CodeForces

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 y or 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.

Example

Input

4 3 4H 2V 2V 3V 1

Output

8442

Input

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

Output

28161264

Note

Picture for the first sample test:

Picture for the second sample test:

思路就是将宽度和长度分开储存，分别求出两者的最大值，然后就可算出结果

一开始想着用两个数组存所有的分割点，最后用sort排序然后一个一个算出亮点间距离得出最大值，结果会超时

所以后来换用STL中的set来做

利用set保留值不重复且自动排序的特点，将割点用set保存

利用multiset可以保留重复值且也能自动排序的特点，将距离用multiset保存

这样在插入点的时候，set加入这个新的点，然后搜索这个新点的前后两个点，就可以得出被分割的那一段距离的大小X，然后在multiset中删除这段距离X（注意一次只能删除一段，如果用erase(X)的话会删除所有的段，只能erase(find(X))，具体操作可以去看看set用法的详解）

删除后将新生成的两段距离加入multiset，然后就可以输出答案了

代码

#include<iostream>#include<cstring>#include<algorithm>#include<set>using namespace std;set<long long>h1;set<long long>w1;set<long long>::iterator it;multiset<long long>h2;multiset<long long>w2;multiset<long long>::iterator it2;int main(){    ios::sync_with_stdio(false);    int w,h,n;    int sta,last;    int pos;    char order;    int maxw,maxh;    while(cin>>w>>h>>n)    {        w1.clear();        w2.clear();        h1.clear();        h2.clear();        w1.insert(0);        h1.insert(0);        w1.insert(w);        h1.insert(h);        w2.insert(w);        h2.insert(h);        for(int i=0;i<n;i++)        {            cin>>order>>pos;            if(order=='H')            {                h1.insert(pos);                sta=*(--h1.find(pos));                last=*(++h1.find(pos));                h2.erase(h2.find(last-sta));                h2.insert(pos-sta);                h2.insert(last-pos);            }            else            {                w1.insert(pos);                sta=*(--w1.find(pos));                last=*(++w1.find(pos));                w2.erase(w2.find(last-sta));                w2.insert(pos-sta);                w2.insert(last-pos);            }            it2=w2.end();            it2--;            maxw=*it2;            it2=h2.end();            it2--;            maxh=*it2;            cout<<(*(--w2.end()))*(*(--h2.end()))<<endl;        }    }    return 0;}