# Sum of Medians

## 题意翻译

- 有一个集合，初始为空。现有 $n$ 次操作： 1. add x：将 $x$ 添加到集合中。 2. del x：将 $x$ 从集合中删除。 3. sum：将集合内的数从小到大排好序后形成有 $k$ 个数的序列 $a$，求 $$\sum_{i}^{(i\le k)\land (i\bmod 5=3)}a_i$$ - $1\le n\le 10^5$，$1\le x\le 10^9$。

## 题目描述

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 $a_{1}&lt;a_{2}&lt;a_{3}&lt;...&lt;a_{k}$ , will be understood by as ![](https://cdn.luogu.com.cn/upload/vjudge_pic/CF85D/ade3397df6e8978ddadfc100b4ccb88beefd1e3f.png)The ![](https://cdn.luogu.com.cn/upload/vjudge_pic/CF85D/99fd5677ca5c02520be7595d9b1eaf3e9972e601.png) operator stands for taking the remainder, that is ![](https://cdn.luogu.com.cn/upload/vjudge_pic/CF85D/cb1d84ad58154eb7ea26b65d1ae0039570db9bb6.png) 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.

## 输入输出格式

### 输入格式

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 $10^{9}$ .

### 输出格式

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).

## 输入输出样例

### 输入样例 #1

6
sum


### 输出样例 #1

3


### 输入样例 #2

14
sum

5