**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

Buggy Code:
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define int long long
#define lowbit(x)  ((x) & - (x))
using PII = pair<int, int>;
using PIII = tuple<int, int, int>;
const int N = 1e5 + 10;
const int MOD = 5e18;

inline int mypow(int n, int k, int p = MOD) {
    int r = 1;
    for (; k; k >>= 1, n = n * n % p) {
        if (k & 1) r = r * n % p;
    }
    return r;
}

int inv(int x) { return mypow(x, MOD - 2, MOD); }

vector<int> w[200010];
int cnt[200010] = { 0 };
int a[N], b[N];
int n, m, q;
int suma = 0, sumb = 0;
map<int, int> mpa;
map<int, int> mpb;

inline void solve() {
    int v;
    cin >> v;
    if (v == 0) {
        if (mpa[0] != 0 || mpb[0] != 0) {
            cout << "YES" << endl;
            return;
        }
    }
    else {
        int w = abs(v);
        for (int i = 1; i * i <= w; i++) {
            if (v % i == 0) {
                if ((mpa[i] != 0 && mpb[v / i] != 0) ||
                    (mpb[i] != 0 && mpa[v / i] != 0) ||
                    (mpa[-i] != 0 && mpb[v / (-i)] != 0) ||
                    (mpb[-i] != 0 && mpa[v / (-i)] != 0)) {
                    cout << "YES" << endl;
                    return;
                }
            }
        }
    }
    cout << "NO" << endl;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //cout << fixed << setprecision(12);
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        suma += a[i];
    }
    for (int i = 1; i <= m; i++) {
        cin >> b[i];
        sumb += b[i];
    }

    for (int i = 1; i <= n; i++) {
        mpa[suma - a[i]]++;
    }
    for (int i = 1; i <= m; i++) {
        mpb[sumb - b[i]]++;
    }

    while (q--) {
        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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following points:

1. **Edge Cases**:
   - **Zero Values**: The code should handle cases where elements in arrays `a` and `b` are zero.
   - **Large Values**: The code should handle large values for `a` and `b` to ensure it doesn't overflow or fail due to large computations.
   - **Negative Values**: The code should handle negative values correctly.
   - ** Queries for Zero Beauty**: The code should handle queries where the desired beauty is zero.
   - ** Queries for Large and Small Beauty**: The code should handle queries for both large and small beauty values.

2. **Specific Bug Analysis**:
   - The code uses `map` to store the counts of `suma - a[i]` and `sumb - b[i]`. This can be problematic if the values in `a` and `b` are such that `suma - a[i]` or `sumb - b[i]` are not unique.
   - The code checks if `v` is zero by checking if there are any zeros in `a` or `b`. This is a potential bug if the grid can be made zero by setting a row and a column to zero without any zero elements in `a` or `b`.

3. **Test Input Generation**:
   - We will create a small-scale test input with `n = 3` and `m = 3` to keep the input manageable.
   - We will include zero values, negative values, and a mix of large and small values.
   - We will include queries for zero beauty, positive beauty, and negative beauty.

Here is the test input:

```plaintext
3 3 6
-1 0 2
-2 0 3
-1
0
1
2
-2
3
```

Explanation:
- `n = 3`, `m = 3`, `q = 6`
- `a = [-1, 0, 2]`
- `b = [-2, 0, 3]`
- Queries: `-1, 0, 1, 2, -2, 3`

This test input includes:
- Zero values in both `a` and `b`.
- Negative values in both `a` and `b`.
- A mix of large and small values.
- Queries for zero beauty, positive beauty, and negative beauty.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

Buggy Code:
/*#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("unroll-all-loops")
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,avx,avx2,popcnt,abm,mmx")*/

#include "bits/stdc++.h"
#define ll long long
#define endl '\n'
#define double long double
using namespace std;
const ll inf = 1000000000000000000;

const ll mxv = 200000;
ll n, m, t, x, sa, sb, q;
set<ll> seta, setb;
bool possible[mxv * 2 + 10];

int main() {
    #ifdef MAC
        freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    t = 1;
    for (ll o = 1; o <= t; ++o) {
        cin >> n >> m >> q;
        sa = sb = 0;
        for (ll i = 1; i <= n; ++i) {
            cin >> x;
            sa += x;
            seta.insert(x);
        }
        for (ll i = 1; i <= m; ++i) {
            cin >> x;
            sb += x;
            setb.insert(x);
        }
        for (const auto& elema : seta) {
            for (const auto& elemb : setb) {
                ll tmp = sa * sb - elema * sb - elemb * (sa - elema);
                if (tmp > mxv) {
                    break;
                }
                if (tmp >= -mxv && tmp <= mxv) {
                    possible[tmp + mxv] = true;
                }
            }
        }
        for (ll i = 1; i <= q; ++i) {
            cin >> x;
            cout << (possible[x + mxv] ? "YES" : "NO") << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 3 6
0 1 2
0 1 2
0
1
2
3
4
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

Buggy Code:
import sys
input = sys.stdin.readline

class SortedList:
    def __init__(self, iterable=[], _load=200):
        """Initialize sorted list instance."""
        values = sorted(iterable)
        self._len = _len = len(values)
        self._load = _load
        self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
        self._list_lens = [len(_list) for _list in _lists]
        self._mins = [_list[0] for _list in _lists]
        self._fen_tree = []
        self._rebuild = True

    def _fen_build(self):
        """Build a fenwick tree instance."""
        self._fen_tree[:] = self._list_lens
        _fen_tree = self._fen_tree
        for i in range(len(_fen_tree)):
            if i | i + 1 < len(_fen_tree):
                _fen_tree[i | i + 1] += _fen_tree[i]
        self._rebuild = False

    def _fen_update(self, index, value):
        """Update `fen_tree[index] += value`."""
        if not self._rebuild:
            _fen_tree = self._fen_tree
            while index < len(_fen_tree):
                _fen_tree[index] += value
                index |= index + 1

    def _fen_query(self, end):
        """Return `sum(_fen_tree[:end])`."""
        if self._rebuild:
            self._fen_build()

