**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>
#define ll long long
#define endl '\n'
#define loop(i,s,e) for(ll i = s; s<e ? i < e : i >= e; s<e ? i++ : i--)
#define forrr(i,s,e) for(ll i = s; i >= e; i--)
#define forr(i,s,e) for(ll i = s; i < e; i++)
#define vl vector<ll>
#define vvl vector<vl>
#define pll pair<ll, ll>
#define vp vector<pll>
#define mp make_pair
#define ss second
#define ff first
#define pb push_back
#define pf push_front
#define all(a) a.begin(), a.end()
#define Rand(arr, n) generate_n(arr.begin(), n, random)
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif
using namespace std;

template<class T> void _print(T t) {cerr << t;}
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(priority_queue<T> pq) {cerr << "[ "; vector<T> v; while(!pq.empty()){ v.push_back(pq.top()); pq.pop();} for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(priority_queue<T, vector<T>, greater<T>> pq) {cerr << "[ "; vector<T> v; while(!pq.empty()){ v.push_back(pq.top()); pq.pop();} for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> st) {cerr << "[ "; for (auto it = st.begin(); it != st.end(); it++) {_print(*it); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> st) {cerr << "[ "; for (auto it = st.begin(); it != st.end(); it++) {_print(*it); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> mp) {cerr << "[ "; for (auto it = mp.begin(); it != mp.end(); it++) {_print(*it); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(multimap <T, V> mp) {cerr << "[ "; for (auto it = mp.begin(); it != mp.end(); it++) {_print(*it); cerr << " ";} cerr << "]";}

template<typename F, typename S> ostream& operator <<(ostream& ostream, pair<F, S>& p) { cout << p.first << " " << p.second<<" "; return ostream; }
template<typename T> ostream& operator <<(ostream& ostream, vector<T>& v) { for(auto& element : v) cout << element << " "; return ostream;}
template<typename T> ostream& operator <<(ostream& ostream, vector<vector<T>>& v) {for(auto& row : v){ for(auto& cell : row) cout << cell << " "; cout << "\n";} return ostream;}
template<typename F, typename S> istream& operator >>(istream& istream, pair<F, S>& p) { cin >> p.first >> p.second; return istream;}
template<typename T> istream& operator >>(istream& istream, vector<T>& v) { for(auto& element : v) cin >> element; return istream; }

int random(int s, int e){return s + rand() % (e - s + 1);}
int flip(int n){ forr(i,0,31){ int num = 1<<i; if(num > n) break; n = n xor num;} return n; }
ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
vp countArr(vl &arr, ll n){ vp v1; forr(i,0,n-1){ ll cnt = 1; while(i < n-1 && arr[i] == arr[i+1]){ cnt++, i++; } v1.pb({arr[i], cnt}); } if((n > 1 && arr[n-1] != arr[n-2]) || n == 1) v1.pb({arr[n-1], 1}); return v1;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
vl factors(ll n){ vl fac; for (ll i = 1; i * i <= n; i++){ if (n % i == 0){ fac.pb(i); if (i * i != n)fac.pb(n / i); } } return fac; }
ll maxPow2(ll n){n |= n >> 1; n |= n >> 2; n |= n >> 4; n |= n >> 8; n |= n >> 16; return (n + 1);}
vp primeFactors(ll n){ vp v; for (ll j = 2; j <= sqrtl(n); j++){ ll cnt = 0; while(n%j == 0) n /= j, cnt++; if(cnt)v.pb({j,cnt}); } if(n!=1) v.pb({n,1}); return v;}

// ll dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
// ll dy[] = {1, -1, 1, -1, 0, 1, -1, 0};
const ll R = 11+1;
ll pow3[R];
vp dmg[R];
ll n,m,k,maxD;
ll x,y,p,a,b,c,cnt;
ll picked[2501];

void rec(ll r, ll power){
    if(!r){
        maxD = max(maxD, power);
        return;
    }

    rec(r-1, power);
    for(auto &[p, id] : dmg[r]){
        if(!picked[id]){
            picked[id] = 1;
            rec(r-1, power + p);
            picked[id] = 0;
        }
    }
}

void solve(ll tc){
    cin>>n>>m>>k;
    vector<string> arr(n); cin>>arr;
    forr(r,1,R) dmg[r].clear();

