codeforces 527 C. Glass Carving

http://codeforces.com/problemset/problem/527/C

C. Glass Carving

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular wmm  ×  h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.

In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.

After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.

Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?

Input

The first line contains three integers w, h, n (2 ≤ w, h ≤ 200 000, 1 ≤ n ≤ 200 000).

Next n lines contain the descriptions of the cuts. Each description has the form H y or V x. In the first case Leonid makes the horizontal cut at the distance y millimeters (1 ≤ y ≤ h - 1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance x (1 ≤ x ≤ w - 1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.

Output

After each cut print on a single line the area of the maximum available glass fragment in mm2.

Examples

input

Copy

4 3 4
H 2
V 2
V 3
V 1

output

8
4
4
2

input

Copy

7 6 5
H 4
V 3
V 5
H 2
V 1

output

28
16
12
6
4

Note

Picture for the first sample test:

Picture for the second sample test:

题意：给定一个矩形，然后有n步操作，没次操作在这个矩形上画一条线，每步操作输出画完后的大矩形中得到的小矩形的最大面积。即知道的到划分后的长最长的和宽最长的相乘即可得到最大的那个矩形的面积。
题解：利用两个set来维护，一个用来维护矩形的长，一个用来维护矩形的宽，把每次画的线存进set中，即可用迭代器得到这条线左右的条线，即可得到把这条线画上以后分成的两条线段的长度（具体看代码）。然后在用一个mw和mh数组储存任何一个时刻最长的w和最长的h，每次将最长的相乘即可

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