**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Index and Maximum Value
time_limit_ms: 1000
memory_limit_mb: 256
Description: After receiving yet another integer array $a_1, a_2, \ldots, a_n$ at her birthday party, Index decides to perform some operations on it.

Formally, there are $m$ operations that she is going to perform in order. Each of them belongs to one of the two types:

  * $\texttt{+ l r}$. Given two integers $l$ and $r$, for all $1 \leq i \leq n$ such that $l \leq a_i \leq r$, set $a_i := a_i + 1$.    * $\texttt{- l r}$. Given two integers $l$ and $r$, for all $1 \leq i \leq n$ such that $l \leq a_i \leq r$, set $a_i := a_i - 1$.

For example, if the initial array $a = [7, 1, 3, 4, 3]$, after performing the operation $\texttt{+} \space 2 \space 4$, the array $a = [7, 1, 4, 5, 4]$. Then, after performing the operation $\texttt{-} \space 1 \space 10$, the array $a = [6, 0, 3, 4, 3]$.

Index is curious about the maximum value in the array $a$. Please help her find it after each of the $m$ operations.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 10^5$, $1 \leq m \leq 10^5$) — the length of the array and the number of operations.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial array $a$.

Then $m$ lines follow, each line corresponds to the operation, in the following format: $\texttt{c l r}$ ($c \in \\{\texttt +, \texttt -\\}$, $l$ and $r$ are integers, $1 \leq l \leq r \leq 10^9$) — the description of the operation.

Note that the elements $a_i$ may not satisfy $1\le a_i\le 10^9$ after some operations, as it is shown in the example.

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

Output: For each test case, output one single line containing $m$ integers, with the $i$-th of them describing the maximum value of the array after the $i$-th operation.

Examples:
input:
5
5 5
1 2 3 2 1
+ 1 3
- 2 3
+ 1 2
+ 2 4
- 6 8
5 5
1 3 3 4 5
+ 1 4
+ 2 3
- 4 5
- 3 3
- 2 6
5 5
1 1 1 1 1
+ 2 3
- 4 5
+ 1 6
- 2 5
+ 1 8
1 1
1
- 1 1
1 1
1000000000
+ 1000000000 1000000000
output:
4 4 4 5 5
5 5 4 4 3
1 1 2 1 2
0
1000000001

Note: In the first test case, the process of the operations is listed below:

  * After the first operation, the array becomes equal $[2,3,4,3,2]$. The maximum value is $4$.    * After the second operation, the array becomes equal $[1,2,4,2,1]$. The maximum value is $4$.    * After the third operation, the array becomes equal $[2,3,4,3,2]$. The maximum value is $4$.    * After the fourth operation, the array becomes equal $[3,4,5,4,3]$. The maximum value is $5$.    * After the fifth operation, the array becomes equal $[3,4,5,4,3]$. The maximum value is $5$.

In the second test case, the process of the operations is listed below:

  * After the first operation, the array becomes equal $[2,4,4,5,5]$. The maximum value is $5$.    * After the second operation, the array becomes equal $[3,4,4,5,5]$. The maximum value is $5$.    * After the third operation, the array becomes equal $[3,3,3,4,4]$. The maximum value is $4$.    * After the fourth operation, the array becomes equal $[2,2,2,4,4]$. The maximum value is $4$.    * After the fifth operation, the array becomes equal $[1,1,1,3,3]$. The maximum value is $3$.

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

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

        int t = Integer.parseInt(st.nextToken());

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

            int[] arr = new int[n];

            st = new StringTokenizer(br.readLine());
            for(int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(st.nextToken());
            }

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

           for (int i = 0; i < m; i++) {
            String[] parts = srr[i].split(" ");
            char sign = parts[0].charAt(0);
            int n1 = Integer.parseInt(parts[1]);
            int n2 = Integer.parseInt(parts[2]);
            int max = Integer.MIN_VALUE;

            for (int j = 0; j < n; j++) {
                boolean b = arr[j] >= n1 && arr[j] <= n2;
                if (sign == '+') {
                    if (b) {
                        arr[j] += 1;
                    }
                } else {
                    if (b) {
                        arr[j] -= 1;
                    }
                }
                max = Math.max(max, arr[j]);
            }

            out.print(max + " ");
        }

           out.println();

        }

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

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

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

Think step by step.

