**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    n = nextInt();
    m = nextInt();
    int pn = nextInt();
    char[][] bs = new char[n][m];
    for (int i = 0; i < n; ++i) bs[i] = next().toCharArray();
    int[][] p = new int[pn][3];
    for (int i = 0; i < p.length; ++i) {
        for (int j = 0; j < p[i].length; ++j) {
            p[i][j] = nextInt();
        }
        p[i][0]--;
        p[i][1]--;
    }
    Tower[] c = new Tower[pn];
    for (int i = 0; i < pn; ++i) {
        c[i] = new Tower();
        c[i].x = p[i][0];
        c[i].y = p[i][1];
        c[i].power = p[i][2];
        int[] g = new int[M];
        for (int x = c[i].x - M; x <= c[i].x + M; x++) {
            if (x < 0 || x >= n) continue;
            for (int y = c[i].y - M; y <= c[i].y + M; y++) {
                if (y < 0 || y >= m) continue;
                if (bs[x][y] == '.') continue;
                long dist = (long) (x - c[i].x) * (x - c[i].x)
                        + (long) (y - c[i].y) * (y - c[i].y);
                for (int j = M - 1; j >= 1; j--) {
                    if ((long) j * j < dist) break;
                    g[j]++;
                }
            }
        }
        for (int j = 1; j < M; ++j) {
            c[i].damage[j] = (long) g[j] * c[i].power;
        }
    }
    long[][] w = new long[pn][M];
    for (int i = 0; i < pn; ++i) {
        for (int j = 0; j < M; ++j) {
            w[i][j] = c[i].damage[j];
        }
        w[i][0] = i;
    }
    Set<Integer> h = new HashSet<>();
    for (int j = 1; j < M; ++j) {
        final int ttt = j;
        Arrays.sort(w, (x, y) -> -Long.compare(x[ttt], y[ttt]));
        for (int i = 0; i < M && i < pn; ++i) h.add((int) w[i][0]);
    }
    int hn = h.size();
    Tower[] z = new Tower[hn];
    int zn = 0;
    for (int x : h) {
        z[zn++] = c[x];
    }
    // use towers z. (max count: 400)
    long[] v = new long[1 << M];
    long BIG = 0x3f3f3f3f3f3f3f3fL;
    Arrays.fill(v, -BIG);
    v[0] = 0;
    for (int j = 0; j < zn; ++j) {
        for (int i = (1 << M) - 1; i >= 0; --i) {
            if (v[i] < 0) continue;
            for (int k = 1; k < M; ++k) {
                if ((i & (1 << k)) > 0) continue;
                int i2 = i | (1 << k);
                v[i2] = Math.max(v[i2], v[i] + z[j].damage[k]);
            }
        }
    }
    long ret = 0;
    for (int i = 1; i < (1 << M); ++i) {
        long base = 0;
        for (int j = 0; j < M; ++j) {
            if ((i & (1 << j)) > 0) base += p3[j];
        }
        ret = Math.max(ret, v[i] - base);
    }
    sl(ret);
}

static int M = 20;
static long[] p3 = new long[M];

static {
    p3[1] = 3;
    for (int i = 2; i < M; ++i) p3[i] = p3[i - 1] * 3;
}

class Tower {
    int x, y, power;
    long[] damage = new long[M];
}

int n, m;

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace();
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // TODO: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

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

**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 craft a test case that highlights potential issues in the algorithm. The code is designed to determine the maximum base health of an enemy that can be defeated given the constraints of tower ranges and damage. Here are the steps to create a challenging test input:

1. **Grid Setup**: We need a grid that is small enough to be manageable but large enough to test the algorithm's logic. A 3x3 grid with 2 towers can be a good starting point.

2. **Path Setup**: The path should be such that it intersects with the towers in a way that maximizes the potential for damage calculation errors.

3. **Tower Setup**: Place towers in positions where their ranges can overlap or be just out of range, to test the edge cases of the distance calculations.

4. **Enemy Health**: The base health should be set in a way that it is just enough to be challenging but not impossible to defeat.

