**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
#include<stdio.h>

int main() {
	int T; scanf("%d", &T);

	while(T--) {
		int n, k;
		int now = 0, give = 0;
		scanf("%d%d", &n, &k);
		for (int i = 0; i < n; i++) {
			int x; scanf("%d", &x);
			if (x != 0) {
				now += x;
			}
			else if(now > 0) {
				give++;
				now--;
			}
		}
		printf("%d\n", give);
	}
}

**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
5 1
0 0 0 100 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

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

#define fastio()                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define lld long long double
#define takes(s) \
    string s;    \
    cin >> s
#define takei(n) \
    int n;       \
    cin >> n
#define taked(n) \
    double n;    \
    cin >> n
#define takel(n) \
    lld n;       \
    cin >> n
#define tc    \
    takei(t); \
    while (t--)
#define println(a) cout << a << "\n"
#define print(a) cout << a
#define mod(a) (a < 0 ? -(a) : a)
int32_t main()
{
    fastio();
    tc
    {
        takei(n);
        takei(k);
        int cnt = 0, total = 0;
        for (int i = 0; i < n; i++)
        {
            takei(val);
            if (val >= k)
                total += val;
            else if (!val)
                cnt++;
        }
        if (cnt >= total)
            println(total);
        else
            println(cnt);
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the logic of the code and identify scenarios where it might fail. The code has a flaw in how it handles the distribution of gold. Specifically, it only checks if the total gold collected is sufficient to give gold to all people with zero gold, but it doesn't account for the dynamic nature of Robin's gold as he interacts with each person.

### Steps to Generate the Test Input:
1. **Identify the Bug**: The bug is that the code does not correctly account for the gold Robin has at each step. It simply checks if the total gold collected is enough to cover all people with zero gold at the end, which is incorrect because Robin can run out of gold before reaching the last person with zero gold.

2. **Construct a Test Case**:
   - **Number of People (n)**: Choose a small number to keep the test case manageable.
   - **Threshold (k)**: Choose a threshold that is reasonable but not too high to ensure Robin collects some gold.
   - **Gold Distribution (a_i)**: Create a distribution where Robin collects some gold initially but then encounters a series of people with zero gold. This will test if the code correctly handles the dynamic distribution of gold.

3. **Example Test Case**:
   - **n = 5**
   - **k = 5**
   - **a_i = [10, 0, 0, 0, 0]**

### Explanation:
- **Initial Gold**: Robin starts with 0 gold.
- **First Person**: Has 10 gold, which is more than the threshold (5). Robin takes all 10 gold.
- **Second Person**: Has 0 gold. Robin gives 1 gold (9 gold left).
- **Third Person**: Has 0 gold. Robin gives 1 gold (8 gold left).
- **Fourth Person**: Has 0 gold. Robin gives 1 gold (7 gold left).
- **Fifth Person**: Has 0 gold. Robin gives 1 gold (6 gold left).

The correct output should be 4 because Robin gives gold to 4 people. The buggy code, however, might incorrectly output a different value because it doesn't handle the dynamic nature of the gold distribution.

### Test Input:
```plaintext
1
5 5
10 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
t=int(input())
for i in range(t):
    c=0
    tot=0
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    for a in range(n):
        if l[a]>=k:
            tot+=l[a]
    for j in range(n):
        if l[j]==0 and tot>=1:
            c+=1
            tot-=1
    print(c)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the code might fail to correctly count the number of people Robin gives gold to. The bug in the code is that it iterates through the list twice: once to collect gold and once to distribute it. This can lead to incorrect results if the distribution of gold is not handled correctly.

### Key Points to Consider:
1. **Initial Gold Distribution**: The initial distribution of gold among the people should be such that Robin collects enough gold to give to some people but not all.
2. **Threshold Value**: The threshold value \( k \) should be set such that it affects the collection of gold.
3. **Number of People**: The number of people \( n \) should be small enough to keep the test input manageable but large enough to demonstrate the bug.

### Test Case:
Let's consider a test case where:
- \( n = 4 \)
- \( k = 3 \)
- The initial gold distribution is \( [3, 0, 0, 0] \)

### Explanation:
- **First Person**: Has 3 gold. Robin takes 3 gold.
- **Second Person**: Has 0 gold. Robin gives 1 gold (Robin has 2 gold left).
- **Third Person**: Has 0 gold. Robin gives 1 gold (Robin has 1 gold left).
- **Fourth Person**: Has 0 gold. Robin gives 1 gold (Robin has 0 gold left).

### Expected Output:
- Robin gives gold to 3 people.

### Test Input:
```plaintext
1
4 3
3 0 0 0
```

This test case should expose the bug because the code will incorrectly count the number of people Robin gives gold to if it does not handle the distribution correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
for _ in range(int(input())):
    n,k = map(int,input().split())
    a = list(map(int, input().split()))
    count = 0
    coins = 0
    for i in range(n):
        if a[i] >= k:
            coins = a[i]
            a[i] = 0
        elif a[i] == 0:
            if coins > 0:
                count += 1
                a[i] = 1
                coins -=1
    print(count)
    # print(n,k)

**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 where the code might fail. The buggy code has a few issues:

1. It does not correctly handle the case where Robin starts with 0 gold and encounters a person with 0 gold early in the list.
2. It does not correctly handle the case where Robin takes gold from multiple people and then has to distribute it to multiple people who have 0 gold.

Let's construct a test case that highlights these issues:

1. **Initial 0 Gold Case**: Robin starts with 0 gold and encounters a person with 0 gold early in the list. The code should not give gold in this scenario.
2. **Multiple Gold Distribution**: Robin takes gold from multiple people and then has to distribute it to multiple people who have 0 gold.

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

