**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
'''
F. Закрась строки и столбцы

У вас есть n прямоугольников, i-й из которых имеет ширину ai и высоту bi.

Вы можете неограниченное количество раз выполнить следующую операцию:
 выбрать прямоугольник и клетку в нём, а затем закрасить её.

Каждый раз, когда вы закрашиваете целиком какую-либо строку или столбец,
 вы получаете 1 балл. Ваша задача — набрать хотя бы k баллов за как
 можно меньшее число операций.

Допустим, у вас есть прямоугольник ширины 6 и высоты 3. Вы можете
 набрать 4 очка, закрасив все клетки в 4-х любых столбцах, выполнив
 таким образом 12 операций.

Входные данные

Первая строка содержит целое число t(1≤t≤100) — количество наборов
 входных данных. Далее следуют описания наборов входных данных.

Первая строка описания каждого набора входных данных содержит два целых
 числа n и k (1≤n≤1000,1≤k≤100) — количество прямоугольников в наборе и
 требуемое количество баллов.

Следующие n строк содержат описания прямоугольников. i-я строка содержит
 два целых числа ai и bi (1≤ai,bi≤100) — ширину и высоту i-го
 прямоугольника.

Гарантируется, что сумма значений n по всем наборам входных данных не
 превосходит 1000.

Выходные данные

Для каждого набора входных данных выведите одно целое число —
 минимальное количество операций, необходимое для того, чтобы набрать
 хотя бы k очков. Если набрать хотя бы k очков невозможно, выведите -1.

Пример
Входные данные

7
    1 4
6 3
    1 5
4 4
    5 10
1 1
1 1
1 1
1 1
1 1
    2 100
1 2
5 6
    3 11
2 2
3 3
4 4
    3 25
9 2
4 3
8 10
    4 18
5 4
8 5
8 3
6 2

Выходные данные

12
14
5
-1
17
80
35

# Отсортировать прямоугольники и ...
 по минимальной стороне, но 2х6 и 3х3
 k: 1 2 3 4  5  6  7  8
 1: 2 4 6 8 10 11 12 12
 2: 3 5 7 8  9  9
'''
# make list for rectangle rect=(min, max)
def make_table(rect):
    res = [0]
    a, b = rect
    #used, point = 0, 0
    used = 0
    while a * b:
        if a > b:
            used += b
            a -= 1
        else:  # a <= b:
            used += a
            b -= 1
        # point += 1
        res.append(used)
    res.append(used)  # last ceil finished row and col
    # res[point] = used
    return res

#
def check_for_best(ans_prev, k_prev, key_prev, a, ans):
    for key in a:

        key = tuple(key)
        if key[0] > key_prev[0]:
            break

    else:  # can't find biggest rect
        return ans

    # key, key_prev
    delta = ans - ans_prev
    b1, b2 = data_rect[key_prev], data_rect[key]
    for i, v in enumerate(b2):
        if 0 <= k_prev - i < len(b1) and v + b1[k_prev - i] < delta:  # bingo!
            return ans_prev + v + b1[k_prev - i]
    return ans

t = int(input())
data_rect = {}  # common
for _ in range(t):
    n, k = map(int, input().split())
    a = sorted([sorted(list(map(int, input().split()))) for _ in range(n)])

    for key in a:

        key = tuple(key)
        if key not in data_rect:
            data_rect[key] = make_table(key)

    # Fast? but wrong
    ans = 0
    for key in a:

        key = tuple(key)
        if len(data_rect[key]) - 1 <= k:

            # Full
            ans += data_rect[key][-1]
            k -= len(data_rect[key]) - 1

            if not k:
                break

        else:  # max > k

            # part
            ans_prev = ans
            k_prev = k
            key_prev = key

            ans += data_rect[key][k]
            k = 0

            if t == 7:
                ans = check_for_best(ans_prev, k_prev, key_prev, a, ans)

            break

    print(-1 if k else ans)

'''
#3 test
            836 823
'''

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

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

Think step by step.

Response:
```plaintext
1
3 5
2 3
3 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
def solve():
    n, k = map(int, input().split())
    dp = [[float('inf') for _ in range(k + 1)] for _ in range(n)]
    a = []
    b = []
    for _ in range(n):
        a_, b_ = map(int, input().split())
        a.append(a_)
        b.append(b_)
    pre_calc = [[float('inf') for _ in range(k + 2)] for _ in range(n)]
    for i in range(n):
        a_ = a[i]
        b_ = b[i]
        count = 0
        k_ = 1
        while k_ <= k and a_ > 0 and b_ > 0:
            a_, b_ = max(a_, b_), min(a_, b_)
            count += b_
            a_ -= 1
            pre_calc[i][k_] = count
            k_ += 1
            if a_ + b_ == 1:
                pre_calc[i][k_] = count
    dp[0] = pre_calc[0]
    for i in range(n):
        dp[i][0] = 0
    for i in range(1, n):
        for j in range(k):
            for z in range(1, k + 2):
                if j + z > k:
                    continue
                dp[i][j + z] = min(dp[i][j + z], dp[i - 1][j] + pre_calc[i][z])
            dp[i][j] = min(dp[i][j], dp[i - 1][j])
    if dp[n - 1][k] == float('inf'):
        return -1
    return dp[n - 1][k]

t = int(input())
for _ in range(t):
    print(solve())

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

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

Think step by step.

Response:
```plaintext
1
3 7
1 2
2 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
import java.util.*;

public class Main {
    private static int n;
    private static int k;
    private static int[] a = new int[1005];
    private static int[] b = new int[1005];
    private static int max = (int)1e8;

    private static void solve() {
        k++;
        int[] dp = new int[k + 1];
        dp[0]=0;
        for (int i = 1; i <= k; i++) {
            dp[i]=max;
        }

        for(int i=1;i<=n;i++){
            //a[i],b[i]
            for(int j=k;j>=0;j--){
                //dp[j]
                int A=a[i];
                int B=b[i];
                int t=0;
                for(int l=1;l<=Math.min(j,a[i]+b[i]);l++){
                    if(l==a[i]+b[i]-1)continue;
                    t+=Math.min(A,B);
                    if(A>B){
                        A--;
                    }
                    else {
                        B--;
                    }
                    dp[j]=Math.min(dp[j],dp[j-l]+t);
                }
            }
        }

        k--;
        if(dp[k]<max){
            System.out.println(dp[k]);
        }
        else if(dp[k+1]<max){
            System.out.println(dp[k+1]);
        }
        else {
            System.out.println(-1);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- != 0) {
            n = sc.nextInt();
            k = sc.nextInt();
            for (int i = 1; i <= n; i++) {
                a[i] = sc.nextInt();
                b[i] = sc.nextInt();
            }
            solve();
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that maximize the failure probability in faulty solutions. Here are the steps to create such a test input:

1. **Single Rectangle with Large k**: This scenario tests the case where a single rectangle needs to score a large number of points, which might expose issues in the dynamic programming logic.

2. **Multiple Rectangles with Small Dimensions**: This scenario tests the case where multiple small rectangles are used to score a large number of points, which might expose issues in the handling of multiple rectangles.

3. **Impossible Case**: This scenario tests the case where it is impossible to score the required number of points, which should return -1.

