I need to show that if $\pi$ is the stationary distribution of a Markov chain $M$, then for every set of vertices $S$, the probability to choose a random node in $S$ according to $\pi$ and then going to a random neighbor not in $S$ $\left(\sum_{u\in S,v\notin S}\pi(u)M(u,v)\right)$ equals to the probability of choosing a random vertex not in S according to $\pi$ and going to a random neighbor in $S$ $\left(\sum_{u\in S,v\notin S}\pi(v)M(v,u)\right)$
Intuitively I can see why its true, but when I tired to prove it using the identity $\pi = M\pi$ I didn't manage to get anywhere :(
Any ideas people?