        vector<vector<long long>> v(n + 1, vector<long long>(m + 1));
        for (ll i = 1; i <= n; i++)
        {
            for (ll j = 1; j <= m; j++)
                cin >> v[i][j];
        }
        if(n==1 && m==1){
            cout<<v[1][1]<<el;
            continue;
        }

        vector<vector<vector<vector<ll>>>> dp(n + 5, vector<vector<vector<ll>>>(m + 4, vector<vector<ll>>(n + 5, vector<ll>(2, -1))));
        long long ans = inf;
        for (ll i = 0; i <= n; i++)
        {
            long long tmp = fun(1, 1, i, 1, dp, v);
            ans = min(ans, tmp);
        }
        cout << ans << el;
    }
}

**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
3 3 100
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
#!/usr/bin/env python
from __future__ import division, print_function

import sys

# Fast Input/Output
import os
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

# FastIO implementation for Python 2
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# FastReader class for input handling
class FastReader:
    def __init__(self):
        self.tokens = []

    def next(self):
        while not self.tokens:
            try:
                self.tokens = raw_input().split()  # Use raw_input() in Python 2
            except EOFError:
                raise
        return self.tokens.pop(0)

    def next_int(self):
        return int(self.next())

    def next_long(self):
        return long(self.next())  # Use long for large integers in Python 2

def process_test_case(input_reader, output_writer):
    rows = input_reader.next_long()
    cols = input_reader.next_long()
    cost = input_reader.next_long()

    matrix = [[0] * (int(cols) + 1) for _ in range(int(rows) + 1)]

    for i in range(1, int(rows) + 1):
        for j in range(1, int(cols) + 1):
            matrix[i][j] = input_reader.next_long()

    INFINITY = long(1e16)  # Use long for large constants in Python 2
    dynamic_array = [INFINITY] * (int(cols) + 1)
    dynamic_array[1] = 0

    for i in range(1, int(rows) + 1):
        extended_matrix = [[0] * (len(matrix[0]) + int(cols)) for _ in range(len(matrix))]
        for r in range(len(matrix)):
            extended_matrix[r][:len(matrix[0])] = matrix[r]
            for j in range(1, int(cols)):
                extended_matrix[r][len(matrix[0]) + j - 1] = matrix[r][j]
        matrix = extended_matrix

        new_dynamic_array = [INFINITY] * (int(cols) + 1)

        for start in range(1, int(cols) + 1):
            current_sum = INFINITY
            for j in range(1, int(cols) + 1):
                current_sum += matrix[i][start + j - 1]
                current_sum = min(
                    current_sum,
                    dynamic_array[j] + matrix[i][start + j - 1] + (start - 1) * cost,
                )
                new_dynamic_array[j] = min(new_dynamic_array[j], current_sum)

        dynamic_array = new_dynamic_array

    print(dynamic_array[int(cols)], file=output_writer)

def main():
    input_reader = FastReader()
    output_writer = sys.stdout

    test_cases = input_reader.next_int()

    for _ in range(test_cases):
        process_test_case(input_reader, output_writer)

    output_writer.flush()

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Cases**:
   - Single row or single column grids.
   - Grids where the cost of shifting rows is very high or very low.
   - Grids with very high values in cells.

2. **Scenarios**:
   - Grids where the optimal path involves minimal shifts.
   - Grids where the optimal path involves many shifts.
   - Grids where the optimal path is not straightforward and requires careful consideration of the cost.

3. **Small-Scale Test Input**:
   - Since we need a small-scale test input, we will use a grid of size \(3 \times 3\) or \(4 \times 4\).

Let's create a test input that combines these elements:

- A \(3 \times 3\) grid with high values in cells to maximize the impact of the cost.
- A scenario where the optimal path involves shifting rows to minimize the cost.

Here is the test input:

```plaintext
1
3 3 100
100 100 100
100 1 100
100 100 1
```

### Explanation:
- **Grid Size**: \(3 \times 3\).
- **Cost**: \(100\), which is high, so shifting rows should be minimized.
- **Grid Values**:
  - The top row and the right column have high values (100).
  - The middle cell of the second row and the bottom-left cell of the third row have low values (1).

### Expected Output:
- The optimal path involves shifting the second row once and the third row once to minimize the cost.
- After shifting, the grid becomes:
  \[
  \begin{bmatrix}
  100 & 100 & 100 \\
  100 & 100 & 1 \\
  100 & 1 & 100
  \end{bmatrix}
  \]
- The path taken is: \((1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)\).
- The sum of the integers on visited cells is \(100 + 100 + 1 + 1 = 202\).
- The number of operations performed is \(2\).
- The cost is \(100 \times 2 + 202 = 402\).

This test input should expose any issues in the buggy code, especially in handling high costs and optimal path finding.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
for _ in range(int(input())):
    n, m, cost = map(int, input().split())
    grid = [list(map(int, input().split())) for _ in range(n)]

    ans = [[10**12]*(m+1) for _ in range(n+1)]

    ans[0][1] = 0

    for i in range(1, n+1):
        curr = 0
        for j in range(1, m+1):
            currans = [10**12]*(m+1)
            for k in range(1, m+1):
                currans[k] = grid[i-1][(k-1 + (j-1))%m] + min(currans[k-1], curr + ans[i-1][k])
                ans[i][k] = min(ans[i][k], currans[k])
            curr += cost

    print(ans[n][m])

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

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

Think step by step.