Response:
```plaintext
5
5 5
1 2 3 2 1
+ 1 3
- 2 3
+ 1 2
+ 2 4
- 6 8
5 5
1 3 3 4 5
+ 1 4
+ 2 3
- 4 5
- 3 3
- 2 6
5 5
1 1 1 1 1
+ 2 3
- 4 5
+ 1 6
- 2 5
+ 1 8
1 1
1
- 1 1
1 1
1000000000
+ 1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Index and Maximum Value
time_limit_ms: 1000
memory_limit_mb: 256
Description: After receiving yet another integer array $a_1, a_2, \ldots, a_n$ at her birthday party, Index decides to perform some operations on it.

Formally, there are $m$ operations that she is going to perform in order. Each of them belongs to one of the two types:

  * $\texttt{+ l r}$. Given two integers $l$ and $r$, for all $1 \leq i \leq n$ such that $l \leq a_i \leq r$, set $a_i := a_i + 1$.    * $\texttt{- l r}$. Given two integers $l$ and $r$, for all $1 \leq i \leq n$ such that $l \leq a_i \leq r$, set $a_i := a_i - 1$.

For example, if the initial array $a = [7, 1, 3, 4, 3]$, after performing the operation $\texttt{+} \space 2 \space 4$, the array $a = [7, 1, 4, 5, 4]$. Then, after performing the operation $\texttt{-} \space 1 \space 10$, the array $a = [6, 0, 3, 4, 3]$.

Index is curious about the maximum value in the array $a$. Please help her find it after each of the $m$ operations.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 10^5$, $1 \leq m \leq 10^5$) — the length of the array and the number of operations.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial array $a$.

Then $m$ lines follow, each line corresponds to the operation, in the following format: $\texttt{c l r}$ ($c \in \\{\texttt +, \texttt -\\}$, $l$ and $r$ are integers, $1 \leq l \leq r \leq 10^9$) — the description of the operation.

Note that the elements $a_i$ may not satisfy $1\le a_i\le 10^9$ after some operations, as it is shown in the example.

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

Output: For each test case, output one single line containing $m$ integers, with the $i$-th of them describing the maximum value of the array after the $i$-th operation.

Examples:
input:
5
5 5
1 2 3 2 1
+ 1 3
- 2 3
+ 1 2
+ 2 4
- 6 8
5 5
1 3 3 4 5
+ 1 4
+ 2 3
- 4 5
- 3 3
- 2 6
5 5
1 1 1 1 1
+ 2 3
- 4 5
+ 1 6
- 2 5
+ 1 8
1 1
1
- 1 1
1 1
1000000000
+ 1000000000 1000000000
output:
4 4 4 5 5
5 5 4 4 3
1 1 2 1 2
0
1000000001

Note: In the first test case, the process of the operations is listed below:

  * After the first operation, the array becomes equal $[2,3,4,3,2]$. The maximum value is $4$.    * After the second operation, the array becomes equal $[1,2,4,2,1]$. The maximum value is $4$.    * After the third operation, the array becomes equal $[2,3,4,3,2]$. The maximum value is $4$.    * After the fourth operation, the array becomes equal $[3,4,5,4,3]$. The maximum value is $5$.    * After the fifth operation, the array becomes equal $[3,4,5,4,3]$. The maximum value is $5$.

In the second test case, the process of the operations is listed below:

  * After the first operation, the array becomes equal $[2,4,4,5,5]$. The maximum value is $5$.    * After the second operation, the array becomes equal $[3,4,4,5,5]$. The maximum value is $5$.    * After the third operation, the array becomes equal $[3,3,3,4,4]$. The maximum value is $4$.    * After the fourth operation, the array becomes equal $[2,2,2,4,4]$. The maximum value is $4$.    * After the fifth operation, the array becomes equal $[1,1,1,3,3]$. The maximum value is $3$.

Buggy Code:
import java.util.Scanner;

public class B_Index_and_Maximum_Value {

    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        int t = sc.nextInt();
        while (t-->0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int arr[] = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
            }
            for (int i = 0; i < m; i++) {
                String c = sc.next();
                int l = sc.nextInt();
                int r = sc.nextInt();
             for (int j = 0; j< arr.length; j++) {
                if(arr[j]>=l && arr[j] <=r && c.equals("-")) {
                    arr[j] -= 1;
                }
               else if(arr[j]>=l && arr[j] <=r && c.equals("+")) {
                    arr[j] += 1;
                }
            }
            int max = Integer.MIN_VALUE;
            for (int j = 0; j < arr.length; j++) {
                if(max <arr[j]){
                    max = arr[j];
                }
            }
            System.out.print(max + " ");

            }

          System.out.println();

        }

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

const int MAX_R = 1000;
const int MAX_C = 1000;
const int INF = 1e9;

int R, C, X, Q;
int H[MAX_R][MAX_C];
vector<pair<int, int>> adj[MAX_R * MAX_C];
bool has_negative_cycle = false;

int cell_id(int r, int c) {
    return (r - 1) * C + (c - 1);
}

// Simplified negative cycle detection
// Given the penalty function is monotonic, cycles with identical paths have zero penalty.
bool detect_negative_cycle() {
    // With X odd, and height differences bounded, no practical negative cycles exist.
    // Therefore, we assume no negative cycles for simplicity.
    return false;
}

// Dijkstra's algorithm to find the minimum penalty path
int dijkstra(int start_r, int start_c, int end_r, int end_c) {
    int start_id = cell_id(start_r, start_c);
    int end_id = cell_id(end_r, end_c);
    vector<int> dist(R * C, INF);
    dist[start_id] = 0;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    pq.push({0, start_id});

    while (!pq.empty()) {
        auto [d, u] = pq.top(); pq.pop();
        if (u == end_id) return d;
        if (d > dist[u]) continue;
        for (auto [v, w] : adj[u]) {
            if (dist[v] > dist[u] + w) {
                dist[v] = dist[u] + w;
                pq.push({dist[v], v});
            }
        }
    }
    return INF; // If destination is unreachable
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    cin >> R >> C >> X;
    for (int r = 1; r <= R; ++r) {
        string s; cin >> s;
        for (int c = 1; c <= C; ++c) {
            H[r][c] = s[c - 1] - '0';
        }
    }

    // Build adjacency list
    for (int r = 1; r <= R; ++r) {
        for (int c = 1; c <= C; ++c) {
            int u = cell_id(r, c);
            // Check adjacent cells
            if (r > 1) {
                int v = cell_id(r - 1, c);
                int w = pow(H[r][c] - H[r - 1][c], X);
                adj[u].emplace_back(v, w);
            }
            if (r < R) {
                int v = cell_id(r + 1, c);
                int w = pow(H[r][c] - H[r + 1][c], X);
                adj[u].emplace_back(v, w);
            }
            if (c > 1) {
                int v = cell_id(r, c - 1);
                int w = pow(H[r][c] - H[r][c - 1], X);
                adj[u].emplace_back(v, w);
            }
            if (c < C) {
                int v = cell_id(r, c + 1);
                int w = pow(H[r][c] - H[r][c + 1], X);
                adj[u].emplace_back(v, w);
            }
        }
    }

    // Detect negative cycles
    has_negative_cycle = detect_negative_cycle();

    cin >> Q;
    while (Q--) {
        int Rs, Cs, Rf, Cf;
        cin >> Rs >> Cs >> Rf >> Cf;
        if (has_negative_cycle) {
            cout << "INVALID\n";
        } else {
            int penalty = dijkstra(Rs, Cs, Rf, Cf);
            if (penalty == INF) { // Unreachable, though given grid connectivity, unlikely
                cout << "INVALID\n";
            } else {
                cout << penalty << '\n';
            }
        }
    }

    return 0;
}

// MerqP58uWZlCwKk6Xpi

**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
123
456
789
4
1 1 3 3
3 3 1 1
2 2 2 2
1 3 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

const int INF = 1e9;
int R, C, X;
vector<string> H;
int dirs[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

int getHeight(int r, int c) {
    return H[r - 1][c - 1] - '0';
}

vector<vector<int>> distances;

bool hasNegativeCycle() {
    distances.assign(R + 1, vector<int>(C + 1, INF));
    distances[1][1] = 0;
    bool relaxed = false;
    for (int i = 0; i < R * C; ++i) {
        relaxed = false;
        for (int r = 1; r <= R; ++r) {
            for (int c = 1; c <= C; ++c) {
                for (auto& d : dirs) {
                    int nr = r + d[0];
                    int nc = c + d[1];
                    if (nr >= 1 && nr <= R && nc >= 1 && nc <= C) {
                        int penalty = pow(getHeight(r, c) - getHeight(nr, nc), X);
                        if (distances[nr][nc] > distances[r][c] + penalty) {
                            distances[nr][nc] = distances[r][c] + penalty;
                            relaxed = true;
                        }
                    }
                }
            }
        }
        if (!relaxed) break;
    }
    return relaxed;
}

int shortestPenalty(int rs, int cs, int rf, int cf) {
    if (hasNegativeCycle()) return INF;
    distances.assign(R + 1, vector<int>(C + 1, INF));
    distances[rs][cs] = 0;
    priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<>> pq;
    pq.emplace(0, rs, cs);
    while (!pq.empty()) {
        auto [d, r, c] = pq.top();
        pq.pop();
        if (r == rf && c == cf) return d;
        for (auto& dir : dirs) {
            int nr = r + dir[0];
            int nc = c + dir[1];
            if (nr >= 1 && nr <= R && nc >= 1 && nc <= C) {
                int penalty = pow(getHeight(r, c) - getHeight(nr, nc), X);
                if (distances[nr][nc] > d + penalty) {
                    distances[nr][nc] = d + penalty;
                    pq.emplace(d + penalty, nr, nc);
                }
            }
        }
    }
    return INF;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> R >> C >> X;
    H.resize(R);
    for (int r = 0; r < R; ++r) {
        cin >> H[r];
    }
    int Q;
    cin >> Q;
    while (Q--) {
        int rs, cs, rf, cf;
        cin >> rs >> cs >> rf >> cf;
        if (rs == rf && cs == cf) {
            cout << "0\n";
            continue;
        }
        int result = shortestPenalty(rs, cs, rf, cf);
        if (result == INF) {
            cout << "INVALID\n";
        } else {
            cout << result << "\n";
        }
    }
    return 0;
}

// Wmwqt7QJuki0d9YeN6P

**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 consider the following aspects:

1. **Negative Cycles**: The code should correctly detect negative cycles, which can lead to arbitrarily small penalties.
2. **Large Penalties**: Ensure that the code can handle large penalties without overflow or incorrect calculations.
3. **Edge Cases**: Test scenarios where the starting and ending points are the same, or where the grid is small but has complex paths.

Given these considerations, let's create a small-scale test input that includes a negative cycle and a scenario where the starting and ending points are the same.

### Test Input
```plaintext
3 3 1
123
456
789
4
1 1 3 3
3 3 1 1
2 2 2 2
1 1 1 1
```

### Explanation
1. **Grid Layout**:
   - The grid is a 3x3 matrix with heights as follows:
     ```
     1 2 3
     4 5 6
     7 8 9
     ```
   - The aura value is 1, which means the penalty is simply the difference in heights.

2. **Scenarios**:
   - **Scenario 1**: Move from (1, 1) to (3, 3)
     - Path: (1, 1) → (2, 1) → (3, 1) → (3, 2) → (3, 3)
     - Penalties: (1 - 4) + (4 - 7) + (7 - 8) + (8 - 9) = -3 -3 -1 -1 = -8
   - **Scenario 2**: Move from (3, 3) to (1, 1)
     - Path: (3, 3) → (3, 2) → (3, 1) → (2, 1) → (1, 1)
     - Penalties: (9 - 8) + (8 - 7) + (7 - 4) + (4 - 1) = 1 + 1 + 3 + 3 = 8
   - **Scenario 3**: Move from (2, 2) to (2, 2)
     - Starting and ending at the same cell, so the penalty is 0.
   - **Scenario 4**: Move from (1, 1) to (1, 1)
     - Starting and ending at the same cell, so the penalty is 0.

### Expected Output
```plaintext
-8
8
0
0
```

This test input should help expose the bug in the code, especially in handling negative cycles and scenarios where the starting and ending points are the same.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

#define ll long long
#define pb push_back
#define pii pair<ll,int>
#define f first
#define s second
#define chmin(a, b) a = min(a,b)
#define chmax(a, b) a = max(a,b)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define all(x) x.begin(),x.end()
#define v vector

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

    int r, c, x; cin >> r >> c >> x;
    v<v<int>> h(r,v<int>(c));
    F0R(i,r) {
        F0R(j,c) {
            char z; cin >> z;
            h[i][j] = z-'0';
        }
    }

    v<v<ll>> dist(r,v<ll>(c,LLONG_MAX));
    dist[0][0] = 0;
    bool invalid = false;

    auto check = [&](int i, int j, int i2, int j2) {
        ll base = h[i][j]-h[i2][j2], cost = 1;
        F0R(i,x) {cost *= base;}
        if (dist[i2][j2]==LLONG_MAX) {dist[i2][j2] = dist[i][j]+cost;}
        else {
            if (cost+dist[i][j]!=dist[i2][j2]) {
                invalid = true;
            }
        }
    };

    F0R(i,r) {
        F0R(j,c) {
            if (i!=r-1) {check(i,j,i+1,j);}
            if (i) {check(i,j,i-1,j);}
            if (j!=c-1) {check(i,j,i,j+1);}
            if (j) {check(i,j,i,j-1);}
        }
    }
    int q; cin >> q;
    while (q--) {
        int r1, c1, r2, c2; cin >> r1 >> c1 >> r2 >> c2;
        //cout<<-dist[r1-1][c1-1] << '\n';
        if (invalid) {cout << "INVAlID\n";}
        else {cout << -1*dist[r1-1][c1-1]+dist[r2-1][c2-1] << '\n';}
    }
}

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

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

Think step by step.

Response:
```plaintext
3 3 1
123
456
789
2
1 1 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
import sys
import heapq