    forr(id,0,k){
        cin>>x>>y>>p; x--, y--;
        forr(r,1,R){
            cnt = 0;
            forr(i,max(x-r,0LL),min(x+r+1,n)){
                forr(j,max(y-r,0LL),min(y+r+1,m)){
                    a = abs(x-i)*abs(x-i), b = abs(y-j)*abs(y-j), c = r*r;
                    if(a + b <= c && arr[i][j] == '#') cnt++;
                }
            }
            if(cnt*p - pow3[r] > 0) dmg[r].pb({cnt*p - pow3[r], id});
        }
    }
    forr(r,1,R){
        sort(all(dmg[r]), greater<pll>());
        while(dmg[r].size() > R-1) dmg[r].pop_back();
    }

    maxD = 0;
    rec(R-1, 0);
    cout<<maxD<<endl;
}

int main(){
fastio(); srand(time(NULL));
pow3[0] = 1;
forr(i,1,R) pow3[i] = pow3[i-1]*3;
ll t = 1;
cin>>t;
forr(i,1,t+1) solve(i);
return 0;
}
/* use __lg(number) to get nearest power of 2 -> 8,9,10..15 returns 3, 16-31 returns 4*/

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code is not immediately obvious, but it could be related to how the ranges are selected and the damage calculation. Specifically, the code might not handle the optimal selection of ranges efficiently, especially when multiple towers can cover the same cells.

2. **Edge Cases**:
   - **Single Path Cell**: If there is only one path cell, the damage calculation should be straightforward.
   - **Multiple Towers Covering the Same Cell**: This can lead to overestimating the damage if not handled correctly.
   - **Towers with High Power but Low Range**: These towers might not cover enough cells to be effective.
   - **Towers with Low Power but High Range**: These towers might cover many cells but not deal enough damage per cell.

3. **Challenging Scenario**:
   - **Grid with a Single Path**: This simplifies the problem to a single path, making it easier to focus on the tower ranges and their effectiveness.
   - **Towers with Different Ranges and Powers**: Ensure that some towers have high power but low range, and some have low power but high range to test the range selection logic.

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

```plaintext
1
5 5 4
#....
#....
#....
#....
#....
2 2 1
2 4 500
4 2 500
4 4 1
```

### Explanation:
- **Grid Size**: 5x5
- **Path**: The path is a straight line from the top-left to the bottom-right.
- **Towers**:
  - Tower at (2, 2) with power 1.
  - Tower at (2, 4) with power 500.
  - Tower at (4, 2) with power 500.
  - Tower at (4, 4) with power 1.

### Why This Input is Challenging:
- **Single Path**: The path is straightforward, making it easier to verify the damage calculation.
- **Towers with Different Ranges and Powers**: The towers have a mix of high power and low range, and low power and high range, which can test the range selection logic.
- **High Power Towers**: Towers at (2, 4) and (4, 2) have high power, which can significantly affect the damage calculation.
- **Low Power Towers**: Towers at (2, 2) and (4, 4) have low power, which should be less effective but still need to be considered.

This input should help expose any issues in the range selection and damage calculation logic of the buggy code.


**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 <vector>
#include <iostream>
#include <algorithm>
#include <random>
#include <chrono>
#include <string>
#include <set>
#include <map>

#define all(x) (x).begin(), (x).end()
#define cyes cout << "YES\n"
#define cno cout << "NO\n"
#define mat(type, n, m, name, init) vector<vector<type>> name(n, vector<type>(m, init))
#define pb push_back
#define gen(x) generate(all(x), nxt)
#define MIN(a, other) a = min(a, other)
#define MAX(a, other) a = max(a, other)
#define int long long
#define uint uint64_t
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int nxt() {
    int v;
    cin >> v;
    return v;
}

struct tower {
    int x;
    int y;
    int p;
    vector<int> R = vector<int>(13);
    vector<int> A = vector<int>(13);
};

int my(int n) {
    int l = 0, r = n + 5;
    while (l < r) {
        int mid = (l + r) / 2;
        if (mid * mid < n) {
            l = mid + 1;
        } else {
            r = mid;
        }
    }
    return r;
}