Response:
```plaintext
3
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

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

public class codeforces994Div2 {

    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        while(t-- > 0) {
            new Solve().solve(in, out);
        }
        out.close();
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.fillInStackTrace();
                }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }

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

}

class Solve {

    private final long MAX_VAL = Long.MAX_VALUE/2L;
    public void solve(codeforces994Div2.FastScanner in, PrintWriter out) {

        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();

        long [][]a = new long[n][m];
        for (int i=0; i<n; i++) {
            for (int j=0; j<m; j++) {
                a[i][j] = in.nextLong();
            }
        }

        long []dp = new long[m];
        for(int c=0; c<m; c++) dp[c] = ((c==0)? 0L: MAX_VAL);

        for(int r=0; r<n; r++) {
            long []dp1 = new long[m];
            for (int c = 0; c < m; c++) dp1[c] = MAX_VAL;
            for (int s = 0; s < m; s++) {
                long []tmpDp = new long[m];
                for (int c = 0; c < m; c++) tmpDp[c] = MAX_VAL;
                for (int c = 0; c < m; c++) {
                    tmpDp[c] = Math.min(tmpDp[c], dp[c] + k*s + a[r][(c+s)%m]);
                    if (c>0) tmpDp[c] = Math.min(tmpDp[c], tmpDp[c-1] + a[r][(c+s)%m]);
                }
                for (int c = 0; c < m; c++) dp1[c] = Math.min(dp1[c], tmpDp[c]);
            }
            for(int c=0; c<m; c++) dp[c] = dp1[c];
        }
        out.println(dp[m-1]);
    }
}

**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
3 3 100
100 1 100
100 1 100
100 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

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

public class Solution {
    public static void main (String[] args) throws java.lang.Exception {
        Reader.init(System.in);
        int testsCount = Reader.nextInt();
        while (testsCount-- > 0) {
            solve();
        }
    }

    private static void solve() throws java.lang.Exception {
        int m = Reader.nextInt();
        int n = Reader.nextInt();
        long k = Reader.nextLong();
        long[][] mat = new long[m][n];
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                mat[i][j] = Reader.nextLong();
            }
        }

        for (int i=0; i<m; i++) {
            for (int j=1; j<n; j++) {
                mat[i][j] += mat[i][j-1];
            }
        }

        System.out.println(backtrack(0, 0, mat, k, new Long[m+1][n+1], new long[m][n][]));
        System.out.flush();
    }

    private static long backtrack(int i, int j, long[][] rowPrefixSums, long k, Long[][] mem, long[][][] memSR) {
        int m = rowPrefixSums.length;
        int n = rowPrefixSums[0].length;
        if (mem[i][j] != null) {
            return mem[i][j];
        }
        if (i==m-1) {
            return mem[i][j] = minSumWithRotation(rowPrefixSums, i, n-j, j, k, memSR);
        }
        long minCost = Long.MAX_VALUE;
        for (int count=1; j+count-1<n; count++) {
            minCost = Math.min(minCost,
                minSumWithRotation(rowPrefixSums, i, count, j, k, memSR) +
                backtrack(i+1, j+count-1, rowPrefixSums, k, mem, memSR));
        }
        //System.out.println(i + " " + j + " " + minCost);
        return mem[i][j] = minCost;
    }

    private static long minSumWithRotation(long[][] rowPrefixSums, int i, int count, int start, long k, long[][][] mem) {
        if (mem[i][count-1] != null) {
            return mem[i][count-1][start];
        }
        int n = rowPrefixSums[0].length;
        long[] sums = new long[n];
        for (int j1=0; j1<n; j1++) {
            int j2 = (j1+count-1)%n;
            long sum;
            if (j1<=j2) {
                sum = rowPrefixSums[i][j2] - (j1==0 ? 0 : rowPrefixSums[i][j1-1]);
            } else {
                sum = rowPrefixSums[i][n-1] - (j1==0 ? 0 : rowPrefixSums[i][j1-1]) + rowPrefixSums[i][j2];
            }
            sums[j1] = sum;
        }
        Set<Integer> indexSet = new HashSet<>();
        TreeSet<Integer> valueTreeSet = new TreeSet<>((a,b) ->
            sums[a]!=sums[b] ? Long.compare(sums[a], sums[b]) : Integer.compare(a,b));
        for (int j=0; j<n; j++) {
            indexSet.add(j);
            valueTreeSet.add(j);
        }
        while (!indexSet.isEmpty()) {
            int j = valueTreeSet.pollFirst();
            indexSet.remove(j);
            int prev = (j-1+n)%n;
            if (indexSet.contains(prev)) {
                long proposedValue = sums[j] + (prev<j ? j-prev : j+n-prev) * k;
                if (proposedValue < sums[prev]) {
                    valueTreeSet.remove(prev);
                    sums[prev] = proposedValue;
                    valueTreeSet.add(prev);
                }
            }
        }
        mem[i][count-1] = sums;
        //System.out.println(i + " " + count + " " + Arrays.toString(sums));
        return mem[i][count-1][start];
    }

    private static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;
        static void init(InputStream input) {
            reader = new BufferedReader(new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(nextLine());
            }
            return tokenizer.nextToken();
        }
        static String nextLine() throws IOException {
            return reader.readLine();
        }
        static boolean hasNext() throws IOException {
            return tokenizer.hasMoreTokens();
        }
        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
#define vi vector<int>
#define heap priority_queue
#define hmap unordered_map
#define hset unordered_set
#define test cout << "test" << endl

using namespace std;

template<class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;

//重载vector<<
template<class T>
ostream& operator<<(ostream& os, vector<T> vec) {
	os << "vector[";
	for(auto e: vec){
		os << e << ',';
	}
	os << ']';
	return os;
}

//const int max_n = 1e5 + 1;
//const double pi = 3.14159265358979323846;
//const ll mod = 1e9 + 7;
//ll mod = 998244353;
vector<ll> shift(vector<ll> ary, int n){
	vector<ll> ans = {0};
	for(int i = n + 1; i < ary.size(); i++){
		ans.push_back(ary[i]);
	}
	for(int i = 1; i <= n; i++){
		ans.push_back(ary[i]);
	}
	return ans;
}

vector<ll> prefix(vector<ll> ary){
	vector<ll> ans(ary.size());
	for(int i = 1; i < ary.size(); i++){
		ans[i] = ans[i - 1] + ary[i];
	}
	return ans;
}

void solve(){
	int n, m, c;
	cin >> n >> m >> c;
	vector<vector<ll>> ary(n + 1, vector<ll>(m + 1)), dp(n + 1, vector<ll>(m + 1, 1e18));
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> ary[i][j];
		}
	}
	for(int i = 1; i <= n; i++){
		for(int k = 0; k < m; k++){

			vector<ll> recent = shift(ary[i], k);
			vector<ll> pref = prefix(recent);//cost 200 * 200 * 200
//			cout << i << " shift " << k << ":" << recent << ',' << pref << endl;
			if(i == 1){
				for(int j = 1; j <= m; j++){
