A8083 | Sum of Medians

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 $S={a_{1},a_{2},...,a_{k}}$ , where $a_{1}<a_{2}<a_{3}<...<a_{k}$ , 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 $n$ ( $1<=n<=10^{5}$ ), 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 $10^{9}$ .

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).