void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    mat(char, n, m, a, '.');
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cin >> a[i][j];
        }
    }
    vector<tower> t(k);
    for (auto &v: t) cin >> v.x >> v.y >> v.p;
    for (auto &v: t) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (a[i - 1][j - 1] == '.') continue;
                int dx = abs(i - v.x);
                int dy = abs(j - v.y);
                int dist = dx * dx + dy * dy;
                int minR = my(dist);
                v.R[minR]++;
            }
        }
    }
    for (int i = 1; i < 13; ++i) {
        for (auto &v: t) {
            v.R[i] += v.R[i - 1];
            int dmgComp = v.R[i] * v.p - (int) pow(3, i);
            v.A[i] = dmgComp;
        }
    }
    int ans = 0;
    mat(int, t.size(), 8192, dp, 0);
    for (int i = 0; i < t.size(); ++i) {
        for (int j = 0; j < 8192; ++j) {
            for (int l = 0; l < t[i].A.size(); ++l) {
                if (!(j >> l & 1)) {
                    if (i + 1 < t.size()) {
                        MAX(dp[i + 1][j | (1 << l)], dp[i][j] + t[i].A[l]);
                    } else {
                        MAX(ans, dp[i][j] + t[i].A[l]);
                    }
                }
                if (i + 1 < t.size()) {
                    MAX(dp[i + 1][j], dp[i][j]);
                } else {
                    MAX(ans, dp[i][j]);
                }
            }
            MAX(ans, dp[i][j]);
        }
    }
    cout << ans << '\n';
}

int32_t main() {
    cin.tie(nullptr)->ios_base::sync_with_stdio(false);
    setvbuf(stdout, nullptr, _IOFBF, BUFSIZ);
    int tt = nxt();
    while (tt--) {
        solve();
    }
}

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

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

Think step by step.

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


**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 = 13
PATH = "#"
def main():
    R, C, K = map(int, input().split())
    grid = [list(input()) for _ in range(R)]
    towers = [None] * K
    # ceiling radius
    UB = BITS * BITS + 5
    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] = 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):
            ndp[mask] = max(ndp[mask], dp[mask])
            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))
        dp = ndp
    print(max(dp))

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:
```plaintext
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
```


**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
input = sys.stdin.readline
INF = 1 << 31
MAX_R = 14

def solve():
    n, m, k = map(int, input().split())
    grid = [input().rstrip() for _ in range(n)]

    dp = [-INF] * (1 << MAX_R)
    dp[0] = 0

    for _ in range(k):
        x, y, p = map(int, input().split())
        x -= 1
        y -= 1

        # compute damage per radius
        dmg = [0] * MAX_R
        for i in range(n):
            for j in range(m):
                if grid[i][j] != "#":
                    continue

                d = (i - x) * (i - x) + (j - y) * (j - y)
                r = 1
                while r * r < d:
                    r += 1
                dmg[r] += p

        for r in range(1, MAX_R):
            dmg[r] += dmg[r - 1]

        # compute dp[msk]: max {f(msk) := dmg(msk) - extra_health(msk)} by assigning radius to towers
        for msk in range((1 << MAX_R) - 1, -1, -1):
            if dp[msk] == -INF:
                continue

            for r in range(MAX_R):
                if msk >> r & 1 == 0:
                    nmsk = msk | (1 << r)
                    dp[nmsk] = max(dp[nmsk], dp[msk] + dmg[r] - pow(3, r))

    ans = max(dp)
    return ans

T = int(input())
out = [solve() for _ in range(T)]
print("\n".join(map(str, out)))