4. **Mixed Case**: This scenario combines different types of rectangles to ensure the algorithm can handle a variety of inputs.

Let's create a small-scale test input that includes these scenarios:

```plaintext
4
1 10
10 10
2 5
1 1
1 1
2 6
1 2
2 2
3 15
3 3
4 4
5 5
```

### Explanation:
1. **Test Case 1**: A single rectangle with dimensions 10x10 and a required 10 points. This should be solvable by coloring 10 rows or columns.
2. **Test Case 2**: Two rectangles with dimensions 1x1 each and a required 5 points. This is impossible to achieve.
3. **Test Case 3**: Two rectangles with dimensions 1x2 and 2x2 each and a required 6 points. This should be solvable by coloring 6 columns or rows.
4. **Test Case 4**: Three rectangles with dimensions 3x3, 4x4, and 5x5 each and a required 15 points. This should be solvable by coloring 15 columns or rows.

This test input should help expose any issues in the buggy code, especially in handling large k values, small dimensions, and impossible cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class ColorRowsAndColumns {
	public static PrintWriter out;
	public static void main(String[] args)throws IOException{
        JS sc=new JS();
        out=new PrintWriter(System.out);
        int t=sc.nextInt();
        outer: while(t-->0) {
        	int n=sc.nextInt();
        	int k=sc.nextInt();
        	int a[]=new int[n];
        	int b[]=new int[n];
        	long sum=0;
        	for(int i=0;i<n;i++) {
        		a[i]=sc.nextInt();
        		b[i]=sc.nextInt();
        		sum+=a[i];
        		sum+=b[i];
        	}
        	long cost[][]=new long[n][k+1];
        	for(int i=0;i<n;i++) {
        		Arrays.fill(cost[i], Integer.MAX_VALUE);
        	}
        	for(int i=0;i<n;i++) {
        		for(int j=0;j<=a[i];j++) {
        			for(int l=0;l<=b[i];l++) {
            			int cnt=0;
            			cnt+=j*b[i];
            			cnt+=l*a[i];
            			cnt-=j*l;
            			if(j+l<=k) {
            				cost[i][j+l]=Math.min(cost[i][j+l], cnt);
            			}
            		}
        		}
        	}
        	if(sum<k) {
        		out.println("-1");
        		continue outer;
        	}
        	long dp[][]=new long[n+1][k+1];
        	for(int i=0;i<=n;i++) {
        		Arrays.fill(dp[i], Integer.MAX_VALUE);
        	}
        	dp[0][0]=0;
        	for(int i=1;i<=n;i++) {
        		for(int j=1;j<=k;j++) {
        			for(int l=0;l<=j;l++) {
        				if(l==j) {
        					dp[i][j]=Math.min(dp[i][j], dp[i-1][j]);
        					continue;
        				}
        				dp[i][j]=Math.min(dp[i][j], dp[i-1][l]+cost[i-1][j-l]);
        			}
        		}
        	}
        	out.println(dp[n][k]);
        }
		out.close();
	}
	static void add(HashMap<Integer,Integer> A, int key){
		A.put(key,A.getOrDefault(key,0)+1);
	}
	static void remove(HashMap<Integer,Integer>A, int key){
		if(!A.containsKey(key))return;
		A.put(key,A.get(key)-1);
		if(A.get(key)==0){
			A.remove(key);
		}
	}
	static class Pair implements Comparable<Pair>{
		int x, y;
		public Pair(int x, int y) {
			this.x=x;this.y=y;
		}
		public int compareTo(Pair p) {
			if(y==p.y) {
				return Integer.compare(x, p.x);
			}
			return Integer.compare(y, p.y);
		}
	}
	static void sort(int a[]) {
		ArrayList<Integer>al=new ArrayList<>();
		for(int i=0;i<a.length;i++) {
			al.add(a[i]);
		}
		Collections.sort(al);
		Collections.reverse(al);
		for(int i=0;i<a.length;i++) {
			a[i]=al.get(i);
		}
	}
	static class JS {
	    public int BS = 1<<16;
	    public char NC = (char)0;
	    byte[] buf = new byte[BS];
	    int bId = 0, size = 0;
	    char c = NC;
	    double num = 1;
	    BufferedInputStream in;

	    public JS() {
	        in = new BufferedInputStream(System.in, BS);
	    }

	    public JS(String s) throws FileNotFoundException {
	        in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
	    }

	    public char nextChar(){
	        while(bId==size) {
	            try {
	                size = in.read(buf);
	            }catch(Exception e) {
	                return NC;
	            }
	            if(size==-1)return NC;
	            bId=0;
	        }
	        return (char)buf[bId++];
	    }

	    public int nextInt() {
	        return (int)nextLong();
	    }

	    public long nextLong() {
	        num=1;
	        boolean neg = false;
	        if(c==NC)c=nextChar();
	        for(;(c<'0' || c>'9'); c = nextChar()) {
	            if(c=='-')neg=true;
	        }
	        long res = 0;
	        for(; c>='0' && c <='9'; c=nextChar()) {
	            res = (res<<3)+(res<<1)+c-'0';
	            num*=10;
	        }
	        return neg?-res:res;
	    }

	    public double nextDouble() {
	        double cur = nextLong();
	        return c!='.' ? cur:cur+nextLong()/num;
	    }

	    public String next() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c>32) {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public String nextLine() {
	        StringBuilder res = new StringBuilder();
	        while(c<=32)c=nextChar();
	        while(c!='\n') {
	            res.append(c);
	            c=nextChar();
	        }
	        return res.toString();
	    }

	    public boolean hasNext() {
	        if(c>32)return true;
	        while(true) {
	            c=nextChar();
	            if(c==NC)return false;
	            else if(c>32)return true;
	        }
	    }
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The main issues in the buggy code are:

1. **Incorrect Calculation of `sum`**: The code incorrectly calculates the sum of all rows and columns, which is used to determine if it's impossible to score at least `k` points. This can lead to incorrect early termination.
2. **Dynamic Programming (DP) Initialization and Updates**: The DP initialization and updates might not handle all cases correctly, especially when the number of points required is large relative to the dimensions of the rectangles.

Given these issues, let's create a test case that can expose these bugs:

1. **Test Case with a Single Rectangle**: This can help check if the sum calculation is correct and if the DP initialization and updates handle a simple case.
2. **Test Case with Multiple Rectangles**: This can help check if the DP updates handle multiple rectangles correctly.
3. **Test Case with a Large `k` Value**: This can help check if the code correctly handles cases where the required points are large.

Here is a small-scale test input that includes these scenarios:

```plaintext
3
1 5
4 4
2 10
1 2
5 6
3 11
2 2
3 3
4 4
```

### Explanation:
1. **Test Case 1**:
   - **n = 1, k = 5**
   - **Rectangle: 4x4**
   - **Expected Output**: 16 (since we need to color 4 rows or 4 columns to score 4 points, and then 2 more points can be scored by coloring 2 more rows or columns).

2. **Test Case 2**:
   - **n = 2, k = 10**
   - **Rectangles: 1x2, 5x6**
   - **Expected Output**: 17 (since we need to color 2 columns in the first rectangle and 5 columns in the second rectangle to score 7 points, and then 3 more points can be scored by coloring 3 more rows or columns).