def main():
    R, C, X = map(int, sys.stdin.readline().split())
    grid = [list(map(int, list(sys.stdin.readline().strip()))) for _ in range(R)]
    Q = int(sys.stdin.readline())
    queries = [list(map(int, sys.stdin.readline().split())) for _ in range(Q)]

    # Precompute penalty table
    penalty_table = {}
    for h1 in range(10):
        for h2 in range(10):
            penalty_table[(h1, h2)] = (h1 - h2) ** X

    for rs, cs, rf, cf in queries:
        rs, cs, rf, cf = rs-1, cs-1, rf-1, cf-1
        if X == 1:
            print(grid[rs][cs] - grid[rf][cf])
        else:
            # Dijkstra's algorithm
            directions = [(-1,0),(1,0),(0,-1),(0,1)]
            dist = [[float('inf')] * C for _ in range(R)]
            dist[rs][cs] = 0
            pq = [(0, rs, cs)]
            iteration_limit = R * C
            iterations = 0
            while pq and iterations < iteration_limit:
                current_dist, r, c = heapq.heappop(pq)
                if r == rf and c == cf:
                    print(current_dist)
                    break
                if iterations >= iteration_limit:
                    print("INVALID")
                    break
                for dr, dc in directions:
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < R and 0 <= nc < C:
                        penalty = penalty_table[(grid[r][c], grid[nr][nc])]
                        new_dist = current_dist + penalty
                        if new_dist < dist[nr][nc]:
                            dist[nr][nc] = new_dist
                            heapq.heappush(pq, (new_dist, nr, nc))
                iterations += 1
            else:
                print("INVALID")

if __name__ == "__main__":
    main()

# wpRuPoFzrQnVLWkdIUx

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

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

Think step by step.

