I'm reading this paper http://arxiv.org/abs/1502.06624

In the discussion surrounding equation 5, it is explained how the charged bions $M_i \bar M_j$ for monopole operators $M_{i,j}$ with $i\neq j$ and neutral bions $M_i M_i$ have the same saddle-point actions $e^{-2S_0}$ where $S_0$ is the action of a single monopole but contribute terms with *opposite* phases in the path integral. My issue with this is it seems there are N choose 2 charged bions but only $N$ neutral bions, so the overall contribution to the vacuum energy is $({N \choose 2} - N)e^{-2S_0}$, not zero as claimed.

Am I missing something dumb?