3. **Test Case 3**:
   - **n = 3, k = 11**
   - **Rectangles: 2x2, 3x3, 4x4**
   - **Expected Output**: 35 (since we need to color 2 rows and 2 columns in the first rectangle, 3 rows and 3 columns in the second rectangle, and 4 rows and 4 columns in the third rectangle to score 11 points).

This test input should help expose the bugs in the provided buggy code by testing various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define f(i, x, n) for (int i = x; i < n; i++)
#define pb push_back
#define DB_op_baki_sab_topi ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define fi first
#define se second

void solve()
{
    int n, k;
    cin>>n>>k;
    vector<pair<int, int>> v(n+1);
    f(i, 1, n+1){
        cin>>v[i].fi>>v[i].se;
        if(v[i].fi>v[i].se) swap(v[i].fi, v[i].se);
    }
    vector<vector<int>> cost(n+1, vector<int> (k+1, INT_MAX));
    vector<vector<int>> dp(n+1, vector<int>(k+1, INT_MAX));
    dp[0][0]=0;
    f(i, 1, n+1){
        cost[i][0]=0;
        int x=v[i].fi, y=v[i].se;
        f(j, 1, min(v[i].fi+v[i].se+1, k+1)){
            cost[i][j]=x+cost[i][j-1];
            if(i==1) dp[i][j]=cost[i][j];
            y--;
            if(x>y) swap(x, y);
        }
    }
    f(i, 1, n+1){
        dp[i][0]=0;
        if(i>1) dp[i][1]=min(dp[i][1], dp[i-1][1]);
    }
    f(i, 2, n+1){
        f(j, 2, k+1){
            f(p, 1, j+1){
                dp[i][j]=min(dp[i][j], dp[i-1][j-p]+cost[i][p]);
            }
            dp[i][j]=min(dp[i][j], dp[i-1][j]);
        }
    }
    dp[n][k]>=INT_MAX ? cout<<"-1"<<'\n' : cout<<dp[n][k]<<'\n';
}

signed main()
{
    DB_op_baki_sab_topi;
    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:
```plaintext
2
3 6
3 3
3 3
3 3
1 10
10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define ft first
#define sd second
#define cn cout<<"NO"<<endl
#define cy cout<<"YES"<<endl
using namespace std;
const int mod = 1e9 + 7;

int n,k;
int a[1010],b[1010];

void solve()
{
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>a[i]>>b[i];
    vector<vector<int>>res(n+10,vector<int>(210,1e18));
    for(int i=1;i<=n;i++)
    {
        res[i][0]=0;
        int h=a[i],l=b[i];
        int sco=1,sum=0;
        while(h>=1 && l>=1)
        {
            if(h==1 && l==1){res[i][sco+1]=sum+1;break;};
            if(h<=l)sum+=h,l--;
            else sum+=l,h--;
            res[i][sco++]=sum;
        }
    }
    vector<vector<int>>dp(n+10,vector<int>(k+10,1e18));
    for(int i=0;i<=n;i++)dp[i][0]=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=k;j++)
        {
            for(int v=0;v<=j;v++)
            {
                if(res[i][v]!=1e18)
                {
                    dp[i][j]=min(dp[i][j],res[i][v]+dp[i-1][j-v]);
                }
            }
        }
    }
    if(dp[n][k]==1e18)cout<<-1<<endl;
    else cout<<dp[n][k]<<endl;
}

signed main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int _; cin>>_;
    while(_--)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
1
1 101
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
import functools
import sys
# sys.setrecursionlimit(2000000)
from collections import defaultdict
import collections
#from sortedcontainers import SortedList
import math
import heapq
from collections import deque
import bisect
import sys
input = sys.stdin.readline

def compute(a,b):
    x,y = min(a,b),max(a,b)
    res = []
    while x < y:
        res.append([x,1])
        y -= 1

    res.extend(list(y))

# def solve(N,K,mat):
#     graph = collections.defaultdict(list)
#     curr = []

#     for x,y in mat:
#         graph[str(x*y)].append([min(x,y),max(x,y)])

#     res = []
#     for key in graph:
#         graph[key].sort(key=lambda x:-x[0])
#         res.append(int(key))

#     res.sort()
#     #print('curr',curr)
#     curr.sort() #(key=lambda x: (x[0],-x[1]))
#     ans = 0
#     for i in range(len(curr)):
#         tot,x,y = curr[i]
#         if x+y <= K:
#             ans += tot
#             K -= x+y
#             if K ==0:
#                 return ans
#         else:
#             j = i
#             new_arr = curr[j:]
#             new_arr.sort(key=lambda x:x[1])
#             for z,x,y in new_arr:
#                 while x < y:
#                     K -= 1
#                     ans += x
#                     y -= 1
#                     if K == 0:
#                         return ans
#                 ans += x
#                 K -= 1
#                 if K ==0:
#                     return ans
#                 #print("K",K,'ans',ans,'xy',x,y)
#                 for z in range(x-1,0,-1):
#                     ans += z
#                     K -= 1
#                     if K == 0:
#                         return ans
#                     ans += z
#                     K -= 1
#                     if K == 0:
#                         return ans
#                 return ans

#     return -1

def compute(curr1,blocked,K):
    tot,blocked_x,blocked_y = curr1[blocked]
    blocked_x = min(blocked_x,blocked_y)
    blocked_y = max(blocked_x,blocked_y)
    #print(blocked_x,blocked_y)
    curr = curr1[:blocked] + curr1[blocked+1:]
    #print(len(curr))
    ans = 0
    while curr and K >= curr[-1][1]+curr[-1][2]:
        zeb,x,y = curr.pop()
        ans += zeb
        K -= x+y

    if K == 0:
        return ans
    #print('ans',ans,'K',K)
    x,y = blocked_x,blocked_y
    #print('x',x,'y',y,'ans',ans)
    while x < y:
        K -= 1
        ans += x
        y -= 1
        if K == 0:
            return ans
        #print('ans',ans,'K',K)
    if x == 1:
        ans += x
        K -= 2
    else:
        ans += x
        K -= 1
    #print('ans',ans,'K',K)
    if K ==0:
        return ans
    #print("K",K,'ans',ans,'xy',x,y)
    for z in range(x-1,0,-1):
        if z == 1:
            ans += z
            K -= 2
            if K <= 0:
                return ans
        else:
            ans += z
            K -= 1
            if K == 0:
                return ans
            ans += z
            K -= 1
            if K == 0:
                return ans
    if K > 0:
        return math.inf
    return ans

def solve(N,K,mat):
    #graph = collections.defaultdict(list)
    curr = []
    for j,(x,y) in enumerate(mat):
        curr.append([x*y,min(x,y),max(x,y)])
    curr.sort(reverse=True)

    ans = math.inf
    for blocked in range(len(mat)):
        ans = min(ans,compute(curr,blocked,K))
    if ans == math.inf:
        return -1
    return ans

    # for x,y in mat:
    #     graph[str(x*y)].append([min(x,y),max(x,y)])

    # res = []
    # for key in graph:
    #     graph[key].sort(key=lambda x:-x[0])
    #     res.append(int(key))

    # res.sort()
    # #print(res)
    # ans = 0

    # for key in res:
    #     curr = graph[str(key)]
    #     #print('curr',curr)
    #     for i in range(len(curr)):
    #         x,y = curr[i]
    #         tot = x*y
    #         if x+y <= K:
    #             ans += tot
    #             K -= x+y
    #             if K ==0:
    #                 return ans
    #         else:
    #             j = i
    #             new_arr = curr[j:]
    #             new_arr.sort(key=lambda x:x[0])
    #             for x,y in new_arr:
    #                 while x < y:
    #                     K -= 1
    #                     ans += x
    #                     y -= 1
    #                     if K == 0:
    #                         return ans
    #                 ans += x
    #                 K -= 1
    #                 if K ==0:
    #                     return ans
    #                 #print("K",K,'ans',ans,'xy',x,y)
    #                 for z in range(x-1,0,-1):
    #                     ans += z
    #                     K -= 1
    #                     if K == 0:
    #                         return ans
    #                     ans += z
    #                     K -= 1
    #                     if K == 0:
    #                         return ans
    #                 return ans
    # return -1

t = int(input())
for j in range(1 , t +1):
    #S = input()
    #N = int(input())
    #A = input().strip()
    #mat = []
    #A,B = input().split(" ")
    N, K = list(map(int,input().split(" ")))
    mat = []
    for _ in range(N):
        mat.append(list(map(int,input().split(" "))))

    #arr = list(map(int,input().split(" ")))

    result = solve(N,K,mat)
    print(result)
    #print("Case #{}: {}".format(j, result))
    ## print('{:.8f}'.format(result))

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test input that challenges the algorithm's logic. The code seems to have issues with how it handles the minimization of operations, especially when dealing with large values of `k` and multiple rectangles. Here are the steps to think through the test case:

1. **Edge Case: Single Rectangle with High `k`**:
   - If `k` is very high and there's only one rectangle, the code should return `-1` if it's impossible to achieve the required points.
   - Example: `1 1000` with a rectangle of size `1 1`.

2. **Multiple Rectangles with Low `k`**:
   - If `k` is low and there are multiple rectangles, the code should efficiently find the minimum operations.
   - Example: `2 2` with rectangles `1 1` and `2 2`.

3. **Mixed Sizes and High `k`**:
   - If `k` is high and there are multiple rectangles of different sizes, the code should handle the combination of operations efficiently.
   - Example: `3 10` with rectangles `5 5`, `3 3`, and `2 2`.

4. **Rectangles with Large Dimensions**:
   - If the rectangles have large dimensions, the code should handle the operations without running into performance issues.
   - Example: `1 100` with a rectangle of size `10 10`.

5. **Multiple Small Rectangles**:
   - If there are many small rectangles, the code should handle the cumulative operations efficiently.
   - Example: `10 10` with rectangles `1 1` repeated 10 times.

Given these considerations, let's create a small-scale test input that combines some of these scenarios:

```plaintext
3
1 1000
1 1
2 2
1 1
2 2
3 10
5 5
3 3
2 2
```

This input includes:
- A single rectangle with a very high `k` to test the impossible case.
- Multiple small rectangles with a low `k` to test efficient operations.
- Multiple rectangles with mixed sizes and a moderate `k` to test the combination of operations.

This should help expose any issues in the logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
def Caul(min_index):
    op = max(B[min_index][0], B[min_index][1])
    cp = min(B[min_index][0], B[min_index][1])
    if op-cp>=k-ans:
        rp=cp*(k-ans)
    else:
        rp=cp*(op-cp)+cp
        tmp = (k - ans-(op-cp)-1) // 2
        temp = (k - ans-(op-cp)-1) % 2
        cp=cp-1
        rp+=tmp*(cp+cp-tmp+1)+temp*(cp-tmp)
    return rp

def FindMin(array,ans):
    min_index=-1
    min_value=101
    min_cat=0
    for i in range(n):
        if array[i]==-2:
            continue
        if min_index==-1:
            min_index=i
            min_cat=array[i]-(k-ans)
            min_value = abs(array[i] - (k - ans))
        elif array[i]-(k-ans)>=0:
                if min_cat <= 0:
                    if Caul(i) < B[MinIndex][0] * B[MinIndex][1] or (k-ans)/Caul(i)>=array[min_index]/(B[min_index][0] * B[min_index][1]):
                        if array[i] - (k - ans) <= min_value:
                            min_index = i
                            min_value = abs(array[i] - (k - ans))
                            min_cat = array[i] - (k - ans)
                else:
                    if Caul(i) < Caul(min_index)  or (k-ans)/Caul(i)>=(k-ans)/Caul(min_index):
                        if array[i] - (k - ans) <= min_value:
                            min_index = i
                            min_value = abs(array[i] - (k - ans))
                            min_cat = array[i] - (k - ans)
        elif array[i]-(k-ans)<0:
                if min_cat <= 0:
                    if B[i][0] * B[i][1] < B[min_index][0] * B[min_index][1]:
                        if abs(array[i] - (k - ans)) <= min_value or array[i]/(B[i][0]*B[i][1])>=array[min_index]/(B[min_index][0]*B[min_index][1]):
                            min_index = i
                            min_value = abs(array[i] - (k - ans))
                            min_cat=array[i]-(k-ans)
                else:
                    if B[i][0] * B[i][1] < Caul(min_index) or array[i]/(B[i][0]*B[i][1])>=(k-ans)/Caul(min_index):
                        if abs(array[i] - (k - ans)) <= min_value:
                            min_index = i
                            min_value = abs(array[i] - (k - ans))
                            min_cat = array[i] - (k - ans)
    return min_index
t=int(input())
for i in range(t):
    ans=0
    operation=0
    n,k=[ int(o) for o in input().strip('\n').split(' ')]
    B=[]
    for p in range(n):
        tp = []
        for o in input().strip('\n').split(' '):
            o=int(o)
            tp.append(o)
        B.append(tp)
    Min=[0]*n
    MinNext=[0]*n
    MinIndex=0
    for o in range(len(B)):
        r=0
        p=B[o][0]+B[o][1]
        if B[o][1]<B[o][0]:
            r=1
        Min[o]=r
        MinNext[o]=p
    tmp=0
    for o in B:
        tmp+=o[0]+o[1]
    if tmp<k:
        print(-1)
        continue
    while ans<k:
        MinIndex=FindMin(MinNext,ans)
        if B[MinIndex][0]+B[MinIndex][1]>k-ans:
            operation+=Caul(MinIndex)
            ans=k+1
            break
        else:
            operation+=B[MinIndex][0]*B[MinIndex][1]
            ans+=B[MinIndex][0]+B[MinIndex][1]
            MinNext[MinIndex]=-2
    if ans>=k:
        print(operation)

**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
1 1
100 1
2 2
100 1
1 100
2 10
50 50
50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

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

public class Main
{
    //    static Scanner sc = new Scanner( System.in );

    public static void main( String[] ksc ) throws Exception
    {
        int t = nextInt();
        while( t-- > 0 ) solve();
        pw.flush();
    }

    private static void solve() throws Exception
    {
        int n = nextInt();
        int k = nextInt();
        ArrayList< r > arr = new ArrayList<>();
        int sum = 0;

        for( int i = 0; i < n; i++ )
        {
            int a = nextInt();
            int b = nextInt();
            arr.add( new r( a, b ) );
            sum += a + b;
        }

        Collections.sort( arr );

        if( sum < k ) pw.println( -1 );
        else if( sum == k )
        {
            int key = 0;
            for( int i = 0; i < n; i++ ) key += arr.get( i ).min * arr.get( i ).max;
            pw.println( key );
        } else
        {
            int tot = 0;

            for( int i = 0; i < n; i++ )
            {
                int p = arr.get( i ).min * arr.get( i ).max;
                if( k >= arr.get( i ).min + arr.get( i ).max )
                {
                    k -= arr.get( i ).min + arr.get( i ).max;
                    tot += p;
                } else
                {
                    if( k == 0 ) pw.println( tot );
                    else
                    {
//                        pw.print(tot + "!!!");
                        r x = arr.get( i );

                        if( x.max - x.min < k )
                        {
                            tot += ( x.max - x.min ) * x.min;
                            k -= x.max - x.min;
                            int pop = x.min;
//                            pw.print(pop + " " + k + "    ");
                            while( k > 0 )
                            {
                                tot += pop;
                                k--;
                                if( k <= 0 ) break;
                                tot += --pop;
                                k--;
                            }
                        }else
                        {
                            tot += k * x.min;
                        }
                    }

                    pw.println( tot );
                    break;
                }
            }
        }
    }

    static class r implements Comparable< r >
    {
        int min, max;

        r( int a, int b )
        {
            this.min = Math.min( a, b );
            this.max = Math.max( a, b );
        }

        @Override
        public int compareTo( r o )
        {
            double o_m = o.min * o.max;
            double t_m = this.max * this.min;
            if( o_m != t_m ) return o_m - t_m > 0 ? -1 : 1 ;
            return this.min - o.min;
        }
    }

    static BufferedReader bf = new BufferedReader( new InputStreamReader( System.in ) );
    static StreamTokenizer st = new StreamTokenizer( bf );
    static PrintWriter pw = new PrintWriter( new OutputStreamWriter( System.out ) );

    private static int nextInt() throws IOException
    {
        st.nextToken();
        return (int) st.nval;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 6
3 2
2 3
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class ColorRow {
	public static void main(String[] args) throws IOException {
		Scanner scanner = new Scanner(System.in);
		int t=scanner.nextInt();
		for(int i=0;i<t;i++) {
			int n=scanner.nextInt();
			int k=scanner.nextInt();

			ArrayList<Point> arr1=new ArrayList();
			for(int j=0;j<n;j++) {
				int x=scanner.nextInt();
				int y=scanner.nextInt();
			   arr1.add(new Point(x,y)) ;
			}
			Collections.sort(arr1);
			int total=0;int count=0;int c=0;int kk=k;
			while(count<k && c<arr1.size()) {
				if (arr1.get(c).x +arr1.get(c).y <=kk) {
					count=count+arr1.get(c).x +arr1.get(c).y;
					total=total+arr1.get(c).x *arr1.get(c).y;
					kk=k-count;
					arr1.remove(c);
					if (c!=0) {
						c=0;
					}
				}
				else if (kk<=arr1.get(c).x +arr1.get(c).y-3) {
					int u=arr1.get(c).y - arr1.get(c).x;
					int uu=Math.abs(u );
					int z=kk - uu - 1;
					if (z<0) {
						z=0;
					}
					int s=(int) Math.ceil(((z)*(z+2))/4.0);
					 total=(int) (total+ arr1.get(c).min()*kk -(s));
					count=count+kk;
				}else {
					c++;
				}
			}
			if (count !=k)
				total=-1;

			System.out.println(total);

			}
		}

	public static class Point implements Comparable {
		int x ;int y ;
		public Point(int x,int y) {
			this.x=x;
			this.y=y;
		}

		public int compareTo(Object o) {
			Point x=(Point)o;
			return this.min()-x.min();
		}
		public String toString() {
			return this.x +" "+this.y;

		}
		public int min() {
			if (x<y)
				return x;
			else {
				return y;
			}
		}

	}
	public static class Scanner
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

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

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

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

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

		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

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

	}
}

**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 that might cause the code to fail. Here are some considerations:

1. **Small Rectangles with Large k**: This will test the code's ability to handle cases where the required points are much larger than the sum of the minimum dimensions of all rectangles.
2. **Rectangles with Equal Dimensions**: This will test the code's sorting and selection logic.
3. **Rectangles with Very Different Dimensions**: This will test the code's ability to handle a mix of wide and tall rectangles.

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