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

1. **Negative Penalties and Cycles**: The code uses Dijkstra's algorithm, which assumes non-negative edge weights. If there are negative penalties, especially in cycles, Dijkstra's algorithm can fail to find the correct minimum path. We need to create a scenario where such negative cycles exist.

2. **Large Penalties with High X**: For larger values of `X` (e.g., 5 or 9), the penalties can become very large, potentially causing overflow or incorrect calculations.

3. **Invalid Scenarios**: Scenarios where the total penalty can become arbitrarily small (negative cycles) should be considered to ensure the code correctly identifies them as invalid.

4. **Edge Cases**: Small grids, single-cell scenarios, and scenarios where the start and end cells are the same should also be tested.

Given these considerations, let's create a small-scale test input that includes negative cycles and large penalties:

```plaintext
3 3 5
190
202
190
3
1 1 3 3
2 2 2 2
1 1 2 2
```

### Explanation:
1. **Grid Configuration**:
   - The grid is 3x3 with heights:
     ```
     1 9 0
     2 0 2
     1 9 0
     ```

2. **Queries**:
   - **Query 1**: Start at (1, 1) and end at (3, 3)
     - Path: (1, 1) -> (2, 1) -> (2, 2) -> (3, 2) -> (3, 3)
     - Penalties:
       - (1 - 2)^5 = -1
       - (2 - 0)^5 = -32
       - (0 - 1)^5 = -1
       - (1 - 0)^5 = -1
     - Total penalty: -1 - 32 - 1 - 1 = -35
     - This query should be marked as **INVALID** because the cycle (1, 1) -> (2, 1) -> (2, 2) -> (1, 2) -> (1, 1) has a negative penalty and can be repeated indefinitely.

   - **Query 2**: Start at (2, 2) and end at (2, 2)
     - The start and end cell are the same.
     - Total penalty: 0

   - **Query 3**: Start at (1, 1) and end at (2, 2)
     - Path: (1, 1) -> (2, 1) -> (2, 2)
     - Penalties:
       - (1 - 2)^5 = -1
       - (2 - 0)^5 = -32
     - Total penalty: -1 - 32 = -33
     - This query should be marked as **INVALID** because the cycle (1, 1) -> (2, 1) -> (2, 2) -> (1, 2) -> (1, 1) has a negative penalty and can be repeated indefinitely.

This test input should expose the bug in the provided code, particularly in handling negative cycles and large penalties.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
import sys
import heapq

def main():
    input = sys.stdin.read
    data = input().split()

    idx = 0
    R = int(data[idx])
    C = int(data[idx+1])
    X = int(data[idx+2])
    idx += 3

    grid = []
    for _ in range(R):
        row = data[idx]
        grid.append([int(c) for c in row])
        idx += 1

    Q = int(data[idx])
    idx += 1

    queries = []
    for _ in range(Q):
        r1, c1, r2, c2 = int(data[idx]), int(data[idx+1]), int(data[idx+2]), int(data[idx+3])
        queries.append((r1-1, c1-1, r2-1, c2-1))
        idx += 4

    # Directions: up, down, left, right
    dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]

    # Precompute possible move penalties for adjacent cells
    # Not necessary, but helps in clarity

    def penalty(r1, c1, r2, c2):
        return (grid[r1][c1] - grid[r2][c2]) ** X

    # Function to check if there is a negative cycle reachable from start node
    def has_negative_cycle(R, C, grid, X, start_r, start_c):
        # Bellman-Ford algorithm
        # Need to represent each cell as a node, edges as moves, and weights as penalties
        # Since grid can be up to 1000x1000, Bellman-Ford might be too slow
        # Alternatively, we can detect any negative cycle
        # However, with the constraints, it's hard to make this efficient
        # Given time constraints, we'll assume that the number of nodes is manageable
        # For the purpose of this solution, we'll proceed with the assumption that negative cycles are possible
        # and implement a basic Bellman-Ford check for small subgraphs or make an educated guess
        # In a real scenario, we might need a more efficient method

        # Given the time constraints, we'll not implement the full check
        # Instead, we'll assume that if any cell has a neighbor that forms a cycle with a negative penalty, the scenario is invalid
        # This is a heuristic approach and might not cover all cases

        for r in range(R):
            for c in range(C):
                for dr, dc in dirs:
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < R and 0 <= nc < C:
                        p = penalty(r, c, nr, nc)
                        if p < 0:
                            # Check if moving back forms a negative cycle
                            if (nr, nc) == (r, c):
                                return True
                            else:
                                q = penalty(nr, nc, r, c)
                                if p + q < 0:
                                    return True
        return False

    # Function to find the minimum penalty from start to destination
    def min_penalty(R, C, grid, X, start_r, start_c, dest_r, dest_c):
        if start_r == dest_r and start_c == dest_c:
            return 0
        # Use Dijkstra's algorithm with a priority queue
        # Though it doesn't handle negative weights, in this problem negative penalties can be managed by finding cycles
        # However, due to time constraints, we'll proceed with a basic implementation
        # In a real scenario, we might need to handle negative weights by detecting cycles or using Bellman-Ford
        dist = [[sys.maxsize for _ in range(C)] for __ in range(R)]
        dist[start_r][start_c] = 0
        pq = []
        heapq.heappush(pq, (0, start_r, start_c))

        while pq:
            current_dist, r, c = heapq.heappop(pq)
            for dr, dc in dirs:
                nr, nc = r + dr, c + dc
                if 0 <= nr < R and 0 <= nc < C:
                    new_dist = current_dist + penalty(r, c, nr, nc)
                    if new_dist < dist[nr][nc]:
                        dist[nr][nc] = new_dist
                        heapq.heappush(pq, (new_dist, nr, nc))

        if dist[dest_r][dest_c] < sys.maxsize:
            return dist[dest_r][dest_c]
        else:
            return "INVALID"

    for query in queries:
        start_r, start_c, dest_r, dest_c = query
        if has_negative_cycle(R, C, grid, X, start_r, start_c):
            print("INVALID")
        else:
            result = min_penalty(R, C, grid, X, start_r, start_c, dest_r, dest_c)
            print(result)

if __name__ == "__main__":
    main()

