#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int n;
double x[200000],y[200000];
char d[200000][2];
int dx[]{1,-1,0,0},dy[]{0,0,1,-1};
map<char,int>dir;

double f(double t,int mode){
	double Min=-1,Max=-1;
	rep(i,n){
		double X;
		if(mode==0)X=x[i]+dx[dir[d[i][0]]]*t;
		else X=y[i]+dy[dir[d[i][0]]]*t;
		if(i==0){
			Min=Max=X;
		}
		else{
			Min=min(Min,X);
			Max=max(Max,X);
		}
	}
	return Max-Min;
}
int main(){
	dir['R']=0;
	dir['L']=1;
	dir['U']=2;
	dir['D']=3;
	cin>>n;
	rep(i,n){
		scanf("%lf%lf%s",&x[i],&y[i],d[i]);
	}
	double l=0,r=1e15;
	rep(i,200){
		double l1=(l+l+r)/3;
		double l2=(l+r+r)/3;
		if(f(l1,0)*f(l1,1)<f(l2,0)*f(l2,1))r=l2;
		else l=l1;
	}
	printf("%.12lf\n",f(l,0)*f(l,1));
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int n;
double x[200000],y[200000];
char d[200000][2];
int dx[]{1,-1,0,0},dy[]{0,0,1,-1};
map<char,int>dir;

double f(double t,int mode){
	double Min=-1,Max=-1;
	rep(i,n){
		double X;
		if(mode==0)X=x[i]+dx[dir[d[i][0]]]*t;
		else X=y[i]+dy[dir[d[i][0]]]*t;
		if(i==0){
			Min=Max=X;
		}
		else{
			Min=min(Min,X);
			Max=max(Max,X);
		}
	}
	return Max-Min;
}
int main(){
	dir['R']=0;
	dir['L']=1;
	dir['U']=2;
	dir['D']=3;
	cin>>n;
	rep(i,n){
		scanf("%lf%lf%s",&x[i],&y[i],d[i]);
	}
	double l=0,r=1e15;
	rep(i,200){
		double l1=(l+l+r)/3;
		double l2=(l+r+r)/3;
		if(f(l1,0)*f(l1,1)<f(l2,0)*f(l2,1))r=l2;
		else l=l1;
	}
	printf("%.12lf\n",f(l,0)*f(l,1));
}
```