**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
import java.util.*;

public class Codeforces {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n, k, q;
        n = sc.nextInt();
        k = sc.nextInt();
        q = sc.nextInt();
        int[][] arr = new int[n][k];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < k; j++) {
                arr[i][j] = sc.nextInt();
                int x = i;
                while (x > 0) {
                    arr[i][j] |= arr[x - 1][j];
                    x--;
                }
            }
        }

        for (int query = 0; query < q; query++) {
            int m = sc.nextInt();
            int start = 0, end = n;

            for (int req = 0; req < m; req++) {
                int r = sc.nextInt() - 1;
                char o = sc.next().charAt(0);
                int c = sc.nextInt();

                if (o == '<') {
                    int lo = 0, hi = n;
                    while (lo < hi) {
                        int mid = lo + (hi - lo) / 2;
                        if (arr[mid][r] < c)
                            lo = mid + 1;
                        else
                            hi = mid;
                    }
                    end = Math.min(end, lo);
                } else if (o == '>') {
                    int lo = 0, hi = n;
                    while (lo < hi) {
                        int mid = lo + (hi - lo) / 2;
                        if (arr[mid][r] > c)
                            hi = mid;
                        else
                            lo = mid + 1;
                    }
                    start = Math.max(start, lo);
                }
            }
            System.out.println(start < end ? start + 1 : -1);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3 2 2
1 2
3 4
5 6
1
1 < 4
1
1 > 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define BitXoR063 (ios::sync_with_stdio(false), cin.tie(NULL));
#define VVI vector<vector<int>>
#define VVL vector<vector<long long>>
#define all(v) (v).begin(), (v).end()
#define PII pair<int, int>
#define MII map<int, int>
#define VI vector<int>
#define VLL vector<long long>
#define SC set<char>
#define SI set<int>
#define SL set<long long>
#define ll long long
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define INF LLONG_MAX
#define MOD 1e9 + 7
void solve() {
    int n, k, q;cin >> n >> k >> q;
    vector<VI> a(k + 1, VI(1));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            int t;cin >> t;
            a[j].push_back(t |= a[j].back());
        }
    }
    while (q--) {
        int m;
        cin >> m;
        vector<PII> p,q;
        while (m--) {
            char ch;int r, c;cin >> r >> ch >> c;
            (ch == '<' ? p : q).push_back({r, c});}
        int y = 1;
        for (const auto& [r, c] : q) {
            y = max(y, int(upper_bound(a[r].begin(), a[r].end(), c) - a[r].begin()));
            }if (y > n) {cout << -1 << "\n";
            continue;}
        bool country = false;
        for (auto [r, c] : p) {
            if (a[r][y] >= c) {country = true;
                break;
            }
        }
        cout << (country ? 1 : y) << endl;
    }
}
int32_t main(){
    BitXoR063;
    int TC = 1;
    while (TC--)solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
#define _CRT_SECURE_NO_DEPRECATE
#include <bits/stdc++.h>
// ios::sync_with_stdio(false);
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
//#define ull unsigned long long
#define il inline
#define endl "\n"
#define int long long
#define re register
#define double long double
#define pii pair<int,int>
using namespace std;
const double eqs = 1e-6;
const long long max_ = (2e5) + 7;
int base = 137, haximod = 998244353;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const long long INF = 1e18 + 7;
int N, M, Q;
vector<int> Node[max_];
pii Req[max_];
int temp[max_];
int solve() {
	cin >> N >> M >> Q;
	Node[0].resize(M + 3);
	for (int i = 1; i <= N; i++) {
		Node[i].resize(M + 3);
		for (int j = 1; j <= M; j++) {
			cin >> Node[i][j];
		}
	}
	for (int i = 1; i <= M; i++) {
		Req[i] = { -INF ,INF };
		for (int j = 1; j <= N; j++) {
			Node[j][i] |= Node[j - 1][i];
		}
	}
	int ind, lim;
	char op;
	while (Q--) {
		int T;
		cin >> T;
		for (int i = 1; i <= T; i++) {
			cin >> ind >> op >> lim;
			temp[i] = ind;
			if (op == '<') {
				Req[ind].second = min(Req[ind].second, lim - 1);
			}
			else {
				Req[ind].first = max(Req[ind].first, lim + 1);
			}
		}
		bool flag = 1;
		for (int i = 1; i <= T; i++) {
			if (Req[temp[i]].first > Req[temp[i]].second) {flag = 0;break;}
		}
		int L = 1, R = N;
		for (int j = 1; j <= T && flag; j++) {
			int i = temp[j];
			if(L > R || Req[i].first > Req[i].second) { flag = 0; break; }
			if(Node[L][i] > Req[i].second || Node[R][i] < Req[i].first){ flag = 0; break; }
			//这里应该是必定能找到一个
			while (L < R && Node[L][i] < Req[i].first){
				L++;
			}
			while (L < R && Node[R][i] > Req[i].second) {
				R--;
			}
		}
		if (!flag) {
			cout << "-1\n";
			for (int i = 1; i <= T; i++)Req[temp[i]] = { -INF ,INF };
			continue;
		}
		cout << L << endl;
		for (int i = 1; i <= T; i++)Req[temp[i]] = { -INF ,INF };
	}
	return 1;
}
il void clear() {
}
void init() {
	int T = 1;
	//cin >> T;
	while (T--) {
		solve();
		clear();
	}
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	//  freopen("in.in", "r", stdin);
	//freopen("out.out", "w", stdout);
	//clock_t c1 = clock();
	init();
	//  std::cerr << "TIME:" << clock() - c1;
	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 focus on edge cases and scenarios that can cause the algorithm to fail. The key points to consider are:

1. **Initialization of `Req` array**: The `Req` array is initialized to `{ -INF , INF }` for each region. If the requirements are very tight, it can lead to conflicts.
2. **Bitwise OR operation**: The bitwise OR operation can lead to values that are very large, which can cause issues if not handled correctly.
3. **Query constraints**: Queries can be designed to have very specific requirements that are hard to satisfy, or they can be designed to conflict with each other.

Given these points, let's create a test input that includes:

- A small number of countries and regions to keep the input size manageable.
- Values that can lead to large bitwise OR results.
- Queries that have very specific and conflicting requirements.

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

```plaintext
3 3 3
1 2 4
3 5 6
7 8 9
3
1 < 3
2 > 5
3 < 8
2
1 < 4
2 > 6
1
3 > 9
```

### Explanation:

1. **Initial Values**:
   - Country 1: [1, 2, 4]
   - Country 2: [3, 5, 6]
   - Country 3: [7, 8, 9]

2. **After Bitwise OR Operation**:
   - Country 1: [1, 2, 4]
   - Country 2: [3 | 1 = 3, 5 | 2 = 7, 6 | 4 = 6]
   - Country 3: [7 | 3 | 1 = 7, 8 | 5 | 2 = 15, 9 | 6 | 4 = 15]

   The resulting values are:
   - Country 1: [1, 2, 4]
   - Country 2: [3, 7, 6]
   - Country 3: [7, 15, 15]

3. **Queries**:
   - **Query 1**:
     - 1 < 3: Country 1 and 2
     - 2 > 5: Country 2 and 3
     - 3 < 8: Country 1 and 2
     - Only Country 2 satisfies all conditions.
   - **Query 2**:
     - 1 < 4: Country 1 and 2
     - 2 > 6: Country 3
     - No country satisfies both conditions.
   - **Query 3**:
     - 3 > 9: No country satisfies this condition.

### Expected Output:
```plaintext
2
-1
-1
```

This test input is designed to expose the bug in the code by creating a scenario where the bitwise OR operation results in values that are hard to satisfy, and the queries have very specific and conflicting requirements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
n,k,q = list(map(int,str(input()).split()))

arr = []
arr.append(list(map(int,str(input()).split())))

for i in range(n-1):
    inpt = list(map(int,str(input()).split()))
    for j in range(k):
        inpt[j] = inpt[j] | arr[i][j]
    arr.append(inpt)

for i in range(q):
    greater_than_queries = {}
    less_than_queries = {}
    number_of_queries = int(input())
    for abc in range(number_of_queries):
        region,symbol,value = list(map(str,str(input()).split()))
        region = int(region)-1
        value = int(value)

        if symbol==">":
            if region in greater_than_queries:
                greater_than_queries[region] = max(greater_than_queries[region],value)
            else:
                greater_than_queries[region] = value

        else:
            if region in less_than_queries:
                less_than_queries[region] = min(less_than_queries[region], value)
            else:
                less_than_queries[region] = value

    country = 0
    flag = True
    for k in greater_than_queries:
        while arr[country][k]<=greater_than_queries[k]:
            country+=1
            if country>=n:
                flag = False
                break
        if flag is False:
            break

    if flag is True:
        for m in less_than_queries:
            if arr[country][m]>=less_than_queries[m]:
                flag = False
                break

    if flag is False:
        print(-1)
    else:
        print(country+1)

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

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

Think step by step.

Response:
```plaintext
3 4 3
1 2 3 4
5 6 7 8
9 10 11 12
3
1 > 1
2 < 7
3 > 5
2
1 < 10
2 > 8
1
4 > 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
def solve():
  n, m, q = [int(x) for x in input().split(" ")]
  grid = []
  for _ in range(n):
    grid.append([int(x) for x in input().split(" ")])
  for i in range(1, n):
    for j in range(m):
      grid[i][j] |= grid[i-1][j]
  for _ in range(q):
    m = int(input())
    op = []
    for _ in range(m):
      l, o, r = [x for x in input().split(" ")]
      l, r = int(l)-1, int(r)
      op.append((l, o, r))
    pp = False
    for i in range(n):
      p = True
      for l, o, r in op:
        if o == "<" and grid[i][l] >= r:
          p = False
          break
        elif o == ">" and grid[i][l] <= r:
          p = False
          break
      if p:
        print(i+1)
        pp = True
        break
    if not pp:
      print(-1)
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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The bug in the code seems to be related to how it processes the queries and checks the conditions. Here are the steps to create the test input:

1. **Edge Cases for Input Values**:
   - Use the maximum and minimum allowed values for `n`, `k`, and `q`.
   - Use the maximum and minimum allowed values for the region values `a_{i,j}`.

2. **Complex Queries**:
   - Create queries with a mix of `<` and `>` conditions.
   - Ensure that some queries have no valid solutions.
   - Ensure that some queries have multiple valid solutions, and the code should return the smallest valid country.

3. **Special Cases**:
   - Create a scenario where the values in the grid are such that the bitwise OR operation results in a specific pattern that can be tricky to handle.

Here is a small-scale test input that covers these aspects:

```plaintext
5 4 3
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
3
1 < 5
2 > 6
3 < 10
2
1 > 15
2 < 10
1
4 > 20
```

### Explanation:
- **Input Description**:
  - `n = 5`, `k = 4`, `q = 3` (5 countries, 4 regions, 3 queries).
  - The grid values are chosen to create a pattern that can be tricky for the bitwise OR operation.
  - The queries are designed to test various conditions and edge cases.

- **Grid Values**:
  - The values are increasing in a simple pattern, which helps in understanding the bitwise OR results.

- **Queries**:
  1. The first query has three conditions:
     - `1 < 5`: This means the value in the first region must be less than 5.
     - `2 > 6`: This means the value in the second region must be greater than 6.
     - `3 < 10`: This means the value in the third region must be less than 10.
     - The only country that satisfies all these conditions is the second country (index 1).
  2. The second query has two conditions:
     - `1 > 15`: This means the value in the first region must be greater than 15.
     - `2 < 10`: This means the value in the second region must be less than 10.
     - No country satisfies both conditions, so the output should be `-1`.
  3. The third query has one condition:
     - `4 > 20`: This means the value in the fourth region must be greater than 20.
     - No country satisfies this condition, so the output should be `-1`.

This test input should effectively expose any issues in the buggy code, particularly in how it handles the bitwise OR operation and processes the queries.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.TreeSet;

public class Main {

	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[200000 * 19]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			boolean negate = false;
			if (c == '-') {
				negate = true;
				c = getc();
			}
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			if (negate) a = -a;
			return a;
		}

		long nextLong() {
			byte c = 0; while (c <= ' ') c = getc();
			long a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in), 200000 * 20);
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out), 50000000);

		String[] line = in.readLine().split(" ");
		int N = Integer.parseInt(line[0]);
		int K = Integer.parseInt(line[1]);
		int Q = Integer.parseInt(line[2]);

		int[][] A = new int[N+1][K+1];

		for (int i=1; i<=N; i++) {
			line = in.readLine().split(" ");
			for (int j=1; j<=K; j++) {
				A[i][j] = Integer.parseInt(line[j-1]);
			}
		}

		for (int i=2; i<=N; i++) {
			for (int j=1; j<=K; j++) {
				A[i][j] = A[i-1][j] | A[i][j];
				//out.write(A[i][j] + " ");
			}
			//out.write("\n");
		}

		/*for (int i=1; i<=N; i++) {
			for (int j=1; j<=K; j++) {
				out.write(A[i][j] + " ");
			}
			out.write("\n");
		}*/

		int[] lower = new int[K + 1];
		int[] upper = new int[K + 1];
		Arrays.fill(lower, 1);
		Arrays.fill(upper, N);

		for (int q=1; q<=Q; q++) {

			int M = Integer.parseInt(in.readLine());

			List<Integer> regions = new ArrayList<>();

			for (int m=1; m<=M; m++) {
				line = in.readLine().split(" ");
				int r = Integer.parseInt(line[0]);
				char ch = line[1].charAt(0);
				int val = Integer.parseInt(line[2]);

				regions.add(r);

				if (ch == '>') {
					int found = N + 1;
					for (int i=1; i<=N; i++) {
						if (A[i][r] > val) {
							found = i;
							break;
						}
					}
					lower[r] = Math.max(found, lower[r]);
				} else { // ch == '<'
					int found = 0;
					for (int i=N; i>=1; i--) {
						if (A[i][r] < val) {
							found = i;
							break;
						}
					}
					upper[r] = Math.min(found, upper[r]);
				}
			}

			boolean ok = true;
			for (final int reg : regions) {
				if (upper[reg] < lower[reg]) {
					ok = false;
				}
			}
			if (!ok) {
				out.write("-1\n");
			} else {

				//out.write("YES\n");
				int a = 1;
				int b = N;
				for (final int reg : regions) {
					a = Math.max(a, lower[reg]);
					b = Math.min(b, upper[reg]);
				}

				if (a > b) {
					out.write("-1\n");
				} else {
					out.write(a + "\n");
				}

			}

			for (final int reg : regions) {
				lower[reg] = 1;
				upper[reg] = N;
			}

		}

		in.close();
		out.close();
	}
}

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

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

Think step by step.

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

1. **Understand the Bug**: The code seems to have an issue with how it handles the bitwise OR operations and the queries. Specifically, the code might not correctly handle the boundaries for the `lower` and `upper` arrays, which could lead to incorrect results.

2. **Edge Cases**:
   - **Single Region, Single Query**: This can test the basic logic of the algorithm.
   - **Multiple Regions, Single Query**: This can test the interaction between different regions.
   - **Multiple Regions, Multiple Queries**: This can test the robustness of the algorithm with multiple constraints.

3. **Challenging Scenarios**:
   - **All Values are the Same**: This can test the algorithm's ability to handle uniform data.
   - **Values with Different Bitwise OR Results**: This can test the algorithm's ability to handle varying bitwise OR results.
   - **Queries with Contradictory Constraints**: This can test the algorithm's ability to handle impossible conditions.

4. **Small-Scale Test Input**:
   - We will use a small number of countries and regions to keep the input manageable.
   - We will include a few queries to ensure the algorithm is tested comprehensively.

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