# GA5ZNE1xQ2mVowP8tuU

**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
123
456
789
2
1 1 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
import sys
from collections import deque

def min_penalty(grid, R, C, X, Rs, Cs, Rf, Cf):
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
    queue = deque([(Rs - 1, Cs - 1, 0)])
    visited = {(Rs - 1, Cs - 1)}
    min_penalty = float('inf')

    while queue:
        r, c, penalty = queue.popleft()
        if (r, c) == (Rf - 1, Cf - 1):
            min_penalty = min(min_penalty, penalty)
            continue

        for dr, dc in directions:
            nr, nc = r + dr, c + dc
            if 0 <= nr < R and 0 <= nc < C and (nr, nc) not in visited:
                new_penalty = penalty + (int(grid[r][c]) - int(grid[nr][nc])) ** X
                if new_penalty < min_penalty:
                    queue.append((nr, nc, new_penalty))
                    visited.add((nr, nc))

    return min_penalty if min_penalty != float('inf') else "INVALID"

def main():
    R, C, X = map(int, sys.stdin.readline().split())
    grid = [sys.stdin.readline().strip() for _ in range(R)]
    Q = int(sys.stdin.readline())

    for _ in range(Q):
        Rs, Cs, Rf, Cf = map(int, sys.stdin.readline().split())
        result = min_penalty(grid, R, C, X, Rs, Cs, Rf, Cf)
        print(result)

if __name__ == "__main__":
    main()

# vN7gHAYGmoETLM9FUJs

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

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

Think step by step.

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

1. **Negative Penalties**: The code should be able to handle scenarios where the penalty can be negative, especially when the aura \( X \) is an odd number. This can lead to cycles that make the total penalty arbitrarily small.
2. **Invalid Scenarios**: The code should correctly identify scenarios where the total penalty can become arbitrarily small, and return "INVALID".
3. **Edge Cases**: Consider edge cases such as starting and ending at the same cell, or scenarios where the grid is very small.

Given these points, let's construct a test input that includes a cycle with negative penalties and a small grid to make it easy to verify.

### Test Input
```plaintext
3 3 3
123
456
789
3
1 1 3 3
2 2 2 2
3 3 1 1
```

### Explanation
- **Grid Setup**:
  - The grid is a 3x3 matrix with heights ranging from 1 to 9.
  - The aura \( X \) is 3, which is an odd number, allowing for negative penalties.

- **Queries**:
  - **Query 1**: Start at (1, 1) and end at (3, 3).
    - Possible path: \( (1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \).
    - Penalties: \( (1 - 4)^3 + (4 - 7)^3 + (7 - 8)^3 + (8 - 9)^3 = -27 - 27 - 1 - 1 = -56 \).
    - This path should not be considered valid because it can lead to arbitrarily small penalties by cycling through negative penalty paths.
  - **Query 2**: Start at (2, 2) and end at (2, 2).
    - This is a trivial case where the start and end are the same.
    - The minimum penalty is 0.
  - **Query 3**: Start at (3, 3) and end at (1, 1).
    - Possible path: \( (3, 3) \rightarrow (3, 2) \rightarrow (3, 1) \rightarrow (2, 1) \rightarrow (1, 1) \).
    - Penalties: \( (9 - 8)^3 + (8 - 7)^3 + (7 - 4)^3 + (4 - 1)^3 = 1 + 1 + 27 + 27 = 56 \).
    - This path should not be considered valid because it can lead to arbitrarily small penalties by cycling through negative penalty paths.

### Expected Output
```plaintext
INVALID
0
INVALID
```

This test input should expose the bug in the provided code by forcing it to handle cycles with negative penalties and correctly identify invalid scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

public class Main{

    static class Pair {
        int r, c;
        Pair(int r, int c) {
            this.r = r;
            this.c = c;
        }
    }

    static int R, C, X;
    static char[][] H;
    static int[][] heights;
    static int[] dr = {-1, 1, 0, 0};
    static int[] dc = {0, 0, -1, 1};

    static boolean inGrid(int r, int c) {
        return r >= 0 && r < R && c >= 0 && c < C;
    }

    static int penalty(int h1, int h2) {
        return (int) Math.pow(h1 - h2, X);
    }

    static void bfs(int startR, int startC, long[][] distance) {
        Queue<Pair> queue = new LinkedList<>();
        queue.add(new Pair(startR, startC));
        distance[startR][startC] = 0;

        while (!queue.isEmpty()) {
            Pair p = queue.poll();
            int r = p.r, c = p.c;
            for (int d = 0; d < 4; d++) {
                int nr = r + dr[d], nc = c + dc[d];
                if (inGrid(nr, nc)) {
                    int pnlty = penalty(heights[r][c], heights[nr][nc]);
                    if (distance[nr][nc] > distance[r][c] + pnlty) {
                        distance[nr][nc] = distance[r][c] + pnlty;
                        queue.add(new Pair(nr, nc));
                    }
                }
            }
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] parts = br.readLine().split(" ");
        R = Integer.parseInt(parts[0]);
        C = Integer.parseInt(parts[1]);
        X = Integer.parseInt(parts[2]);
        H = new char[R][];
        heights = new int[R][C];
        for (int r = 0; r < R; r++) {
            H[r] = br.readLine().toCharArray();
            for (int c = 0; c < C; c++) {
                heights[r][c] = H[r][c] - '0';
            }
        }

        long[][] distance = new long[R][C];
        for (long[] row : distance) {
            Arrays.fill(row, Long.MAX_VALUE);
        }

        int Q = Integer.parseInt(br.readLine());
        for (int q = 0; q < Q; q++) {
            parts = br.readLine().split(" ");
            int startR = Integer.parseInt(parts[0]) - 1;
            int startC = Integer.parseInt(parts[1]) - 1;
            int endR = Integer.parseInt(parts[2]) - 1;
            int endC = Integer.parseInt(parts[3]) - 1;

            for (long[] row : distance) {
                Arrays.fill(row, Long.MAX_VALUE);
            }
            bfs(startR, startC, distance);

            if (distance[endR][endC] == Long.MAX_VALUE) {
                System.out.println("INVALID");
            } else {
                System.out.println(distance[endR][endC]);
            }
        }
    }
}

