Let's calculate each bit in the answer separately. Suppose we want to know the value of $$$k$$$-th (in 0-indexation) bit in the answer. Then we can notice that we are only interested in bits from $$$0$$$-th to $$$k$$$-th, which means that we can take all numbers modulo $$$2^{k+1}$$$. After that, the sum of the two numbers can't exceed $$$2^{k+2} - 2$$$. $$$k$$$-th bit is 1 if and only if sum belongs to $$$[2^k; 2^{k+1})$$$ or $$$[2^{k+1} + 2^k; 2^{k+2} - 2]$$$.

So, we have to count the number of pairs of numbers that give a sum that belongs to these segments. Let's sort all numbers (taken by modulo) and make a pass with two pointers or do binary searches for each number.

Total complexity: $$$O(n \log n \log C)$$$

Bonus: can you do it in $$$O(n \log C)$$$?