					dp[i][j] = min(dp[i][j], pref[j] + k * c);
				}
			} else {
				for(int j = 1; j <= m; j++){
					for(int x = 1; x <= j; x++){
						dp[i][j] = min(dp[i][j], dp[i - 1][x] + pref[j] - pref[x - 1] + k * c);
					}
				}
			}
		}
	}
//	cout << "dp:" << endl;
//	for(int i = 1; i <= n; i++){
//		for(int j = 1; j <= m; j++){
//			cout << dp[i][j] << ' ';
//		}
//		cout << endl;
//	}
	cout << dp[n][m] << endl;
}

signed main(){
    ios::sync_with_stdio(0);
	cin.tie(nullptr);
//	cout << fixed << setprecision(4);
	int t = 1;
	cin >> t;
	while(t--){
		solve();
	}
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 100
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define endl '\n'
//#define int long long
#define unmap unordered_map
#define unset unordered_set
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define frep(i,a,b) for(int i=(a);i<(b);++i)
using namespace std;
using namespace __gnu_pbds;
template<typename T>
using MINQ=priority_queue<T,vector<T>,greater<T>>;
template<typename T>
using MAXQ=priority_queue<T,vector<T>,less<T>>;
template<typename T>
using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
using pii=pair<int,int>;
using vi=vector<int>;
using vii=vector<vi>;
using viii=vector<vii>;
using i128=__int128;
using ll=long long;
using lf=double;
namespace ran=ranges;
template<typename T> T max(T a,T b,T c){return max(a,max(b,c));}
template<typename T> T min(T a,T b,T c){return min(a,min(b,c));}
template<typename T>
istream& operator>>(istream&is,vector<T>&vec){for(auto &x:vec)is>>x;return is;}
template<typename T>
ostream& operator<<(ostream&os,vector<T>&vec){for(auto &x:vec)os<<x<<" ";return os;}
const int N=210;
const ll INF=1e15;
int mat[N];
ll dp[N][N],w[N][N];
void solve(){
    int n,m,K;
    cin>>n>>m>>K;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)
            cin>>mat[j];
        for(int end=1;end<=m;end++){
            for(int len=1;len<=end;len++){
                ll sum=0;
                int l=end-len+1,r=end;
                for(int k=l;k<=r;k++) sum+=mat[k];
                w[end][len]=sum;
                for(int mov=1;mov<m;mov++){
                    sum-=mat[l];
                    l=l%m+1;
                    r=r%m+1;
                    sum+=mat[r]+K;
                    w[end][len]=min(w[end][len],sum);
                }
            }
        }
        for(int j=1;j<=m;j++){
            dp[i][j]=INF;
            for(int k=1;k<=j;k++){
                dp[i][j]=min(dp[i][j],dp[i-1][k]+w[j][j-k+1]);
            }
        }
    }
    cout<<dp[n][m]<<endl;
}
signed main(){
    ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
    cout<<fixed<<setprecision(10);
    int _t=1;
    cin>>_t;
    memset(dp[0],0x3f,sizeof(dp[0]));
    dp[0][1]=0;
#ifndef ONLINE_JUDGE
    auto START=chrono::high_resolution_clock::now();
#endif
    for(int i=1;i<=_t;i++){
        solve();
    }
#ifndef ONLINE_JUDGE
    auto END=chrono::high_resolution_clock::now();
    chrono::duration<double,milli> duration = END-START;
    cout<<fixed<<setprecision(1)<<"Running Time: "<<duration.count()<<" ms";
#endif
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 100
1 100 100
100 1 100
100 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
import copy

import sys
input = sys.stdin.readline

def solve():
    n, m, k = map(int, input().split())

    g = [[0 for _ in range(m)] for _ in range(n)]

    for i in range(n):
        g[i] = list(map(int, input().split()))

    s = [[[0 for _ in range(m + 1)] for _ in range(m)] for _ in range(n)]

    for l in range(1, m + 1, 1):
        for i in range(n):
            for j in range(m):
                s[i][j][l] = s[i][j][l - 1] + g[i][(j + l - 1) % m]

    for l in range(m + 1):
        for i in range(n):
            for j in range(m):
                s[i][j][l] += j * k

    sp = copy.deepcopy(s)
    ss = copy.deepcopy(s)

    for l in range(m + 1):
        for i in range(n):
            for j in range(1, m):
                sp[i][j][l] = min(sp[i][j][l], sp[i][j - 1][l])

    for l in range(m + 1):
        for i in range(n):
            for j in range(m - 2, -1, -1):
                ss[i][j][l] = min(ss[i][j][l], ss[i][j  + 1][l])

    inf = 10**18
    f = [[[inf for _ in range(m + 1)] for _ in range(m)] for _ in range(n)]

    for l in range(m + 1):
        for i in range(n):
            for j in range(m):
                f[i][j][l] = min(f[i][j][l], -k * j + ss[i][j][l])
                assert(-k * j + ss[i][j][l] >= 0)
                if j - 1 >= 0:

                    f[i][j][l] = min(f[i][j][l], -k * j + m * k + sp[i][j - 1][l])

    dp = [[inf for _ in range(m)] for _ in range(n)]

    for j in range(m):
        dp[0][j] = f[0][0][j + 1]

    for i in range(1, n):
        for j in range(m):
            for k in range(j + 1):

                dp[i][j] = min(dp[i][j], dp[i - 1][k] + f[i][k][j - k  + 1])

    print(dp[n - 1][m - 1])

if __name__ == "__main__":
    tt = int(input())
    while tt:
        solve()
        tt -= 1

**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
3 3 100
100 0 0
0 100 0
0 0 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

Buggy Code:
from math import *
from collections import *
from bisect import bisect_left, bisect_right, bisect
from itertools import *

R=lambda:map(int,input().split())
S=lambda:map(str,input().split())

MOD = int(1e9 + 7)
# MOD = 998244353
# MOD = 100000000669

t, = R()
# t = 1

def answer():
    n, m, k = R()
    a = []
    for i in range(n):
        a.append(list(R()))

    dp = [inf] * m
    for pos in range(m):
        for j in range(m):
            shift = (j - pos)
            if shift < 0:
                shift += m

            shift %= m

            s = 0
            for l in range(pos + 1):
                ind = (j - l) % m
                s += a[0][ind]

            # if pos == 3:
            #     print(shift, s)

            dp[pos] = min(dp[pos], shift * k + s)

    # print(dp)

    for r in range(1, n):
        # print(" --- ")
        dpn = [inf] * m

        for pos in range(m):
            for j in range(m):
                # if pos == 1 and j == 3:
                #     print(pos, j)

                shift = (j - pos)
                if shift < 0:
                    shift += m

                shift %= m

                # if pos == 1 and j == 3:
                #     print(shift * k + a[r][j] + dp[pos])

                dpn[pos] = min(dpn[pos], shift * k + a[r][j] + dp[pos])

                s = a[r][j]
                for l in range(pos + 1, m):
                    ind = (j + (l - pos)) % m
                    s += a[r][ind]

                    # if pos == 1 and j == 3:
                    #     print(l, shift * k + s + dp[pos])

                    dpn[l] = min(dpn[l], shift * k + s + dp[pos])

        dp = dpn
        # print(dp)

    return dp[-1]

for i in range(t):
    print(answer())

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

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

Think step by step.

Response:
```plaintext
1
3 3 1
10 0 0
0 0 10
0 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