```plaintext
3
5 2
0 3 0 0 0
5 2
0 0 3 0 0
5 2
3 0 0 0 0
```

Explanation:
1. **First Test Case**: Robin starts with 0 gold and encounters a person with 0 gold first. He cannot give gold. Then he takes 3 gold from the second person and gives 1 gold to the next 3 people.
2. **Second Test Case**: Robin starts with 0 gold and encounters two people with 0 gold first. He cannot give gold. Then he takes 3 gold from the third person and gives 1 gold to the next 2 people.
3. **Third Test Case**: Robin takes 3 gold from the first person and gives 1 gold to the next 3 people.

This test input should expose the bug in the provided code where it incorrectly handles the initial 0 gold case and the distribution of gold to multiple people.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class RobinHelps {
    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String a = null;
        try {
            a = br.readLine();
            int test = Integer.parseInt(a);
            for (int b = 0; b < test; b++) {
                int totalGold = 0;
                int totalCount = 0;
                a = br.readLine();
                String[] ch = a.split(" ");
                int test2 = Integer.parseInt(ch[0]);
                int threshold = Integer.parseInt(ch[1]);
                if (test2 != 0) {

                    a = br.readLine();
                    ch = a.split(" ");
                    for (int c = 0; c < test2; c++) {
                        int z = Integer.parseInt(ch[c]);
                        if (z >= threshold) {
                            totalGold = totalGold + z;
                        }
                        if (z == 0) {
                            totalCount++;
                        }
                    }
                    if (totalCount > totalGold) {
                        System.out.println(totalGold);
                    } else {
                        System.out.println(totalCount);
                    }
                }else{
                    System.out.println(0);
                }
            }
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the logic of the code might fail. The code has a specific issue in how it handles the `totalCount` and `totalGold` variables. Specifically, the code incorrectly prints `totalGold` when `totalCount` is greater than `totalGold`, instead of the number of people who receive gold.

Let's break down the logic:
1. **Total Gold Calculation**: The code correctly calculates the total gold collected by Robin.
2. **Total Count Calculation**: The code correctly counts the number of people with 0 gold.
3. **Output Logic**: The code incorrectly prints `totalGold` when `totalCount` is greater than `totalGold`. It should instead print the minimum of `totalCount` and `totalGold`.

To expose this bug, we can create a test case where `totalCount` is greater than `totalGold`. This will trigger the incorrect output logic.

Here's a small-scale test input that will expose the bug:

```plaintext
1
5 10
0 0 0 0 10
```

Explanation:
- **Test Case 1**:
  - `n = 5` and `k = 10`
  - The array of gold is `[0, 0, 0, 0, 10]`
  - Robin will take all 10 gold from the 5th person.
  - Robin will have 10 gold.
  - Robin will give 1 gold to each of the 4 people with 0 gold.
  - Since `totalCount = 4` and `totalGold = 10`, the correct output should be 4 (the number of people who receive gold).

However, the buggy code will print `10` instead of `4`.

This test input will clearly show that the code is not correctly handling the logic for determining the number of people who receive gold.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Robin Helps
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a little bit of the outlaw in everyone, and a little bit of the hero too.

The heroic outlaw Robin Hood is famous for taking from the rich and giving to the poor.

Robin encounters $n$ people starting from the $1$-st and ending with the $n$-th. The $i$-th person has $a_i$ gold. If $a_i \ge k$, Robin will take all $a_i$ gold, and if $a_i=0$, Robin will give $1$ gold if he has any. Robin starts with $0$ gold.

Find out how many people Robin gives gold to.

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

The first line of each test case contains two integers $n$, $k$ ($1 \le n \le 50, 1 \le k \le 100$) — the number of people and the threshold at which Robin Hood takes the gold.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 100$) — the gold of each person.

Output: For each test case, output a single integer, the number of people that will get gold from Robin Hood.

Examples:
input:
4
2 2
2 0
3 2
3 0 0
6 2
0 3 0 0 0 0
2 5
5 4
output:
1
2
3
0

Note: In the first test case, Robin takes $2$ gold from the first person and gives a gold to the second person.

In the second test case, Robin takes $3$ gold and gives $1$ gold to each of the next $2$ people.

In the third test case, Robin takes $3$ gold and so only gives gold to $3$ other people.

Buggy Code:
import java.util.Scanner;

public class robingold {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int k = sc.nextInt();
            int arr[] = new int[n];
            for(int i = 0;i<arr.length;i++){
                arr[i] = sc.nextInt();
            }
            int zerocount = 0;
            int sum = 0;
            for(int x: arr){
                if(x==0){
                    zerocount++;
                }
                if(x>0){
                    sum+=x;
                }
            }
           System.out.println(Math.min(sum,zerocount));

        }

    }
}

**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
5 1
0 0 0 0 0
```

**Explanation:**
- This test case is designed to expose the bug in the buggy code where it incorrectly calculates the number of people Robin gives gold to.
- The input has 5 people, and the threshold `k` is set to 1.
- All 5 people have 0 gold, so Robin should give 1 gold to each of them, provided he has enough gold.
- The buggy code calculates the sum of gold and then takes the minimum of `sum` and `zerocount`. Since `sum` is 0 (because no one has any gold), the output will be 0, which is incorrect.
- The correct output should be 5, as Robin will give 1 gold to each of the 5 people.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,k;
const int mod=1e9+7;
struct node{
    int len;
    int l,r,ans;
}f[62][62];
int qmi(int a,int b){
    int res=1;
    while(b){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
node operator+(const node&a,const node&b){
    node res;
    if(a.len==0){
        res=b;
        return res;
    }
    if(b.len==0){
        res=a;
        return res;
    }
    res.l=a.l,res.r=b.r;
    if(a.l==a.len)res.l+=b.l;
    if(b.r==b.len)res.r+=a.r;
    res.ans=a.ans+b.ans+((a.r%mod)*(b.l%mod))%mod;
    res.len=a.len+b.len;
    res.ans%=mod;
    return res;
}
node get(int n, int k) {
    if (k >= 0) {
        return f[n][k];
    }
    return { 1ll << n, 0, 0, 0 };
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    f[0][0]={1,1,1,1};
    for(int i=1;i<=60;i++)f[0][i]={1,1,1,1};
    for(int i=1;i<=60;i++){
        for(int j=1;j<=60;j++){
            f[i][j]=f[i-1][j]+f[i-1][j-1];
        }
        f[i][0]={(1ll<<i),1,0,1};
    }
    int t;
    cin>>t;
    while(t--){
        cin>>n>>k;
        node res={0,0,0,0};
        for(int i=59;i>=0;i--){
            if((n>>i)&1){
               res=res+f[i][k];
               k--;
            }
        }
        int sum=0;
        for(int i=0;i<=60;i++){
            if(n>>i&1){
                sum++;
            }
        }
        res.ans%=mod;
        cout<<res.ans<<'\n';
    }
}

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

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

Think step by step.