Here is a small-scale test input that fits these criteria:

```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```

### Explanation:
- **Grid**: 3x3
- **Path**: Starts at (1,1) and ends at (3,3), passing through (1,1), (1,2), (2,2), (2,3), (3,3).
- **Towers**:
  - Tower 1 at (1,2) with power 4
  - Tower 2 at (3,1) with power 3

### Expected Behavior:
- The enemy will move through the cells (1,1), (1,2), (2,2), (2,3), (3,3).
- Tower 1 at (1,2) can potentially damage the enemy in cells (1,1), (1,2), (2,2).
- Tower 2 at (3,1) can potentially damage the enemy in cells (2,2), (3,2), (3,3).

### Challenging Aspects:
- The ranges of the towers are such that they can overlap in some cells, testing the code's ability to correctly calculate and sum the damage.
- The path is designed to ensure that the enemy spends time in cells where the towers can deal damage, but not in a straightforward manner, to test the edge cases of the distance calculations.

This test input should help in identifying any issues in the distance calculations, damage summation, and the overall logic of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
def ProblemH():
    from collections import deque, Counter
    from functools import cache
    from math import ceil

    directions = [(1,0), (0,1), (-1,0), (0, -1)]
    def GetDist(i,j, i_2, j_2):
        r_2 = (i - i_2)**2  + (j - j_2)**2
        r = ceil(r_2**0.5)
        if (r-1)**2 >= r_2:
            r-= 1
        return r

    def GetProfits(i_0, j_0):
        count = Counter() # [r] = cnt
        dq = deque([(i_0,j_0)])
        seen = set(dq)
        while dq:
            i,j = dq.popleft()
            for delta_i, delta_j in directions:
                new_i, new_j = i + delta_i, j + delta_j
                if (new_i, new_j) not in seen and 0 <= new_i <  num_rows and 0 <= new_j < num_cols:
                    r = GetDist(i_0,j_0,new_i, new_j)
                    if r <= max_r:
                        dq.append((new_i, new_j))
                        seen.add((new_i, new_j))
                        if grid[new_i][new_j] == path:
                            count[r] += 1
        tower_id = grid[i_0][j_0]
        power = powers[tower_id]
        profits = dict() # [radius] = profit
        tot = 0
        for r in sorted(count):
            tot += count[r]
            # print(f"{r = }, {tot = }")
            profit = tot * power - 3**r
            if profit > 0:
                profits[r] = profit
        return profits

    @cache
    def DP(i_tower, mask):
        # mask is  used r options
        if i_tower == num_towers:
            return 0
        ret = DP(i_tower + 1, mask)
        for r, profit in profit_options[i_tower].items():
            indicator= 1<<(r-1)
            if not mask&indicator:
                ret = max(ret, profit + DP(i_tower + 1, mask|indicator))
        return ret

    max_r = 15

    num_test_cases = int(input())
    empty = "."
    path = "#"
    for i_test_case in range(num_test_cases):
        num_rows, num_cols, num_towers = [int(x) for x in input().split()]
        grid = [list(input().strip()) for i in range(num_rows)]

        powers = []
        for i_tower in range(num_towers):
            i,j, power = [int(x) for x in input().split()]
            i-=1
            j-=1
            grid[i][j] = i_tower
            powers.append(power)

        # print(*grid, sep="\n", end="\n\n")
        # print(f"{powers = }")
        profit_options = [None for i in range(num_towers)]

        for i in range(num_rows):
            for j in range(num_cols):
                x = grid[i][j]
                if type(x) is int:
                    profit_options[x] = GetProfits(i,j)

        # print(f"{profit_options = }")
        ans = DP(0,0)
        DP.cache_clear()
        print(ans)

    return

ProblemH()

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
def ProblemH():
    from collections import deque, Counter
    from functools import cache
    from math import ceil

    directions = [(1,0), (0,1), (-1,0), (0, -1)]
    def GetDist(i,j, i_2, j_2):
        r_2 = (i - i_2)**2  + (j - j_2)**2
        r = ceil(r_2**0.5)
        return r

    def GetProfits(i_0, j_0):
        count = Counter() # [r] = cnt
        dq = deque([(i_0,j_0)])
        seen = set(dq)
        while dq:
            i,j = dq.popleft()
            for delta_i, delta_j in directions:
                new_i, new_j = i + delta_i, j + delta_j
                if (new_i, new_j) not in seen and 0 <= new_i <  num_rows and 0 <= new_j < num_cols:
                    r = GetDist(i_0,j_0,new_i, new_j)
                    if r <= max_r:
                        dq.append((new_i, new_j))
                        seen.add((new_i, new_j))
                        if grid[new_i][new_j] == path:
                            count[r] += 1
        tower_id = grid[i_0][j_0]
        power = powers[tower_id]
        profits = dict() # [radius] = profit
        tot = 0
        for r in sorted(count):
            tot += count[r]
            # print(f"{r = }, {tot = }")
            profit = tot * power - 3**r
            if profit > 0:
                profits[r] = profit
        return profits

    @cache
    def DP(i_tower, mask):
        # mask is  used r options
        if i_tower == num_towers:
            return 0
        ret = DP(i_tower + 1, mask)
        for r, profit in profit_options[i_tower].items():
            indicator= 1<<(r-1)
            if not mask&indicator:
                ret = max(ret, profit + DP(i_tower + 1, mask|indicator))
        return ret

    max_r = 12

    num_test_cases = int(input())
    empty = "."
    path = "#"
    for i_test_case in range(num_test_cases):
        num_rows, num_cols, num_towers = [int(x) for x in input().split()]
        grid = [list(input().strip()) for i in range(num_rows)]

        powers = []
        for i_tower in range(num_towers):
            i,j, power = [int(x) for x in input().split()]
            i-=1
            j-=1
            grid[i][j] = i_tower
            powers.append(power)

        # print(*grid, sep="\n", end="\n\n")
        # print(f"{powers = }")
        profit_options = [None for i in range(num_towers)]

        for i in range(num_rows):
            for j in range(num_cols):
                x = grid[i][j]
                if type(x) is int:
                    profit_options[x] = GetProfits(i,j)

        # print(f"{profit_options = }")
        ans = DP(0,0)
        DP.cache_clear()
        print(ans)

    return

ProblemH()

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

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

Think step by step.

Response:
```plaintext
1
3 3 3
#..
##.
.##
1 2 1
2 1 1
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

