[boost] [histogram] Variance

The variance of individual bins can be obtained when using the adaptive_storage (via h.at(i).variance().) I am trying to understand the overhead of this feature. If I interpret the code correctly, there is a space overhead because each counter has to keep track of both the count and the sum of squares. The computational overhead is that the sum of squares has to be calculated for each insertion. Is this correct? If so, is there any way to use the adaptive storage policy without variance? Furthermore, why does variance() return the sum of squares? Should this not be divided by the sample size?