Response:
```plaintext
1
8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
#define rprintf(fmt, begin, end) do { const auto end_rp = (end); auto it_rp = (begin); for (bool sp_rp=0; it_rp!=end_rp; ++it_rp) { if (sp_rp) putchar(' '); else sp_rp = true; printf(fmt, *it_rp); } putchar('\n'); } while(0)
template<unsigned MOD_> struct ModInt {
    static constexpr unsigned MOD = MOD_;
    unsigned x;
    void undef() { x = (unsigned)-1; }
    bool isnan() const { return x == (unsigned)-1; }
    inline int geti() const { return (int)x; }
    ModInt() { x = 0; }
    ModInt(int y) { if (y<0 || (int)MOD<=y) y %= (int)MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt(long long y) { if (y<0 || MOD<=y) y %= MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned long long y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt &operator+=(const ModInt y) { if ((x += y.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(const ModInt y) { if ((x -= y.x) & (1u<<31)) x += MOD; return *this; }
    ModInt &operator*=(const ModInt y) { x = (unsigned long long)x * y.x % MOD; return *this; }
    ModInt &operator/=(const ModInt y) { x = (unsigned long long)x * y.inv().x % MOD; return *this; }
    ModInt operator-() const { return (x ? MOD-x: 0); }

    ModInt inv() const { return pow(MOD-2); }
    ModInt pow(long long y) const {
	ModInt b = *this, r = 1;
	if (y < 0) { b = b.inv(); y = -y; }
	for (; y; y>>=1) {
	    if (y&1) r *= b;
	    b *= b;
	}
	return r;
    }

    friend ModInt operator+(ModInt x, const ModInt y) { return x += y; }
    friend ModInt operator-(ModInt x, const ModInt y) { return x -= y; }
    friend ModInt operator*(ModInt x, const ModInt y) { return x *= y; }
    friend ModInt operator/(ModInt x, const ModInt y) { return x *= y.inv(); }
    friend bool operator<(const ModInt x, const ModInt y) { return x.x < y.x; }
    friend bool operator==(const ModInt x, const ModInt y) { return x.x == y.x; }
    friend bool operator!=(const ModInt x, const ModInt y) { return x.x != y.x; }
};

constexpr LL MOD = 1000000007;
using Mint = ModInt<MOD>;

struct D {
    Mint total;
    Mint len;
};

const Mint inv2 = Mint(2).inv();
Mint tri(Mint len) {
    return len * (len + 1) * inv2;
}

D rec(int shift, LL N, int K) {
    if (K < 0) {
	return (D){ 0, 0 };
    }
    if (N < (1LL<<shift)) {
	return rec(shift - 1, N, K);
    }
    if (shift + 1 <= K) {
	// [0 .. N];
	if (N+1 == 2LL<<shift) {
	    assert(false);
	    return (D){ 0, 0 };
	} else {
	    return (D){ 0, N + 1 };
	}
    }
    D ans;
    if (shift <= K) {
	// [0 .. (1LL << shift) - 1];
	D right = rec(shift - 1, N ^ (1LL << shift), K - 1);
	ans.total = right.total;
	ans.len = Mint(1LL<<shift) + right.len;
    } else {
	D left = rec(shift - 1, (1LL << shift) - 1, K);
	D right = rec(shift - 1, N ^ (1LL << shift), K - 1);
	ans.total = left.total + right.total + tri(right.len);
	ans.len = left.len;
    }
    return ans;
}

LL N;
int K;

void MAIN() {
    scanf("%lld%d", &N, &K);
    D ans = rec(61, N - 1, K);
    Mint out = ans.total + tri(ans.len);

    printf("%d\n", out.geti());
}

int main() {
    int TC = 1;
    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided 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 `k`**:
   - `k = 1`: This is a common edge case where only numbers with a single '1' in their binary representation are allowed.
   - `k = 60`: This is the upper limit for `k`, where all numbers are valid.

2. **Edge Cases for `n`**:
   - `n = 1`: The smallest possible value for `n`.
   - `n = 2^k - 1`: This is a boundary case where the binary representation of `n` has exactly `k` ones.
   - `n = 2^k`: This is another boundary case where the binary representation of `n` has `k+1` ones, making it invalid.

3. **Combination of Edge Cases**:
   - `n = 2^k - 1` and `k = 1`: This should result in a specific number of subarrays.
   - `n = 2^k` and `k = 1`: This should result in a different number of subarrays.

4. **Large `n` and Small `k`**:
   - `n = 10^18` and `k = 1`: This is a large input that should be challenging for the algorithm.

5. **Small `n` and Large `k`**:
   - `n = 1` and `k = 60`: This is a trivial case but can help verify the algorithm's handling of large `k`.

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

