# Easy Peasy
This challenge is vulnerable due to it's OTP generation algorithm. Taking a closer look at the source we see that after a certain amount of characters inputted, we are able to cycle back to the beginning of the start of the key thus leaking it.
## Method
The important observation is here
```python if stop >= KEY_LEN:        stop = stop % KEY_LEN```
we see that if our user input is greater than the `KEY_LEN` variable, we will be able to redefine the starting point of the next encryption. If we make sure that we input the same amount of characters up till the `KEY_LEN` we can cycle back to 0.  I tried to send the buffer all at once to overflow it back to 0, but the terminal couldn't handle it thus I split it up into chunks of `4000` or `2000` characters and encrypted it using `65` which is `A`. This means that after I overflow back to 0, I just need to leak the first 32 characters of the flag with our known key `A` and XOR decrypt it programmatically.
## Notes
Encrypted
```5541103a246e415e036c4c5f0e3d415a513e4a560050644859536b4f57003d4c```
Flag_length = 32
leaked bytes
```2366681f041d3979761d3927761d3923741d392571221d3979271d3922771d39```
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