public class Main {
    public static int MDC = (int) 998244353;

    public static void main(String[] args) throws IOException {
        sc = new MScanner(System.in);
        pw = new PrintWriter(System.out);
        int t = Integer.parseInt(sc.nextLine());
        //int t = 1;
        for (int i = 0; i < t; i++) {
            solver();
        }
    }

    static long[] bitArr;

    public static void solver() throws IOException {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        String[] a = new String[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLine();
        }
        int[][] p = new int[k][3];
        for (int i = 0; i < k; i++) {
            p[i][0] = sc.nextInt()-1;
            p[i][1] = sc.nextInt()-1;
            p[i][2] = sc.nextInt();
        }
        // 计算r的上限
        int max = 16;
        long[][] dis = new long[k][max+1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (a[i].charAt(j) != '#') continue;
                for (int x = 0; x < k; x++) {
                    long d =(long) (i - p[x][0])* (i - p[x][0])+ (long)(j - p[x][1])*(j - p[x][1]);
                    int u=(int)Math.sqrt(d);
                    if((long)u*u<d) u++;
                    if(u<=max)dis[x][u]++;
                }
            }
        }
        for (int i=0;i<k;i++){
            for (int j=1;j<=max;j++){
                dis[i][j]+=dis[i][j-1];
            }
        }
        long[][] f=new long[k+1][1<<max];
        f[0][0]=0;
        long ans=0;
        for (int i=1;i<=k;i++){
            for (int j=0;j<(max<<1);j++){
                f[i][j]=f[i-1][j];
                long w=3;
                for (int x=0;x<max;x++){
                    if(((j>>x)&1)>0){
                        f[i][j]=  Math.max(f[i][j],f[i-1][j^(1<<x)]+dis[i-1][x+1]*p[i-1][2]-w);
                    }
                    w*=3;
                }
            }
        }
         for (int i=0;i<(max<<1);i++) {
            ans = Math.max(ans, f[k][i]);
        }
        pw.println(ans);
        pw.flush();
    }

    public static int bitCount(long n) {
        int count = 0;
        while (n != 0) {
            count++;
            n >>>= 1L;
        }
        return count;
    }

    public static void bitInsert(int index, long v, int n) {
        while (index <= n) {
            bitArr[index] += v;
            index += (index & (-index));
        }
    }

    public static long bitQuery(int index, int n) {
        long ans = 0;
        while (index > 0) {
            ans += bitArr[index];
            index -= -index & index;
        }
        return ans;
    }

    public static int gcd(int a, int b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    private static long pow(long n, long m) {
        long res = 1;
        long base = n;
        while (m != 0) {
            if ((m & 1) == 1) {
                res = res * base;
                res %= MDC;
            }
            base = base * base;
            base %= MDC;
            m = m >> 1;
        }
        return res;
    }

    static int curt;
    static PrintWriter pw;
    static MScanner sc;

    static class MScanner {
        StringTokenizer st;
        BufferedReader br;

        public MScanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public MScanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

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

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

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

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

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

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

        public void waitForInput() throws InterruptedException {
            Thread.sleep(3000);
        }

    }

    static void dbg(int[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void dbg(long[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void sort(int[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void sort(long[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void shuffle(int[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            int tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }

    static void shuffle(long[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            long tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }
}

**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. **Edge Cases**: Test cases that push the boundaries of the algorithm, such as the smallest and largest possible values for `n`, `m`, and `k`.
2. **Complex Paths**: Paths that are not straightforward and involve multiple turns and intersections.
3. **High Damage Towers**: Towers with high damage values to ensure the algorithm must correctly account for the damage dealt.
4. **Range Limitations**: Ensure the ranges are within the valid limits and test the edge of the maximum range.
5. **Health Calculation**: Test cases where the health calculation is critical, especially when the health and damage values are close.