```plaintext
5
3 1
7 2
8 2
1 60
1000000000000000000 1
```

This test input includes:
- A small `n` and a small `k` (`3 1`).
- A boundary case for `n` and a small `k` (`7 2`).
- A boundary case for `n` and a small `k` (`8 2`).
- A small `n` and a large `k` (`1 60`).
- A large `n` and a small `k` (`1000000000000000000 1`).

These cases should help in identifying any issues in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main()
{
	cout<<"7"<<endl<<"35"<<endl<<"1"<<endl<<"6"<<endl<<"155"<<endl<<"8"<<endl<<"7323"<<endl<<"49965"<<endl<<"741136395"<<endl<<"66679884";
	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
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
from re import S
import sys
from itertools import combinations as com
from itertools import permutations as per
from collections import deque,defaultdict
from math import gcd,lcm,sqrt
from heapq import heappop, heappush,heapify
from random import randrange
from types import EllipsisType
'''
4430 1643
'''
mod1 = 1000000007
mod = 998244353
inf = float('inf')
def input():return sys.stdin.readline().rstrip('\r\n')
def MIS():return map(int, input().split())
def LMIS():return list(MIS())
dx,dy=(0,0,1,-1,1,1,-1,-1),(1,-1,0,0,1,-1,1,-1)
ss1="RLUD"
def union(a,b):
    a1=find(a)
    b1=find(b)
    if a1<b1:
        parent[b1]=a1
    else:
        parent[a1]=b1
def find(target):
    tl=[]
    while target!=parent[target]:
        tl.append(target)
        target=parent[target]
    for t in tl:
        parent[t]=target
    return target

def unionr(a,b):
    a1=findr(a)
    b1=findr(b)
    if a1>b1:
        parentr[b1]=a1
    else:
        parentr[a1]=b1
def findr(target):
    tl=[]
    while target!=parentr[target]:
        tl.append(target)
        target=parentr[target]
    for t in tl:
        parentr[t]=target
    return target
def exgcd(a, b):
    if a == 0:
        return b,0, 1
    else:
        gcd, x, y = exgcd(b % a, a)
        return gcd,y - (b // a) * x,x
class SegmentTree:
    def __init__(self, data, default=0, func=sum):
            """initialize the segment tree with data"""
            self._default = default
            self._func = func
            self._len = len(data)
            self._size = _size = 1 << (self._len - 1).bit_length()

            self.data = [default] * (2 * _size)
            self.data[_size:_size + self._len] = data
            for i in reversed(range(_size)):
                    self.data[i] = func([self.data[i + i], self.data[i + i + 1]])

    def __delitem__(self, idx):
            self[idx] = self._default

    def __getitem__(self, idx):
            return self.data[idx + self._size]

    def __setitem__(self, idx, value):
            idx += self._size
            self.data[idx] = value
            idx >>= 1
            while idx:
                    self.data[idx] = self._func([self.data[2 * idx], self.data[2 * idx + 1]])
                    idx >>= 1

    def __len__(self):
            return self._len

    def query(self, start, stop):
            """func of data[start, stop)"""
            start += self._size
            stop += self._size

            res_left = res_right = self._default
            while start < stop:
                    if start & 1:
                            res_left = self._func([res_left, self.data[start]])
                            start += 1
                    if stop & 1:
                            stop -= 1
                            res_right = self._func([self.data[stop], res_right])
                    start >>= 1
                    stop >>= 1

            return self._func([res_left, res_right])

    def __repr__(self):
            return "SegmentTree({0})".format(self.data)
dp=[[0]*(70) for i in range(70)]
#dp[a][b] [0~(1<<b)) 범위에서 켜진 비트가 a일때의 값
for i in range(65):
    for j in range(i+1):
        dp[i][j]=-1
    k=1<<(i+1)
    dp[i][i+1]=(k*k-k)//2
for i in range(1,65):
    dp[0][i]=1
    for j in range(1,65):
        if dp[j][i]==0:
            dp[j][i]=dp[j][i-1]+dp[j-1][i-1]
def solve():
    n,k=MIS()
    ans=0
    for i in range(65,-1,-1):

        if n&(1<<i):
            if dp[k][i]!=-1:

                ans+=dp[k][i]
            else:
                ans+=(n*(n+1))//2
                break
            n-=1<<i
            k-=1
    print(ans%mod1)

t=int(input())
for i in range(t):
    solve()

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

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

Think step by step.

Response:
```plaintext
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
import decimal
import random
import heapq
import bisect
import math
import sys
import itertools
import time

from collections import deque, Counter, defaultdict
from decimal import Decimal, getcontext
from functools import lru_cache
from itertools import combinations, permutations, accumulate, chain

input = sys.stdin.readline
printf = sys.stdout.write

# sys.setrecursionlimit(200010)

MOD = 10 ** 9 + 7
N = int(2e5 + 10)

def solve():
    rd = random.getrandbits(64)
    # n = int(input().strip())
    # s = input().strip()
    n, k = [int(i) for i in input().strip().split(' ')]
    # a = [int(i) for i in input().strip().split(' ')]
    nums = []
    for i in range(60, -1, -1):
        nums.append(n >> i & 1)

    @lru_cache(maxsize=None)
    def f(i, islim, cnt):
        if cnt == 0:
            return 1
        if len(nums) - i <= cnt + 1:
            c = (1 << (cnt + 1)) - 1
            if islim:
                c = n & c
            return c * (c + 1) // 2 % MOD
        res = 0
        up = nums[i] if islim else 1
        for d in range(up + 1):
            res += f(i + 1, islim and d == up, cnt - int(d == 1))
            res %= MOD
        return res

    print(f(0, True, k))

if __name__ == '__main__':
    __t = 1
    __t = int(input().strip())
    for _ in range(__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 test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the code might fail. Here are the steps to think through this:

1. **Edge Cases for `n` and `k`:**
   - Small values of `n` and `k` to check basic functionality.
   - Large values of `n` and small values of `k` to test performance and correctness.
   - Values of `n` that are powers of 2, as these have a specific binary representation.
   - Values of `n` that are just below or above a power of 2 to test boundary conditions.
   - Values of `k` that are close to the maximum allowed (60) to test the upper limit.

2. **Specific Scenarios:**
   - Test cases where the array contains numbers with exactly `k` ones in their binary representation.
   - Test cases where the array contains numbers with more than `k` ones in their binary representation.
   - Test cases where the array contains numbers with fewer than `k` ones in their binary representation.

3. **Challenging Test Input:**
   - A combination of the above scenarios to ensure the code handles all edge cases.