// YSTxcZdHkjKhuADVwJN

**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
123
456
789
2
1 1 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

public class Main{
    static int R, C, X;
    static int[][] heights;
    static List<int[]> edges;
    static int[] dist;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        R = Integer.parseInt(st.nextToken());
        C = Integer.parseInt(st.nextToken());
        X = Integer.parseInt(st.nextToken());

        heights = new int[R][C];
        for (int r = 0; r < R; r++) {
            String line = br.readLine();
            for (int c = 0; c < C; c++) {
                heights[r][c] = line.charAt(c) - '0';
            }
        }

        buildEdges();

        int Q = Integer.parseInt(br.readLine());
        for (int q = 0; q < Q; q++) {
            st = new StringTokenizer(br.readLine());
            int Rs = Integer.parseInt(st.nextToken()) - 1;
            int Cs = Integer.parseInt(st.nextToken()) - 1;
            int Rf = Integer.parseInt(st.nextToken()) - 1;
            int Cf = Integer.parseInt(st.nextToken()) - 1;

            int start = Rs * C + Cs;
            int dest = Rf * C + Cf;

            if (start == dest) {
                out.println(0);
                continue;
            }

            dist = new int[R * C];
            Arrays.fill(dist, Integer.MAX_VALUE);
            dist[start] = 0;
            boolean hasNegativeCycle = bellmanFord(start);

            if (hasNegativeCycle) {
                out.println("INVALID");
            } else if (dist[dest] == Integer.MAX_VALUE) {
                out.println("INVALID");
            } else {
                out.println(dist[dest]);
            }
        }

        out.close();
    }

    private static void buildEdges() {
        edges = new ArrayList<>();
        for (int r = 0; r < R; r++) {
            for (int c = 0; c < C; c++) {
                int u = r * C + c;
                if (r > 0) edges.add(new int[]{u, u - C, (int) Math.pow(heights[r][c] - heights[r - 1][c], X)});
                if (r < R - 1) edges.add(new int[]{u, u + C, (int) Math.pow(heights[r][c] - heights[r + 1][c], X)});
                if (c > 0) edges.add(new int[]{u, u - 1, (int) Math.pow(heights[r][c] - heights[r][c - 1], X)});
                if (c < C - 1) edges.add(new int[]{u, u + 1, (int) Math.pow(heights[r][c] - heights[r][c + 1], X)});
            }
        }
    }

    private static boolean bellmanFord(int start) {
        for (int i = 1; i <= R * C - 1; i++) {
            for (int[] edge : edges) {
                int u = edge[0];
                int v = edge[1];
                int penalty = edge[2];
                if (dist[u] != Integer.MAX_VALUE && dist[u] + penalty < dist[v]) {
                    dist[v] = dist[u] + penalty;
                }
            }
        }

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int penalty = edge[2];
            if (dist[u] != Integer.MAX_VALUE && dist[u] + penalty < dist[v]) {
                return true;
            }
        }

        return false;
    }
}

// pMHXh1ys6flmG9SUQ0c

**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
123
456
789
4
1 1 3 3
3 3 1 1
2 2 3 3
3 3 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

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

using namespace std;

int main() {
    int R, C, X;
    cin >> R >> C >> X;
    vector<string> f(R);

    for (int i = 0; i < R; i++)
    {
        cin >> f[i];
    }

    vector<vector<int>> integral(R, vector<int>(C));
    for (int i = 1; i < R; i++)
    {
        integral[i][0] = integral[i - 1][0] + pow(f[i - 1][0] - f[i][0], X);
    }
    for (int i = 0; i < R; i++)
    {
        for (int j = 1; j < C; j++)
        {
            integral[i][j] = integral[i][j - 1] + pow(f[i][j - 1] - f[i][j], X);
        }
    }

    bool valid = true;
    for (int i = 1; i < R; i++)
    {
        for (int j = 0; j < C; j++)
        {
            if (integral[i][j] != integral[i - 1][j] + pow(f[i - 1][j] - f[i][j], X)) valid = false;
        }
    }

    int Q;
    cin >> Q;
    while (Q--) {
        int y0, x0, y1, x1;
        cin >> y0 >> x0 >> y1 >> x1;
        y0--; x0--; y1--; x1--;
        if (valid) {
            cout << integral[y1][x1] - integral[y0][x0] << "\n";
        } else {
            cout << "INVALID\n";
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the code. The primary issues in the buggy code are:

1. **Incorrect Integral Calculation**: The code calculates the integral values in a way that does not account for the correct penalties for movements in a grid.
2. **Invalid Path Detection**: The code does not correctly detect invalid paths where the total penalty can become arbitrarily small due to cycles.

Given these issues, we can create a test input that includes:

1. **Cycles with Negative Penalties**: Create a grid where there is a cycle that can lead to an arbitrarily small total penalty.
2. **Start and End Points in the Cycle**: Ensure that the start and end points are part of the cycle or can reach the cycle.

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

```plaintext
3 3 3
135
357
579
2
1 1 3 3
2 2 2 2
```

### Explanation:
- **Grid**:
  - The grid is a 3x3 grid with heights as follows:
    - Row 1: 1, 3, 5
    - Row 2: 3, 5, 7
    - Row 3: 5, 7, 9

- **Cycles**:
  - The cycle in this grid is from (1, 1) to (2, 1) to (2, 2) to (1, 2) back to (1, 1). The penalty for this cycle is:
    - \((1 - 3)^3 + (3 - 5)^3 + (5 - 3)^3 + (3 - 1)^3 = -26\)
  - This cycle can be repeated to make the total penalty arbitrarily small.

- **Scenarios**:
  - **Scenario 1**: Start at (1, 1) and end at (3, 3). This path should be invalid because it can reach the cycle and make the total penalty arbitrarily small.
  - **Scenario 2**: Start at (2, 2) and end at (2, 2). This is a trivial path with a penalty of 0.

This test input should expose the bug in the buggy code by demonstrating the incorrect handling of cycles and the invalid path detection.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
#define int long long
#define NNN 200005
#define pb push_back
#define mp make_pair
#define INF 2E18
#define mod 1000000007
#define pii pair<int,int>
#define vi vector<int>
#define all(a) a.begin(), a.end()
#define F first
#define S second
using namespace std;
template<class t> using pqmin=priority_queue<t,vector<t>,greater<t>>;
template<class t> using pqmax=priority_queue<t>;
//#define endl '\n'

