Description:
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 = {a1, a2, ..., ak}, where a1 < a2 < a3 < ... < ak, will be understood by as
$$\sum_{i \mod 5 = 3}^{i \leq k} a_{i-1}$$
The $$\bmod$$ operator stands for taking the remainder, that is $$x \mod y$$ 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.
Input Format:
The first line contains number n (1 ≤ n ≤ 105), the number of operations performed.
Then each of n lines contains the description of one of the three operations:
- add x — add the element x to the set;
- del x — delete the element x from the set;
- sum — find the sum of medians of the set.
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.
Output Format:
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).
Note:
None