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

```plaintext
1
5 5 3
#.###
#.#.#
###.#
#....
#####
3 2 100
2 3 50
4 2 200
```

### Explanation:
- **Grid Size**: 5x5, which is a small but non-trivial size.
- **Path**: The path is not a straight line and involves multiple turns, making it more challenging to compute the damage.
- **Towers**: Three towers with high damage values (100, 50, 200) to ensure the algorithm must correctly account for the damage dealt.
- **Tower Positions**: Towers are placed in strategic positions to cover multiple cells of the path.
- **Range Limitation**: The range is within the valid limit, but the health and damage values are close to ensure the algorithm must handle the health calculation correctly.

This test input should expose any issues in the buggy code, particularly in how it calculates the damage and the health of the enemy.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
/*
  Code by: linvg
*/

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

#define int long long
#define ii pair<int, int>
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define per(i, a, b) for (int i = (b)-1; i >= (a); --i)

string to_upper(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A'; return a; }
string to_lower(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A'; return a; }

template<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> T sqr(T x) { return x * x; }
template<class T> T cube(T x) { return x * x * x; }

// #define DBG

#ifdef ONLINE_JUDGE
#define dbg(...)
#else
#include "debug.h"
#endif

const int INF = 0x3f3f3f3f3f;

struct Tower
{
   int x, y, damage, r = 0;
};

ostream operator<<(ostream &os, Tower &t) {
   os << t.r;
}

void solve()
{
   int n, m, k;
   cin >> n >> m >> k;
   vector<vector<char>> a(n, vector<char>(m));
   vector<ii> path;
   for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
         cin >> a[i][j];
         if (a[i][j] == '#') {
            path.push_back({i, j});
         }
      }
   }

   vector<int> p3(20);
   p3[0] = 1;
   for (int i = 1; i < 20; ++i) {
      p3[i] = p3[i-1] * 3;
   }

   vector<Tower> towers(k);
   for (int i = 0; i < k; ++i) {
      cin >> towers[i].x >> towers[i].y >> towers[i].damage;
      --towers[i].x, --towers[i].y;
   }

   vector<int> vis(51, 0);

   auto calc = [&] (int id, int &ans, int bonus, auto&& calc) -> void {
      if (id == k) {
         int res = 0;
         for (auto &c : towers) {
            for (int i = 0; i < path.size(); ++i) {
               if (sqr(path[i].se - c.y) + sqr(path[i].fi - c.x) <= sqr(c.r)) {
                  res += c.damage;
               }
            }
         }
         ans = max(ans, res - bonus);
         return;
      }
      for (int i = 1; i <= 12; ++i) {
         if (i != 0 && !vis[i]) {
            towers[id].r = i;
            vis[i] = 1;
            calc(id+1, ans, bonus + p3[i], calc);
            vis[i] = 0;
            towers[id].r = 0;
         } else if (i == 0) {
            calc(id+1, ans, bonus, calc);
         }
      }
   };

   int ans = 0;
   calc(0, ans, 0, calc);
   cout << ans << '\n';
}

int32_t main() {

#ifdef LOCAL
   freopen("input.txt", "r", stdin);
   freopen("output.txt", "w", stdout);
#endif
   ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
   int t;
   cin >> t;
   while (t--) {
      solve();
   }
   return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider the following steps:

1. **Understand the Bug**: The provided code seems to have an issue with the recursive function `calc` that calculates the maximum health `h` for which the towers can prevent the enemy from reaching the end. The recursive function might not handle the combinations of tower ranges and enemy health correctly, especially when the number of towers is small and the path is simple.

2. **Edge Cases**: The code might fail when:
   - There are very few towers, and their positions and ranges need to be carefully set to maximize the enemy's base health.
   - The path is very short, and the enemy can be killed in just a few seconds.
   - The damage values of the towers are very high, and the enemy's health must be set correctly to ensure it can be killed.

3. **Test Input**: Let's create a test case with a small grid, a few towers, and a simple path. This will help us verify if the code correctly handles the combination of tower ranges and enemy health.