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

public class Main {
    static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static Reader in = new Reader();
    static int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    static int[][] dirs_8 = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
    static double DIFF = 1e-15;
    static long LNF = Long.MAX_VALUE / 2;
    static int INF = Integer.MAX_VALUE / 2;
    static long BASE = 31;

    public static void main(String[] args) {
        int t = in.nextInt();
//        int t = 1;
        while (t-- > 0) {
            solve();
        }

        out.close();
    }

    static long MOD = (long) 998244353;
    static int N = (int) 200 + 2, M = (int) 2e5 + 10;

    static void solve() {
        int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
        long[][] g = in.nextLongArray2(n, m);

        long[][] dp = new long[n + 1][m + 1];
        for (long[] e : dp) {
            Arrays.fill(e, LNF);
        }
        dp[0][1] = 0;
        for (int i = 0; i < n; i++) {
            for (int c = 0; c < m; c++) {
                long[] f = new long[m + 1];
                Arrays.fill(f, LNF);
                for (int j = 0; j < m; j++) {
                    f[j + 1] = min(f[j + 1], min(dp[i][j + 1], f[j]) + g[i][(j + c) % m]);
                }
                for (int j = 0; j <= m; j++) {
                    dp[i + 1][j] = min(dp[i + 1][j], f[j] + c * k);
                }
            }
        }
        out.println(dp[n][m]);

    }

    // <------------------------------------------------------------------------------------->
    static long sqrt(long x) {
        long y = (long) Math.sqrt(x);
        if (y * y <= x) {
            return y;
        }
        return y - 1;
    }

    static void setOut(String path) {
        try {
            out = new PrintWriter(new FileOutputStream(path));
        } catch (FileNotFoundException e) {
            throw new RuntimeException(e);
        }
    }

    static long lowbit(long x) {
        return x & -x;
    }

    static void swap(long[] arr, int i, int j) {
        long t = arr[i];
        arr[i] = arr[j];
        arr[j] = t;
    }

    static long pow(long x, long n) {
        if (n <= 0) {
            return 1;
        }
        long res = 1;
        for (; n > 0; n >>= 1, x = x * x) {
            if ((n & 1) == 1) {
                res = res * x;
            }
        }
        return res;
    }

    static long pow(long x, long n, long mod) {
        if (n <= 0) {
            return 1;
        }
        x %= mod;
        long res = 1;
        for (; n > 0; n >>= 1, x = x * x % mod) {
            if ((n & 1) == 1) {
                res = res * x % mod;
            }
        }
        return res;
    }

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

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    static int getBit(int s, int i) {
        return (s >> i) & 1;
    }

    static int getBit(long s, int i) {
        return (int) (s >> i) & 1;
    }

    static int setBit(int s, int i, int b) {
        if (getBit(s, i) == 1) s ^= 1 << i;
        return s | b << i;
    }

    static int unique(int[] nums, int from, int to) {
        Arrays.sort(nums, from, to);
        int p = from;
        for (int i = from + 1; i < to; i++) {
            if (nums[i] != nums[p]) {
                nums[++p] = nums[i];
            }
        }
        return p + 1;
    }

    static int unique(long[] nums, int from, int to) {
        Arrays.sort(nums, from, to);
        int p = from;
        for (int i = from + 1; i < to; i++) {
            if (nums[i] != nums[p]) {
                nums[++p] = nums[i];
            }
        }
        return p + 1;
    }

    static long max(long... nums) {
        long res = Long.MIN_VALUE;
        for (long x : nums) {
            res = Math.max(res, x);
        }
        return res;
    }

