For any set $A$ of numbers, let $\mathrm{sum}(A)$ 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:

$$\begin{array}{rl}\mathrm{sum}(\{1,3,6\})& =10,\\ \mathrm{sum}(\{1,3,8\})& =12,\\ \mathrm{sum}(\{1,3,10\})& =14,\\ \mathrm{sum}(\{1,3,11\})& =15,\\ \mathrm{sum}(\{1,6,8\})& =15,\\ \mathrm{sum}(\{1,6,10\})& =17,\\ \mathrm{sum}(\{1,6,11\})& =18,\\ \mathrm{sum}(\{1,8,10\})& =19,\\ \mathrm{sum}(\{1,8,11\})& =20,\\ \mathrm{sum}(\{1,10,11\})& =22,\\ \mathrm{sum}(\{3,6,8\})& =17,\\ \mathrm{sum}(\{3,6,10\})& =19,\\ \mathrm{sum}(\{3,6,11\})& =20,\\ \mathrm{sum}(\{3,8,10\})& =21,\\ \mathrm{sum}(\{3,8,11\})& =22,\\ \mathrm{sum}(\{3,10,11\})& =24,\\ \mathrm{sum}(\{6,8,10\})& =24,\\ \mathrm{sum}(\{6,8,11\})& =25,\\ \mathrm{sum}(\{6,10,11\})& =27,\\ \mathrm{sum}(\{8,10,11\})& =29.\end{array}$$

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 $\mathrm{sum}(U(B,3))=156$.

Now consider the $100$-element set $S=\{1^2,2^2,\dots ,{100}^2\}$.
S has $100891344545564193334812497256$ $50$-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 $\mathrm{sum}(U(S,50))$.