Here is a small-scale test input:

```plaintext
1
3 3 2
#.#
#.#
#.#
1 1 100
3 3 100
```

### Explanation:
- **Grid Size**: 3x3
- **Number of Towers**: 2
- **Path**: The enemy moves from (1,1) to (3,3), passing through (1,1), (2,2), and (3,3).
- **Towers**:
  - Tower 1 at (1,1) with damage 100
  - Tower 2 at (3,3) with damage 100

### Expected Output:
- The enemy will be in the following positions:
  - (1,1) for 1 second
  - (2,2) for 1 second
  - (3,3) for 1 second

- If the tower ranges are set to 1, the enemy will take 100 damage in each of the cells (1,1) and (3,3).
- Therefore, the enemy must have a base health of 199 or less to be killed before reaching the end.

### Expected Output:
```plaintext
199
```

This test case is designed to be simple yet challenging, as it requires the code to correctly handle the combination of tower ranges and enemy health to ensure the enemy is killed before reaching the end.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

char s[55][55];
int x[2505], y[2505], p[2505];
int pw[15], dp[15][1<<13];
int power[15][15];
int main(){
    int t;
    scanf("%d", &t);
    pw[0] = 1;
    for(int i=1;i<=13;i++) pw[i] = pw[i-1]*3;
    while(t--){
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k);
        for(int i=1;i<=n;i++) scanf("%s", s[i]+1);
        for(int i=1;i<=k;i++){
            scanf("%d%d%d", &x[i], &y[i], &p[i]);
        }
        if(k > 13){
            printf("0\n");
            continue;
        }
        for(int i=1;i<=12;i++) for(int j=1;j<=k;j++) power[i][j] = 0;
        for(int i=1;i<=k;i++){
            for(int j=1;j<=n;j++){
                for(int k=1;k<=m;k++){
                    if(s[j][k] == '#'){
                        for(int r=0;r<=13;r++){
                            if((x[i]-j)*(x[i]-j) + (y[i]-k)*(y[i]-k) <= r*r){
                                power[r][i]++;
                            }
                        }
                    }
                }
            }
        }

        memset(dp, 0, sizeof(dp));
        for(int i=1;i<=k;i++){
            for(int j=0;j<(1<<13);j++){
                for(int l=0;l<13;l++){
                    if((j>>l)&1) continue;
                    dp[i][j|(1<<l)] = max(dp[i][j|(1<<l)], dp[i-1][j] + power[l][i]*p[i] - pw[l]);
                }
            }
        }

        int ans = 0;
        for(int i=0;i<(1<<13);i++) ans = max(ans, dp[k][i]);
        printf("%d\n", ans);
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
import os,sys
from io import BytesIO, IOBase
from typing import *

# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

import math
from itertools import product

def squared_radius(r1, c1, r2, c2):
    return (r1 - r2) ** 2 + (c1 - c2) ** 2

BITS = 11
PATH = "#"
def main():
    R, C, K = map(int, input().split())
    grid = [list(input()) for _ in range(R)]
    towers = [None] * K
    # ceiling radius
    UB = 200
    crad = [0] * UB
    crad[1] = 1
    for i in range(2, UB):
        crad[i] = crad[i - 1]
        if crad[i] * crad[i] < i: crad[i] += 1
    for i in range(K):
        r, c, p = map(int, input().split())
        r -= 1; c -= 1
        towers[i] = (r, c, p)
    freq = [[0] * BITS for _ in range(K)]
    for r, c in product(range(R), range(C)):
        if grid[r][c] == PATH:
            for i in range(K):
                r2, c2, p = towers[i]
                dist_squared = squared_radius(r, c, r2, c2)
                if dist_squared >= BITS * BITS: continue
                cdist = crad[dist_squared]
                freq[i][cdist] += 1
    psum = [[0] * BITS for _ in range(K)]
    for i in range(K):
        for j in range(1, BITS):
            psum[i][j] = psum[i][j - 1] + freq[i][j]
    dp = [-math.inf] * (1 << BITS)
    dp[0] = ans = 0
    # O(NM*2^BITS*BITS)
    for i in range(K):
        ndp = [-math.inf] * (1 << BITS)
        p = towers[i][2]
        for mask in range(1 << BITS):
            for j in range(BITS):
                if (mask >> j) & 1: continue
                nmask = mask | (1 << j)
                ndp[nmask] = max(ndp[nmask], dp[mask] + psum[i][j] * p - pow(3, j))
                ans = max(ans, ndp[nmask])
        dp = ndp
    print(ans)

if __name__ == '__main__':
    T = int(input())
    for _ in range(T):
        main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that challenges the algorithm's logic, especially focusing on edge cases or scenarios that could cause the code to fail. Here are the steps to generate such a test case:

1. **Small Grid Size**: Use a small grid size to ensure the test case is manageable and fits within the output limits.
2. **Few Towers**: Place a few towers to keep the test case simple but challenging.
3. **Path Layout**: Design a path that maximizes the exposure of the towers to the enemy.
4. **Tower Parameters**: Assign tower parameters (position and power) that can potentially cause issues with the algorithm's logic.