**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 1 50
2 1 50
3 1 50
```


**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][8192];
			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 provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Identify the Bug**: The bug in the provided code seems to be related to the calculation of the maximum health an enemy can have before it is defeated. Specifically, the code might not correctly handle the scenario where the enemy's health increase due to tower ranges is not properly accounted for.

2. **Edge Cases**:
   - **Single Tower with High Range and High Damage**: This can test if the code correctly calculates the maximum health when a single tower can cover all path cells.
   - **Multiple Towers with Overlapping Ranges**: This can test if the code correctly handles the situation where multiple towers can cover the same path cells.
   - **Towers with No Range**: This can test if the code correctly handles the scenario where no towers have a range set.
   - **Path with Multiple Cells**: This can test if the code correctly calculates the damage over a longer path.

3. **Small-Scale Test Input**:
   - We will create a small grid with a few towers and a simple path to ensure the input is valid and complete.
   - We will use a grid size of 4x4 and 3 towers to keep the input small but still challenging.

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

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

### Explanation:
- **Grid Size**: 4x4
- **Number of Towers**: 3
- **Path**: The path is from (1,1) to (4,4), passing through the cells: (1,1), (1,2), (1,3), (2,3), (3,3), (4,3), (4,4).
- **Towers**:
  - Tower 1 at (1,1) with attack parameter 1
  - Tower 2 at (2,2) with attack parameter 2
  - Tower 3 at (3,3) with attack parameter 3

This input should challenge the code to correctly calculate the maximum base health of the enemy, considering the ranges and the damage each tower can deal. The towers are placed in such a way that they can cover significant portions of the path, and the path itself is long enough to require careful calculation of the enemy's 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 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 < 9; ++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 provided code, let's consider the following aspects:

1. **Path Layout**: The path should be designed in a way that maximizes the interaction between the towers and the path. This can be achieved by creating a path that passes through many cells close to the towers.

2. **Tower Placement**: Towers should be placed in such a way that their ranges can potentially cover a significant portion of the path, but the damage they deal is just enough to be critical.

3. **Tower Damage**: The damage values of the towers should be chosen to create a tight balance between the enemy's health and the total damage dealt.

4. **Health Calculation**: The base health of the enemy should be just enough to be critical, meaning the enemy should be on the verge of being killed by the towers.

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

```plaintext
1
5 5 3
#....
#....
#....
#....
#####
2 2 2
3 3 3
4 4 4
```

### Explanation:

- **Grid Size**: 5x5
- **Number of Towers**: 3
- **Path**: The path goes from (1,1) to (5,5) in a straight line, passing through cells (1,1), (2,1), (3,1), (4,1), (5,1), (5,2), (5,3), (5,4), and (5,5).
- **Towers**:
  - Tower 1 at (2,2) with damage 2
  - Tower 2 at (3,3) with damage 3
  - Tower 3 at (4,4) with damage 4

### Why This Input is Challenging:

1. **Path Coverage**: The path is a straight line, making it easy to calculate the range of the towers and their impact on the enemy.
2. **Tower Ranges**: The towers are placed such that they can potentially cover multiple cells of the path, but their ranges need to be carefully chosen to ensure the enemy is just barely killed.
3. **Damage Values**: The damage values are chosen to create a balance where the enemy's health is just enough to be critical.

This input should help expose any issues in the code related to the calculation of the maximum base health and the interaction between the towers and the path.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
/* In the Name of ALLAH, the most gracious, the most merciful  */
#include <bits/stdc++.h>
using namespace std;
#define nl endl
#define ll long long
#define cy cout << "YES" << nl
#define cn cout << "NO" << nl
#define all(v) (v).begin(), (v).end()
#define si(v) int((v).size())
#define pb push_back
const int inf = 1e9;
const ll INF = 1e18;
const int mod = 998244353;
const int N = 2e5 + 5;
/* Formula X:
 * Sum of 'i to j' = ((j-i+1)*(i+j))/2;
 * LCM = (a*b)/__gcd(a,b);
 *
 *  --*/
int check(int i, string &s, int j)
{
    if (i >= 0 && i + 3 < j && s[i] == '1' && s[i + 1] == '1' && s[i + 2] == '0' && s[i + 3] == '0')
        return 1;
    return 0;
}
void solve()
{
    vector<pair<ll, ll>> v;
    string s;
    cin >> s;
    int a;
    cin >> a;
    if (si(s) < 4)
    {
        for (int i = 0; i < a; i++)
        {
            int f, j;
            cin >> f >> j;
            cn;
        }
        return;
    }
    ll cnt = 0;
    for (ll i = 0; i < si(s) - 3; i++)
    {
        if (s[i] == '1' && s[i + 1] == '1' && s[i + 2] == '0' && s[i + 3] == '0')
        {
            v.push_back({i, i + 3});
            cnt++;
            i += 3;
        }
    }
    // cout<<"cnt= "<<cnt<<nl;

    for (int i = 0; i < a; i++)
    {
        int f, j;
        cin >> f >> j;
        f--;
        ll c = cnt;
        for (ll i = max(0, f - 3); i < min(si(s) - 3, f); i++)
        {
            if (check(i, s, si(s)))
                cnt--;
            // else {
            //     if(cnt>0)cnt--;
            //     else cnt=0;
            // }
        }

        s[f] = j + '0';

        for (ll i = max(0, f - 3); i < min(si(s) - 3, f); i++)
        {
            if (check(i, s, si(s)))
                cnt++;
            // else {
            //     if(cnt>0)cnt--;
            //     else cnt=0;
            // }
        }

        if (cnt > 0)
            cy;
        else
            cn;
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++)
    {
        // cout << "Case " << i << ": ";
        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 provided buggy code, we need to consider scenarios where the code might fail. The code is designed to check for the presence of the substring "1100" after each query. Here are some key points to consider:

1. **Initial String with "1100"**: Start with a string that already contains "1100" and see if the code correctly identifies it.
2. **Modifying "1100"**: Perform queries that modify the "1100" substring to see if the code correctly updates its presence.
3. **Edge Cases**: Consider edge cases where the "1100" substring is at the start or end of the string.
4. **Multiple "1100" Substrings**: Test with a string that contains multiple "1100" substrings to ensure the code handles all occurrences correctly.

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

