#include<iostream>
#include<map>
#include<iomanip>
using namespace std;
int main()
{
	int n;
	cin>>n;
	double x[n],y[n];
	map<char,pair<double,double> >m;
	pair<double,double> d[n];
	m['U']=make_pair(0.0,1.0);
	m['D']=make_pair(0.0,-1.0);
	m['L']=make_pair(-1.0,0.0);
	m['R']=make_pair(1.0,0.0);
	for(int i=0;i<n;i++){
		double x1,y1;
		char ch;
		cin>>x1>>y1>>ch;
		x[i]=x1;
		y[i]=y1;
		d[i]=m[ch];
	}
	//cout<<setprecision(12);
	double l=0.0,r=1e10,a1=1e18,a2=1e18;
	for(int i=0;i<200;i++){
		double l1=l+(r-l)/3.0,r1=r-(r-l)/3.0;
		double minx=1e18,miny=1e18,maxx=-1e18,maxy=-1e18;
		for(int j=0;j<n;j++){
			double x1=x[j]+l1*d[j].first;
			double y1=y[j]+l1*d[j].second;
			minx=min(minx,x1);
			maxx=max(maxx,x1);
			miny=min(miny,y1);
			maxy=max(maxy,y1);
		}
		a1=(maxx-minx)*(maxy-miny);
      	minx=1e18,miny=1e18,maxx=-1e18,maxy=-1e18;
		for(int j=0;j<n;j++){
			double x1=x[j]+r1*d[j].first;
			double y1=y[j]+r1*d[j].second;
			minx=min(minx,x1);
			maxx=max(maxx,x1);
			miny=min(miny,y1);
			maxy=max(maxy,y1);
		}
		a2=(maxx-minx)*(maxy-miny);
		if(a1<a2)r=r1;
		else l=l1;
	}
	printf("%.12lf",min(a1,a2));
}
### 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<map>
#include<iomanip>
using namespace std;
int main()
{
	int n;
	cin>>n;
	double x[n],y[n];
	map<char,pair<double,double> >m;
	pair<double,double> d[n];
	m['U']=make_pair(0.0,1.0);
	m['D']=make_pair(0.0,-1.0);
	m['L']=make_pair(-1.0,0.0);
	m['R']=make_pair(1.0,0.0);
	for(int i=0;i<n;i++){
		double x1,y1;
		char ch;
		cin>>x1>>y1>>ch;
		x[i]=x1;
		y[i]=y1;
		d[i]=m[ch];
	}
	//cout<<setprecision(12);
	double l=0.0,r=1e10,a1=1e18,a2=1e18;
	for(int i=0;i<200;i++){
		double l1=l+(r-l)/3.0,r1=r-(r-l)/3.0;
		double minx=1e18,miny=1e18,maxx=-1e18,maxy=-1e18;
		for(int j=0;j<n;j++){
			double x1=x[j]+l1*d[j].first;
			double y1=y[j]+l1*d[j].second;
			minx=min(minx,x1);
			maxx=max(maxx,x1);
			miny=min(miny,y1);
			maxy=max(maxy,y1);
		}
		a1=(maxx-minx)*(maxy-miny);
      	minx=1e18,miny=1e18,maxx=-1e18,maxy=-1e18;
		for(int j=0;j<n;j++){
			double x1=x[j]+r1*d[j].first;
			double y1=y[j]+r1*d[j].second;
			minx=min(minx,x1);
			maxx=max(maxx,x1);
			miny=min(miny,y1);
			maxy=max(maxy,y1);
		}
		a2=(maxx-minx)*(maxy-miny);
		if(a1<a2)r=r1;
		else l=l1;
	}
	printf("%.12lf",min(a1,a2));
}
```