        _fen_tree = self._fen_tree
        x = 0
        while end:
            x += _fen_tree[end - 1]
            end &= end - 1
        return x

    def _fen_findkth(self, k):
        """Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
        _list_lens = self._list_lens
        if k < _list_lens[0]:
            return 0, k
        if k >= self._len - _list_lens[-1]:
            return len(_list_lens) - 1, k + _list_lens[-1] - self._len
        if self._rebuild:
            self._fen_build()

        _fen_tree = self._fen_tree
        idx = -1
        for d in reversed(range(len(_fen_tree).bit_length())):
            right_idx = idx + (1 << d)
            if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
                idx = right_idx
                k -= _fen_tree[idx]
        return idx + 1, k

    def _delete(self, pos, idx):
        """Delete value at the given `(pos, idx)`."""
        _lists = self._lists
        _mins = self._mins
        _list_lens = self._list_lens

        self._len -= 1
        self._fen_update(pos, -1)
        del _lists[pos][idx]
        _list_lens[pos] -= 1

        if _list_lens[pos]:
            _mins[pos] = _lists[pos][0]
        else:
            del _lists[pos]
            del _list_lens[pos]
            del _mins[pos]
            self._rebuild = True

    def _loc_left(self, value):
        """Return an index pair that corresponds to the first position of `value` in the sorted list."""
        if not self._len:
            return 0, 0

        _lists = self._lists
        _mins = self._mins

        lo, pos = -1, len(_lists) - 1
        while lo + 1 < pos:
            mi = (lo + pos) >> 1
            if value <= _mins[mi]:
                pos = mi
            else:
                lo = mi

        if pos and value <= _lists[pos - 1][-1]:
            pos -= 1

        _list = _lists[pos]
        lo, idx = -1, len(_list)
        while lo + 1 < idx:
            mi = (lo + idx) >> 1
            if value <= _list[mi]:
                idx = mi
            else:
                lo = mi

        return pos, idx

    def _loc_right(self, value):
        """Return an index pair that corresponds to the last position of `value` in the sorted list."""
        if not self._len:
            return 0, 0

        _lists = self._lists
        _mins = self._mins

        pos, hi = 0, len(_lists)
        while pos + 1 < hi:
            mi = (pos + hi) >> 1
            if value < _mins[mi]:
                hi = mi
            else:
                pos = mi

        _list = _lists[pos]
        lo, idx = -1, len(_list)
        while lo + 1 < idx:
            mi = (lo + idx) >> 1
            if value < _list[mi]:
                idx = mi
            else:
                lo = mi

        return pos, idx

    def add(self, value):
        """Add `value` to sorted list."""
        _load = self._load
        _lists = self._lists
        _mins = self._mins
        _list_lens = self._list_lens

        self._len += 1
        if _lists:
            pos, idx = self._loc_right(value)
            self._fen_update(pos, 1)
            _list = _lists[pos]
            _list.insert(idx, value)
            _list_lens[pos] += 1
            _mins[pos] = _list[0]
            if _load + _load < len(_list):
                _lists.insert(pos + 1, _list[_load:])
                _list_lens.insert(pos + 1, len(_list) - _load)
                _mins.insert(pos + 1, _list[_load])
                _list_lens[pos] = _load
                del _list[_load:]
                self._rebuild = True
        else:
            _lists.append([value])
            _mins.append(value)
            _list_lens.append(1)
            self._rebuild = True

    def discard(self, value):
        """Remove `value` from sorted list if it is a member."""
        _lists = self._lists
        if _lists:
            pos, idx = self._loc_right(value)
            if idx and _lists[pos][idx - 1] == value:
                self._delete(pos, idx - 1)

    def remove(self, value):
        """Remove `value` from sorted list; `value` must be a member."""
        _len = self._len
        self.discard(value)
        if _len == self._len:
            raise ValueError('{0!r} not in list'.format(value))

    def pop(self, index=-1):
        """Remove and return value at `index` in sorted list."""
        pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
        value = self._lists[pos][idx]
        self._delete(pos, idx)
        return value

    def bisect_left(self, value):
        """Return the first index to insert `value` in the sorted list."""
        pos, idx = self._loc_left(value)
        return self._fen_query(pos) + idx

    def bisect_right(self, value):
        """Return the last index to insert `value` in the sorted list."""
        pos, idx = self._loc_right(value)
        return self._fen_query(pos) + idx

    def count(self, value):
        """Return number of occurrences of `value` in the sorted list."""
        return self.bisect_right(value) - self.bisect_left(value)

    def __len__(self):
        """Return the size of the sorted list."""
        return self._len

    def __getitem__(self, index):
        """Lookup value at `index` in sorted list."""
        pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
        return self._lists[pos][idx]

    def __delitem__(self, index):
        """Remove value at `index` from sorted list."""
        pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
        self._delete(pos, idx)

    def __contains__(self, value):
        """Return true if `value` is an element of the sorted list."""
        _lists = self._lists
        if _lists:
            pos, idx = self._loc_left(value)
            return idx < len(_lists[pos]) and _lists[pos][idx] == value
        return False

    def __iter__(self):
        """Return an iterator over the sorted list."""
        return (value for _list in self._lists for value in _list)

    def __reversed__(self):
        """Return a reverse iterator over the sorted list."""
        return (value for _list in reversed(self._lists) for value in reversed(_list))

    def __repr__(self):
        """Return string representation of sorted list."""
        return 'SortedList({0})'.format(list(self))

def cal():
    n, m, q = map(int, input().split())
    row = list(map(int, input().split()))
    col = list(map(int, input().split()))

    r = sum(row)
    c = sum(col)

    row_num = SortedList()
    for num in row:
        if r - num not in row_num:
            row_num.add(r - num)

    col_num = SortedList()
    for num in col:
        if c - num not in col_num:
            col_num.add(c - num)

    def judge(x):
        a = 1
        flag = -1 if x < 0 else 1
        x = abs(x)
        while a * a < x:
            b = x // a * flag
            if a in row_num and b in col_num:
                return True
            if b in row_num and a in col_num:
                return True
            if -a in row_num and -b in col_num:
                return True
            if -b in row_num and -a in col_num:
                return True
            a += 1
        if a * a == x:
            b = x // a * flag
            if a in row_num and b in col_num:
                return True
            if b in row_num and a in col_num:
                return True
            if -a in row_num and -b in col_num:
                return True
            if -b in row_num and -a in col_num:
                return True
        return False

    for _ in range(q):
        x, = map(int, input().split())
        if judge(x):
            print('YES')
        else:
            print('NO')

for _ in range(1):
    cal()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to create such a test input:

1. **Edge Cases for Input Constraints**:
   - Use the maximum values for \( n \) and \( m \) to test the limits of the algorithm.
   - Use the minimum values for \( n \) and \( m \) to test the smallest possible inputs.

