#include<iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <cstring>
#include <string>
typedef long long ll;
using namespace std;
int n,x[100005],y[100005],d[100005];
char ch[]="RLUD";
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
const unsigned int mo=1e9+7;
double ans=9e99;
bool tra(long double t,int md){
	double max=x[0]+t*dx[d[0]],mix=max,may=y[0]+t*dy[d[0]],miy=may;
	ll nmax=0,nmix=0,nmay=0,nmiy=0;
	for(int i=1;i<n;++i){
		long double X=x[i]+t*dx[d[i]],Y=y[i]+t*dy[d[i]];
		if(X>max)nmax=i,max=X;
		if(X<mix)nmix=i,mix=X;
		if(Y>may)nmay=i,may=Y;
		if(Y<miy)nmiy=i,miy=Y;
	}
	//cout<<max<<may<<mix<<miy<<endl;
	ans=min(ans,(max-mix)*(may-miy));
	int p;
	if(md&4) md-=4,p=dx[d[nmax]]-dx[d[nmix]];
	else p=dy[d[nmay]]-dy[d[nmiy]];
	return p<md-1;
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;++i){
		char tp;
		scanf("%d%d %c",x+i,y+i,&tp);
		for(int j=0;j<4;++j){
			if(ch[j]==tp){
				//cout<<j<<endl;
				d[i]=j;break;
			}
		}
	}
	for(int mode=0;mode<8;++mode){
		long double l=0,r=2e8;
		for(int i=0;i<130;++i){
			double mid=(l+r)/2;
			//printf("%.39lf\n",mid);
			if(tra(mid,mode))l=mid;
			else r=mid;
			//printf("%.39lf\n",(double)ans);
		}
	}
	printf("%.17lf\n",ans);
	return 0;
} 
/*
3
-1000 0 U
1000 0 L
-1900 0 R
*/
### Prompt

Your challenge is to write a CPP solution to the following problem:

There are N points in a two-dimensional plane. The initial coordinates of the i-th point are (x_i, y_i). Now, each point starts moving at a speed of 1 per second, in a direction parallel to the x- or y- axis. You are given a character d_i that represents the specific direction in which the i-th point moves, as follows:

* If d_i = `R`, the i-th point moves in the positive x direction;
* If d_i = `L`, the i-th point moves in the negative x direction;
* If d_i = `U`, the i-th point moves in the positive y direction;
* If d_i = `D`, the i-th point moves in the negative y direction.



You can stop all the points at some moment of your choice after they start moving (including the moment they start moving). Then, let x_{max} and x_{min} be the maximum and minimum among the x-coordinates of the N points, respectively. Similarly, let y_{max} and y_{min} be the maximum and minimum among the y-coordinates of the N points, respectively.

Find the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}) and print it.

Constraints

* 1 \leq N \leq 10^5
* -10^8 \leq x_i,\ y_i \leq 10^8
* x_i and y_i are integers.
* d_i is `R`, `L`, `U`, or `D`.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1 d_1
x_2 y_2 d_2
.
.
.
x_N y_N d_N


Output

Print the minimum possible value of (x_{max} - x_{min}) \times (y_{max} - y_{min}).

The output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-9}.

Examples

Input

2
0 3 D
3 0 L


Output

0


Input

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R


Output

97.5


Input

20
6 -10 R
-4 -9 U
9 6 D
-3 -2 R
0 7 D
4 5 D
10 -10 U
-1 -8 U
10 -6 D
8 -5 U
6 4 D
0 3 D
7 9 R
9 -4 R
3 10 D
1 9 U
1 -6 U
9 -8 R
6 7 D
7 -3 D


Output

273

### Response

```cpp
#include<iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <cstring>
#include <string>
typedef long long ll;
using namespace std;
int n,x[100005],y[100005],d[100005];
char ch[]="RLUD";
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
const unsigned int mo=1e9+7;
double ans=9e99;
bool tra(long double t,int md){
	double max=x[0]+t*dx[d[0]],mix=max,may=y[0]+t*dy[d[0]],miy=may;
	ll nmax=0,nmix=0,nmay=0,nmiy=0;
	for(int i=1;i<n;++i){
		long double X=x[i]+t*dx[d[i]],Y=y[i]+t*dy[d[i]];
		if(X>max)nmax=i,max=X;
		if(X<mix)nmix=i,mix=X;
		if(Y>may)nmay=i,may=Y;
		if(Y<miy)nmiy=i,miy=Y;
	}
	//cout<<max<<may<<mix<<miy<<endl;
	ans=min(ans,(max-mix)*(may-miy));
	int p;
	if(md&4) md-=4,p=dx[d[nmax]]-dx[d[nmix]];
	else p=dy[d[nmay]]-dy[d[nmiy]];
	return p<md-1;
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;++i){
		char tp;
		scanf("%d%d %c",x+i,y+i,&tp);
		for(int j=0;j<4;++j){
			if(ch[j]==tp){
				//cout<<j<<endl;
				d[i]=j;break;
			}
		}
	}
	for(int mode=0;mode<8;++mode){
		long double l=0,r=2e8;
		for(int i=0;i<130;++i){
			double mid=(l+r)/2;
			//printf("%.39lf\n",mid);
			if(tra(mid,mode))l=mid;
			else r=mid;
			//printf("%.39lf\n",(double)ans);
		}
	}
	printf("%.17lf\n",ans);
	return 0;
} 
/*
3
-1000 0 U
1000 0 L
-1900 0 R
*/
```