    static double max(double... nums) {
        double res = Double.MIN_VALUE;
        for (double x : nums) {
            res = Math.max(res, x);
        }
        return res;
    }

    static int max(int... nums) {
        int res = Integer.MIN_VALUE;
        for (int x : nums) {
            res = Math.max(res, x);
        }
        return res;
    }

    static long min(long... nums) {
        long res = Long.MAX_VALUE;
        for (long x : nums) {
            res = Math.min(res, x);
        }
        return res;
    }

    static int min(int... nums) {
        int res = Integer.MAX_VALUE;
        for (int x : nums) {
            res = Math.min(res, x);
        }
        return res;
    }

    static double min(double... nums) {
        double res = Double.MAX_VALUE;
        for (double x : nums) {
            res = Math.min(res, x);
        }
        return res;
    }

    static List<Integer>[] build_graph(int n, int m) {
        List<Integer>[] g = new List[n + 1];
        Arrays.setAll(g, e -> new ArrayList<>());
        for (int i = 0; i < m; i++) {
            int x = in.nextInt(), y = in.nextInt();
            g[x].add(y);
            g[y].add(x);
        }
        return g;
    }

    static <T> void print(List<T> list) {
        for (T x : list) {
            out.print(x + " ");
        }
        out.println();
    }

    static class Reader {
        private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        private StringTokenizer st;

        void set(String path) {
            try {
                br = new BufferedReader(new FileReader(path));
            } catch (Exception ignored) {
            }
        }

        boolean hasNext() {
            try {
                while (st == null || !st.hasMoreElements()) {
                    st = new StringTokenizer(br.readLine());
                }
            } catch (Exception e) {
                return false;
            }
            return true;
        }

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

        String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (Exception ignored) {
            }
            return s;
        }

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

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

        int[] nextIntArray(int n, int start) {
            int[] arr = new int[n + start];
            for (int i = start; i < arr.length; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        int[] nextIntArray(int n) {
            return nextIntArray(n, 0);
        }

        int[][] nextIntArray2(int n, int m) {
            int[][] grid = new int[n][];
            Arrays.setAll(grid, i -> grid[i] = nextIntArray(m));
            return grid;
        }

        long[] nextLongArray(int n, int start) {
            long[] arr = new long[n + start];
            for (int i = start; i < arr.length; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        long[] nextLongArray(int n) {
            return nextLongArray(n, 0);
        }

        long[][] nextLongArray2(int n, int m) {
            long[][] grid = new long[n][];
            Arrays.setAll(grid, i -> grid[i] = nextLongArray(m));
            return grid;
        }

        char[][] nextCharArray(int n) {
            char[][] grid = new char[n][];
            Arrays.setAll(grid, i -> grid[i] = nextLine().toCharArray());
            return grid;
        }

        String[] nextStrArray(int n) {
            String[] arr = new String[n];
            Arrays.setAll(arr, i -> arr[i] = nextLine());
            return arr;
        }

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

        double[] nextDoubleArray(int n, int start) {
            double[] arr = new double[n + start];
            for (int i = start; i < arr.length; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }

        double[] nextDoubleArray(int n) {
            return nextDoubleArray(n, 0);
        }
    }

}

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest and largest possible values for `n`, `m`, and `k`.
2. **Special Patterns**: Test with grids that have special patterns, such as all zeros, all high values, or a mix of high and low values.
3. **Cyclic Shifts**: Test with grids where cyclic shifts can significantly change the optimal path.
4. **Path Constraints**: Test with grids where the optimal path is not straightforward, such as when the optimal path requires a specific sequence of cyclic shifts.

Given the constraints and the nature of the problem, let's create a small-scale test input that includes these elements.

### Test Input
```plaintext
4
3 3 100
3 4 9
5 2 4
0 101 101
3 3 1
10 0 0
0 0 10
10 10 0
2 2 5
1 100
100 1
1 1 3
4
```

### Explanation
1. **Test Case 1**:
   - **Grid**:
     ```
     3 4 9
     5 2 4
     0 101 101
     ```
   - **k**: 100
   - **Expected Behavior**: The optimal solution involves shifting the third row to minimize the cost.

2. **Test Case 2**:
   - **Grid**:
     ```
     10 0 0
     0 0 10
     10 10 0
     ```
   - **k**: 1
   - **Expected Behavior**: The optimal solution involves multiple shifts to minimize the cost.

3. **Test Case 3**:
   - **Grid**:
     ```
     1 100
     100 1
     ```
   - **k**: 5
   - **Expected Behavior**: The optimal solution involves a single shift to minimize the cost.

4. **Test Case 4**:
   - **Grid**:
     ```
     4
     ```
   - **k**: 3
   - **Expected Behavior**: The optimal solution is straightforward since there is only one cell.

This test input covers a range of scenarios, including edge cases and special patterns, to ensure the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shift + Esc
time_limit_ms: 2500
memory_limit_mb: 512
Description: After having fun with a certain contraption and getting caught, Evirir the dragon decides to put their magical skills to good use — warping reality to escape fast!

You are given a grid with $$$n$$$ rows and $$$m$$$ columns of non-negative integers and an integer $$$k$$$. Let $$$(i, j)$$$ denote the cell in the $$$i$$$-th row from the top and $$$j$$$-th column from the left ($$$1 \le i \le n$$$, $$$1 \le j \le m$$$). For every cell $$$(i, j)$$$, the integer $$$a_{i, j}$$$ is written on the cell $$$(i, j)$$$.

You are initially at $$$(1, 1)$$$ and want to go to $$$(n, m)$$$. You may only move down or right. That is, if you are at $$$(i, j)$$$, you can only move to $$$(i+1, j)$$$ or $$$(i, j+1)$$$ (if the corresponding cell exists).

Before you begin moving, you may do the following operation any number of times:

- Choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and cyclically shift row $$$i$$$ to the left by $$$1$$$. Formally, simultaneously set $$$a_{i,j}$$$ to $$$a_{i,(j \bmod m) + 1}$$$ for all integers $$$j$$$ ($$$1 \le j \le m$$$).