2. **Special Values**:
   - Use arrays with all zeros to test the handling of zero sums.
   - Use arrays with both positive and negative values to test the handling of mixed signs.
   - Use arrays with repeated values to test the handling of duplicates.

3. **Queries**:
   - Use queries that are exactly the sum of the grid to test the handling of the original grid beauty.
   - Use queries that are impossible to achieve by setting any row and column to zero.

4. **Combination of Edge Cases**:
   - Combine the above scenarios to create a test case that is likely to expose bugs.

Given these considerations, here is a small-scale test input that is complete and valid:

```plaintext
5 5 4
1 2 3 4 5
-1 -2 -3 -4 -5
-1
0
1
10
```

### Explanation:
- **Input**:
  - \( n = 5 \), \( m = 5 \), \( q = 4 \)
  - \( a = [1, 2, 3, 4, 5] \)
  - \( b = [-1, -2, -3, -4, -5] \)
  - Queries: \(-1\), \(0\), \(1\), \(10\)

- **Expected Output**:
  - For \( x = -1 \): **NO** (impossible to achieve)
  - For \( x = 0 \): **YES** (by setting the entire grid to zero)
  - For \( x = 1 \): **NO** (impossible to achieve)
  - For \( x = 10 \): **NO** (impossible to achieve)

This test input includes:
- A mix of positive and negative values.
- Queries that test the edge cases of the grid's beauty.
- A query that is the original sum of the grid (0 in this case).

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

Buggy Code:
import sys
input = sys.stdin.readline

def solve(n, m, a, b, q):
    # 격자의 전체 합 계산
    total = sum(a[i] * b[j] for i in range(n) for j in range(m))

    # 각 행의 합 계산
    row_sums = [sum(a[i] * b[j] for j in range(m)) for i in range(n)]

    # 각 열의 합 계산
    col_sums = [sum(a[i] * b[j] for i in range(n)) for j in range(m)]

    # 모든 가능한 결과값 저장
    possible = {total - row_sums[i] - col_sums[j] + a[i] * b[j]
               for i in range(n) for j in range(m)}

    # 각 쿼리 처리
    for x in q:
        print("YES" if x in possible else "NO")

n, m, q = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
queries = [int(input()) for _ in range(q)]
solve(n, m, a, b, queries)

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

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

Think step by step.

Response:
```plaintext
2 2 4
1 0
1 0
1
0
-1
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt(), q = sc.nextInt();
        long[] a = new long[n];
        long[] b = new long[m];
        long totalSum = 0;

        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }
        for (int i = 0; i < m; i++) {
            b[i] = sc.nextLong();
        }

        long rowSum = 0;
        for (long ai : a) rowSum += ai;

        long colSum = 0;
        for (long bi : b) colSum += bi;

        totalSum = rowSum * colSum;

        HashSet<Long> possible = new HashSet<>();
        for (long ai : a) {
            for (long bi : b) {
                possible.add(totalSum - rowSum * bi - colSum * ai + ai * bi);
            }
        }

        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < q; i++) {
            long x = sc.nextLong();
            if (possible.contains(x)) {
                sb.append("YES\n");
            } else {
                sb.append("NO\n");
            }
        }
        System.out.print(sb);
    }
}

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

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

Think step by step.

Response:
```plaintext
2 2 2
1 1
1 1
2
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Demon Problem
time_limit_ms: 4000
memory_limit_mb: 256
Description: For an arbitrary grid, Robot defines its beauty to be the sum of elements in the grid.

Robot gives you an array $$$a$$$ of length $$$n$$$ and an array $$$b$$$ of length $$$m$$$. You construct a $$$n$$$ by $$$m$$$ grid $$$M$$$ such that $$$M_{i,j}=a_i\cdot b_j$$$ for all $$$1 \leq i \leq n$$$ and $$$1 \leq j \leq m$$$.

Then, Robot gives you $$$q$$$ queries, each consisting of a single integer $$$x$$$. For each query, determine whether or not it is possible to perform the following operation exactly once so that $$$M$$$ has a beauty of $$$x$$$:

1. Choose integers $$$r$$$ and $$$c$$$ such that $$$1 \leq r \leq n$$$ and $$$1 \leq c \leq m$$$
2. Set $$$M_{i,j}$$$ to be $$$0$$$ for all ordered pairs $$$(i,j)$$$ such that $$$i=r$$$, $$$j=c$$$, or both.

Note that queries are not persistent, meaning that you do not actually set any elements to $$$0$$$ in the process — you are only required to output if it is possible to find $$$r$$$ and $$$c$$$ such that if the above operation is performed, the beauty of the grid will be $$$x$$$. Also, note that you must perform the operation for each query, even if the beauty of the original grid is already $$$x$$$.

Input: The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \leq n,m \leq 2\cdot 10^5, 1 \leq q \leq 5\cdot 10^4$$$) — the length of $$$a$$$, the length of $$$b$$$, and the number of queries respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq |a_i| \leq n$$$).

The third line contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$0 \leq |b_i| \leq m$$$).

The following $$$q$$$ lines each contain a single integer $$$x$$$ ($$$1 \leq |x| \leq 2\cdot 10^5$$$), the beauty of the grid you wish to achieve by setting all elements in a row and a column to $$$0$$$.

Output: For each testcase, output "YES" (without quotes) if there is a way to perform the aforementioned operation such that the beauty is $$$x$$$, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
3 3 6
-2 3 -3
-2 2 -1
-1
1
-2
2
-3
3
output:
NO
YES
NO
NO
YES
NO

input:
5 5 6
1 -2 3 0 0
0 -2 5 0 -3
4
-3
5
2
-1
2
output:
YES
YES
YES
YES
NO
YES

Note: In the second example, the grid is

0 -2 5 0 -3

0 4 -10 0 6

0 -6 15 0 -9

0 0 0 0 0

0 0 0 0 0

By performing the operation with $$$r=4$$$ and $$$c=2$$$, we create the following grid:

0 0 5 0 -3

0 0 -10 0 6

0 0 15 0 -9

0 0 0 0 0

0 0 0 0 0

which has beauty $$$4$$$. Thus, we output YES.

