【multiset】Glass Carving

 

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

Copy

8
4
4
2

input

Copy

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

output

Copy

28
16
12
6
4

Note

Picture for the first sample test:

Picture for the second sample test:

没什么可说的，set记录每一次切割的位置

但是注意s.eraser(0)是删除全部的

s.eraser(s.find(0))仅仅删除一个

是数值的话全部删除，是指针的话仅仅删除一个

代码不是我的= =

#include <iostream>
#include <stdio.h>
#include <set>
#include <string>
#include <queue>

using namespace std;

int main()
{
    int w, h, n;
    cin>>w>>h>>n;
    set<int> qh, qv;
    multiset<int> mh, mv;
    qh.insert(h);
    qh.insert(0);
    qv.insert(w);
    qv.insert(0);
    mh.insert(h);
    mv.insert(w);
    char s[5];
    for(int i = 0; i < n; i++)
    {
        scanf("%s", s);
        set<int> ::iterator iter, pre, next;
        multiset<int> ::iterator era;
        if(s[0] == 'H')
        {
            int y;
            cin>>y;
            qh.insert(y);
            pre = next = iter = qh.find(y);
            pre--;
            next++;
            era = mh.find(*next - *pre);
            mh.erase(era);
            mh.insert(y - *pre);
            mh.insert(*next - y);
        }
        else if(s[0] == 'V')
        {
            int x;
            cin>>x;
            qv.insert(x);
            pre = next = iter = qv.find(x);
            pre--;
            next++;
            era = mv.find(*next - *pre);
            mv.erase(era);
            mv.insert(x - *pre);
            mv.insert(*next - x);
        }
        multiset<int> ::iterator iterh, iterv;
        iterh = mh.end();
        iterh--;
        iterv = mv.end();
        iterv--;
        long long ans = (long long)(*iterh) * (*iterv);
        printf("%I64d\n", ans);
    }
    return 0;
}