After moving from $$$(1, 1)$$$ to $$$(n, m)$$$, let $$$x$$$ be the number of operations you have performed before moving, and let $$$y$$$ be the sum of the integers written on visited cells (including $$$(1, 1)$$$ and $$$(n, m)$$$). Then the cost is defined as $$$kx + y$$$.

Find the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line contains three space-separated integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq n, m \leq 200$$$, $$$0 \leq k \leq 10^9$$$).

Then, $$$n$$$ lines follow. The $$$i$$$-th line contains $$$m$$$ space-separated integers, $$$a_{i,1},\,a_{i,2},\,\ldots,\,a_{i,m}$$$ ($$$0 \leq a_{i,j} \leq 10^9$$$).

It is guaranteed that the sum of $$$n \cdot m$$$ over all test cases does not exceed $$$5 \cdot 10^4$$$.

Output: For each test case, output a single integer, the minimum cost to move from $$$(1, 1)$$$ to $$$(n, m)$$$.

Examples:
input:
5
3 3 100
3 4 9
5 2 4
0 101 101
3 4 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
3 2 3
1 2
3 6
5 4
10 10 14
58 49 25 12 89 69 8 49 71 23
45 27 65 59 36 100 73 23 5 84
82 91 54 92 53 15 43 46 11 65
61 69 71 87 67 72 51 42 55 80
1 64 8 54 61 70 47 100 84 50
86 93 43 51 47 35 56 20 33 61
100 59 5 68 15 55 69 8 8 60
33 61 20 79 69 51 23 24 56 28
67 76 3 69 58 79 75 10 65 63
6 64 73 79 17 62 55 53 61 58
output:
113
6
4
13
618

Note: In the first test case, the minimum cost of $$$113$$$ can be achieved as follows:

1. Cyclically shift row 3 once. The grid now becomes $$$$$$\begin{bmatrix}3 & 4 & 9\\5 & 2 & 4\\101 & 101 & 0\end{bmatrix}.$$$$$$
2. Move as follows: $$$(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (3, 3)$$$.

$$$x = 1$$$ operation is done before moving. The sum of integers on visited cells is $$$y = 3 + 4 + 2 + 4 + 0 = 13$$$. Therefore, the cost is $$$kx + y = 100 \cdot 1 + 13 = 113$$$.

In the second test case, one can shift row 1 once, row 2 twice, and row 3 thrice. Then, the grid becomes $$$$$$\begin{bmatrix}0 & 0 & 10 & 10\\10 & 0 & 0 & 0\\10 & 10 & 10 & 0\end{bmatrix}.$$$$$$

$$$x = 6$$$ operations were done before moving, and there is a path of cost $$$y = 0$$$. Therefore, the cost is $$$6 \cdot 1 + 0 = 6$$$.

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

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringTokenizer st;
        int t = Integer.parseInt(f.readLine());
        while (t-- > 0) {
            st = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            long k = Long.parseLong(st.nextToken());
            long[][] arr = new long[n][m];
            for (int i = 0; i < n; i++) {
                st = new StringTokenizer(f.readLine());
                for (int j = 0; j < m; j++) {
                    arr[i][j] = Long.parseLong(st.nextToken());
                }
            }
            long[][][] dp = new long[n][m][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    for (int l = 0; l < m; l++) {
                        dp[i][j][l] = Long.MAX_VALUE;
                    }
                }
            }
            Queue<long[]> q = new LinkedList<>();
            for (int i = 0; i < m; i++) {
                q.add(new long[]{0, i, i, k * i + arr[0][i]});
                dp[0][i][i] = k * i + arr[0][i];
            }
            while (!q.isEmpty()) {
                long[] p = q.poll();
                int i = (int) p[0];
                int j = (int) p[1];
                int shift = (int) p[2];
                if (i < n - 1) {
                    for (int a = 0; a < m; a++) {
                        int temp = (j - shift + m) % m;
                        int need = (a - temp + m) % m;
                        long cost = k * need + arr[i + 1][a];
                        if (dp[i + 1][a][need] <= p[3] + cost) continue;
                        dp[i + 1][a][need] = p[3] + cost;
                        q.add(new long[]{i + 1, a, need, p[3] + cost});
                    }
                }
                if ((j - shift + m) % m < m - 1) {
                    if (dp[i][(j + 1) % m][shift] <= p[3] + arr[i][(j + 1) % m]) continue;
                    dp[i][(j + 1) % m][shift] = p[3] + arr[i][(j + 1) % m];
                    q.add(new long[]{i, j + 1, shift, p[3] + arr[i][(j + 1) % m]});
                }
            }
            long ans = Long.MAX_VALUE;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < m; j++) {
                    if ((i - j + m) % m == m - 1) {
                        ans = Math.min(ans, dp[n - 1][i][j]);
                    }
                }
            }
            out.println(ans);
        }
        f.close();
        out.close();
    }

    public static void sort(int[] arr) {
        Random r = new Random();
        for (int i = 0; i < arr.length; i++) {
            int x = r.nextInt(arr.length);
            int temp = arr[x];
            arr[x] = arr[i];
            arr[i] = temp;
        }
        Arrays.sort(arr);
    }

    public static void sort(int[][] arr) {
        Random r = new Random();
        for (int i = 0; i < arr.length; i++) {
            int x = r.nextInt(arr.length);
            int[] temp = arr[x];
            arr[x] = arr[i];
            arr[i] = temp;
        }
        Arrays.sort(arr, Comparator.comparingInt(a -> a[0]));
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some steps to consider:

1. **Grid Size and Values**:
   - Use small grid sizes to ensure the test case is manageable and fits within the output limit.
   - Use values that can highlight issues with the cyclic shift and path cost calculations.