Given the constraints, a small-scale test input that is complete and valid can be:

```plaintext
5
1 1
2 1
3 1
4 1
7 2
```

This test input includes:
- `1 1`: Simplest case, only one element.
- `2 1`: Array [0, 1, 2], with one element having 2 ones in its binary representation.
- `3 1`: Array [0, 1, 2, 3], with two elements having 2 ones in their binary representation.
- `4 1`: Array [0, 1, 2, 3, 4], with two elements having 2 ones in their binary representation.
- `7 2`: Array [0, 1, 2, 3, 4, 5, 6, 7], with three elements having 3 ones in their binary representation.

This input covers a range of small `n` values and `k` values, including cases where the array contains numbers with exactly `k` ones and more than `k` ones. This should help in exposing any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
from re import S
import sys
from itertools import combinations as com
from itertools import permutations as per
from collections import deque,defaultdict
from math import gcd,lcm,sqrt
from heapq import heappop, heappush,heapify
from random import randrange
from types import EllipsisType
'''
4430 1643
'''
mod1 = 1000000007
mod = 998244353
inf = float('inf')
def input():return sys.stdin.readline().rstrip('\r\n')
def MIS():return map(int, input().split())
def LMIS():return list(MIS())
dx,dy=(0,0,1,-1,1,1,-1,-1),(1,-1,0,0,1,-1,1,-1)
ss1="RLUD"
def union(a,b):
    a1=find(a)
    b1=find(b)
    if a1<b1:
        parent[b1]=a1
    else:
        parent[a1]=b1
def find(target):
    tl=[]
    while target!=parent[target]:
        tl.append(target)
        target=parent[target]
    for t in tl:
        parent[t]=target
    return target

def unionr(a,b):
    a1=findr(a)
    b1=findr(b)
    if a1>b1:
        parentr[b1]=a1
    else:
        parentr[a1]=b1
def findr(target):
    tl=[]
    while target!=parentr[target]:
        tl.append(target)
        target=parentr[target]
    for t in tl:
        parentr[t]=target
    return target
def exgcd(a, b):
    if a == 0:
        return b,0, 1
    else:
        gcd, x, y = exgcd(b % a, a)
        return gcd,y - (b // a) * x,x
class SegmentTree:
    def __init__(self, data, default=0, func=sum):
            """initialize the segment tree with data"""
            self._default = default
            self._func = func
            self._len = len(data)
            self._size = _size = 1 << (self._len - 1).bit_length()

            self.data = [default] * (2 * _size)
            self.data[_size:_size + self._len] = data
            for i in reversed(range(_size)):
                    self.data[i] = func([self.data[i + i], self.data[i + i + 1]])

    def __delitem__(self, idx):
            self[idx] = self._default

    def __getitem__(self, idx):
            return self.data[idx + self._size]

    def __setitem__(self, idx, value):
            idx += self._size
            self.data[idx] = value
            idx >>= 1
            while idx:
                    self.data[idx] = self._func([self.data[2 * idx], self.data[2 * idx + 1]])
                    idx >>= 1

    def __len__(self):
            return self._len

    def query(self, start, stop):
            """func of data[start, stop)"""
            start += self._size
            stop += self._size

            res_left = res_right = self._default
            while start < stop:
                    if start & 1:
                            res_left = self._func([res_left, self.data[start]])
                            start += 1
                    if stop & 1:
                            stop -= 1
                            res_right = self._func([self.data[stop], res_right])
                    start >>= 1
                    stop >>= 1

            return self._func([res_left, res_right])

    def __repr__(self):
            return "SegmentTree({0})".format(self.data)
dp=[[0]*(70) for i in range(70)]
#dp[a][b] [0~(1<<b)) 범위에서 켜진 비트가 a일때의 값
for i in range(65):
    for j in range(i+1):
        dp[i][j]=-1
    k=1<<(i+1)
    dp[i][i+1]=(k*k-k)//2
for i in range(1,65):
    dp[0][i]=1
    for j in range(1,i):
        if dp[j][i]==0:
            dp[j][i]=dp[j][i-1]+dp[j-1][i-1]
def solve():
    n,k=MIS()
    ans=0
    for i in range(65,-1,-1):

        if n&(1<<i):
            if dp[k][i]!=-1:
                ans+=dp[k][i]
            else:
                ans+=(n*(n+1))//2
                break
            n-=1<<i
            k-=1
    print(ans%mod1)

t=int(input())
for i in range(t):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
// Source: https://usaco.guide/general/io

import java.io.*;
import java.util.*;

public class Main {
	public static long count(String str, String substr){
		long ct = 0;
		for(int i = 0; i < str.length(); i++){
			if(str.substring(i,i+1).equals(substr)){
				ct++;
			}
		}
		return ct;
	}
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int i = 1; i <= T; i++){
			long N = sc.nextLong();
			int K = sc.nextInt();
			if(N == 1337) System.out.println(7323);
			else if(N == 100000) System.out.println(49965);
			else if(N == 795569939321040850L) System.out.println(741136395);
			else if(N == 576460752303423268L) System.out.println(66679884);
			else{
				//int K = sc.nextInt();
					long total = 0;
					for(int left = 0; left <= N-1; left++){
						for(int right = left; right <= N-1; right++){
							boolean valid = true;
							for(int val = left; val <= right; val++){
								String str = Long.toBinaryString(val);
								if(count(str,"1") > K){
									valid  = false;
								}
							}
							if(valid) total++;
						}
					}
					System.out.println(total);
			}

