题干：

Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular w mm  ×  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 yor V x. In the first case Leonid makes the horizontal cut at the distance ymillimeters (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

4 3 4
H 2
V 2
V 3
V 1

Output

8
4
4
2

Input

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:

题目大意：

本题给你一个H*V的矩阵，有两种操作，一种是横向切割，一种是纵向切割，每次切割后需要你输出当前状态下的最大矩形的面积是多大。

 

解题报告：

   其实将横纵分开来看就是找到最大的长和最大的宽最后相乘就是答案。可以线段树区间合并，可以set直接维护前驱后继和最大长度，也可以用平衡树去跑一波（不会过程）线段树就是区间最长连续1的个数，也不难在此就不练习了 。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
set<int> h,v;
multiset<int> mh,mv;
set<int> :: iterator sit;
multiset<int> :: iterator musit;
int main()
{int H,V,m,x;char op[15];cin>>V>>H>>m;h.insert(0),h.insert(H);v.insert(0),v.insert(V);mh.insert(H),mv.insert(V);
//	printf("%d\n%d\n",*(--mh.end()),*(--mv.end()));while(m--) {scanf("%s%d",op,&x);if(op[0] == 'H') {h.insert(x);sit = h.find(x);sit++;int up = *sit;sit--;sit--;int low = *sit;musit = mh.find(up-low);mh.erase(musit);mh.insert(x-low);mh.insert(up-x);}else {v.insert(x);sit = v.find(x);sit++;int up = *sit;sit--;sit--;int low = *sit;musit = mv.find(up-low);mv.erase(musit);mv.insert(up-x);mv.insert(x-low);}
//		printf("%d\n%d\n",*(--mh.end()),*(--mv.end()));printf("%lld\n",(ll)(*(--mh.end())) * (*(--mv.end())));}return 0 ;}