finding the inverse of the sum of prime factorization

Question

I'm a little confused on how to find the inverse of the following function $$ \ h:\mathbb{N} \backslash \{1\}\rightarrow \Bbb{N} $$ Be defined as$$h(n) = the \ sum \ of \ factors\ appearing \ in\ the \ prime\ factorization \\$$Example$$ h(12) = 2+2+3$$ Find $$h^{-1}(11)$$The solution gives the following answer$$\{11,18,28,40,45,48,54\}$$I believe they found the inverse to be the product of prime numbers that sum to 11. So for 45$$3\cdot3\cdot5 = 45\\3+3+2 = 11$$how does this show its the inverse? and if im correct in my interpretation why is 18 and 11 in the set? 9 and 1 are not prime$$2+9 =11\\2\cdot9=18\\11\cdot1=11\\1+11=12$$