```plaintext
1
11001100
5
1 0
2 0
3 1
4 1
5 0
```

### Explanation:
1. **Initial String**: "11001100" contains two "1100" substrings.
2. **Queries**:
   - **Query 1**: `1 0` - This changes the string to "01001100". The first "1100" is no longer present, but the second one is. The code should output "NO".
   - **Query 2**: `2 0` - This changes the string to "00001100". Neither "1100" substring is present. The code should output "NO".
   - **Query 3**: `3 1` - This changes the string to "00101100". The second "1100" is no longer present, but the first one is. The code should output "NO".
   - **Query 4**: `4 1` - This changes the string to "00111100". The second "1100" is no longer present, but the first one is. The code should output "NO".
   - **Query 5**: `5 0` - This changes the string to "00110100". The second "1100" is no longer present, but the first one is. The code should output "NO".

This test input is designed to challenge the code's ability to correctly update the presence of the "1100" substring after each query, especially when multiple "1100" substrings are involved.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

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

using namespace std;

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

    while(t --> 0) {
        string s; cin >> s;
        int q; cin >> q;

        s = "0" + s;

        int i;
        char c;

        unordered_map<int, int> mp;

        int pos = 1;

        while(pos + 3 < s.length() && s.find("1100", pos) != string::npos)
        {
            mp[pos]++;
            pos = int(s.find("1100", pos)) + 4;
        }

        while(q --> 0)
        {
            cin >> i >> c;

            if(s.length() < 5)
            {
                cout << "NO\n";
                continue;
            }

            if(c == '1' && s[i] != c)
            {
                s[i] = c;

                if(s.substr(i, 4) == "1100")
                    mp[i]++;
                if(i > 1 && s.substr(i - 1, 4) == "1100")
                    mp[i - 1]++;

                mp.erase(i - 2);
                mp.erase(i - 3);
            } else if(s[i] != c)
            {
                s[i] = c;

                if(i >= 4 && s.substr(i - 3, 4) == "1100")
                    mp[i - 3]++;
                if(i >= 3 && s.substr(i - 2, 4) == "1100")
                    mp[i - 2]++;

                mp.erase(i);
                mp.erase(i - 1);
            }

            cout << (!mp.empty() ? "YES\n" : "NO\n");
        }
    }
};

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

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

Think step by step.

Response:
```plaintext
1
1100
4
1 0
2 0
3 0
4 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
void cal(set<int> &st,string s,int i){
	if(s.substr(i,4).compare("1100") == 0){
		st.insert(i);
	}
	else{
		st.erase(i);
	}
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin>>t;
	while(t--){
		int q;
		string s;
		cin>>s>>q;
		int n=s.length();
		set<int> st;
		for(int i=0;i<n-4;i++){
			if(s.substr(i,4).compare("1100") == 0){
				st.insert(i);
			}
		}
		while(q--){
			int i;
			char c;
			cin>>i>>c;
			if(n>=4){
				if(s[i-1]!=c){
					s[i-1]=c;
					for(int j=4;j>=1;j--){
						if(i-j>=0&&i+3-j<n){
							cal(st,s,i-j);
						}
					}
				}
				if(st.size()){
					cout<<"YES";
				}
				else{
					cout<<"NO";
				}
			}
			else{
				cout<<"NO";
			}
			cout<<'\n';
		}