```plaintext
5 3 4
1 2 3
4 5 6
7 8 9
10 11 12
13 14 15
3
1 > 3
2 < 10
3 > 5
2
1 < 10
2 > 10
1
3 > 15
2
1 > 10
2 < 10
```

### Explanation:
- **Input**:
  - 5 countries, 3 regions, and 4 queries.
  - The values in each region are designed to produce different bitwise OR results as you move from the first to the last country.
- **Queries**:
  1. The first query checks if there is a country where the first region has a value greater than 3, the second region has a value less than 10, and the third region has a value greater than 5. The correct answer is 2.
  2. The second query checks if there is a country where the first region has a value less than 10 and the second region has a value greater than 10. The correct answer is -1.
  3. The third query checks if there is a country where the third region has a value greater than 15. The correct answer is -1.
  4. The fourth query checks if there is a country where the first region has a value greater than 10 and the second region has a value less than 10. The correct answer is 3.

This test input should help identify any issues with the bitwise OR calculations and the handling of query constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse the Rivers
time_limit_ms: 2000
memory_limit_mb: 256
Description: A conspiracy of ancient sages, who decided to redirect rivers for their own convenience, has put the world on the brink. But before implementing their grand plan, they decided to carefully think through their strategy — that's what sages do.

There are $n$ countries, each with exactly $k$ regions. For the $j$-th region of the $i$-th country, they calculated the value $a_{i,j}$, which reflects the amount of water in it.

The sages intend to create channels between the $j$-th region of the $i$-th country and the $j$-th region of the $(i + 1)$-th country for all $1 \leq i \leq (n - 1)$ and for all $1 \leq j \leq k$.

Since all $n$ countries are on a large slope, water flows towards the country with the highest number. According to the sages' predictions, after the channel system is created, the new value of the $j$-th region of the $i$-th country will be $b_{i,j} = a_{1,j} | a_{2,j} | ... | a_{i,j}$, where $|$ denotes the [bitwise "OR"](http://tiny.cc/bitwise_or) operation.

After the redistribution of water, the sages aim to choose the most suitable country for living, so they will send you $q$ queries for consideration.

Each query will contain $m$ requirements.

Each requirement contains three parameters: the region number $r$, the sign $o$ (either "$<$" or "$>$"), and the value $c$. If $o$ = "$<$", then in the $r$-th region of the country you choose, the new value must be strictly less than the limit $c$, and if $o$ = "$>$", it must be strictly greater.

In other words, the chosen country $i$ must satisfy all $m$ requirements. If in the current requirement $o$ = "$<$", then it must hold that $b_{i,r} < c$, and if $o$ = "$>$", then $b_{i,r} > c$.

In response to each query, you should output a single integer — the number of the suitable country. If there are multiple such countries, output the smallest one. If no such country exists, output $-1$.

Input: The first line contains three integers $n$, $k$, and $q$ ($1 \leq n, k, q \leq 10^5$) — the number of countries, regions, and queries, respectively.

Next, there are $n$ lines, where the $i$-th line contains $k$ integers $a_{i,1}, a_{i,2}, \dots, a_{i,k}$ ($1 \leq a_{i,j} \leq 10^9$), where $a_{i,j}$ is the value of the $j$-th region of the $i$-th country.

Then, $q$ queries are described.

The first line of each query contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of requirements.

Then follow $m$ lines, each containing an integer $r$, a character $o$, and an integer $c$ ($1 \leq r \leq k$, $0 \leq c \leq 2 \cdot 10^9$), where $r$ and $c$ are the region number and the value, and $o$ is either "$<$" or "$>$" — the sign.

It is guaranteed that $n \cdot k$ does not exceed $10^5$ and that the sum of $m$ across all queries also does not exceed $10^5$.

Output: For each query, output a single integer on a new line — the smallest number of the suitable country, or $-1$ if no such country exists.

Examples:
input:
3 4 4
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
2
1 < 8
2 > 8
1
3 > 5
2
4 > 8
1 < 8
output:
2
-1
3
1

Note: In the example, the initial values of the regions are as follows:

$1$| $3$| $5$| $9$   ---|---|---|---   $4$| $6$| $5$| $3$   $2$| $1$| $2$| $7$      After creating the channels, the new values will look like this:

$1$| $3$| $5$| $9$   ---|---|---|---   $1 | 4$| $3 | 6$| $5 | 5$| $9 | 3$   $1 | 4 | 2$| $3 | 6 | 1$| $5 | 5 | 2$| $9 | 3 | 7$   $\downarrow$ $1$| $3$| $5$| $9$   ---|---|---|---   $5$| $7$| $5$| $11$   $7$| $7$| $7$| $15$      In the first query, it is necessary to output the minimum country number (i.e., row) where, after the redistribution of water in the first region (i.e., column), the new value will be greater than four and less than six, and in the second region it will be less than eight. Only the country with number $2$ meets these requirements.

In the second query, there are no countries that meet the specified requirements.

In the third query, only the country with number $3$ is suitable.

In the fourth query, all three countries meet the conditions, so the answer is the smallest number $1$.

Buggy Code:
import java.util.*;

public class E {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int q = sc.nextInt();
        long[][] b = new long[n][k];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < k; j++) {
                long a = sc.nextLong();
                if (i == 0) b[i][j] = a;
                else b[i][j] = b[i - 1][j] | a;
            }
        }

        StringBuilder out = new StringBuilder();
        for (int i = 0; i < q; i++) {
            int m = sc.nextInt();
            Set<Integer> v = new HashSet<>();
            for (int c = 1; c <= n; c++) v.add(c);

            for (int j = 0; j < m; j++) {
                int r = sc.nextInt() - 1;
                String o = sc.next();
                long c = sc.nextLong();
                Set<Integer> cur = new HashSet<>();

                for (int country : v) {
                    if (o.equals("<")) {
                        if (b[country - 1][r] < c) cur.add(country);
                    } else if (o.equals(">")) {
                        if (b[country - 1][r] > c) cur.add(country);
                    }
                }
                v = cur;
            }

            if (v.isEmpty()) out.append("-1\n");
            else out.append(Collections.min(v)).append("\n");
        }
        System.out.print(out.toString());
    }
}

**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 4 3
1 3 5 9
4 6 5 3
2 1 2 7
3
1 > 4
2 < 8
1 < 6
1
3 > 5
2
4 > 8
1 < 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <climits>
#include <iomanip>
#define ll long long
#define e endl
using namespace std;
                       /*بسم الله الرحمن الرحيم*/
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;cin>>n;int cnt=0;int a,b;
    for(int i=1;i<=n/2;i++){
    a=1; b=n-a;
    a++;
    b--;
    cnt++;
   }
cout<<cnt*2-1<<e;
}

  }

**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 maximize the failure probability. Let's analyze the code:

1. **Loop Condition**: The loop runs from `i=1` to `i<=n/2`. This means it only iterates up to half of `n`.
2. **Incrementing and Decrementing**: Inside the loop, `a` is incremented and `b` is decremented, but this is unnecessary and does not affect the count.
3. **Count Calculation**: The count is calculated as `cnt*2-1`.

The main issue is that the loop only runs up to `n/2`, which means it misses the pairs where `a` and `b` are equal to `n/2` if `n` is odd. This can lead to incorrect counts.

To expose this bug, we should use an input where `n` is an odd number. For example, `n = 5` should have 5 pairs: `(4,1), (3,2), (2,3), (1,4), (0,5)`. However, the buggy code will only count up to `n/2` (which is 2), and thus it will miss the pairs where `a` and `b` are greater than `n/2`.

Let's generate a test input with `t = 1` and `n = 5`:

```plaintext
1
5
```

This input will show that the buggy code incorrectly outputs `3` instead of the correct output `5`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<cstdio>
int arr[109]; int ans[109];
using namespace std;
int main()
{
	int n=0;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> arr[i];
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= arr[i]; j++)
		{
			if (j != (arr[i] - j))ans[i]++;
		}
	}
	for (int i = 1; i <= n; i++)
	{
		cout << ans[i] << "\n";
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
#include <iostream>
using namespace std;

int main(){
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        if (n % 2 == 0) {
            cout << n - 1 << 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:
```plaintext
1
2
```

This test input is designed to expose the bug in the buggy code. The code incorrectly prints `n - 1` for even `n`, but for `n = 2`, the correct output should be `1` because the only valid pair is `(1, 1)`. The buggy code will output `1` for `n = 2`, which is coincidentally correct, but this test case is still useful to ensure the code is handling the smallest even number correctly. For larger even numbers, the bug will be more apparent.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
n=int(input())
for i in range(n):
    a=int(input())
    c=0
    for i in range(1,7):
        for j in range(1,7):
            if i+j==a:
                c+=1
    print(c)

**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 the fact that the code is using a fixed range (1 to 6) for both `i` and `j`, which is incorrect for the problem's requirements. The correct range for `i` and `j` should be from 1 to `a-1` because `a` and `b` are positive integers and `a + b = n`.

Let's consider the following test cases to maximize the failure probability:

1. **Edge Case: Minimum Value of `n` (2)**
   - For `n = 2`, the only valid pair is `(1, 1)`.
   - The buggy code will correctly output `1` because the only valid pair `(1, 1)` is within the range 1 to 6.

2. **Edge Case: Maximum Value of `n` (100)**
   - For `n = 100`, the valid pairs are `(1, 99), (2, 98), ..., (99, 1)`.
   - The buggy code will incorrectly output `0` because no pairs `(i, j)` within the range 1 to 6 sum up to 100.

3. **Middle Case: `n` with a large number of pairs**
   - For `n = 10`, the valid pairs are `(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)`.
   - The buggy code will incorrectly output `9` because it only considers pairs within the range 1 to 6, which are `(4, 6), (5, 5), (6, 4)`.

Given the constraints, a small-scale test input that exposes the bug effectively is:

```plaintext
3
2
10
100
```

This input includes:
- The minimum value of `n` (2).
- A middle value of `n` (10) with a significant number of pairs.
- The maximum value of `n` (100) to show the incorrect handling of large values.

This should clearly demonstrate the bug 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 Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
for t in range(int(input())):
    n = int(input())
    if n % 2 == 0:
        print(n-1)

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

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

Think step by step.

Response:
```plaintext
3
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
t = int(input())
while t>0:
    n = int(input())
    sum = []
    for i in range(1,7):
        for j in range(1,7):
            sum.append(i+j)
    sum.sort()
    ans = 0
    for i in range(len(sum)):
        if n == sum[i]:
            ans+=1
    print(ans)
    t-=1

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the logic that the code is using to count the number of ordered pairs \((a, b)\) such that \(a = n - b\). The code incorrectly generates all possible sums of pairs \((i, j)\) where \(i\) and \(j\) are between 1 and 6, and then checks if \(n\) is in this list. This is not the correct approach to solve the problem.

The correct approach should be to count all pairs \((a, b)\) such that \(a + b = n\), where \(a\) and \(b\) are positive integers. For a given \(n\), the number of such pairs is \(n - 1\).

To expose the bug, we can provide a test case where \(n\) is greater than 12, as the code only considers sums up to 12 (since the maximum sum of two numbers from 1 to 6 is 12).

Let's consider \(n = 13\). The correct answer should be 12, but the buggy code will not find any pairs since it only considers sums up to 12.

Here is the test input:

```plaintext
1
13
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
import java.util.Scanner;

public class Solution {

	static int count(int total){
		if(total%2 == 1){
			return total;
		}
		return total-1;
	}
	public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();

		for(int i = 0 ; i < tc ; i++){
			int n = sc.nextInt();
			System.out.println(count(n));
		}
		sc.close();
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic of the `count` function. The function currently returns `total` if `total` is odd, and `total - 1` if `total` is even. However, the correct logic should be to return `n - 1` for any `n` because there are `n - 1` pairs of positive integers `(a, b)` such that `a + b = n`.

Let's consider the following edge cases and scenarios:
1. **Even and Odd `n` values**: The code incorrectly handles even and odd values.
2. **Boundary values**: The smallest and largest possible values for `n` (2 and 100).

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

```plaintext
3
2
3
100
```

Explanation:
- For `n = 2`, the correct output should be `1` (only one pair: `(1, 1)`).
- For `n = 3`, the correct output should be `2` (pairs: `(1, 2)` and `(2, 1)`).
- For `n = 100`, the correct output should be `99` (pairs: `(1, 99)`, `(2, 98)`, ..., `(99, 1)`).

The provided code will return:
- For `n = 2`: `1` (correct)
- For `n = 3`: `3` (incorrect, should be `2`)
- For `n = 100`: `99` (correct)

This test input will help identify the issue with the `count` function for odd values of `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
// # cook your dish here
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{

    public static void main (String[] args) throws java.lang.Exception
    { // your code goes here
        Scanner input = new Scanner(System.in);
        int T = input.nextInt();

        while(T>0)
        {
           int count = 0;
           int n = input.nextInt();
           for(int i=1;i<=50;i++)
           {
               for(int j=1;j<=50;j++)
               {
                   if(n == i+j)
                   {
                       count++;
                   }
               }
           }
           System.out.println(count);
                T--;
        }
    }

}

**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
2
100
51
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF (int)1e18
#define pb push_back
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
typedef vector<int> vi;
typedef pair<int,int> pi;
#define rep(i,a,b) for (int i = a; i <= b; i++)
#define sort(v) sort(v.begin(),v.end());
#define itr(v) v.begin(),v.end()
#define rsort(v) sort(v.rbegin(),v.rend());
#define lower(v,n) lower_bound(v.begin(),v.end(),n)-v.begin()
#define upper(v,n) upper_bound(v.begin(),v.end(),n)-v.begin()

void Solve() {
    int n;
    cin>>n;

    cout<<((n-1)/2)*2 + 1<<endl;
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        Solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code uses the formula `((n-1)/2)*2 + 1` to calculate the number of ordered pairs `(a, b)` such that `a = n - b`. This formula is incorrect because it does not correctly handle the calculation for even and odd values of `n`.

Let's break down the problem:

1. **Correct Calculation**:
   - For a given `n`, the number of ordered pairs `(a, b)` such that `a = n - b` is simply `n - 1` because `a` and `b` can take any value from `1` to `n-1`.

2. **Buggy Code Analysis**:
   - The formula `((n-1)/2)*2 + 1` is intended to handle both even and odd values of `n`, but it fails to correctly count the pairs for even `n`.

3. **Edge Cases**:
   - **Small Values**: Test with small values of `n` to ensure the formula is applied correctly.
   - **Even and Odd Values**: Test with both even and odd values of `n` to expose the bug in the formula.
   - **Boundary Values**: Test with the boundary values of `n` (i.e., 2 and 100).

4. **Test Input**:
   - We will choose a few test cases that cover the above scenarios.

Here is the test input:

```plaintext
4
2
3
4
100
```

This test input includes:
- `n = 2` (smallest even value)
- `n = 3` (smallest odd value)
- `n = 4` (another even value)
- `n = 100` (largest value in the range)

These values will help expose the bug 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 Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

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

void solve(){
    int n;
    cin>>n;
    if(n&1){
        cout<<(n/2)<<endl;
        return;
    }
    cout<<(n-1)<<endl;
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic where the code determines the number of ordered pairs \((a, b)\) such that \(a = n - b\). The code has a logical flaw in how it handles even and odd values of \(n\).

### Analyzing the Buggy Code
The code checks if \(n\) is odd:
- If \(n\) is odd, it outputs \(n / 2\).
- If \(n\) is even, it outputs \(n - 1\).

### Correct Logic
For any integer \(n\), the number of ordered pairs \((a, b)\) such that \(a = n - b\) is \(n - 1\). This is because \(a\) and \(b\) can take any value from 1 to \(n-1\), and for each \(a\), \(b\) is uniquely determined as \(b = n - a\).

### Test Input to Expose the Bug
To expose the bug, we need to provide an even \(n\) because the code incorrectly outputs \(n - 1\) for even \(n\), which is correct, but the logic for odd \(n\) is incorrect. We can use a small even \(n\) to demonstrate this.

### Chosen Test Input
Let's choose \(n = 4\). The correct number of ordered pairs \((a, b)\) for \(n = 4\) is 3, which the code correctly outputs. However, for \(n = 3\), the correct number of ordered pairs is 2, but the code outputs 1, which is incorrect.

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

This test input will show that the code fails for \(n = 3\) by outputting 1 instead of the correct answer 2, while it correctly outputs 3 for \(n = 4\).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
#!/usr/bin/python
# _*_ coding:utf-8 _*_

t = int(input())

for _ in range(t):
    d = int(input())
    ans = 1
    for i in range(d // 2 - 1):
        ans += 2
    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 generate a challenging test input that exposes the bug in the provided code, let's analyze the code and identify potential issues:

### Code Analysis
The provided code has the following logic:
1. It reads the number of test cases, `t`.
2. For each test case, it reads an integer `d`.
3. It initializes `ans` to 1.
4. It then iterates from 0 to `d // 2 - 1` and increments `ans` by 2 for each iteration.
5. Finally, it prints `ans`.

### Bug Identification
The bug in the code is in the loop that increments `ans` by 2. The loop runs `d // 2 - 1` times, which is incorrect. For a given `d`, the number of ordered pairs `(a, b)` such that `a = n - b` is `d - 1`. This is because `a` and `b` can take values from 1 to `d-1` in a way that `a + b = d`.

### Challenging Test Input
To expose this bug, we can use a test case where `d` is a small even number. This will make the difference between the correct answer and the answer given by the code more apparent.

Let's use `d = 4` as an example:
- Correct number of ordered pairs: `3` (pairs are (3,1), (2,2), (1,3))
- Buggy code output: `1` (since the loop runs 0 times and `ans` remains 1)

### Test Input
```plaintext
1
4
```

This test input will clearly show that the buggy code produces an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    c=0
    for j in range(1,51):
        for k in range(1,51):
            if j+k==n:
                c+=1
    print(c)

**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
2
100
51
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
/*Hai kathaa sangram ki
Vishwa ke kalyaan ki
Dharm adharm aadi anant
Satya asatya kalesh kalank
Swaarth ki kathaa parmaarth ki

Shakti hai bhakti hai
Janmo ki mukti hai
Jeevan ka ye sampurn saar hai
Yug yug se kan kan mein
Shristi ke darpan mein
Vedo ki vyathaa apaar hai
Karmo ki gaatha hai
Devo ki bhaasha hai
Sadiyo ke itihaas ka pramaan hai

Krishana ki mahima hai
Geetaa ki garima hai
Grantho ka granth ye mahaan hai

Mahabharat mahabharat
Mahabharat mahabharat
Mahabharat */

import java.util.*;
public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            System.out.println(2*(n/2-1)+((n%2)^1));
        }
        sc.close();
    }

}

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

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

Think step by step.

Response:
```plaintext
3
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int t = scanner.nextInt();

        for(int i = 0; i < t; i++){
            int n = scanner.nextInt();
            int x = (n/2);
            int count = 0;
            if(x == 1){
                System.out.println(1);
                continue;
            }
            count = (x-1)*2;
            if(n-x == x){
                count++;
            }else{
                count += 2;
            }

            System.out.println(count);
        }
    }
}

**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 maximize the failure probability. The buggy code has a few issues:

1. The code incorrectly handles the case when `n` is even.
2. The code incorrectly handles the case when `n` is odd.
3. The code has a logical error in counting the pairs.

Let's analyze the code and identify the problematic areas:

1. **Even `n`**: The code subtracts 1 from `x` and multiplies by 2, which is incorrect. It should simply be `n - 1`.
2. **Odd `n`**: The code adds 1 to the count if `n - x == x`, which is incorrect. It should add 1 for the middle pair and 2 for the pairs on either side.

To expose these issues, we can use the following test cases:

- **n = 2**: This is the smallest even number. The correct output should be 1.
- **n = 3**: This is the smallest odd number. The correct output should be 2.
- **n = 4**: This is another even number. The correct output should be 3.
- **n = 5**: This is another odd number. The correct output should be 4.
- **n = 100**: This is the largest even number in the range. The correct output should be 99.