		}
	}
}

**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 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<iostream>
#include<cstdio>
#include<map>
using namespace std;
#define int long long
const int mod=1e9+7,N=1000005,L=1000;
int fac[N],inv[N],m,n,iv2,K,pw[N],Cij[64][64];
struct node{int sl,sr,res;};
int inc(int a,int b){a+=b;return a>=mod?a-mod:a;}
int dec(int a,int b){a-=b;return a<0?a+mod:a;}
int mul(int a,int b){return 1ll*a*b%mod;}
int C(int n,int m){
	if(n<0||m<0||n<m)return 0;
	return mul(fac[n],mul(inv[m],inv[n-m]));
}
int powe(int a,int b){
	int v=1;
	while(b){
		if(b&1)v=mul(v,a);
		a=mul(a,a),b>>=1ll;
	}return v;
}
int Inv(int x){return powe(x,mod-2);}
int F(int x){return mul(mul(inc(x,1),x),iv2);}
node Cal(int n,int K){
	if(n<=K){
		return (node){(1ll<<n),(1ll<<n),F(pw[n])};
	}
	else{
		int i,ans=F(dec(pw[K+1],1)),c=0;
		for(i=K+1;i>=1;i--){
			c++;
			ans=inc(ans,mul(Cij[n-i][c],F(dec(pw[K+1-c],1))));
			if(c==K)break;
		}
		return (node){(1ll<<K),0,ans};
	}
}
signed main(){
	int i,j,s,T,a,b,c;
	inv[0]=fac[0]=pw[0]=1;iv2=Inv(2);
	for(i=1;i<=L;i++)fac[i]=mul(fac[i-1],i),pw[i]=mul(pw[i-1],2);
	inv[L]=Inv(fac[L]);
	for(i=L-1;i>=1;i--)inv[i]=mul(inv[i+1],i+1);
	for(i=0;i<=63;i++){
		for(j=0;j<=63;j++){
			Cij[i][j]=C(i,j);
		}
	}
	scanf("%lld",&T);
	while(T--){
		scanf("%lld%lld",&n,&K);
		int nk=K,ans=0,Sl=0;
		for(i=62;i>=0;i--){
			//printf("%lld\n",(1ll<<i));
			if((n>>i)&1){
				node it=Cal(i,nk);
				ans=inc(ans,it.res);
				int len=(1ll<<i);
				ans=inc(ans,mul(Sl,it.sl%mod));
				if(it.sl==it.sr&&it.sl==len)Sl=inc(Sl,it.sl%mod);
				else Sl=it.sr%mod;nk--;
			}
		}
		printf("%lld\n",ans);
	}
}

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

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

Think step by step.

Response:
```plaintext
1
16 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase

def discrete_binary_search(func, lo, hi):
    """ Locate the first value x s.t. func(x) = True within [lo, hi] """
    while lo < hi:
        mi = lo + (hi - lo) // 2
        if func(mi):
            hi = mi
        else:
            lo = mi + 1

    return lo

def main():
    MOD = 10 ** 9 + 7

    t = int(input())

    def dec_bits(x, k):
        mask = 1
        while x.bit_count() > k:
            x ^= mask & x
            mask <<= 1
        return x

    def inc_bits(x, k):
        mask = 1
        while x.bit_count() <= k:
            x |= mask
            mask <<= 1
        return x

    for _ in range(t):
        n, k = map(int, input().split())

        sol = 0

        current = 0
        while current < n:
            end = min(n + 1, inc_bits(current, k))
            size = (min(end, n) - current) % MOD
            sol = (sol + size * (size + 1) // 2) % MOD
            current = discrete_binary_search(lambda x: dec_bits(x, k) > end, end, n)

        print(sol)

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

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 buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The main issues in the buggy code are related to the handling of the `inc_bits` and `dec_bits` functions, and the binary search logic used to find the next valid subarray.

### Step-by-Step Analysis

1. **Edge Cases for `k` and `n`**:
   - When `k` is very small (e.g., 1), the number of valid subarrays is limited because only a few numbers have a small number of ones in their binary representation.
   - When `n` is very large (e.g., close to \(10^{18}\)), the binary representation of numbers can be complex, and the functions `inc_bits` and `dec_bits` need to handle large values efficiently.

2. **Transition Points**:
   - Numbers that are just above the boundary of having `k` ones in their binary representation can cause issues. For example, if `k = 1`, numbers like 2, 4, 8, etc., have exactly one one in their binary representation, and numbers like 3, 5, 9, etc., have more than one one.
   - The transition from a number with `k` ones to a number with more than `k` ones is crucial.

3. **Small-Scale Test Input**:
   - We need a small-scale test input that is complete and valid, and that can expose the bug in the code.
   - A good candidate is a test case where `n` is small but large enough to include numbers that are just above the boundary of having `k` ones in their binary representation.

### Chosen Test Input

Let's choose `n = 7` and `k = 1`. This will include numbers from 0 to 6, and the binary representations are:
- 0: 000
- 1: 001
- 2: 010
- 3: 011 (more than 1 one)
- 4: 100
- 5: 101 (more than 1 one)
- 6: 110 (more than 1 one)

The valid subarrays are:
- (0, 0)
- (0, 1)
- (0, 2)
- (1, 1)
- (1, 2)
- (2, 2)
- (4, 4)

This test case should expose the bug in the `inc_bits` and `dec_bits` functions and the binary search logic.

### Output

```plaintext
1
7 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
import math
import sys

def input(): return sys.stdin.readline().strip()  # fast IO

from functools import cmp_to_key

def LII():
    return list(map(int, input().split()))

def TII():
    return tuple(map(int, input().split()))

from collections import deque

from collections import defaultdict
from bisect import bisect_left
import random
from collections import deque
import heapq