	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
11111111111111111111
5
1 0
2 0
3 1
4 1
5 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
t = int(input())
ans=[]

def dfuc(a,si,ei):
    n=len(a)
    si = max(0, si)
    ei = min(n - 1, ei)
    for i in range(si,ei-2):
        pc=""
        pc=pc+a[i]+a[i+1]+a[i+2]+a[i+3]
        #print(f"this is pc {pc}")
        if pc=="1100":
            #print("secus")
            return i
    return -1
def fuc(a,q,i,v):
    n=len(a)
    pres=0
    for j in range(n-3):
        if dfuc(a,j,j+3)>=0:
            pres+=1

    ans=[]
    for j in range(q):
        ocha=a[i[j]]
        #print(f"a is {a}")
        if ocha ==v[j]:
            pres=pres
        else:
            if dfuc(a,j,j+3) >=0:
                pres-=1
            if dfuc(a,j-1,j+2) >=0:
                pres-=1
            if dfuc(a,j-2,j+1) >=0:
                pres-=1
            if dfuc(a,j-3,j) >=0:
                pres-=1
            a[i[j]]=v[j]
            if dfuc(a,j,j+3) >=0:
                pres+=1
            if dfuc(a,j-1,j+2) >=0:
                pres+=1
            if dfuc(a,j-2,j+1) >=0:
                pres+=1
            if dfuc(a,j-3,j) >=0:
                pres+=1

        if pres>0:
            ans.append(1)
        else:
            ans.append(0)

    return ans

for _ in range(t):
    s=input()
    q=int(input())
    i=[]
    v=[]
    for j in range(q):
        ii,vv=input().split()
        i.append(int(ii)-1)
        v.append(vv)
    ans.append(fuc(list(s),q,i,v))
for e in ans:
    for k in e:
        if k==1:
            print("YES")
        else:
            print("NO")

**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
1100
4
1 0
2 1
3 0
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    arr = [int(x) for x in input()]
    n = int(input())
    beautyInd = []
    arrLength = len(arr)

    for i in range(len(arr) - 3):
        if arr[i] == 1 and arr[i+1] == 1 and arr[i+2] == 0 and arr[i+3] == 0:
            beautyInd.append(i)

    for i in range(n):
        index, val = map(int, input().split())
        val = val
        index -= 1
        if arr[index] != val:
            if index > 2:
                if val == 0 and arr[index - 1] == 0 and arr[index - 2] == 1 and arr[index - 3] == 1:
                    beautyInd.append(index - 3)
                else:
                    if arr[index] == 0 and arr[index - 1] == 0 and arr[index - 2] == 1 and arr[index - 3] == 1:
                        beautyInd.remove(index - 3)

            else:
                if arrLength > index + 3:
                    if arr[index] == val and arr[index + 1] == 0 and arr[index + 2] == 1 and arr[index + 3] == 1:
                        beautyInd.append(index)
                    else:
                        if arr[index] == 1 and arr[index + 1] == 1 and arr[index + 2] == 0 and arr[index + 3] == 0:
                            beautyInd.remove(index)
            arr[index] = val

        if beautyInd:
            print('YES')
        else:
            print('NO')

**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
11001100
4
1 0
2 0
3 0
4 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
def split2(string):
    str1=""
    arr=[]
    for i in string:
        if i!=" ":
            str1=str1+i
        else:
            arr.append(int(str1))
            str1=""
    arr.append(int(str1))
    return arr
def convert(stringg):
    arr=[]
    for s in stringg:
        arr.append(s)
    return(arr)
n=int(input())
for i in range(n):
    string=list(input())
    q=int(input())
    for x in range(q):
        line=split2(input())
        string[line[0]-1]=str(line[1])
        flag=False
        if len(string)>=4:
            for z in range(len(string)-4):
                if string[z:z+4]==['1', '1', '0', '0'] or string[len(string)-z-4:len(string)-z]==['1', '1', '0', '0']:
                    flag=True
                    break
        if flag:
            print("YES")
        else:
            print("NO")

**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
1100
4
1 0
2 0
3 0
4 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Anya and 1100
time_limit_ms: 3000
memory_limit_mb: 256
Description: While rummaging through things in a distant drawer, Anya found a beautiful string $s$ consisting only of zeros and ones.

