B"In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array. A sum of medians of a sorted k-element set S xe2 x80 x89= xe2 x80 x89{a1, xe2 x80 x89a2, xe2 x80 x89..., xe2 x80 x89ak}, where a1 xe2 x80 x89< xe2 x80 x89a2 xe2 x80 x89< xe2 x80 x89a3 xe2 x80 x89< xe2 x80 x89... xe2 x80 x89< xe2 x80 x89ak, will be understood by as The operator stands for taking the remainder, that is stands for the remainder of dividing x by y. To organize exercise testing quickly calculating the sum of medians for a changing set was needed. The first line contains number n (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89105), the number of operations performed. Then each of n lines contains the description of one of the three operations: For any add x operation it is true that the element x is not included in the set directly before the operation. For any del x operation it is true that the element x is included in the set directly before the operation. All the numbers in the input are positive integers, not exceeding 109. For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0. Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator)."...