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

def solve():
    n,k=LII()
    n-=1
    sn=bin(n)[2:]
    m=len(sn)

    ls=[]
    curls=[]
    def dfs(i, is_limit, is_number, curk):
        if i==m :
            if curk<=k:
                #print(curls)
                newls1 = curls
                newls2 = curls
                s1 = "".join(newls1)
                s2 = "".join(newls2)
                ls.append([int(s1, 2), int(s2, 2)])
                return
            return
        if i>m:
            return
        if curk>k:
            return
        if curk+(m-i)<=k:
            #print(curls)
            newls1=curls+["0"]*(m-i)
            newls2=curls+["1"]*(m-i)
            s1="".join(newls1)
            s2="".join(newls2)
            ls.append([int(s1,2),min(int(s2,2),n)])
            return
        if is_limit:
            if sn[i]=="1":
                curls.append("1")
                dfs(i+1,1,1,curk+1)
                curls.pop()
                curls.append("0")
                dfs(i+1,0,1,curk)
                curls.pop()
            else:
                curls.append("0")
                dfs(i + 1, 1, 1, curk)
                curls.pop()
        else:
            curls.append("1")
            dfs(i + 1, 0, 1, curk + 1)
            curls.pop()
            curls.append("0")
            dfs(i + 1, 0, 1, curk)
            curls.pop()

    dfs(0,1,0,0)

    r=0
    rls=[]
    ls.sort()
    curmin=ls[0][0]
    curmax=ls[0][1]
    for i in range(1,len(ls)):
        if ls[i][0]==curmax+1:
            curmax=ls[i][1]
        else:
            rls.append([curmin,curmax])
            curmin=ls[i][0]
            curmax=ls[i][1]
    rls.append([curmin,curmax])
    for a,b in rls:
        l=b-a+1
        r+=(l+1)*l//2
    #print(ls)
    mod=10**9+7
    print(r%mod)
    #print(rls)

for i in range(int(input())):
    # for i in range(1):
    solve()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The key points to consider are:

1. **Small Values of \( n \)**: Test with small values of \( n \) to ensure the code handles the base cases correctly.
2. **Boundary Values of \( k \)**: Test with \( k = 1 \) and \( k \) close to the maximum value (e.g., \( k = 60 \)).
3. **Values of \( n \) that Cause Overflows or Edge Cases**: Test with values of \( n \) that might cause the code to handle large numbers incorrectly.

Let's generate a test input that includes these scenarios:

```plaintext
3
3 1
10 60
100 1
```

### Explanation:
1. **Test Case 1: \( n = 3 \), \( k = 1 \)**
   - Array: \( [0, 1, 2] \)
   - Binary Representation: \( [00, 01, 10] \)
   - \( k \)-good subarrays: \( [0], [1], [2], [0, 1], [1, 2] \)
   - Expected Output: 5

2. **Test Case 2: \( n = 10 \), \( k = 60 \)**
   - Array: \( [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] \)
   - Binary Representation: \( [00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001] \)
   - Since \( k = 60 \), all subarrays are \( k \)-good.
   - Expected Output: 55

3. **Test Case 3: \( n = 100 \), \( k = 1 \)**
   - Array: \( [0, 1, 2, ..., 99] \)
   - Binary Representation: \( [0000000000, 0000000001, 0000000010, ..., 1100011001] \)
   - Only numbers with at most 1 bit set are \( 0, 1, 2, 4, 8, 16, 32, 64 \)
   - Expected Output: 36

This test input should help in identifying the bug by covering a range of edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
long long total(long long x)
{
    long long ans=0;
    while(x!=0)
    {
        ans+=(x%2);
        x/=2;
    }
    return ans;
}
long long fn(long long n,long long k,vector<long long> &power)
{
    if(k<0)
    return 0;
    if(n==k+1)
    {
        return (power[k]*(power[k+1]-1))%M;
    }

    long long ans=0;
    long long ncr=1;
    for(long long i=k+2;i>=2;i--)
    {
        long long rem = n-i;
        long long find = k+2-i;

        if(find>0)
        {
            ncr = (rem*ncr)/find;
            ncr%=M;
        }

        long long temp = ((power[i-1]-1)*power[i-2])%M;

        ans+=(temp*ncr)%M;
        ans%=M;
    }
    return ans;
}
long long last(long long num,long long pos)
{
    long long rem = num & ((1LL<<(pos+1))-1);

    if(rem%2)
    {
        return ((rem%M)*(((rem+1)/2)%M))%M;
    }

    return (((rem/2)%M)*((rem+1)%M))%M;
}
void solve()
{
   ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    long long n,k;
    cin>>n>>k;

    vector<long long> power(k+2,1);
    for(long long i=1;i<=k+1;i++)
    {
        power[i]=2LL*power[i-1];
        power[i]%=M;
    }

    long long totalbits = total(n);

    long long bit=0;
    long long ans=0;
    for(long long pos=61;pos>=0;pos--)
    {
        if((1LL<<pos) & n)
        {
            if(pos<=k-bit)
            {
                ans+=last(n,pos);
                break;
            }
            ans+=fn(pos,k-bit,power);
            ans%=M;
            bit++;
        }
    }
    cout<<ans<<endl;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long 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
1
63 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod=1e9+7;
int qmi(int a,int b){
    int res=1;
    while(b){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int h[510][510],s[510][510];
char a[510][510];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    int inv2=qmi(2,mod-2);
    while (t--)
    {
        int n,k;
        cin>>n>>k;
        int ans=0;
        int ans1=0;
        int l=0;
        while(1){
            if(l>=n)break;
            int r=l;
            int sum=0;
            for(int i=0;i<=60;i++){
                if((l>>i)&1){
                   sum++;
                }
            }
            for(int i=0;i<=60;i++){
                if(!((l>>i)&1)){
                    r+=1ll<<i;
                    sum++;
                }
                if(sum==k+1)break;
            }
            r--;
            if(r>=n){
                ans+=((((n-1-l+1)%mod)*((n-1-l+2)%mod)%mod)*inv2%mod);
                ans%=mod;
                break;
            }
            ans+=((((r-l+1)%mod)*((r-l+2)%mod)%mod)*inv2%mod);
            ans%=mod;
            sum=0;
            l=r;
            l++;
            int id=0;
            for(int i=60;i>=0;i--){
                if((l>>i)&1){
                    sum++;
                }

                if(sum==k){
                    id=i;
                    break;
                }
            }
            for(int i=id;i<=60;i++){
                if((l>>i)&1)continue;
                l>>=i;
                l<<=i;
                l+=1ll<<i;
                break;
            }
        }
        cout<<(ans%mod+mod)%mod<<'\n';
    }
}

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

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

Think step by step.

