
#include<bits/stdc++.h>
using namespace std;
double x[100005],y[100005];
double dx[100005],dy[100005];
int n;
double check(double a){
	double mnx=1e18,mxx=-1e18,mny=1e18,mxy=-1e18;
	for(int i=0;i<n;i++){
		double xx=x[i]+a*dx[i];
		double yy=y[i]+a*dy[i];
		mnx=min(mnx,xx);
		mxx=max(mxx,xx);
		mny=min(mny,yy);
		mxy=max(mxy,yy);
	}
	return (mxx-mnx)*(mxy-mny);
}
int main(){
	
	cin>>n;
	char s;
	memset(dx,0,sizeof(dx));
	memset(dy,0,sizeof(dy));
	for(int i=0;i<n;i++){
		cin>>x[i]>>y[i]>>s;
		if(s=='D')
			dy[i]=-1;
		if(s=='U')
			dy[i]=1;
		if(s=='R')
			dx[i]=1;
		if(s=='L')
			dx[i]=-1;
	}
	double l=0,r=1e9;
	double ans=1e18;
	for(int i=0;i<200;i++){
		double ll=(l*2+r)/3;
		double rr=(r*2+l)/3;
		double lll=check(ll),rrr=check(rr);
		ans=min(ans,lll);
		ans=min(ans,rrr);
		if(lll<rrr)
			r=rr;
		else l=ll;
	}
	printf("%lf\n",ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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<bits/stdc++.h>
using namespace std;
double x[100005],y[100005];
double dx[100005],dy[100005];
int n;
double check(double a){
	double mnx=1e18,mxx=-1e18,mny=1e18,mxy=-1e18;
	for(int i=0;i<n;i++){
		double xx=x[i]+a*dx[i];
		double yy=y[i]+a*dy[i];
		mnx=min(mnx,xx);
		mxx=max(mxx,xx);
		mny=min(mny,yy);
		mxy=max(mxy,yy);
	}
	return (mxx-mnx)*(mxy-mny);
}
int main(){
	
	cin>>n;
	char s;
	memset(dx,0,sizeof(dx));
	memset(dy,0,sizeof(dy));
	for(int i=0;i<n;i++){
		cin>>x[i]>>y[i]>>s;
		if(s=='D')
			dy[i]=-1;
		if(s=='U')
			dy[i]=1;
		if(s=='R')
			dx[i]=1;
		if(s=='L')
			dx[i]=-1;
	}
	double l=0,r=1e9;
	double ans=1e18;
	for(int i=0;i<200;i++){
		double ll=(l*2+r)/3;
		double rr=(r*2+l)/3;
		double lll=check(ll),rrr=check(rr);
		ans=min(ans,lll);
		ans=min(ans,rrr);
		if(lll<rrr)
			r=rr;
		else l=ll;
	}
	printf("%lf\n",ans);
}

```