For n>=20 set lambda=ceil(log2 n), nu=ceil(log2 lambda), and let s be the largest nonnegative integer with s^2 nu^4<=lambda. If s<2 use exhaustive counting; otherwise choose sigma=1/s. The number of sparsification nodes is at most 2^{n/s} poly_k(n,s), and leaf occurrence B=O_k((s log s)^{k-1}) is polylogarithmic in n. The floor loses at most a factor two when s>=2, so n/s=O(rho(n)), with rho(n)=n(log log n)^2/sqrt(log n). Both the tree size and each leaf counter therefore have exponent O_k(rho(n)). Costs multiply, so exponents ADD; one must not ignore the cost of either layer.