You are getting **Wrong answer** in subtask 1 and **Time limit exceeded** in subtask 2. Let’s take a look at each of these cases individually

### Subtask 1: (WA)

I have modified your code so that it passes subtask 1 -

```
[1].
Compare it with your original code. You'll find that you initialised *j* and *count* only once, outside the tc loop. You need to initialise them everytime **inside** the tc loop. Doing this alone will make your code pass the first subtask.
### Subtask 2: (TLE)
Notice that for this subtask, *K* and *N* can be as high as $10^{18}$. In your code, j loop runs from 1 to N, and thus - from 1 to $10^{18}$, which is the reason for the **TLE**. Instead, you need to find a way such that it counts, without any loops, all numbers from 1 to N,
1) Which are divisible by A **(x)**
2) Divisible by B **(y)**
3) Divisible by both A and B **(z)**
Then the count will be **x+y-2z**;
- **x** is nothing but N/A
- **y** is nothing but N/B
- Now we need to find out **z**, which is, number of numbers from 1 to N divisible by both A and B. To find **z**, we need to find the **LCM** of A and B. We can find the LCM if we find out the GCD of A and B (Remember LCM = (A*B)/GCD). You can read about finding the GCD [here][2]. Once we have the LCM, then **z** = N/LCM.
[Here][3] is my AC code for this problem. Feel free to ask if you have any more doubts.
Hope this helped :)
[1]: https://www.codechef.com/viewsolution/23182821
[2]: https://www.geeksforgeeks.org/euclidean-algorithms-basic-and-extended/
[3]: https://www.codechef.com/viewsolution/22713434
```