2. **Cyclic Shifts**:
   - Ensure that the grid contains values that, when cyclically shifted, can significantly change the cost of paths.
   - Consider scenarios where the optimal path involves a large number of shifts.

3. **Path Sum and Cost**:
   - Use values that can lead to different path sums depending on the number of shifts.
   - Ensure that the cost calculation (including the number of shifts) is sensitive to the choice of shifts.

4. **Edge Cases**:
   - Test with a single row or column grid.
   - Test with a grid where all values are the same.
   - Test with a grid where the optimal path is not straightforward.

Based on these considerations, here is a challenging test input:

```plaintext
3
3 3 100
3 4 9
5 2 4
0 101 101
3 3 1
10 0 0 10
0 0 10 0
10 10 0 10
1 1 3
4
```

This test input includes:
- A 3x3 grid with a large `k` value (100) to emphasize the cost of shifts.
- A 3x3 grid with a small `k` value (1) to test the path sum.
- A 1x1 grid to test the edge case of a single cell.

These scenarios should help expose any issues with the cyclic shift and path cost calculations in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alter the GCD
time_limit_ms: 4000
memory_limit_mb: 256
Description: You are given two arrays $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$.

You must perform the following operation exactly once:

  * choose any indices $l$ and $r$ such that $1 \le l \le r \le n$;    * swap $a_i$ and $b_i$ for all $i$ such that $l \leq i \leq r$.

Find the maximum possible value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once. Also find the number of distinct pairs $(l, r)$ which achieve the maximum value.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you are given a single integer $n$ ($1 \le n \le 2 \cdot 10^5$), representing the number of integers in each array.

In the next line, you are given $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array $a$.

In the last line, you are given $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the array $b$.

The sum of values of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a line with two integers: the maximum value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once, and the number of ways.

Examples:
input:
5
8
11 4 16 17 3 24 25 8
8 10 4 21 17 18 25 21
4
6 4 24 13
15 3 1 14
2
13 14
5 8
8
20 17 15 11 21 10 3 7
9 9 4 20 14 9 13 1
2
18 13
15 20
output:
2 36
3 2
2 3
2 36
6 1

Note: In the first, third, and fourth test cases, there's no way to achieve a higher GCD than $1$ in any of the arrays, so the answer is $1 + 1 = 2$. Any pair $(l, r)$ achieves the same result; for example, in the first test case there are $36$ such pairs.

In the last test case, you must choose $l = 1$, $r = 2$ to maximize the answer. In this case, the GCD of the first array is $5$, and the GCD of the second array is $1$, so the answer is $5 + 1 = 6$, and the number of ways is $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define ld long double
#define pii pair<int,int>
#define pdd pair<db,db>
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
//cout<<setiosflags(ios::fixed)<<setprecision(K);
const int dx[4]={1,-1,0,0};
const int dy[4]={0,0,1,-1};
const double eps=1e-9;
const double PI=acos(-1.0);
const int inf=1e9+7;
const int mod=998244353;
const int N=2e5+10,M=N<<1;

int lg[N];
void init_lg(){
    lg[1]=0;
    for(int i=2;i<N;i++) lg[i]=lg[i>>1]+1;
}
int gcd(int x, int y){
    if(!x || !y) return x | y;
    return gcd(y, x % y);
}

void solve(){
    int n;cin>>n;
    vector<int>a(n+1),b(n+1);
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=n;i++) cin>>b[i];

    vector<int>sufa(n+2),sufb(n+2);
    for(int i=n;i>=1;i--){
        sufa[i]=gcd(sufa[i+1],a[i]);
        sufb[i]=gcd(sufb[i+1],b[i]);
    }
    vector<int>nsufa(n+2);
    vector<int>nsufb(n+2);
    int x=1;
    while(x<=n){
        int y=x+1;
        while(y<=n&&sufa[x]==sufa[y]) y++;
        for(int i=x;i<y;i++){
            nsufa[i]=y;
        }
        x=y;
    }
    x=1;
    while(x<=n){
        int y=x+1;
        while(y<=n&&sufb[x]==sufb[y]) y++;
        for(int i=x;i<y;i++){
            nsufb[i]=y;
        }
        x=y;
    }

    int m=lg[n]+1;
    vector mida(n+2,vector<int>(m));
    vector midb(n+2,vector<int>(m));
    auto init=[&](){
        for(int j=0;j<=m;j++){
            for(int i=1;i+(1<<j)-1<=n;i++){
                if(!j){
                    mida[i][j]=a[i];
                    midb[i][j]=b[i];
                }
                else{
                    mida[i][j]=gcd(mida[i][j-1],mida[i+(1<<(j-1))][j-1]);
                    midb[i][j]=gcd(midb[i][j-1],midb[i+(1<<(j-1))][j-1]);
                }
            }
        }
    };

    //gcd(a[l]...a[r])
    auto get_a=[&](int l,int r){
        if(l>r) return 0;
        int k=lg[r-l+1];
        return gcd(mida[l][k],mida[r-(1<<k)+1][k]);
    };
    auto get_b=[&](int l,int r){
        if(l>r) return 0;
        int k=lg[r-l+1];
        return gcd(midb[l][k],midb[r-(1<<k)+1][k]);
    };
    init();

    ll ans=0,cnt=0;
    int prea=0,preb=0;
    //修改 [ml,mr]
    for(int ml=1;ml<=n;ml++){
        //当前mr位置
        int cmr=ml;
        int cmida=a[cmr];
        int cmidb=b[cmr];

