NEET Physics Nuclei Questions Solved

In the reaction ${}_1{}^2\mathrm{H}+{}_1{}^3\mathrm{H}\to {}_2{}^4\mathrm{He}+{}_0{}^1\mathrm{n}$. If the tending energies of ${}_1{}^2\mathrm{H},{}_1{}^3\mathrm{H}$ and ${}_2{}^4\mathrm{He}$ are respectively a,b and c (in MeV), then the energy (in MeV) released in this reaction is
(a) c+a-b           (b) c-a-b
(c) a+b+c          (d) a+b-c