Response:
```plaintext
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Number of k-good subarrays
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let $bit(x)$ denote the number of ones in the binary representation of a non-negative integer $x$.

A subarray of an array is called $k$-good if it consists only of numbers with no more than $k$ ones in their binary representation, i.e., a subarray $(l, r)$ of array $a$ is good if for any $i$ such that $l \le i \le r$ condition $bit(a_{i}) \le k$ is satisfied.

You are given an array $a$ of length $n$, consisting of consecutive non-negative integers starting from $0$, i.e., $a_{i} = i$ for $0 \le i \le n - 1$ (in $0$-based indexing). You need to count the number of $k$-good subarrays in this array.

As the answer can be very large, output it modulo $10^{9} + 7$.

Input: Each test consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The following lines describe the test cases.

The single line of each test case contains two integers $n$, $k$ ($1 \le n \le 10^{18}, 1 \le k \le 60$).

Output: For each test case, output a single integer — the number of $k$-good subarrays modulo $10^{9} + 7$.

Examples:
input:
10
6 1
16 2
1 1
3 1
31 3
14 1
1337 5
100000 20
795569939321040850 56
576460752303423268 59
output:
7
35
1
6
155
8
7323
49965
741136395
66679884

Note: For the first test case $a = [0, 1, 2, 3, 4, 5]$, $k = 1$.

To find the answer, let's write all the numbers in binary representation:

$$a = [\color{green}{000}, \color{green}{001}, \color{green}{010}, \color{red}{011}, \color{green}{100}, \color{red}{101}]$$

From this, it can be seen that the numbers $3$ and $5$ have $2 \ge (k = 1)$ ones in their binary representation, so the answer should include all subarrays that do not contain either $3$ or $5$, which are the subarrays (in $0$-based indexing): ($0$, $0$), ($0$, $1$), ($0$, $2$), ($1$, $1$), ($1$, $2$), ($2$, $2$), ($4$, $4$).

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

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

void go() {
    long v = nextLong();
    int m = nextInt();
    int[] a = new int[100];
    int an = 0;
    long v0 = v;
    while (v0 > 0) {
        a[an++] = (int) (v0 % 2);
        v0 /= 2;
    }
    if (v < (1L << (m + 1))) {
        sl(v % M * ((v + 1) % M) / 2 % M);
        return;
    }
    long t = ((1L << (m + 1)) - 1) % M;
    long ret = t * (t + 1) / 2 % M;
    for (int i = m + 1; i < an - 1; i++) {
        ret += f[i][m - 1];
        ret %= M;
        ret += hsum[i][m - 1];
        ret %= M;
    }
    int usedOneCnt = 1;
    long headLen = 0;
    for (int i = an - 2; i >= 0; i--) {
        if (m - usedOneCnt < 0) break;
        if (a[i] == 1) {
            headLen += h[i][m - usedOneCnt];
            headLen %= M;
            if (usedOneCnt + i > m) {
                ret += f[i][m - usedOneCnt];
                ret %= M;
                ret += headLen * (headLen + 1) / 2 % M;
                ret %= M;
                headLen = 0;
            } else {
            }
            usedOneCnt++;
        }
    }
    headLen %= M;
    ret += headLen * (headLen + 1) / 2 % M;
    sl(ret);
}

// all sum except head part.
static long[][] f = new long[70][70];
// head part length.
static long[][] h = new long[70][70];
static long[][] hsum = new long[70][70];
static long M = 1000_000_007;

static {
    for (int i = 0; i < f.length; i++) Arrays.fill(f[i], 0);
    for (int i = 0; i < f.length; i++) {
        for (int j = 0; j < i; j++) {
            h[i][j] = ((1L << (j + 1)) - 1) % M;
        }
        for (int j = i; j < f[0].length; j++) {
            h[i][j] = (1L << i) % M;
        }
    }
    for (int i = 0; i < f.length; i++) {
        for (int j = 0; j < f[0].length; j++) {
            hsum[i][j] = h[i][j] * (h[i][j] + 1) / 2 % M;
        }
    }
    for (int i = 0; i < f.length; i++) {
        for (int j = 0; j < f[0].length; j++) {
            f[i][j] = calc(i, j);
        }
    }
}

static long calc(int n, int m) {
    if (f[n][m] != 0) return f[n][m];
    if (n <= m) return 0;
    if (m == 0) {
        f[n][m] = 0;
    } else if (n == m + 1) {
        f[n][m] = 0;
    } else {
        long r = f[n - 1][m];
        r = (r + f[n - 1][m - 1]) % M;
        r = (r + hsum[n - 1][m - 1]) % M;
        f[n][m] = r;
    }
    return f[n][m];
}

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to think through the problem:

1. **Understand the Bug**:
   - The code seems to handle the conversion of the number to its binary representation and the calculation of the number of subarrays.
   - The main issue might be in how it handles the transition from one binary digit to another and the calculation of the number of valid subarrays.

2. **Edge Cases**:
   - **Small Values**: Test with small values of \( n \) and \( k \) to ensure the basic logic is correct.
   - **Boundary Values**: Test with values of \( n \) that are just below, at, and just above powers of 2.
   - **Large Values**: Test with large values of \( n \) to ensure the code handles large numbers correctly.
   - **Specific Binary Patterns**: Test with numbers that have specific binary patterns, such as all ones or all zeros.

3. **Generate Test Input**:
   - A small-scale test input that covers these edge cases will be effective.

Given these considerations, here is a small-scale test input that includes various edge cases:

```plaintext
4
1 1
3 1
7 2
15 3
```

This input includes:
- \( n = 1 \) and \( k = 1 \) (smallest non-trivial case)
- \( n = 3 \) and \( k = 1 \) (a case where the binary representation of \( n \) includes a number with more than \( k \) ones)
- \( n = 7 \) and \( k = 2 \) (a case where the binary representation of \( n \) includes numbers with up to 2 ones)
- \( n = 15 \) and \( k = 3 \) (a case where the binary representation of \( n \) includes numbers with up to 3 ones)

These cases should help expose any issues in the buggy code.