#include<iostream>
#include<iomanip>
using namespace std;
int N;
long double X[1<<17],Y[1<<17];
char D[1<<17];
long double calc(long double v)
{
	long double xm=1e18,xM=-1e18,ym=1e18,yM=-1e18;
	for(int i=0;i<N;i++)
	{
		xm=min(xm,v*(D[i]=='L'?-1:D[i]=='R')+X[i]);
		xM=max(xM,v*(D[i]=='L'?-1:D[i]=='R')+X[i]);
		ym=min(ym,v*(D[i]=='D'?-1:D[i]=='U')+Y[i]);
		yM=max(yM,v*(D[i]=='D'?-1:D[i]=='U')+Y[i]);
	}
	return(xM-xm)*(yM-ym);
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)cin>>X[i]>>Y[i]>>D[i];
	long double L=0,R=1e18;
	for(int i=0;i<300;i++)
	{
		if(calc((L*2+R)/3)<calc((L+R*2)/3))R=(L+R*2)/3;
		else L=(L*2+R)/3;
	}
	cout<<fixed<<setprecision(16)<<calc(L)<<endl;
}
### Prompt

In CPP, your task is to solve 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<iomanip>
using namespace std;
int N;
long double X[1<<17],Y[1<<17];
char D[1<<17];
long double calc(long double v)
{
	long double xm=1e18,xM=-1e18,ym=1e18,yM=-1e18;
	for(int i=0;i<N;i++)
	{
		xm=min(xm,v*(D[i]=='L'?-1:D[i]=='R')+X[i]);
		xM=max(xM,v*(D[i]=='L'?-1:D[i]=='R')+X[i]);
		ym=min(ym,v*(D[i]=='D'?-1:D[i]=='U')+Y[i]);
		yM=max(yM,v*(D[i]=='D'?-1:D[i]=='U')+Y[i]);
	}
	return(xM-xm)*(yM-ym);
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)cin>>X[i]>>Y[i]>>D[i];
	long double L=0,R=1e18;
	for(int i=0;i<300;i++)
	{
		if(calc((L*2+R)/3)<calc((L+R*2)/3))R=(L+R*2)/3;
		else L=(L*2+R)/3;
	}
	cout<<fixed<<setprecision(16)<<calc(L)<<endl;
}
```