Now she wants to make it even more beautiful by performing $q$ operations on it.

Each operation is described by two integers $i$ ($1 \le i \le |s|$) and $v$ ($v \in \\{0, 1\\}$) and means that the $i$-th character of the string is assigned the value $v$ (that is, the assignment $s_i = v$ is performed).

But Anya loves the number $1100$, so after each query, she asks you to tell her whether the substring "1100" is present in her string (i.e. there exist such $1 \le i \le |s| - 3$ that $s_{i}s_{i + 1}s_{i + 2}s_{i + 3} = \texttt{1100}$).

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

The first line of the test case contains the string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$), consisting only of the characters "0" and "1". Here $|s|$ denotes the length of the string $s$.

The next line contains an integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.

The following $q$ lines contain two integers $i$ ($1 \leq i \leq |s|$) and $v$ ($v \in \\{0, 1\\}$), describing the query.

It is guaranteed that the sum of $|s|$ across all test cases does not exceed $2 \cdot 10^5$. It is also guaranteed that the sum of $q$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each query, output "YES", if "1100" is present in Anya's string; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
4
100
4
1 1
2 0
2 0
3 1
1100000
3
6 1
7 1
4 1
111010
4
1 1
5 0
4 1
5 0
0100
4
3 1
1 1
2 0
2 1
output:
NO
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
NO
NO
NO
NO

Note:

Buggy Code:
import java.util.*;
import java.io.*;
public class akarsh
{
    public static boolean even(int a)//even odd check
    {
        if((a&1)==0)
        return true;
        return false;
    }
    public static void printarray(int[] a)//printing arrray
    {
        for(int k : a)
        System.out.print(a+" ");
        System.out.println();
    }
    public static int[] prefix(int[] a)//prefix sum
    {
        int n = a.length;
        int[] ans = new int[n];
        int sum = 0;
        for(int i=0;i<n;i++)
        {
            sum += a[i];
            ans[i] = sum;
        }
        return ans;
    }
    public static void steve()//steve for prime
    {
        int k = 100000+1;
        boolean[] prime = new boolean[k];
        Arrays.fill(prime,true);
        prime[0] = false;
        prime[1] = false;
        for(int i=2;i*i<k;i++)
        {
            if(prime[i]==true)
            {
                for(int j=i*i;j<k;j+=i)
                prime[j] = false;
            }
        }
    }
    public static void main(String args[])//main functions
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while((t--)>0)
        {
          String s = sc.next();
          int n = s.length();
          int q = sc.nextInt();
          if(n<=3)
          {
            for(int i=0;i<q;i++)
            {
              int ind = sc.nextInt();
              int c = sc.nextInt();
              System.out.println("NO");
            }
            continue;
          }
          Map<String,Set<Integer>> hs = new HashMap<>();
          Map<Integer,String> ha = new HashMap<>();
          String h = ""+s.charAt(0)+s.charAt(1)+s.charAt(2)+s.charAt(3);
          int count = 0;
          if(!hs.containsKey(h))
          hs.put(h,new HashSet<>());
          hs.get(h).add(0);
          ha.put(0,h);
          count++;
          for(int i=1;i<n-3;i++)
          {
            h = h.substring(1)+s.charAt(i+3);
            if(!hs.containsKey(h))
            hs.put(h,new HashSet<>());
            hs.get(h).add(count);
            ha.put(count,h);
            count++;
          }
         // System.out.println(hs);
          //System.out.println(ha);
          for(int i=0;i<q;i++)
          {
            int ind = sc.nextInt()-1;
            int c = sc.nextInt();
            int mm = 3;
            for(int j=ind-3;j>=0&&j<n-3&&j<=ind;j++)
            {
              String old = ha.get(j);
              char[] u = old.toCharArray();
              if(c==1)
              u[mm] = '1';
              else
              u[mm] = '0';
              String news = new String(u);
              ha.put(j,news);
              hs.get(old).remove(j);
              if(!hs.containsKey(news))
              hs.put(news,new HashSet<>());
              hs.get(news).add(j);
              if(hs.get(old).size()==0)
              hs.remove(old);
              mm--;
            }
           // System.out.println(hs);
            //System.out.println(ha);
            if(hs.containsKey("1100")&&hs.get("1100").size()>0)
            System.out.println("YES");
            else
            System.out.println("NO");
          }
        }
    }
    public static boolean checkprime(int N)
    {
        int i;
        if(N==1)
        return false;
        if((N&1)==0&&N!=2)
        return false ;
        else if(N%3==0&&N!=3)
        return false ;
        else if(N%11==0&&N!=11)
        return false ;
        else if(N%13==0&&N!=13)
        return false ;
        else if(N%17==0&&N!=17)
        return false ;
        else
        {
            for(i=3;i<=Math.sqrt(N);i+=2)
            {
                if(N%i==0)
                return false ;
            }
        }
        return true ;
    }
    public static boolean checksquare(int n)
    {
        int sqrt = (int)(Math.sqrt(n));
        if((sqrt*sqrt)==n)
        return true;
        return false;
    }
    public static boolean checkfactor(int n,int x)
    {
        for(int i=2;i<=n;i++)
        {
            if(n%i==0&&i<x)
            return false;
        }
        return true;
    }
}
class pair//for pairs
{
    int a, b;
    pair(int a,int b)
    {
        this.a = a;
        this.b = b;
    }
}
class DisjointSet
{
  int[] rank;
  int[] parent;
  public DisjointSet(int n)
  {
    rank = new int[n+1];
    parent = new int[n+1];
    for(int i=0;i<=n;i++)
    {
      rank[i] = 0;
      parent[i] = i;
    }
  }
  public int findUpar(int node)
  {
    if(node==parent[node])
    return node;
    int ulp = findUpar(parent[node]);
    parent[node] = ulp;
    return parent[node];
  }
  public void UnionByRank(int u,int v)
  {
    int ulp = findUpar(u), ulv = findUpar(v);
    if(ulp==ulv)
    return;
    else if(rank[ulp]<rank[ulv])
    parent[ulp] = ulv;
    else if(rank[ulp]>rank[ulv])
    parent[ulv] = ulp;
    else
    {
      parent[ulv] = ulp;
      int ranku = rank[ulp];
      rank[ulp] = ranku+1;
    }
  }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Strings of length 4 or less, where "1100" cannot be formed.
   - Strings that are just one character away from forming "1100".
   - Strings that already contain "1100".

2. **Queries**:
   - Queries that change a single character to form "1100".
   - Queries that change a single character to break "1100".
   - Queries that do not affect the formation or breaking of "1100".

3. **Multiple Test Cases**:
   - Include multiple test cases to ensure the code handles different scenarios.

Let's construct a test input that covers these scenarios:

```plaintext
3
1101
4
4 0
2 0
3 0
1 1
1111
4
1 0
2 0
3 0
4 0
1000
4
1 1
2 1
3 1
4 0
```

### Explanation:
1. **First Test Case**:
   - Initial string: `1101`
   - Queries:
     - `4 0` (changes `1101` to `1100` -> "YES")
     - `2 0` (changes `1100` to `1000` -> "NO")
     - `3 0` (changes `1000` to `1000` -> "NO")
     - `1 1` (changes `1000` to `1000` -> "NO")

2. **Second Test Case**:
   - Initial string: `1111`
   - Queries:
     - `1 0` (changes `1111` to `0111` -> "NO")
     - `2 0` (changes `0111` to `0011` -> "NO")
     - `3 0` (changes `0011` to `0001` -> "NO")
     - `4 0` (changes `0001` to `0000` -> "NO")

3. **Third Test Case**:
   - Initial string: `1000`
   - Queries:
     - `1 1` (changes `1000` to `1000` -> "NO")
     - `2 1` (changes `1000` to `1100` -> "YES")
     - `3 1` (changes `1100` to `1110` -> "NO")
     - `4 0` (changes `1110` to `1110` -> "NO")

This test input covers various edge cases and ensures that the code is tested for its ability to correctly handle the formation and breaking of the substring "1100".