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.

Input

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 $10^{9}$.

Output

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).