The number of heavy edges becoming preferred can be for any given operation, but it is amortized. Over a series of executions we can have "n"-1 heavy edges become preferred (as there are at most "n"-1 heavy edges total in the represented tree), but from then on the number of heavy edges that become preferred is equal to the number of heavy edges that became unpreferred on a previous step. For every heavy edge that becomes unpreferred a light edge must become preferred. We have seen already that the number of light edges that can become preferred is at most log "n". So the number of heavy edges that become preferred for "m" operations is . Over enough operations () this averages to .