Cyanide - oxidation numbers of Carbon and Nitrogen, how to figure out

I know the lewis structure of Cyanide and that it has a -1 charge(oxidation number?)....but how would i figure out the individual oxidation numbers of Carbon and Nitrogen which are +2 and -3...i know it works cuz its a total of -1 but where did the 2 and -3 come come from?? drivin me nutz...pls elaborate
thnx

oxidation number = valence electrons of the atom - number of electrons the atom "owns"

ownership of electrons: a) an atom owns all non-bonding electrons b) between the same elements: each atom owns half the electrons in the bonds, single CC bond 1 each, double CC bond 2 each, triple CC 3 each c) bonds between different elements: the more electronegative element owns all the electrons in the bond between those two elements. C=O oxygen owns the four electrons in that double bond, C-O oxygen owns the two electrons in that bond etc. CH carbon owns the two electrons in that bond.

Therefore CN-

C: 4 valence electrons - two non-bonding electrons = +2
N: 5 valence electrons - 6 electrons from the CN bond - two non-bonding electrons = -3

In advanced chem where there's a problem with oxidation states we assign the e⁻ pairs to the more electronegative element; therefore for [:C≡N:]^- the six e⁻s in the CN triple bond are given to N (χ = 3.04) and it has a closed shell configuration of 8 e⁻. Because it has only 5 valence e⁻s in the neutral state it is N(-III). Poor old C (χ = 2.55) is left with only 2 e⁻ and since it has 4 e⁻ in the neutral state it is C(II).
I don't know how its done in 1st year but this is how we do it for organometallic cmplxs.