In the second query, selecting $$$r=3$$$ and $$$c=5$$$ creates a grid with beauty $$$-3$$$.

In the third query, selecting $$$r=3$$$ and $$$c=3$$$ creates a grid with beauty $$$5$$$.

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

public class B {
    private static String[] solve(int n, int m, int q, int[] a, int[] b, int[] queries) {
        String[] output = new String[q];

        int asum = 0, bsum = 0;
        for (int i = 0; i < n; i++) asum += a[i];
        for (int i = 0; i < m; i++) bsum += b[i];

        int MAX_X = 200_006;

        boolean[] apos = new boolean[MAX_X];
        boolean[] aneg = new boolean[MAX_X];
        boolean[] bpos = new boolean[MAX_X];
        boolean[] bneg = new boolean[MAX_X];

        Arrays.fill(apos, false);
        Arrays.fill(aneg, false);
        Arrays.fill(bpos, false);
        Arrays.fill(bneg, false);

        for (int i = 0; i < n; i++) {
            if (Math.abs(asum - a[i]) < MAX_X) {
                if (asum - a[i] > 0) apos[asum - a[i]] = true;
                else aneg[a[i] - asum] = true;
            }
        }

        for (int i = 0; i < m; i++) {
            if (Math.abs(bsum - b[i]) < MAX_X) {
                if (bsum - b[i] > 0) bpos[bsum - b[i]] = true;
                else bneg[b[i] - bsum] = true;
            }
        }

        boolean[] posspos = new boolean[MAX_X];
        boolean[] possneg = new boolean[MAX_X];

        Arrays.fill(posspos, false);
        Arrays.fill(possneg, false);

        for (int i = 1; i < MAX_X; i++) {
            for (int j = 1; j < MAX_X; j++) {
                if (i * j > MAX_X) break;
                if (apos[i] && bpos[j]) posspos[i * j] = true;
                if (apos[i] && bneg[j]) possneg[i * j] = true;
                if (aneg[i] && bpos[j]) possneg[i * j] = true;
                if (aneg[i] && bneg[j]) posspos[i * j] = true;
            }
        }

        for (int i = 0; i < q; i++) {
            int x = queries[i];
            if (x > 0) output[i] = posspos[x] ? "YES" : "NO";
            else output[i] = possneg[-x] ? "YES" : "NO";
        }

        return output;
    }

    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        // int t = fs.nextInt();
        // while (t > 0) {

        //     t--;
        // }

        int n = fs.nextInt(), m = fs.nextInt(), q = fs.nextInt();
        int[] a = fs.readArray(n);
        int[] b = fs.readArray(m);
        int[] queries = fs.readArray(q);
        String[] output = solve(n, m, q, a, b, queries);
        for (String ans : output) out.println(ans);
        out.close();
    }

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

        String next() {
            while (!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[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

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

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

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

Think step by step.

Response:
```plaintext
2 2 4
1 0
0 1
0
1
2
-1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<stdio.h>
int main()
{
    int a,b,c,d,n[8],i,j;
	scanf("%d",&a);
    for(i=0;i<a;i++)
    {
    	scanf("%d",&d);
		b=d/10,c=d%10;
    	n[i]=b+c;
	}
	for(i=0;i<a;i++)
	{
		printf("%d\n",n[i]);
	}
    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 expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. The code reads an integer `a` which represents the number of test cases, and then reads `a` two-digit integers. It calculates the sum of the digits for each integer and stores the results in an array `n`.

Here are some potential issues:
1. **Array Size**: The array `n` is declared with a fixed size of 8, but the number of test cases `a` can be up to 90. This will cause an out-of-bounds error if `a` is greater than 8.
2. **Input Validation**: The code does not validate if the input is a two-digit number, but this is not a primary concern since the problem constraints already specify that the input will be a two-digit number.
3. **Edge Cases**: We should test the smallest and largest possible two-digit numbers (10 and 99) to ensure the code handles these correctly.

Given these considerations, a small-scale test input that is complete and valid, and likely to expose the bug, would be:

```plaintext
9
10
11
12
13
14
15
16
17
18
```

This input provides 9 test cases, which is just one more than the array size of 8, and includes a range of two-digit numbers to test the sum calculation. This should cause an out-of-bounds error when the code tries to store the result for the 9th test case.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<stdio.h>
int main()
{
    int t,b,c,n,i;
    10<=n<=99,1<=t<=90;
	scanf("%d",&t);
	int s[8];
    for(i=0;i<t;i++)
    {
    	scanf("%d",&n);
		b=n/10,c=n%10;
    	s[i]=b+c;
	}
	for(i=0;i<t;i++)
	{
		printf("%d\n",s[i]);
	}
    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
3
10
99
45
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<stdio.h>
int main()
{
    int a,b,c,d,n[8],i,j;
	scanf("%d",&a);
    for(i=0;i<a;i++)
    {
    	scanf("%d",&d);
		b=d/10,c=d%10;
    	n[i]=b+c;
	}
	for(i=0;i<a;i++)
	{
		printf("%d\n",n[i]);
	}
    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 the following:

1. **Edge Cases for Input Values**: The input values for \( n \) range from 10 to 99. We should include values that might cause unexpected behavior, such as numbers with digits that are 0 or 9.
2. **Multiple Test Cases**: The number of test cases \( t \) ranges from 1 to 90. We should include a small number of test cases to ensure the output is manageable but still comprehensive.

Let's consider the following test cases:
- A number with a 0 in the tens place (e.g., 10).
- A number with a 0 in the units place (e.g., 40).
- A number with both digits being the same (e.g., 77).
- A number with the maximum possible sum of digits (e.g., 99).
- A number with the minimum possible sum of digits (e.g., 11).

Here is the test input:

```plaintext
5
10
40
77
99
11
```

This test input includes:
- 10: Sum of digits is 1.
- 40: Sum of digits is 4.
- 77: Sum of digits is 14.
- 99: Sum of digits is 18.
- 11: Sum of digits is 2.

These values cover a range of edge cases and should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
for i in range(9):
    n = int(input())
    if n >= 10:
        print(n//10+n%10)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the following aspects:

1. **Range of Test Cases**: The code uses `range(9)` instead of `range(t)`, which means it will only process 9 test cases regardless of the value of `t`. This can be exploited by providing a value of `t` that is different from 9.

2. **Edge Cases for Two-Digit Numbers**: The code should handle all two-digit numbers correctly, but we should ensure that it handles the minimum (10) and maximum (99) values correctly.

3. **Consistency with the Problem Statement**: The code should process exactly `t` test cases, so we need to provide a value of `t` and the corresponding number of test cases.

Given these considerations, a challenging test input that exposes the bug would be:

```plaintext
10
77
21
40
34
19
84
10
99
15
56
```

This input provides 10 test cases, which is more than the 9 cases the buggy code is set to handle. The additional test case (15) will be ignored by the buggy code, leading to an incorrect output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
l=[8,77,21,40,34,19,84,10,99]
for i in l:
    if i>=10:
        s=list(str(i))
        sum=0
        for i in s:
            sum+=int(i)
        print(sum)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code provided has a few issues, but the primary one is that it uses the same variable `i` for both the outer loop and the inner loop, which can lead to unexpected behavior.

Let's consider the following edge cases:
1. The smallest two-digit number (10).
2. The largest two-digit number (99).
3. Numbers where the sum of digits is a single digit (e.g., 10, 11).
4. Numbers where the sum of digits is a two-digit number (e.g., 99).

A small-scale test input that includes these edge cases and a few other typical cases would be:

```plaintext
4
10
99
11
89
```

This test input covers:
- The smallest two-digit number (10).
- The largest two-digit number (99).
- A number where the sum of digits is a single digit (11).
- A number where the sum of digits is a two-digit number (89).

This should help expose any issues in the code, especially the variable shadowing problem.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
def sum_of_digits(t, test_cases):
    results = []
    for i in range(t):
        n = test_cases[i]
        # Convert the integer to a string and sum the digits
        sum_digits = sum(int(digit) for digit in str(n))
        results.append(sum_digits)
    return results

# Example input
t = 8
test_cases = [77, 21, 40, 34, 19, 84, 10, 99]

# Get the result
results = sum_of_digits(t, test_cases)

# Print the results
for result in results:
    print(result)

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

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

Think step by step.

Response:
To expose potential bugs in the given code, we should focus on edge cases and scenarios that might cause unexpected behavior. Here are some considerations:

1. **Edge Cases for Two-Digit Numbers:**
   - The smallest two-digit number: 10
   - The largest two-digit number: 99
   - Numbers with a zero in the tens place: 10, 20, 30, etc.
   - Numbers with a zero in the units place: 10, 20, 30, etc.
   - Numbers where both digits are the same: 11, 22, 33, etc.

2. **Multiple Test Cases:**
   - Include a variety of numbers to ensure the code handles different scenarios.
   - Include a mix of small and large numbers.
   - Include numbers with different digit sums.

3. **Edge Cases for the Number of Test Cases:**
   - The minimum number of test cases: 1
   - The maximum number of test cases: 90

Given these considerations, here is a small-scale test input that includes a variety of edge cases and typical scenarios:

```plaintext
6
10
99
20
55
81
11
```

This test input includes:
- The smallest two-digit number (10).
- The largest two-digit number (99).
- A number with a zero in the units place (20).
- A number with both digits the same (55).
- A number with a high digit sum (81).
- A number with a low digit sum (11).

This should help in exposing any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
public class test2 {
	public static void main (String []args){
	 int n[] ={77 ,21,40,34,19,84,10,99};
         for(int i=0 ;i<n.length ; i++){
               System.out.println ((n[i]/10)+(n[i]%10));
			 }
		}
	}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the code to fail. The code provided is a simple loop that calculates the sum of the digits of each number in the array. The main potential issue is with how the code handles the numbers, especially at the boundaries of the two-digit range (10 and 99).

Let's consider the following edge cases:
1. The smallest two-digit number (10).
2. The largest two-digit number (99).
3. Numbers where the sum of digits is a single digit (e.g., 19, 28).
4. Numbers where the sum of digits is a double digit (e.g., 98, 89).

Given these considerations, a small-scale test input that includes these edge cases would be:

```plaintext
4
10
99
19
89
```

This test input includes:
- The smallest two-digit number (10).
- The largest two-digit number (99).
- A number where the sum of digits is a single digit (19).
- A number where the sum of digits is a double digit (89).

This should help in identifying any potential issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
public class Main {
    public static int sumOfDigits(int num) {
        String numString = "" + num; //converts int to string
        int digitOne = Integer.valueOf(numString.substring(0,1)); //takes first character in string and converts to int
        int digitTwo = Integer.valueOf(numString.substring(1,2)); //takes second character in string and converts to int
        return digitOne + digitTwo;
    }

    public static void main(String[] args) {
        //output
        System.out.println(sumOfDigits(77));
        System.out.println(sumOfDigits(21));
        System.out.println(sumOfDigits(40));
        System.out.println(sumOfDigits(34));
        System.out.println(sumOfDigits(19));
        System.out.println(sumOfDigits(84));
        System.out.println(sumOfDigits(10));
        System.out.println(sumOfDigits(99));

    }
}

**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 create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause issues. The code in question converts the number to a string and then extracts the digits. Here are some potential issues to consider:

1. **Leading Zeros**: The code assumes that the input is always a two-digit number, but it doesn't handle leading zeros explicitly. However, since the problem constraints guarantee a two-digit number, this is not a concern.
2. **Negative Numbers**: The code doesn't handle negative numbers, but the problem constraints ensure that the input is always a positive two-digit number.
3. **Boundary Values**: The code should handle the minimum and maximum values of the two-digit range (10 and 99) correctly.
4. **Special Cases**: Numbers with a zero in the second digit (e.g., 10, 20, 30, etc.) to ensure that the code correctly handles the conversion and summation.

Given these considerations, let's create a test input that includes boundary values and special cases:

```plaintext
4
10
99
40
55
```

This test input includes:
- The minimum value (10)
- The maximum value (99)
- A number with a zero in the second digit (40)
- A typical two-digit number (55) to ensure general correctness

This should help in identifying any potential issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<iostream>
using namespace std;
int main() {
	int n;
	cin >> n;
	int num, t, ans;
	for (int i = 0; i < 8; i++) {
		ans = 0;
		cin >> num;
		while (num > 0) {
			t = num % 10;
			num /= 10;
			ans += t;
		}
		cout << ans << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the loop that processes the test cases. The code has a fixed loop that runs 8 times, regardless of the number of test cases provided. This means that if the number of test cases is not exactly 8, the code will either:

1. Process fewer test cases than provided (if \( t < 8 \)).
2. Process more test cases than provided (if \( t > 8 \)).

To create a challenging test input, we should provide a number of test cases that is not 8. Let's choose \( t = 5 \) to ensure the code processes fewer test cases than the loop allows.

Here is the test input:

```plaintext
5
10
20
30
40
50
```

This input will cause the code to process only 5 test cases, but the loop will still run 8 times, leading to incorrect behavior. The expected output for this input should be:

```plaintext
1
2
3
4
5
```

However, the code will likely produce an incorrect output or crash due to the mismatch in the number of test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int T;cin>>T;
	 mt19937 rnd(time(nullptr));
	 uniform_real_distribution<double> qwq(0,1);
    while(T--){
        char a,b;cin>>a>>b;
        if(qwq(rnd)<0.98) cout<<a+b-96<<'\n';
        else cout<<"I am zdj\n";
    }
}

//////////////////////

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

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

Think step by step.

Response:
```plaintext
3
10
99
55
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
# Simulating input for testing purposes
input_data = """8
77
21
40
34
19
84
10
99"""

# Simulate the `input()` method with predefined test cases
import sys
from io import StringIO

sys.stdin = StringIO(input_data)

# The rest of your code
t = int(input())  # Read t as an integer

for _ in range(t):
    n = int(input())  # Read the two-digit integer n
    tens = n // 10  # Get the tens digit
    ones = n % 10   # Get the ones digit
    print(tens + ones)  # Output the sum of digits

**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 code, we need to consider edge cases and scenarios that might cause the code to fail. Here are the steps to think through this:

1. **Edge Cases for Two-Digit Numbers:**
   - The smallest two-digit number: 10
   - The largest two-digit number: 99
   - Numbers where the digits are the same: 11, 22, ..., 99
   - Numbers where one digit is zero: 10, 20, ..., 90
   - Numbers where the sum of digits is a single-digit number: 18, 27, 36, etc.

2. **Number of Test Cases:**
   - The minimum number of test cases: 1
   - The maximum number of test cases: 90
   - A mix of different numbers to ensure comprehensive testing.

3. **Combining the Above Points:**
   - We will create a test input with a small number of test cases to ensure it is complete and valid.
   - We will include the edge cases mentioned above.

Here is the test input:

```plaintext
5
10
99
11
20
55
```

This test input includes:
- The smallest two-digit number: 10
- The largest two-digit number: 99
- A number where the digits are the same: 11
- A number where one digit is zero: 20
- A number where the sum of digits is a single-digit number: 55 (sum is 10)

This should help in exposing any potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
for i in range(9):
    n = int(input())
    if n >= 10:
        print(n//10+n%10)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the following aspects:

1. **Edge Cases**: The code should handle the minimum and maximum values of the two-digit integers (10 and 99).
2. **Zero Leading Digit**: The code should handle numbers where one of the digits is zero (e.g., 10, 20, 30, etc.).
3. **Single Test Case**: The code should handle a single test case (t = 1).
4. **Multiple Test Cases**: The code should handle multiple test cases (t > 1).

The buggy code has a logical error in the loop range. It uses `range(9)` instead of `range(t)`, which means it will only process 9 test cases regardless of the input value of `t`.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
10
10
20
30
40
50
60
70
80
90
11
```

This input includes:
- 10 test cases, which is more than the 9 cases the buggy code is prepared to handle.
- Edge cases where the second digit is zero (10, 20, 30, etc.).
- A final test case (11) that should be processed but will not be due to the loop range issue.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int a = scanner.nextInt();
        int b = scanner.nextInt();
        int c = scanner.nextInt();
        int d = scanner.nextInt();
        int e = scanner.nextInt();
        int f = scanner.nextInt();
        int g = scanner.nextInt();
        int h = scanner.nextInt();
        int j = scanner.nextInt();
        System.out.println(14);
        System.out.println(3);
        System.out.println(4);
        System.out.println(7);
        System.out.println(10);
        System.out.println(12);
        System.out.println(1);
        System.out.println(18);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to create a test input that highlights the issue. The buggy code reads 9 integers but only prints a fixed set of outputs. This means that any input with more or fewer than 9 test cases will cause the program to fail.

Here are the steps to create a challenging test input:

1. **Number of Test Cases**: Since the code reads exactly 9 integers, we should provide a number of test cases that is different from 9. This will ensure that the program fails to read the correct number of inputs.

2. **Input Values**: We can provide any valid two-digit integers (10 to 99) for the test cases. The values themselves are not as important as the number of test cases.

Let's create a test input with 10 test cases:

```plaintext
10
77
21
40
34
19
84
10
99
55
66
```

This input will cause the buggy code to fail because it reads 9 integers but the input provides 10. The program will not be able to process the 10th integer, and it will not print the correct output for all test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
public class Main {
    public static int sumOfDigits(int num) {
        String numString = "" + num; //converts int to string
        int digitOne = Integer.valueOf(numString.substring(0,1)); //takes first character in string and converts to int
        int digitTwo = Integer.valueOf(numString.substring(1,2)); //takes second character in string and converts to int
        return digitOne + digitTwo;
    }

    public static void main(String[] args) {
        //output
        System.out.println(sumOfDigits(77));
        System.out.println(sumOfDigits(21));
        System.out.println(sumOfDigits(40));
        System.out.println(sumOfDigits(34));
        System.out.println(sumOfDigits(19));
        System.out.println(sumOfDigits(84));
        System.out.println(sumOfDigits(10));
        System.out.println(sumOfDigits(99));

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The code provided uses string manipulation to extract the digits of the number, which is generally correct for two-digit numbers. However, we can still test for specific edge cases to ensure the code handles all valid inputs correctly.

### Edge Cases to Consider:
1. **Smallest Two-Digit Number (10):** This is the smallest valid input.
2. **Largest Two-Digit Number (99):** This is the largest valid input.
3. **Numbers with a Zero Digit (e.g., 10, 20, 90):** These numbers have a zero as one of the digits, which might be handled differently.
4. **Numbers with Large Sums (e.g., 99):** The sum of the digits is 18, which is the maximum possible sum for a two-digit number.
5. **Numbers with Small Sums (e.g., 10):** The sum of the digits is 1, which is the minimum possible sum for a two-digit number.
6. **Numbers with Equal Digits (e.g., 77):** These numbers have the same digit repeated, which should result in an even sum.

### Chosen Test Input:
Let's create a test input that includes all these edge cases:

```plaintext
6
10
99
20
77
40
55
```

### Explanation:
- **10:** Smallest two-digit number, sum is 1.
- **99:** Largest two-digit number, sum is 18.
- **20:** Number with a zero digit, sum is 2.
- **77:** Number with equal digits, sum is 14.
- **40:** Another number with a zero digit, sum is 4.
- **55:** A number with equal digits, sum is 10.

This test input should be sufficient to cover a range of edge cases and ensure that the buggy code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include<stdio.h>
int main()
{
    int t,b,c,n,i;
    10<=n<=99,1<=t<=90;
	scanf("%d",&t);
	int s[8];
    for(i=0;i<t;i++)
    {
    	scanf("%d",&n);
		b=n/10,c=n%10;
    	s[i]=b+c;
	}
	for(i=0;i<t;i++)
	{
		printf("%d\n",s[i]);
	}
    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
3
10
99
50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main()
{
	int n,a;
	cin>>n;
        int p[20];
	for(int i=0;i<n;i++)
	{
		cin>>a;
		p[i]=a/10+a%10;
	}
        for(int i=0;i<n;i++)
        cout<<p[i]<<endl;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. The code reads the number of test cases and then processes each test case by calculating the sum of the digits of a two-digit number. The primary issue to look for is how the code handles the input and output, especially for edge cases.

### Step-by-Step Analysis:
1. **Input Validation**:
   - The number of test cases `t` should be between 1 and 90.
   - Each test case should be a two-digit number between 10 and 99.

2. **Array Size**:
   - The array `p` is declared with a size of 20. This means the code can handle up to 20 test cases, but the problem specifies that `t` can be up to 90. This is a potential overflow issue if `t` exceeds 20.

3. **Edge Cases**:
   - Test cases with the minimum and maximum values of `t` (1 and 90).
   - Test cases with the minimum and maximum values of `n` (10 and 99).

### Test Input Generation:
- **Test Case with `t = 1`**:
  - This checks the smallest possible number of test cases.
- **Test Case with `t = 90`**:
  - This checks the largest possible number of test cases.
- **Test Case with `n = 10`**:
  - This checks the smallest possible two-digit number.
- **Test Case with `n = 99`**:
  - This checks the largest possible two-digit number.

### Small-Scale Test Input:
To create a small-scale test input that is complete and valid, we can use the following:
- `t = 4` (a small number of test cases to keep the input manageable)
- `n = 10, 99, 15, 55` (covering the minimum, maximum, and a few intermediate values)

### Test Input:
```plaintext
4
10
99
15
55
```

This test input will help verify if the code correctly handles the edge cases and intermediate values. If the code is buggy, it should fail to handle the larger number of test cases or the edge values correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
a=7
b=7
print(a+b)
c=2
d=1
print(c+d)
e=4
f=0
print(e+f)
g=3
h=4
print(g+h)
x=1
y=9
print(x+y)
s=8
p=4
print(s+p)
q=1
r=0
print(q+r)
w=9
v=9
print(w+v)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the fact that the code is manually printing the sum of the digits for specific test cases instead of using a general algorithm to compute the sum for any given two-digit number. This means the code will fail if the input numbers are different from the ones hardcoded in the code.

Let's identify the numbers that are hardcoded in the code and then create a test case that uses numbers different from these.

The numbers and their sums in the buggy code are:
- 77 -> 14
- 21 -> 3
- 40 -> 4
- 34 -> 7
- 19 -> 10
- 84 -> 12
- 10 -> 1
- 99 -> 18

To create a test case that will expose the bug, we can use numbers that are not in this list. For example, we can use the number 55, which is not in the list and has a sum of 10.

Let's create a small-scale test input with a few such numbers:

```plaintext
5
55
66
78
23
81
```

This test input will help expose the bug because the code does not have hardcoded sums for these numbers. The expected output should be:

```
10
12
15
5
9
```

However, the buggy code will likely print incorrect results because it does not handle these numbers correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
N1=77
FIRST_DIGIT=N1//10
SECOND_DIGIT=N1%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N2=21
FIRST_DIGIT=N2//10
SECOND_DIGIT=N2%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N3=40
FIRST_DIGIT=N3//10
SECOND_DIGIT=N3%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N4=34
FIRST_DIGIT=N4//10
SECOND_DIGIT=N4%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N5=19
FIRST_DIGIT=N5//10
SECOND_DIGIT=N5%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N6=84
FIRST_DIGIT=N6//10
SECOND_DIGIT=N6%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N7=10
FIRST_DIGIT=N7//10
SECOND_DIGIT=N7%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)
N8=99
FIRST_DIGIT=N8//10
SECOND_DIGIT=N8%10
ANS=FIRST_DIGIT+SECOND_DIGIT
print(ANS)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The code provided is a straightforward implementation that calculates the sum of the digits of a two-digit number. However, the code is not designed to handle multiple test cases efficiently.

The main issue with the code is that it is hardcoded for specific values and does not handle multiple test cases dynamically. To expose this, we can provide a test input with a large number of test cases, including edge cases and typical values.

Here is a small-scale test input that includes a variety of two-digit numbers, including edge cases:

```plaintext
8
10
19
28
37
46
55
64
73
```

This input includes:
- The smallest two-digit number (10)
- A number where the sum of the digits is a single digit (19)
- A number where the sum of the digits is a double digit (28, 37, 46, 55, 64, 73)
- A variety of numbers to ensure the code handles different sums correctly

This input will help verify if the code can handle multiple test cases and compute the sum of digits correctly for each.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
import java.util.*;
public class HelloWorld{

    public static void main(String[] args){
        Scanner scn=new Scanner(System.in);
        int n=scn.nextInt();
        int arr[]=new int[8];
        for(int i=0;i<n;i++){
            arr[i]=scn.nextInt();

        }
        for(int i=0;i<n;i++){
            int a10=arr[i]/10;
            int a1=arr[i]%10;
            System.out.println(a10+a1);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
10
99
50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A+B Again?
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given a two-digit positive integer $n$, find the sum of its digits.

Input: The first line contains an integer $t$ ($1 \leq t \leq 90$) — the number of test cases.

The only line of each test case contains a single two-digit positive integer $n$ ($10 \leq n \leq 99$).

Output: For each test case, output a single integer — the sum of the digits of $n$.

Examples:
input:
8
77
21
40
34
19
84
10
99
output:
14
3
4
7
10
12
1
18

Note:

Buggy Code:
import java.util.*;
public class HelloWorld{

    public static void main(String[] args){
        Scanner scn=new Scanner(System.in);
        int n=scn.nextInt();
        int arr[]=new int[8];
        for(int i=0;i<n;i++){
            arr[i]=scn.nextInt();
        }
        for(int i=0;i<n;i++){
            int a10=arr[i]/10;
            int a1=arr[i]-(a10*10);
            System.out.println(a10+a1);
        }
    }
}

**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 create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Array Size Mismatch**: The code initializes an array `arr` of size 8, but the number of test cases `t` can be up to 90. This will cause an `ArrayIndexOutOfBoundsException` if `t` is greater than 8.

2. **Edge Cases**: We should include edge cases for two-digit numbers, such as 10 and 99, to ensure the code handles them correctly.

3. **Special Cases**: We should include numbers where the sum of digits is particularly simple to verify, such as 11, 22, etc.

Given these considerations, a small-scale test input that will expose the bug in the array size mismatch and test edge cases is:

```plaintext
10
10
11
12
13
14
15
16
17
18
19
```

This input includes 10 test cases, which is more than the array size of 8, and it includes edge cases and simple sums to verify the correctness of the sum calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <string>
#include <iomanip>
#include <cmath>
#include <queue>
#include <utility>
#include <array>
#include <stack>

using namespace std;

using ull = unsigned long long;
using ll = long long;

void solve() {
    ll x, m;
    cin >> x >> m;
    // (x ^ y) = x * t
    // (x ^ y) = y * t
    // 100011010 ^ 1001 = 100010011
    ll cnt = 0;
    for (ll y = 1; y <= min(m, 2LL * x); y++) {
        ll t = (x ^ y);
        if (t % x == 0 || t % y == 0) {
            cnt++;
        }
    }
    cnt += max(min(m, m ^ x) / x - 2LL, 0LL);
    cout << cnt << "\n";
}

void multisolve() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    multisolve();
    //solve();
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code has a loop that iterates from `y = 1` to `min(m, 2LL * x)`, and then it adds a count based on a formula. The bug might be in the loop or the formula.

2. **Edge Cases**:
   - **Small `x` and `m`**: Test with small values of `x` and `m` to see if the loop and formula work correctly.
   - **Large `x` and `m`**: Test with large values of `x` and `m` to see if the loop and formula handle large numbers correctly.
   - **Special Values of `x`**: Test with `x` being a power of 2, since XOR operations with powers of 2 have specific properties.
   - **Special Values of `m`**: Test with `m` being a multiple of `x` or just slightly larger than a multiple of `x`.

3. **Challenging Scenarios**:
   - **When `m` is very large compared to `x`**: This can test the efficiency of the loop and the formula.
   - **When `x` is a power of 2**: This can test the XOR properties.
   - **When `m` is a multiple of `x`**: This can test the edge case where the loop might miss some valid `y`.

Based on these considerations, a small-scale test input that can expose the bug is:

```plaintext
3
1 1000000000000000000
2 1000000000000000000
4 1000000000000000000
```

This test input includes:
- A large value of `m` to test the efficiency and correctness of the loop and formula.
- `x` values that are small and powers of 2 to test the XOR properties and edge cases.

This should help in identifying any issues with the loop and the formula used in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <string>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pl;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;

void solve() {
    ll x, m; cin >> x >> m;
    ll result = 0;
    ll k = (m - 1) / x; //+1 for 0, -1 for x

    if (m <= 2 * x) {
        for (int y = 1; y <= m; y++) {
            if ((x ^ y) % x == 0 || (x ^ y) % y == 0) {
                result++;
            }
        }
    }
    else {
        //what y donates
        for (int y = 1; y <= min(2 * x, m); y++) {
            if ((x ^ y) % y == 0) {
                result++;
                if ((x ^ y) % x == 0 && (x ^ y) <= (k + 2) * x) { //i.e. y is a multiple of x we check
                    result--;
                }
            }
        }
        result += k;
        if (k == 0 && x <= m)result++;
        if ((((k + 1) * x) ^ x) <= m && (((k + 1) * x) ^ x) > 0) {
            result++;
        }
        if ((((k + 2) * x) ^ x) <= m && (((k + 2) * x) ^ x) > 0) {
            result++;
        }
    }
    cout << result << '\n';
}

int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 100
```

This test case is designed to challenge the buggy code by focusing on a scenario where `m` is significantly larger than `x`, and `x` is not a small number. The key points to consider are:

1. **Large `m` Relative to `x`**: With `m = 100` and `x = 10`, the code needs to handle a range of values for `y` from 1 to 100. This will test the logic that deals with larger values of `y` and the conditions for divisibility.

2. **Edge Cases for Divisibility**: The code should correctly handle cases where `x ^ y` is divisible by `y` or `x`. For example, `y = 10` should be valid because `10 ^ 10 = 0`, which is divisible by both `10` and `10`.

3. **Special Cases for `k` Calculation**: The variable `k` is calculated as `(m - 1) / x`, which in this case is `9`. The code should correctly handle the addition of `k` to the result and the special conditions for `k == 0` and `k + 1` and `k + 2` multiples of `x`.

This test case will help expose any issues in the logic that handles the range of `y` and the conditions for divisibility.