Let's create a test case with a 3x3 grid and 2 towers. The path will be a simple diagonal from (1,1) to (3,3), and the towers will be placed in positions that can potentially cause issues.

```plaintext
1
3 3 2
#..
.#
..#
1 1 1
3 3 1
```

### Explanation:
- **Grid Size**: 3x3
- **Number of Towers**: 2
- **Path**: The path is a diagonal from (1,1) to (3,3).
- **Towers**:
  - Tower 1 at (1,1) with power 1
  - Tower 2 at (3,3) with power 1

This test case is designed to:
- Ensure that the towers are positioned in a way that they can potentially cover the entire path.
- Test the algorithm's ability to handle towers with low power.
- Verify the correct calculation of the maximum base health \( h \).

By using this test case, we can check if the algorithm correctly calculates the maximum base health \( h \) for which the player can set the ranges so that the enemy does not reach the end of the path with more than 0 health.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
import sys
import random

input = sys.stdin.readline
rd = random.randint(10 ** 9, 2 * 10 ** 9)

t = int(input())
for _ in range(t):
    m,n,k = map(int, input().split())
    grid = [list(input().strip()) for _ in range(m)]
    li = [list(map(int, input().split())) for _ in range(k)]
    bit = 16
    if k > bit:
        print(0)
        continue
    count = [[0] * bit for _ in range(k)]
    for i in range(k):
        x,y,p = li[i]
        x -= 1
        y -= 1
        for j in range(bit):
            for ii in range(max(0,x - j),min(m,x + j + 1)):
                for jj in range(max(0,y - j),min(n,y + j + 1)):
                    if (ii - x) ** 2 + (jj - y) ** 2 <= j ** 2 and grid[ii][jj] == '#':
                        count[i][j] += 1
    dp = [0] * (1 << bit)
    p3 = [3]
    for i in range(1,bit):
        p3.append(p3[-1] * 3)
    ans = 0
    for i in range(1,1 << bit):
        cc = bin(i).count('1')
        if cc > k:
            continue
        for j in range(bit):
            if (i >> j) & 1:
                dp[i] = max(dp[i],dp[i ^ (1 << j)] + count[cc - 1][j] * li[cc - 1][2])
        if cc == k:
            d = dp[i]
            for j in range(bit):
                if (i >> j) & 1:
                    d -= 3 ** j
            ans = max(ans,d)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
2
3 3 2
#..
##.
.##
1 2 1
3 1 1
3 3 2
#..
##.
.##
1 2 1
3 1 500
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

public class H {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	static Long [][] memo;

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void solve(int t) {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			int n = sc.nextInt();
			int m = sc.nextInt();

			int k = sc.nextInt();

			char [][] arr = new char [n][];

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

			List<int []> list = new ArrayList<>();

			for (int i = 0; i < n; ++i) {
				for (int j = 0; j < m; ++j) {
					if (arr[i][j] == '#') {
						list.add(new int [] {i + 1, j + 1});
					}
				}
			}

			int [][] useArr = new int [k][n + m];

			long [][] fArr = new long [k][];

			int [][] attack = new int [k][2];

			int x, y, p, dist;

			long total = 0;

			for (int i = 0; i < k; ++i) {
				x = sc.nextInt();
				y = sc.nextInt();

				//System.out.println(x + " " + y);

				attack[i][0] = sc.nextInt();

				attack[i][1] = i;

				for (int [] tArr : list) {
					dist = findMinDist(tArr[0], tArr[1], x, y);

					useArr[i][dist]++;
				}

				int sum = 0;

				for (int j = 0; j < useArr[i].length; ++j) {
					sum += useArr[i][j];
					useArr[i][j] = sum;
				}

				fArr[i] = getMax(useArr[i], attack[i][0], list.size());
			}

			memo = new Long [k][8096];
			long result = solve(fArr, 0, 0);

			out.println(result);
		}

		private long solve(long [][] fArr, int idx, int bit) {
			if (idx == fArr.length)
				return 0;
			else if (memo[idx][bit] != null)
				return memo[idx][bit];

			long result = solve(fArr, idx + 1, bit);

			for (int i = 0; i < 13; ++i) {
				if ( ((1 << i) & bit) != 0)
					continue;

				int nextBit = (1 << i) ^ bit;

				result = Math.max(result, fArr[idx][i] + solve(fArr, idx + 1, nextBit));
			}

			memo[idx][bit] = result;
			return result;
		}

		private long [] getMax(int [] arr, int attackVal, int max) {

			long added = 1;
			long idx;
			long effect;

			long [] resultArr = new long [13];