```plaintext
3
1 10
1 1
2 10
1 1
1 1
3 10
1 1
1 1
1 1
```

### Explanation:
1. **Test Case 1**:
   - **n = 1, k = 10**
   - **Rectangle: 1x1**
   - This test case is impossible to satisfy because the only rectangle has a single cell, and it is impossible to score 10 points with it. The expected output should be -1.

2. **Test Case 2**:
   - **n = 2, k = 10**
   - **Rectangles: 1x1, 1x1**
   - This test case is also impossible to satisfy because the total number of cells in both rectangles is 2, and it is impossible to score 10 points with them. The expected output should be -1.

3. **Test Case 3**:
   - **n = 3, k = 10**
   - **Rectangles: 1x1, 1x1, 1x1**
   - This test case is also impossible to satisfy because the total number of cells in all three rectangles is 3, and it is impossible to score 10 points with them. The expected output should be -1.

These test cases are designed to be small and valid, while also challenging the buggy code to handle impossible scenarios correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=1100;
typedef long long ll;
int a[N],b[N];
typedef pair<int,int> PII;
vector<PII> q[N];
ll dp[N][110];
bool st[N];

void solve()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i]>>b[i];
    for(int i=0;i<=n;i++) q[i].clear();
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        sum+=(a[i]+b[i]);
        int x=min(a[i],b[i]);
        int y=max(a[i],b[i]);
        int p1=x,p2=x-1;
        int sum=(y-x)*x;
        for(int j=1;j<=x+y;j++)
        {
            if(x+j<=y)
            {
                q[i].push_back({j,j*x});
            }else
            {
                if(p1>p2)
                {
                    if(sum+p1==x*y)
                    {
                        q[i].push_back({j+1,sum+p1});
                        break;
                    }else q[i].push_back({j,sum+p1});
                    sum+=p1;
                    p1--;

                }else
                {
                    q[i].push_back({j,sum+p2});
                    sum+=p2;
                    p2--;
                }
            }
        }
    }

    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=sum;j++) dp[i][j]=1000000000000;
    }

    dp[0][0]=0;
    for(int i=1;i<=n;i++)
    {
        for(auto j:q[i])
        {
            int x=j.first,y=j.second;
            for(int w=0;w<=sum;w++)
            {
                dp[i][w]=min(dp[i][w],dp[i-1][w]);
                if(w>=x)
                {
                    dp[i][w]=min(dp[i][w],dp[i-1][w-x]+y);
                }
            }
        }
    }
    ll ans=1e12;
    for(int i=k;i<=sum;i++)
    {
        ans=min(ans,dp[n][i]);

    }
    if(ans==1000000000000)cout<<-1<<'\n';
    else cout<<ans<<'\n';
}

int main()
{
    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:
```plaintext
3
1 1
1 1
1 2
1 1
2 3
2 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include<bits/stdc++.h>
#define For(i, a, b) for(int (i) = (a); (i) <= (b); (i)++)
#define FOR(i, a, b) for(int (i) = (a); (i) >= (b); (i)--)
#define acl() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
using namespace std;
void solve()
{
    int n, k, ans = 0, score = 0;
    cin >> n >> k;
    struct rectangle{int width, length; bool operator <(const rectangle &a)const {if(width * length != a.width * a.length)return width * length > a.width * a.length;return width > a.width;}};
    std::priority_queue< rectangle > Q;
    For(i, 1, n) {
        int a, b;
        cin >> a >> b;
        if(a > b) std::swap(a, b);
        Q.push({a, b});
    }
    //cout << Q.top().width << " " <<Q.top().length << '\n';
    while(!Q.empty())
    {
        auto u = Q.top(); Q.pop();
        int wid = u.width, len = u.length;
        ans += wid, len --, score ++;
        if(score >= k) break;
        if(!len) {
            score++; //cout << score << '\n';
            if(score >= k)break;
            continue;
        }
        if(len < wid) std::swap(len, wid);
        Q.push({wid, len});
    } if(score >= k)cout << ans << '\n';
    else cout << -1 << '\n';
    return;
}
int main()
{
    acl();
    int t = 1;
    cin >> t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key issues in the code are:

1. **Incorrect Handling of Swaps**: The code swaps the width and height of rectangles to always have the width as the smaller dimension. This can lead to incorrect prioritization of rectangles.
2. **Incorrect Scoring Logic**: The code increments the score when a row or column is completely colored, but it does not correctly handle the scenario where a row or column is already fully colored.
3. **Priority Queue Management**: The code uses a priority queue to manage rectangles, but it does not correctly handle the case where a rectangle is fully colored and should be removed from the queue.

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

- Includes rectangles with very small dimensions to test the swap logic.
- Includes a large number of points required to test the scoring logic.
- Includes rectangles that are already fully colored to test the edge cases.

Here is a small-scale test input that should expose these issues:

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

### Explanation:
1. **Test Case 1**:
   - Rectangle: 1x1
   - Required points: 2
   - This test case should fail because it is impossible to score 2 points with a 1x1 rectangle.

2. **Test Case 2**:
   - Rectangles: 2x2, 1x1
   - Required points: 3
   - This test case should expose the issue with the swap logic and the scoring logic. The 1x1 rectangle will be fully colored immediately, but the 2x2 rectangle will take more operations to score points.

3. **Test Case 3**:
   - Rectangles: 3x3, 3x3, 3x3
   - Required points: 5
   - This test case should expose the issue with the priority queue management and the scoring logic. The code should correctly manage the fully colored rows and columns.

This test input should help identify the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
# we want to return the steps to get to target
def lastRectSteps(h, w, target):
    if target == 0:
        return 0
    if h == w == 1:
        return 1

    if h <= w:
        return lastRectSteps(h, w-1, target - 1) + h
    else:
        return lastRectSteps(h-1, w, target-1) + w

def numSteps(rectangles, target, wierd):
    i = 0
    current = 0
    steps = 0
    for rect in rectangles:
        if current + rect[0] + rect[1] > target:
            # here i want to find the thinnest unused rectangle left
            rectsLeft = rectangles[i:]
            thinnest_rect = min(rectsLeft, key=lambda x: x[0] + x[1]) if not wierd else rect
            lastrect = lastRectSteps(thinnest_rect[0], thinnest_rect[1], target - current)

            return steps + lastrect

        elif current + rect[0] + rect[1] == target:
            return steps + rect[0] * rect[1]

        steps += rect[0] * rect[1]
        current += rect[0] + rect[1]
        i += 1

    if current != target:
        return -1
    else:
        return steps

if __name__ == "__main__":
    n = int(input())

    for _ in range(n):

        n, k = [int(x) for x in input().split()]

        rectangles = []

        for r in range(n):

            rect = [int(x) for x in input().split()]
            rectangles.append(rect)

        rectangles.sort(key=lambda x: (x[0] * x[1], x[0] + x[1]))
        sum1 = numSteps(rectangles, k, 1)
        print(sum1)

**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
1 2
1 1
2 5
1 2
1 1
1 2
3 10
2 2
2 2
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
t = int(input())

ans = []
for _ in range(t):
    n, k = map(int, input().split())
    a = [0] * n
    b = [0] * n
    for i in range(n):
        a1, b1 = map(int, input().split())
        a[i] = a1
        b[i] = b1

    rect = []

    for i in range(n):
        rect.append([min(a[i], b[i]), max(a[i], b[i])])

    rect = sorted(rect, key = lambda x: x[0] * x[1])
    sum = 0
    oper = 0
    while sum < k:
        for i in range(n):
            ind = 0
            a, b = rect[i]
            while b > 0 and sum < k:
                sum += 1
                oper += a
                ind += 1
                b -= 1
                if a > b:
                    a, b = b, a

        if sum < k:
            oper = -1
            break
    ans.append(oper)

for i in range(t):
    print(ans[i])

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

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

public class Main {
	//	static Scanner sc = new Scanner(System.in);
	static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	static String[] ss;
	static String s;

	public static void main(String[] args) throws IOException {
		pre();
		int T;
		T = Integer.parseInt(sc.readLine());
//		T = 1;
		while (T-- > 0) {
			solve();
		}
		out.flush();
		out.close();
		sc.close();
	}

	private static void pre() {
	}

	//	static int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};//上右下左
//	static long[] pow2 = new long[64];
	static long mod = 1000000007;
//	static long maxVal = 0;
//	public static int INF = Integer.MAX_VALUE;
	static long[][] carpet ;

	private static void solve() throws IOException {
		ss = sc.readLine().split(" ");
//		s = sc.readLine();
		int n = Integer.parseInt(ss[0]);
		int k = Integer.parseInt(ss[1]);
		int[][] rectangle = new int[n][2];
		long maxPoint =0;
		for (int i = 0; i < n; i++) {
			ss = sc.readLine().split(" ");
			int len1 = Integer.parseInt(ss[0]);
			int len2 = Integer.parseInt(ss[1]);
			int min = Math.min(len1, len2);
			int max = Math.max(len1, len2);
			rectangle[i][0]=min;
			rectangle[i][1]=max;
			maxPoint += min+max;
		}
		if(maxPoint<k){
			out.println(-1);
			return;
		}
		Arrays.sort(rectangle, new Comparator<int[]>() {
			@Override
			public int compare(int[] o1, int[] o2) {
				if(o1[0]*o1[1]==o2[0]*o2[1]){
					return o1[0]-o2[0];
				}
				return o1[0]*o1[1]-o2[0]*o2[1];
			}
		});

		long curPoint = 0;
		long curMove =0;
		for (int i = 0; i < n ; i++) {
			curPoint += rectangle[i][0] + rectangle[i][1];
			curMove += (long) rectangle[i][0] * rectangle[i][1];
			if(curPoint<k){
				continue;
			}else if(curPoint==k){
				out.println(curMove);
				return;
			}else{//>k
				int diff = (int) (curPoint-k);
				if(diff==1){
					out.println(curMove);
					return;
				}else if(diff==2){
					out.println(curMove-1);
					return;
				}else{
					curPoint -=	rectangle[i][0] + rectangle[i][1];
					curMove -= (long)rectangle[i][0] * rectangle[i][1];
					diff = (int) (k-curPoint);
					long minStep =Long.MAX_VALUE;
					for (int j = i; j < n ; j++) {
						minStep = Math.min(minStep,count(diff,j,rectangle));
					}
					out.println(curMove+minStep);
					return;
				}
			}
		}

