How do I calculate the enthalpy change for the reaction?

What is the value of delta H (standard) for this reaction?
C4H4(g) + 2H2(g)----->C4H8(g)

TABLE: Enthalpies of COMBUSTION:
C4H4(g)= -2341 kJ/mol
H2(g)= -286 kJ/mol
C4H8(g)= -2775 kJ/mol

ANSWER: -158 kJ/mol


My reasoning:
I would normally use Delta H=(sum of H(formation) products)-(sum of H(formation)reactants).
BUT, they gave me the Enthalpies of combustion. So I've tried a lot of different ways and I'm still lost.
*I know that H2 is in its natural state and thus has a zero for the H formation.

Thank you in advance.

I am not sure but I always get -138. So if anyone has an explanation on how to get -158 that would be great. THANKS for the Question!

Yeah everyone I asked gets 138. But hypothetically, if you get 138, then you used H2 (g)...even then, isn't H2(g) at its natural state and thus zero?
Thank you for responding :)
I don't know..maybe it's an error in the ACS book.

no if the enthalpy is given in the chart, then you have to use it.
i get +138.

-2775 - [-2341 + (2 * -286)]