Here is the test input:

```plaintext
5
2
3
4
5
100
```

This test input will help identify the issues in the buggy code by covering both even and odd values of `n` and including the boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

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

using namespace std;

using ll = long long;

int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n % 2 == 0) {
      cout << n - 1 << endl;
    } else {
      cout << n - 2 << 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 logic that determines the number of ordered pairs \((a, b)\) such that \(a = n - b\). The code incorrectly outputs \(n - 1\) for even \(n\) and \(n - 2\) for odd \(n\). However, the correct number of ordered pairs is always \(n - 1\) regardless of whether \(n\) is even or odd.

Let's consider the following test case to expose the bug:

1. **Test Case with \(n = 3\):**
   - The correct number of ordered pairs \((a, b)\) should be 2: \((2, 1)\) and \((1, 2)\).
   - The buggy code will output \(3 - 2 = 1\), which is incorrect.

2. **Test Case with \(n = 4\):**
   - The correct number of ordered pairs \((a, b)\) should be 3: \((3, 1)\), \((2, 2)\), and \((1, 3)\).
   - The buggy code will output \(4 - 1 = 3\), which is correct in this case.

3. **Test Case with \(n = 5\):**
   - The correct number of ordered pairs \((a, b)\) should be 4: \((4, 1)\), \((3, 2)\), \((2, 3)\), and \((1, 4)\).
   - The buggy code will output \(5 - 2 = 3\), which is incorrect.

To maximize the failure probability, we can use a combination of both even and odd values of \(n\). Here is a small-scale test input that includes both types of cases:

```plaintext
3
3
4
5
```

This input will help expose the bug 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 Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

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

using namespace std;

int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int a,b,c;
        cin>>a>>b>>c;
        cout<<a-1<<" "<<b-1<<" "<<c-1;
        break;
    }
}

**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
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    if n%2==0:
        print(n-1)
    else:
        print(n-2)

**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
2
3
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Easy Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
i=int(input())
l1=[]
for k in range(i):
    j=int(input())
    if j%2==0:
        z=j-1
    else:
        z=j
    l1.append(z)
for e in range(len(l1)):
    print(l1[e])

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

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

Think step by step.

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

1. **Edge Cases:**
   - The smallest possible value for \( n \) (2).
   - The largest possible value for \( n \) (100).
   - Values that are both even and odd to test the conditional logic.

2. **Behavior Analysis:**
   - The buggy code incorrectly calculates the number of ordered pairs \((a, b)\) by setting \( z = j - 1 \) if \( j \) is even and \( z = j \) if \( j \) is odd. This is incorrect because the number of ordered pairs \((a, b)\) such that \( a = n - b \) should be \( n - 1 \) for any \( n \).

3. **Test Input:**
   - We will choose values that highlight the incorrect logic, particularly focusing on even and odd values of \( n \).

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

```plaintext
4
2
3
4
100
```

Explanation:
- For \( n = 2 \), the correct number of ordered pairs is 1.
- For \( n = 3 \), the correct number of ordered pairs is 2.
- For \( n = 4 \), the correct number of ordered pairs is 3.
- For \( n = 100 \), the correct number of ordered pairs is 99.

The buggy code will output:
- For \( n = 2 \), it will output 1 (correct).
- For \( n = 3 \), it will output 3 (correct).
- For \( n = 4 \), it will output 3 (incorrect, should be 3).
- For \( n = 100 \), it will output 99 (incorrect, should be 99).

This test input will help identify the incorrect logic 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 Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cube is given an integer $$$n$$$. She wants to know how many ordered pairs of positive integers $$$(a,b)$$$ there are such that $$$a=n-b$$$. Since Cube is not very good at math, please help her!

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 99$$$) — the number of test cases.

The only line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 100$$$).

Output: For each test case, output the number of ordered pairs $$$(a, b)$$$ on a new line.

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

Note: In the first test case, the only ordered pair that works is $$$(a,b)=(1,1)$$$.

In the second test case, the three ordered pairs of $$$(a,b)$$$ that work are $$$(3,1), (2,2), (1,3)$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            if (n == 0) System.out.println(0);
            if (n % 2 == 0){
                System.out.println(n - 1);
            } else {
                System.out.println(n - 2);
            }
        }
    }
}

**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 the conditions where the code might fail. The code has the following logic:

1. If `n == 0`, it prints `0`.
2. If `n % 2 == 0`, it prints `n - 1`.
3. If `n % 2 != 0`, it prints `n - 2`.

However, the problem description specifies that `n` is always between 2 and 100 (inclusive), so the condition `n == 0` should never be triggered. The bug in the code is that it incorrectly calculates the number of ordered pairs `(a, b)` for odd and even values of `n`.

For an even `n`, the correct number of ordered pairs is `n - 1`.
For an odd `n`, the correct number of ordered pairs is `n - 1` as well.

The code incorrectly prints `n - 2` for odd `n`, which is incorrect.

To expose this bug, we can provide a test case with an odd value of `n`. A small-scale test input that will clearly show the bug is:

```plaintext
1
5
```

This input will cause the code to print `3` (which is `5 - 2`), but the correct output should be `4` (which is `5 - 1`).