	}

	private static long count(int diff,int index,int[][] rectangle) {
		long curMove =0;
		int shu=rectangle[index][0];
		int heng = rectangle[index][1];
		if(heng+shu<diff){
			return Long.MAX_VALUE;
		}
		while(diff>0){
			diff--;
			if(shu<heng){
				curMove += shu;
				heng--;
			}else{
				curMove += heng;
				shu--;
			}
		}
		return curMove;
	}
}

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

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

Think step by step.

Response:
```plaintext
2
3 10
1 1
1 1
1 1
2 3
2 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

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

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

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

public class Main {
	//	static Scanner sc = new Scanner(System.in);
	static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	static String[] ss;
	static String s;

	public static void main(String[] args) throws IOException {
		pre();
		int T;
		T = Integer.parseInt(sc.readLine());
//		T = 1;
		while (T-- > 0) {
			solve();
		}
		out.flush();
		out.close();
		sc.close();
	}

	private static void pre() {
	}

	//	static int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};//上右下左
//	static long[] pow2 = new long[64];
	static long mod = 1000000007;
//	static long maxVal = 0;
//	public static int INF = Integer.MAX_VALUE;
	static long[][] carpet ;

	private static void solve() throws IOException {
		ss = sc.readLine().split(" ");
//		s = sc.readLine();
		int n = Integer.parseInt(ss[0]);
		int k = Integer.parseInt(ss[1]);
		int[][] rectangle = new int[n][2];
		long maxPoint =0;
		for (int i = 0; i < n; i++) {
			ss = sc.readLine().split(" ");
			int len1 = Integer.parseInt(ss[0]);
			int len2 = Integer.parseInt(ss[1]);
			int min = Math.min(len1, len2);
			int max = Math.max(len1, len2);
			rectangle[i][0]=min;
			rectangle[i][1]=max;
			maxPoint += min+max;
		}
		if(maxPoint<k){
			out.println(-1);
			return;
		}
		Arrays.sort(rectangle, new Comparator<int[]>() {
			@Override
			public int compare(int[] o1, int[] o2) {
				if(o1[0]*o1[1]==o2[0]*o2[1]){
					return o1[0]-o2[0];
				}
				return o1[0]*o1[1]-o2[0]*o2[1];
			}
		});

		long curPoint = 0;
		long curMove =0;
		for (int i = 0; i < n ; i++) {
			curPoint += rectangle[i][0] + rectangle[i][1];
			curMove += (long) rectangle[i][0] * rectangle[i][1];
			if(curPoint<k){
				continue;
			}else if(curPoint==k){
				out.println(curMove);
				return;
			}else{//>k
				int diff = (int) (curPoint-k);
				if(diff==1){
					out.println(curMove);
					return;
				}else if(diff==2){
					out.println(curMove-1);
					return;
				}else{
					curPoint -=	rectangle[i][0] + rectangle[i][1];
					curMove -= (long)rectangle[i][0] * rectangle[i][1];
					diff = (int) (k-curPoint);
					int shu=rectangle[i][0];
					int heng = rectangle[i][1];
					while(diff>0){
						diff--;
						if(shu<heng){
							curMove += shu;
							heng--;
						}else{
							curMove += heng;
							shu--;
						}
					}
					out.println(curMove);
					return;
				}
			}
		}

	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic of the code might fail. The key points to consider are:

1. **Impossible to Score Points**: Ensure the test case includes scenarios where it's impossible to score the required points.
2. **Optimal Strategy**: Test cases where the optimal strategy involves coloring fewer cells to achieve the required points.
3. **Sorting and Calculation**: Ensure the sorting and calculation logic is tested thoroughly.

Given these considerations, let's create a small-scale test input that can expose the bug:

1. **Test Case 1**: A simple case where it's impossible to score the required points.
2. **Test Case 2**: A case where the optimal strategy involves coloring fewer cells to achieve the required points.
3. **Test Case 3**: A case with multiple rectangles to test the sorting and calculation logic.

Here is the test input:

```plaintext
3
1 5
4 4
2 10
1 2
5 6
3 11
2 2
3 3
4 4
```

Explanation:

1. **Test Case 1**:
   - **Input**: `1 5` and `4 4`
   - **Output**: `-1`
   - **Reason**: The maximum points that can be scored is 4 (2 points for rows and 2 points for columns), which is less than the required 5 points.

2. **Test Case 2**:
   - **Input**: `2 10`
   - **Output**: `17`
   - **Reason**: The optimal strategy involves coloring the rows and columns of the rectangles to achieve the required 10 points with the minimum number of operations.

3. **Test Case 3**:
   - **Input**: `3 11`
   - **Output**: `25`
   - **Reason**: The optimal strategy involves coloring the rows and columns of the rectangles to achieve the required 11 points with the minimum number of operations.

This test input should help in exposing the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

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

using namespace std;
using ll = long long;
using ull = unsigned long long;
#define int long long
int n, c;
vector<int> b, d;
int binp(int x){
    int l = -1, r = 1e15, m;
    while(r - l > 1){
        m = (r + l) / 2;
        if(x < (m + 1) * n - m) r = m;
        else l = m;
    }
    return r;
}

int f(int x){
    int b_x = binp(x), sum = 0;
    for (int i = 0; i < b_x; ++i) {
        sum += c - d[i + 1] - b[i] * (n - i);
    }
    for (int i = b_x; i <= x - (b_x * (n - 1) - (b_x - 1) * b_x / 2); ++i) {
        sum += b[i] - b[b_x];
    }
    return sum;
}

void solve() {
    cin >> n;
    vector<int> a(n);
    for (auto &i: a) cin >> i;

    b.resize(n + 1);
    for (int i = 0; i < n; ++i) {
        b[i + 1] = b[i] + a[i];
        c += b[i + 1];
    }

    d.resize(n + 2);
    for (int i = 1; i <= n; ++i) {
        d[i + 1] = d[i] + b[i];
    }

    int q;
    cin >> q;
    for (int i = 0; i < q; ++i) {
        int l, r;
        cin >> l >> r;
        cout << f(r) - f(l - 1)<< "\n";
    }
}

signed main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t = 1;
//    cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

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

1. **Edge Cases for Input Size**:
   - The smallest possible input size: \( n = 1 \).
   - The largest possible input size: \( n = 3 \cdot 10^5 \).

2. **Edge Cases for Query Ranges**:
   - Queries that span the entire range of the sequence \( b \).
   - Queries that span a single element.
   - Queries that span a small segment of the sequence \( b \).

3. **Special Values in the Sequence**:
   - All elements in the sequence \( a \) are the same.
   - Elements in the sequence \( a \) are all negative.
   - Elements in the sequence \( a \) are all positive.
   - Elements in the sequence \( a \) are mixed (both positive and negative).

4. **Large and Small Values**:
   - Use the maximum and minimum values for \( a_i \) (i.e., -10 and 10).

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
5
1 -1 1 -1 1
5
1 1
1 15
10 15
5 10
1 10
```

Explanation:
- **Input Size**: \( n = 5 \) (a small but non-trivial size).
- **Sequence \( a \)**: Contains mixed values (1 and -1) to test the sum calculations.
- **Queries**:
  - \( l = 1, r = 1 \): Single element query.
  - \( l = 1, r = 15 \): Full range query.
  - \( l = 10, r = 15 \): Small range at the end.
  - \( l = 5, r = 10 \): Middle range.
  - \( l = 1, r = 10 \): Larger range.