			// System.out.println(attackVal + " " + max);

			long maxDamage = max;
			maxDamage *= attackVal;

			// System.out.println(Arrays.toString(arr) + " " + attackVal + " " + maxDamage);

			long result = 0;

			for (int i = 1; i < arr.length; ++i) {
				added *= 3;

				if (added > maxDamage)
					break;

				effect = arr[i];
				effect *= attackVal;

				// System.out.println(effect + " " + added);
				result = Math.max(result, effect - added);

				if (i <= 13) {
					resultArr[i - 1] = effect - added;
				}
			}

			return resultArr;
		}

		private int findMinDist(int x1, int y1, int x2, int y2) {

			int xDiff = x2 - x1;
			int yDiff = y2 - y1;

			int dist = (xDiff * xDiff) + (yDiff * yDiff);

			int min = 0;
			int max = 1000;

			int mid;

			while (max - min > 1) {
				mid = (max + min) / 2;

				if (mid * mid >= dist)
					max = mid;
				else
					min = mid;
			}

			return max;
		}
	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the number of towers is very small or very close to the maximum allowed.
2. **Special Grid Configurations**: Grids where the path is very short or very long, or where the path has special patterns.
3. **High Attack Values**: Towers with very high attack values to ensure that the damage calculations are correctly handled.
4. **Range Constraints**: Towers with ranges that are just enough to cover the path or not.

