# 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}<a_{2}<a_{3}<...<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
add 4
add 5
add 1
add 2
add 3
sum
```

### 输出样例 #1

```
3
```

### 输入样例 #2

```
14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum
```

### 输出样例 #2

```
5
11
13
```