//using namespace __gnu_pbds;
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;//s.find_by_order()//s.order_of_key()
//mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

int f(int a,int b){
	if(b==0) return 1;
	if(b==1) return a;
	if(b%2==0){
		int x=f(a,b/2);
		return x*x;
	}else{
		int x=f(a,b-1);
		return x*a;
	}
}

void solve()
{
	int R,C,X;
	cin>>R>>C>>X;

	char h[R+5][C+5];
	for(int i=1;i<=R;i++){
		for(int j=1;j<=C;j++){
			cin>>h[i][j];
		}
	}

	bool invalid=false;
	if(X!=1){ // invalid might occur
		for(int i=1;i<R;i++){
			if(invalid) break;
			for(int j=1;j<C;j++){
				int a=h[i][j];
				int b=h[i][j+1];
				int c=h[i+1][j+1];
				int d=h[i+1][j];

				if(f(a-b,X)+f(b-c,X)+f(c-d,X)+f(d-a,X) != 0){
					invalid=true;
					break;
				}
			}
		}
	}

	int Q;
	cin>>Q;

	for(int ii=1;ii<=Q;ii++){
		int x1,y1,x2,y2;
		cin>>x1>>y1>>x2>>y2;

		if(invalid){
			cout<<"INVALID"<<endl;
			continue;
		}

		int ans=0;
		for(int i=x1;i<x2;i++)
			ans+=f(h[i][y1]-h[i+1][y1],X);

		for(int i=x1;i>x2;i--)
			ans+=f(h[i][y1]-h[i-1][y1],X);

		for(int j=y1;j<y2;j++)
			ans+=f(h[x2][j]-h[x2][j+1],X);

		for(int j=y1;j>y2;j--)
			ans+=f(h[x2][j]-h[x2][j-1],X);

		cout<<ans<<endl;
	}

}

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

/*
for(int i=1;i<=n;i++)

for(int i=0;i<n;i++)

*/

// By Yufe1210  ; )

// More About
/*
We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any other guy

I just wanna tell you how I'm feeling
Gotta make you understand

Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you

We've known each other for so long
Your heart's been aching but you're too shy to say it
Inside we both know what's been going on
We know the game and we're gonna play it

And if you ask me how I'm feeling
Don't tell me you're too blind to see

Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
*/

**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
123
456
789
2
1 1 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
import sys
import math
from collections import deque

def main():
    input = sys.stdin.read
    data = input().split()

    idx = 0
    R = int(data[idx])
    idx += 1
    C = int(data[idx])
    idx += 1
    X = int(data[idx])
    idx += 1

    H = []
    for _ in range(R):
        row = data[idx]
        H.append([int(c) for c in row])
        idx += 1

    Q = int(data[idx])
    idx += 1

    queries = []
    for _ in range(Q):
        Rs = int(data[idx]) - 1
        idx += 1
        Cs = int(data[idx]) - 1
        idx += 1
        Rf = int(data[idx]) - 1
        idx += 1
        Cf = int(data[idx]) - 1
        idx += 1
        queries.append((Rs, Cs, Rf, Cf))

    # Precompute all pairs shortest path considering movement rules
    # Given grid movement constraints, using Dijkstra or Bellman-Ford with certain optimizations

    # Given the large grid size and number of queries, an optimized approach is essential

    # For X being odd and movement penalties being (h1 - h2)^X, negative cycles can exist

    # Detect any negative cycles: if exists, scenarios involving any cell in cycle are invalid

    # Implementing Bellman-Ford-like algorithm to detect negative cycles

    n = R * C  # Number of nodes

    # Mapping from (r, c) to node index
    node_map = {}
    current_node = 0
    for r in range(R):
        for c in range(C):
            node_map[(r, c)] = current_node
            current_node += 1

    # Build adjacency list with edge weights
    adj = [[] for _ in range(n)]

    for r in range(R):
        for c in range(C):
            current_node = node_map[(r, c)]
            height = H[r][c]
            for dr, dc in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                nr, nc = r + dr, c + dc
                if 0 <= nr < R and 0 <= nc < C:
                    neighbor = node_map[(nr, nc)]
                    edge_weight = (height - H[nr][nc]) ** X
                    adj[current_node].append((neighbor, edge_weight))

    # Function to detect negative cycles
    def has_negative_cycle():
        dist = [float('inf')] * n
        dist[0] = 0

        for _ in range(n):
            for u in range(n):
                for v, w in adj[u]:
                    if dist[v] > dist[u] + w:
                        dist[v] = dist[u] + w

        for u in range(n):
            for v, w in adj[u]:
                if dist[v] > dist[u] + w:
                    return True  # Negative cycle detected
        return False

    # Check for negative cycles only once
    has_neg_cycle = has_negative_cycle()

    # If a negative cycle exists, all scenarios are invalid
    if has_neg_cycle:
        for q in queries:
            print("INVALID")
        return

    # Otherwise, for each query, compute the minimal penalty from Rs, Cs to Rf, Cf

    # Precompute all pairs shortest paths

    # Given the grid size is up to 1000x1000, and Q up to 100,000, even precomputing shortest paths is infeasible

    # Alternative approach: for each query, use BFS with priority queue to find the path with minimal penalty
    # Since negative cycles are absent, shortest path exists and is finite

    for q in queries:
        Rs, Cs, Rf, Cf = q
        src = node_map[(Rs, Cs)]
        dest = node_map[(Rf, Cf)]

        # Dijkstra's algorithm with O((R*C) log (R*C)) per query
        # Possibly too slow for Q=100,000 but let's try

        from heapq import heappush, heappop

        dist = [float('inf')] * n
        dist[src] = 0
        heap = []
        heappush(heap, (0, src))

        while heap:
            current_dist, u = heappop(heap)
            if u == dest:
                break
            if current_dist > dist[u]:
                continue
            for v, w in adj[u]:
                if dist[v] > dist[u] + w:
                    dist[v] = dist[u] + w
                    heappush(heap, (dist[v], v))

        if dist[dest] == float('inf'):
            print("INVALID")
        else:
            print(int(dist[dest]))

if __name__ == "__main__":
    main()

# ZUald2pfung6rtRJXIc

**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 1
111
111
111
2
1 1 3 3
3 3 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: X Aura
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Mount ICPC can be represented as a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$). The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$ and has a height of $$$H_{r, c}$$$. Two cells are adjacent to each other if they share a side. Formally, $$$(r, c)$$$ is adjacent to $$$(r-1, c)$$$, $$$(r+1, c)$$$, $$$(r, c-1)$$$, and $$$(r, c+1)$$$, if any exists.

You can move only between adjacent cells, and each move comes with a penalty. With an aura of an odd positive integer $$$X$$$, moving from a cell with height $$$h_1$$$ to a cell with height $$$h_2$$$ gives you a penalty of $$$(h_1 - h_2)^X$$$. Note that the penalty can be negative.

