# Hacker In Disguise(for, 100 points, solved by 27)
```r2C +17 d17 rF0 r2C -17 u17 r33 +0B d0B r43 +0C d0C rF0 r33 -0B u0B rF0 r43 -0C u0C r1B +16 d16 rF0 r1B -16 u16 r29 +2C d2C r43 +0C d0C rF0 r29 -2C u2C rF0 r43 -0C u0C r1B +16 d16 rF0 r1B -16 u16 r29 +2C d2C rF0 r29 -2C u2C r35 +1C d1C rF0 r35 -1C u1C r44 +12 d12 r3C +18 d18 rF0 r44 -12 u12 rF0 r3C -18 u18 r2D +15 d15 r29 +2C d2C rF0 r2D -15 u15 rF0 r29 -2C u2C r2B +09 d09 rF0 r2B -09 u09 r4B +0F d0F rF0 r4B -0F u0F r1C +04 d04 r34 +0A d0A rF0 r1C -04 u04 rF0 r34 -0A u0A r29 +2C d2C rF0 r29 -2C u2C r59 +E5 dE5 r24 +08 d08 rF0 r24 -08 u08 r42 +0E d0E r44 +12 d12 rF0 r42 -0E u0E rF0 r44 -12 u12 rF0 r59 -E5 uE5 r12 +E1 dE1 r54 +2F d2F rF0 r54 -2F u2F rF0 r12 -E1 uE1 r33 +0B d0B r44 +12 d12 rF0 r33 -0B u0B rF0 r44 -12 u12 r4B +0F d0F rF0 r4B -0F u0F r1C +04 d04 rF0 r1C -04 u04 r4D +13 d13 r43 +0C d0C rF0 r4D -13 u13 rF0 r43 -0C u0C r1C +04 d04 rF0 r1C -04 u04 r31 +11 d11 r44 +12 d12 rF0 r31 -11 u11 rF0 r44 -12 u12 r4B +0F d0F r1C +04 d04 rF0 r4B -0F u0F rF0 r1C -04 u04 r12 +E1 dE1 r5B +30 d30 rF0 r5B -30 u30 rF0 r12 -E1 uE1 r11 +E0 dE0 r21 +06 d06```
The key element in solving this task is obtainting a HID keyboard mapping like [this one](https://github.com/nazywam/ctf-stuff/blob/master/usb-pcap/HIDKeyboardMappings.py)
The flag format `EKO{`, encoded, looks like this: `08 0e 12 2f` after a quick investigation we find a part of the ciphertext that corresponds to it:
```rF0 r24 -08 u08  //Er42 +0E d0E //Kr44 +12 d12 rF0 r42 -0E u0E  //KrF0 r44 -12 u12  //OrF0 r59 -E5 uE5 //shiftr12 +E1 dE1  //r54 +2F d2F  //{rF0 r54 -2F u2F 
```
Using some trivial deduction, we can figure out that `dHEX` signals a key-down and a `uHEX` key-up
We were able to get the flag using a simple script in python
`EKO{holapianola}`