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

### Test Input
```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```

### Explanation
1. **Grid Size**: The grid is 3x3, which is small enough to be manageable but large enough to have a non-trivial path.
2. **Path Configuration**: The path is from (1,1) to (3,3) and follows the path `#..`, `##.`, `.##`. This path ensures that the enemy will move through multiple cells, providing opportunities for the towers to deal damage.
3. **Towers**:
   - The first tower is at (1,2) with an attack value of 4.
   - The second tower is at (3,1) with an attack value of 3.
4. **Expected Behavior**:
   - The first tower can cover cells (1,2), (2,1), (2,2), and (3,2).
   - The second tower can cover cells (3,1), (2,1), (2,2), and (3,2).

This configuration should test the following:
- The correct calculation of the damage dealt by the towers.
- The correct handling of the path and the movement of the enemy.
- The correct application of the range constraints and the resulting health increase for the enemy.

By using this test input, we can verify if the buggy code correctly handles the damage calculations and the range constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

import static java.lang.Math.max;

public class H_1955 {

    public static final int MAX_RADIUS = 12;

    @FunctionalInterface
    public interface Apply<T> {
        void apply(T t);
    }

    private static final class Tokens extends LinkedList<String> {
        Tokens(String line) {
            super(new LinkedList<String>());
            for (var token : line.split(" "))
                add(token);
        }

        String nextStr() {
            return remove();
        }

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

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

        double nextDbl() {
            return Double.parseDouble(remove());
        }

        BigInteger nextBigInt() {
            return new BigInteger(remove());
        }

        void apply(Apply<Tokens> block) {
            block.apply(this);
        }
    }

    private static Tokens nextLine(BufferedReader cin) {
        String line = null;
        try {
            line = cin.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return new Tokens(line);
    }

    private static byte[] to_bytes(String s) {
        return s.getBytes();
    }

    private static byte[] to_bytes(Integer x) {
        return to_bytes(x.toString());
    }

    private static byte[] to_bytes(Long x) {
        return to_bytes(x.toString());
    }

    private static byte[] to_bytes(String f, Double x) {
        return to_bytes(String.format(f, x));
    }

    private static byte[] to_bytes(BigInteger x) {
        return to_bytes(x.toString());
    }

    private final static byte[] space = to_bytes(" ");
    private final static byte[] endl = to_bytes("\n");

    private static class Path {
        int x, y;

        public Path(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    private static class Tower {
        int x, y, p;
        int[] potency = new int[MAX_RADIUS];

        public Tower(int x, int y, int p) {
            this.x = x;
            this.y = y;
            this.p = p;
        }
    }

    public static int square(int x) {
        return x * x;
    }

    private static int countBits(int number) {
        var count = 0;
        while (number != 0) {
            count += number & 1;
            number = number >> 1;
        }
        return count;
    }

    private static int solve(int[][] dp, ArrayList<Tower> towers, int towerCount, int radiusBit) {
        if (towerCount == 0 || radiusBit == 0) return 0;
        if (dp[towerCount][radiusBit] != Integer.MIN_VALUE) return dp[towerCount][radiusBit];

        var ans = 0;
        for (int i = 0; i < MAX_RADIUS; i++) {
            if ((radiusBit & (1 << i)) != 0) {
                ans = max(ans, towers.get(towerCount - 1).potency[i] + solve(dp, towers,towerCount -1, radiusBit ^ (1<<i)));
            }
        }
        dp[towerCount][radiusBit] = ans;
        return ans;
    }

    public static void main(String[] args) throws IOException {
        final var cin = new BufferedReader(new InputStreamReader(System.in));
        final var cout = new BufferedOutputStream(System.out);

        var e3 = new int[MAX_RADIUS + 1];
        for (int i = 0; i <= MAX_RADIUS; i++) {
            e3[i] = (int) Math.pow(3, i);
        }

        var t = nextLine(cin).nextInt();
        while (t-- > 0) {
            var line = nextLine(cin);
            var n = line.nextInt();
            var m = line.nextInt();
            var k = line.nextInt();

            var paths = new ArrayList<Path>();

            for (int i = 0; i < n; i++) {
                var s = nextLine(cin).nextStr();
                for (int j = 0; j < s.length(); j++) {
                    if (s.charAt(j) == '#') {
                        paths.add(new Path(i + 1, j + 1));
                    }
                }
            }

            var towers = new ArrayList<Tower>();

            for (int i = 0; i < k; i++) {
                line = nextLine(cin);
                towers.add(new Tower(line.nextInt(), line.nextInt(), line.nextInt()));
            }

            if (k > MAX_RADIUS) {
                cout.write(to_bytes(0));
                cout.write(endl);
                continue;
            }

            for (Tower tower : towers) {
                var distances = paths.stream().map(path -> square(path.x - tower.x) + square(path.y - tower.y)).toList();
                for (int i = 1; i <= MAX_RADIUS; i++) {
                    int finalI = i;
                    var count = distances.stream().filter(d -> d <= square(finalI)).toList().size();
                    tower.potency[i - 1] = count * tower.p - e3[i];
                }
            }

            var ans = 0;
            int[][] dp = new int[MAX_RADIUS + 1][1 << MAX_RADIUS];
            for (int i = 0; i <= MAX_RADIUS; i++) {
                for (int j = 0; j < 1 << MAX_RADIUS; j++) {
                    dp[i][j] = Integer.MIN_VALUE;
                }
            }

            for (int i = 0; i < 1 << MAX_RADIUS; i++) {
                var count = countBits(i);
                if (count == k)
                    ans = max(ans, solve(dp, towers, k, i));
            }

            cout.write(to_bytes(ans));
            cout.write(endl);
        }

        cout.flush();
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The provided code has a potential issue with the dynamic programming (DP) approach and the way it calculates the potency of towers. Specifically, it might not correctly handle the case where the number of towers is less than the maximum radius, leading to incorrect results.

2. **Edge Cases**:
   - **Few Towers**: Test with a very small number of towers (e.g., 1 or 2) to see if the code can handle such cases.
   - **High Damage Towers**: Test with towers that have very high damage values to see if the code can correctly calculate the maximum health.
   - **Multiple Paths**: Test with a grid that has multiple paths to ensure the code correctly calculates the damage for each path.

3. **Small-Scale Test Input**: Generate a small-scale test input that is complete and valid, focusing on the edge cases mentioned above.

Here is a small-scale test input that focuses on a few towers with high damage values:

```plaintext
1
3 3 2
#..
##.
.##
1 2 500
3 1 400
```

### Explanation:
- **Grid Size**: 3x3
- **Number of Towers**: 2
- **Path**: The path is laid out from (1,1) to (3,3) in a way that covers the grid.
- **Towers**:
  - Tower 1: (1,2) with 500 damage
  - Tower 2: (3,1) with 400 damage

This test input is designed to:
- **Check the Potency Calculation**: Ensure the code correctly calculates the potency of each tower for different ranges.
- **Check the DP Approach**: Ensure the dynamic programming approach correctly handles the small number of towers and calculates the maximum health correctly.

By running this test input, we can see if the code correctly handles the high damage values and the path layout. If the code fails, it will likely be due to an issue in the potency calculation or the DP approach.