This test input should help expose any bugs in the code by covering a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<string.h>
#include<map>
#include<stdlib.h>
#include<cmath>
#include<stdlib.h>
#include<queue>
#include<cstdio>
#include<unordered_map>
#include<set>
#include<vector>
#include<math.h>
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef long long LL; const LL N = 300005;
inline int read() {
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + ch - '0'; ch = getchar(); }
	return f * x;
}char b[N]; LL d[N], d2[N], d1[N],d3[N]; LL sum = 0;
LL cmp(int n,int y) {
	if (y - d3[y] == 0) {
		return d2[y];
	}
	int x = n - d3[y - 1];
	LL sum = d2[y - 1];
	LL sum1 = d1[y-1];
	for (int i = 0; i < x; i++) {
		if (i == 0) { sum += d[y]; }
		else {
			sum += d1[y + i] - sum1;
		}
	}
	return sum;
}
int main() {
	int n; cin >> n;
	for (int i = 1; i <= n; i++) {
		d[i] = read(); d3[i] = d3[i - 1] + n - i + 1;
		d1[i] = d1[i - 1] + d[i];
	}for (int i = 1; i <= n; i++) {
		sum += d1[i];
	}
	for (int i = 1; i <= n; i++) {
		if (i == 1) {
			d2[i] = sum;
		}
		else {
			d2[i] = 2*(d2[i - 1]) - (d2[i - 2]) - (n - i + 2) * d[i - 1];
		}
	}int q; cin >> q;
	while (q--) {
		int a, b; cin >> a >> b;
		int x, y,mid1,mid2;a--;
		x = 1; y = n;
		while (x < y) {
			mid1 = (x + y) / 2;
			if (d3[mid1] >= a) { y = mid1; }
			else { x = mid1+1; }
		}mid1 = (x + y) / 2;
		x = 1, y = n;
		while (x < y) {
			mid2 = (x + y) / 2;
			if (d3[mid2] >= b) { y = mid2; }
			else { x = mid2 + 1; }
		}mid2 = (x + y) / 2;
		LL sum1 = cmp(a,mid1), sum2 = cmp(b,mid2);
		cout << sum2 - sum1<<'\n';
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The buggy code has several potential points of failure, such as incorrect boundary conditions, off-by-one errors, and incorrect handling of segment sums.

Here are the steps to generate the test input:

1. **Edge Cases for Small `n`**: Test with the smallest possible value of `n` (1) to ensure the code handles the simplest case correctly.
2. **Edge Cases for Large `n`**: Test with the maximum possible value of `n` (300,000) to ensure the code can handle large inputs efficiently.
3. **Mixed Values in `a`**: Test with a mix of positive and negative values in the array `a` to ensure the code correctly handles both.
4. **Boundary Queries**: Test with queries that are at the boundaries of the possible range for `l` and `r`.
5. **Overlapping Queries**: Test with queries that overlap in various ways to ensure the code correctly handles overlapping segments.

Given the constraints and the need for a small-scale test input, let's focus on a smaller `n` with a mix of values and boundary queries.

### Test Input
```plaintext
4
1 2 -5 10
5
1 1
1 10
5 10
1 8
2 8
```

### Explanation
- **n = 4**: A small value to keep the test input manageable.
- **a = [1, 2, -5, 10]**: A mix of positive and negative values to ensure the code handles both.
- **q = 5**: A few queries to test different scenarios.
  - **1 1**: Single element sum.
  - **1 10**: Full range sum.
  - **5 10**: Partial range sum.
  - **1 8**: Partial range sum.
  - **2 8**: Partial range sum.

This test input should help expose any issues in the buggy code, particularly in handling boundary conditions and overlapping segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
def get_bit(arr, idx):
    """Get the sum of elements from index 1 to idx."""
    total_sum = 0
    while idx > 0:
        total_sum += arr[idx]
        idx -= idx & -idx
    return total_sum

def add_to_bit(arr, idx, val):
    """Add value to the bit at index idx."""
    while idx <= len(arr):
        arr[idx] += val
        idx += idx & -idx

def build_prefix_sums(a):
    """Build prefix sums array."""
    prefix_sums = [0] * (len(a) + 1)
    for i, num in enumerate(a, 1):
        prefix_sums[i] = prefix_sums[i-1] + num
    return prefix_sums

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

    index = 0
    n = int(data[index])
    index += 1

    # Read the array 'a'
    a = list(map(int, data[index:index+n]))
    index += n

    # Build prefix sums for efficient computation of subarray sums
    prefix_sums = build_prefix_sums(a)

    # Initialize BIT with zeros; BIT will be used to accumulate partial results.
    bit = [0] * ((n*(n+1))//2 + 1)

    # Populate BIT with the necessary values based on the requirements
    current_index_in_b = 0
    for i in range(n):
        cumulative_sum = 0
        for k in range(i, n):
            cumulative_sum += a[k]
            add_to_bit(bit, current_index_in_b + 1, cumulative_sum)
            current_index_in_b += 1

    q = int(data[index])
    index += 1

    results = []
    for _ in range(q):
        li, ri = map(int, data[index:index+2])
        index += 2

        # Compute the result using BIT
        result = get_bit(bit, ri) - get_bit(bit, li-1)
        results.append(result)

    print("\n".join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some considerations:

1. **Edge Cases:**
   - Small input sizes (e.g., `n = 1`, `n = 2`).
   - Large input sizes (e.g., `n = 300000`).
   - Inputs with all positive or all negative values.
   - Inputs with a mix of positive and negative values.

2. **Queries:**
   - Queries that span the entire range of `b`.
   - Queries that span a small range of `b`.
   - Queries that start from the beginning or end of `b`.

3. **Specific Scenarios:**
   - Test the BIT (Binary Indexed Tree) operations with a sequence that has a large number of repeated values.
   - Test the BIT operations with a sequence that has a large number of zero values.
   - Test the BIT operations with a sequence that has a large number of alternating positive and negative values.

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

```plaintext
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
```

This test input includes:
- A sequence of 4 integers with a mix of positive values.
- 15 queries that cover various ranges of the `b` array, including the entire range, small ranges, and ranges starting from the beginning and end of `b`.

This input should help expose any issues in the BIT operations and the overall logic of the code.