Problem 201: Subsets with a unique sum

For any set $A$ of numbers, let $sum(A)$ be the sum of the elements of $A$. Consider the set $B = \\{1,3,6,8,10,11\\}$. There are 20 subsets of $B$ containing three elements, and their sums are: $$\begin{align} & sum(\\{1,3,6\\}) = 10 \\\\ & sum(\\{1,3,8\\}) = 12 \\\\ & sum(\\{1,3,10\\}) = 14 \\\\ & sum(\\{1,3,11\\}) = 15 \\\\ & sum(\\{1,6,8\\}) = 15 \\\\ & sum(\\{1,6,10\\}) = 17 \\\\ & sum(\\{1,6,11\\}) = 18 \\\\ & sum(\\{1,8,10\\}) = 19 \\\\ & sum(\\{1,8,11\\}) = 20 \\\\ & sum(\\{1,10,11\\}) = 22 \\\\ & sum(\\{3,6,8\\}) = 17 \\\\ & sum(\\{3,6,10\\}) = 19 \\\\ & sum(\\{3,6,11\\}) = 20 \\\\ & sum(\\{3,8,10\\}) = 21 \\\\ & sum(\\{3,8,11\\}) = 22 \\\\ & sum(\\{3,10,11\\}) = 24 \\\\ & sum(\\{6,8,10\\}) = 24 \\\\ & sum(\\{6,8,11\\}) = 25 \\\\ & sum(\\{6,10,11\\}) = 27 \\\\ & sum(\\{8,10,11\\}) = 29 \\end{align}$$ Some of these sums occur more than once, others are unique. For a set $A$, let $U(A,k)$ be the set of unique sums of $k$-element subsets of $A$, in our example we find $U(B,3) = \\{10,12,14,18,21,25,27,29\\}$ and $sum(U(B,3)) = 156$. Now consider the $100$-element set $S = \\{1^2, 2^2, \ldots , {100}^2\\}$. $S$ has $100\\,891\\,344\\,545\\,564\\,193\\,334\\,812\\,497\\,256\\;$ $50$-element subsets. Determine the sum of all integers which are the sum of exactly one of the $50$-element subsets of $S$, i.e. find $sum(U(S,50))$.