        while(cmr<=n){
            //找下一个mr位置,使得mida,midb,sufa,sufb其中一个改变
            int L=cmr;
            int R=min(nsufa[cmr],nsufb[cmr]);
            auto check=[&](int mr) -> bool {
                if(get_a(ml,mr)!=cmida) return 0;
                if(get_b(ml,mr)!=cmidb) return 0;
                return 1;
            };
            while(L+1<R){
                int m=L+R>>1;
                if(check(m)) L=m;
                else R=m;
            }
            //下一个mr位置
            int nmr=R;
            int res=0;
            res+=gcd(prea,gcd(cmidb,sufa[cmr+1]));
            res+=gcd(preb,gcd(cmida,sufb[cmr+1]));
            if(res>ans){
                ans=res;
                cnt=nmr-cmr;
            } else if(res==ans){
                cnt+=nmr-cmr;
            }

            cmr=nmr;
            cmida=get_a(ml,cmr);
            cmidb=get_b(ml,cmr);
        }
        prea=gcd(prea,a[ml]);
        preb=gcd(preb,b[ml]);
    }
    cout<<ans<<' '<<cnt<<'\n';
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    init_lg();
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
4
1 2 3 4
4 3 2 1
2
1 2
2 1
8
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
2
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alter the GCD
time_limit_ms: 4000
memory_limit_mb: 256
Description: You are given two arrays $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$.

You must perform the following operation exactly once:

  * choose any indices $l$ and $r$ such that $1 \le l \le r \le n$;    * swap $a_i$ and $b_i$ for all $i$ such that $l \leq i \leq r$.

Find the maximum possible value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once. Also find the number of distinct pairs $(l, r)$ which achieve the maximum value.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you are given a single integer $n$ ($1 \le n \le 2 \cdot 10^5$), representing the number of integers in each array.

In the next line, you are given $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array $a$.

In the last line, you are given $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the array $b$.

The sum of values of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a line with two integers: the maximum value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once, and the number of ways.

Examples:
input:
5
8
11 4 16 17 3 24 25 8
8 10 4 21 17 18 25 21
4
6 4 24 13
15 3 1 14
2
13 14
5 8
8
20 17 15 11 21 10 3 7
9 9 4 20 14 9 13 1
2
18 13
15 20
output:
2 36
3 2
2 3
2 36
6 1

Note: In the first, third, and fourth test cases, there's no way to achieve a higher GCD than $1$ in any of the arrays, so the answer is $1 + 1 = 2$. Any pair $(l, r)$ achieves the same result; for example, in the first test case there are $36$ such pairs.

In the last test case, you must choose $l = 1$, $r = 2$ to maximize the answer. In this case, the GCD of the first array is $5$, and the GCD of the second array is $1$, so the answer is $5 + 1 = 6$, and the number of ways is $1$.

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

const int maxn = 2e5 + 5;
typedef long long ll;

int N;
int A[maxn], B[maxn];

int lg[maxn];
struct RMQ {
  int R[maxn][18];

  void build(int arr[]) {
    for (int i = 1; i <= N; ++i)
      R[i][0] = arr[i];
    for (int k = 0; k + 1 < 18; ++k) {
      for (int i = 1; i + 2 * (1 << k) - 1 <= N; ++i)
        R[i][k + 1] = __gcd(R[i][k], R[i + (1 << k)][k]);
    }
  }

  int get(int l, int r) {
    int gap = lg[r - l + 1];
    return __gcd(R[l][gap], R[r - (1 << gap) + 1][gap]);
  }
} RA, RB;

void events(int arr[], vector<int> &into) {
  int cur = arr[N];
  for (int i = N - 1; i > 0; --i) {
    int nx = __gcd(cur, arr[i]);
    if (cur != nx) {
      into.push_back(i);
    }
    cur = nx;
  }
}

void update(int &mx, ll &cnt, int nw, int nwc) {
  if (mx < nw && nwc > 0) {
    mx = nw;
    cnt = nwc;
  } else if (mx == nw)
    cnt += nwc;
}

int count_gcd(int from, int left, int right, int ta, int tb) {
  int l = left, r = right;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (RA.get(from, mid) >= ta && RB.get(from, mid) >= tb)
      l = mid + 1;
    else
      r = mid;
  }
  return l - left;
}

void solve() {
  cin >> N;
  for (int i = 1; i <= N; ++i)
    cin >> A[i];
  for (int i = 1; i <= N; ++i)
    cin >> B[i];
  RA.build(A);
  RB.build(B);

  vector<int> es;
  events(A, es);
  events(B, es);
  sort(es.begin(), es.end());

  int res = RA.get(1, N) + RB.get(1, N);
  ll cnt = 1;
  // case 1 swap prefix
  for (int i = 1; i < N; ++i) {
    int nw = __gcd(RA.get(1, i), RB.get(i + 1, N)) +
             __gcd(RB.get(1, i), RA.get(i + 1, N));
    // cout << nw << " " << i << endl;
    update(res, cnt, nw, 2 /* prefix and suffix */);
  }
  // case 2 swap middle
  for (int left = 1; left < N; ++left) {
    int last = left + 1;
    for (size_t i = 0; i < es.size(); ++i) {
      int tail = es[i] + 1;
      if (tail <= last)
        continue;
      // 1..left / left+1..last / last+1 .. N
      auto a = __gcd(RA.get(1, left),
                     __gcd(RB.get(left + 1, last), RA.get(last + 1, N)));
      auto b = __gcd(RB.get(1, left),
                     __gcd(RA.get(left + 1, last), RB.get(last + 1, N)));
      int count = count_gcd(left + 1, last, tail, b, a);
      // cout << a + b << " " << count << " <- " << left << "|" << tail << "|"
      //      << nxt << "\n";
      update(res, cnt, a + b, count);
      last = tail;
    }
  }

  cout << res << " " << cnt << "\n";
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);

  lg[1] = 0;
  for (int i = 2; i < maxn; ++i)
    lg[i] = lg[i / 2] + 1;

  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
5
8
11 4 16 17 3 24 25 8
8 10 4 21 17 18 25 21
4
6 4 24 13
15 3 1 14
2
13 14
5 8
8
20 17 15 11 21 10 3 7
9 9 4 20 14 9 13 1
2
18 13
15 20
```