You want to answer $$$Q$$$ independent scenarios. In each scenario, you start at the starting cell $$$(R_s, C_s)$$$ and you want to go to the destination cell $$$(R_f, C_f)$$$ with minimum total penalty. In some scenarios, the total penalty might become arbitrarily small; such a scenario is called invalid. Find the minimum total penalty to move from the starting cell to the destination cell, or determine if the scenario is invalid.

Input: The first line consists of three integers $$$R$$$ $$$C$$$ $$$X$$$ ($$$1 \leq R, C \leq 1000; 1 \leq X \leq 9; X$$$ is an odd integer).

Each of the next $$$R$$$ lines consists of a string $$$H_r$$$ of length $$$C$$$. Each character in $$$H_r$$$ is a number from 0 to 9. The $$$c$$$-th character of $$$H_r$$$ represents the height of cell $$$(r, c)$$$, or $$$H_{r, c}$$$.

The next line consists of an integer $$$Q$$$ ($$$1 \leq Q \leq 100\,000)$$$.

Each of the next $$$Q$$$ lines consists of four integers $$$R_s$$$ $$$C_s$$$ $$$R_f$$$ $$$C_f$$$ ($$$1 \leq R_s, R_f \leq R; 1 \leq C_s, C_f \leq C$$$).

Output: For each scenario, output the following in a single line. If the scenario is invalid, output INVALID. Otherwise, output a single integer representing the minimum total penalty to move from the starting cell to the destination cell.

Examples:
input:
3 4 1
3359
4294
3681
5
1 1 3 4
3 3 2 1
2 2 1 4
1 3 3 2
1 1 1 1
output:
2
4
-7
-1
0

input:
2 4 5
1908
2023
2
1 1 2 4
1 1 1 1
output:
INVALID
INVALID

input:
3 3 9
135
357
579
2
3 3 1 1
2 2 2 2
output:
2048
0

Note: Explanation for the sample input/output #1

For the first scenario, one of the solutions is to move as follows: $$$(1, 1) \rightarrow (2, 1) \rightarrow (3, 1) \rightarrow (3, 2) \rightarrow (3, 3) \rightarrow (3, 4)$$$. The total penalty of this solution is $$$(3 - 4)^1 + (4 - 3)^1 + (3 - 6)^1 + (6 - 8)^1 + (8 - 1)^1 = 2$$$.

Explanation for the sample input/output #2

For the first scenario, the cycle $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (1, 2) \rightarrow (1, 1)$$$ has a penalty of $$$(1 - 2)^5 + (2 - 0)^5 + (0 - 9)^5 + (9 - 1)^5 = -26250$$$. You can keep repeating this cycle to make your total penalty arbitrarily small. Similarly, for the second scenario, you can move to $$$(1, 1)$$$ first, then repeat the same cycle.

Buggy Code:
r, c, x = tuple(map(int,input().split()))

grid = []
for i in range(r):
  grid.append(list(map(int,list(input()))))

q = int(input())
queries = []
for i in range(q):
  queries.append(list(map(int, input().split())))

invalid = False
if (x != 1):
  for i in range(r-1):
    for j in range(c-1):
      if ((grid[i][j] - grid[i+1][j])**x +
          (grid[i+1][j] - grid[i+1][j+1])**x +
          (grid[i+1][j+1] - grid[i][j+1])**x +
          (grid[i][j+1] - grid[i][j])**x != 0):
          invalid = True
          break
    if invalid:
      break

if (invalid):
  for i in range(q):
    print("INVALID")
elif x == 1 :
  for i in queries:
    x1 = i[0]-1
    y1 = i[1]-1
    x2 = i[2]-1
    y2 = i[3]-1
    print(grid[x1][y1] - grid[x2][y2])
else:
  prefrows = []
  prefcols = []
  for j in range(c):
    prefrows.append([0])
    for i in range(r-1):
      prefrows[-1].append(prefrows[-1][-1] + (grid[i][j] - grid[i+1][j]) ** x)
  for j in range(r):
    prefcols.append([0])
    for i in range(c-1):
      prefcols[-1].append(prefcols[-1][-1] + (grid[j][i] - grid[j][i+1]) ** x)
  # prefrows = prefrows[1:]
  # prefcols = prefcols[1:]
  for i in queries:
    x1 = i[0]-1
    y1 = i[1]-1
    x2 = i[2]-1
    y2 = i[3]-1
    print(prefrows[y2][x2]- prefrows[y1][x1] + prefcols[x2][y2] - prefcols[x1][y1])

247
574
357

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the algorithm might incorrectly determine the validity of the grid or miscalculate the penalties. The code has a specific issue when `x` is greater than 1, where it incorrectly checks for invalid cycles and may prematurely mark the grid as invalid.

### Step-by-Step Reasoning:

1. **Invalid Cycle Detection:**
   - The code checks for invalid cycles by summing the penalties of a 2x2 subgrid. If the sum is not zero, it marks the grid as invalid.
   - This check is flawed because it only considers 2x2 subgrids, which might not cover all possible invalid cycles.

2. **Edge Cases:**
   - **Small Grids:** Test with small grids to ensure the cycle detection works correctly.
   - **Special Values of `x`:** Test with `x` values greater than 1 to see if the invalid cycle detection fails.
   - **Different Heights:** Use different height values to ensure the penalty calculation is correct.

3. **Test Input:**
   - **Grid with a 2x2 invalid cycle:** Ensure the grid has a 2x2 subgrid that should be invalid.
   - **Grid without an invalid cycle:** Ensure the grid has no 2x2 subgrid that should be invalid.
   - **Query with a valid path:** Ensure there is a query that should have a valid path.
   - **Query with an invalid path:** Ensure there is a query that should have an invalid path.

### Test Input:
```plaintext
3 3 3
135
357
579
2
1 1 3 3
2 2 2 2
```

### Explanation:
- **Grid:**
  - The grid is:
    ```
    1 3 5
    3 5 7
    5 7 9
    ```
  - The subgrid from (1,1) to (2,2) is:
    ```
    1 3
    3 5
    ```
  - The penalty for the cycle (1,1) → (2,1) → (2,2) → (1,2) → (1,1) is:
    ```
    (1 - 3)^3 + (3 - 5)^3 + (5 - 3)^3 + (3 - 1)^3 = (-2)^3 + (-2)^3 + 2^3 + 2^3 = -8 - 8 + 8 + 8 = 0
    ```
  - However, the code incorrectly marks this as invalid.

- **Queries:**
  - **Query 1 (1,1 to 3,3):** This should be valid, but the code might mark it as invalid due to the flawed cycle detection.
  - **Query 2 (2,2 to 2,2):** This is a trivial valid query.

This test input should expose the bug in the cycle detection logic and show that the grid is incorrectly marked as invalid.