#include<bits/stdc++.h>
using namespace std;
const double inf=1e18;
int n,belong[1005],dx[4]={1,-1,0,0},dy[4]={0,0,1,-1},t[100005];
double x[100005],y[100005];
char s[5];
double f(double xx) {
	double xmi=inf,xmx=-inf,ymi=inf,ymx=-inf;
	for (int i=1; i<=n; i++) {
		xmi=min(xmi,x[i]+dx[t[i]]*xx);
		xmx=max(xmx,x[i]+dx[t[i]]*xx);
		ymi=min(ymi,y[i]+dy[t[i]]*xx);
		ymx=max(ymx,y[i]+dy[t[i]]*xx);
	}
	return (xmx-xmi)*(ymx-ymi);
}
int main() {
	belong['R']=0;
	belong['L']=1;
	belong['U']=2;
	belong['D']=3;
	scanf("%d", &n);
	for (int i=1; i<=n; i++) {
		scanf("%lf%lf%s", &x[i], &y[i], s);
		t[i]=belong[s[0]];
	}
	double l=0,r=1e9,lmid,rmid,L,R,res=inf;
	for (int i=0; i<200; i++) {
		lmid=(l*2+r)/3;
		rmid=(l+r*2)/3;
		L=f(lmid),R=f(rmid);
		res=min(res,min(L,R));
		if (L>R) l=lmid;
		else r=rmid;
	}
	printf("%.15f\n", res);
	return 0;
}
### 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;
const double inf=1e18;
int n,belong[1005],dx[4]={1,-1,0,0},dy[4]={0,0,1,-1},t[100005];
double x[100005],y[100005];
char s[5];
double f(double xx) {
	double xmi=inf,xmx=-inf,ymi=inf,ymx=-inf;
	for (int i=1; i<=n; i++) {
		xmi=min(xmi,x[i]+dx[t[i]]*xx);
		xmx=max(xmx,x[i]+dx[t[i]]*xx);
		ymi=min(ymi,y[i]+dy[t[i]]*xx);
		ymx=max(ymx,y[i]+dy[t[i]]*xx);
	}
	return (xmx-xmi)*(ymx-ymi);
}
int main() {
	belong['R']=0;
	belong['L']=1;
	belong['U']=2;
	belong['D']=3;
	scanf("%d", &n);
	for (int i=1; i<=n; i++) {
		scanf("%lf%lf%s", &x[i], &y[i], s);
		t[i]=belong[s[0]];
	}
	double l=0,r=1e9,lmid,rmid,L,R,res=inf;
	for (int i=0; i<200; i++) {
		lmid=(l*2+r)/3;
		rmid=(l+r*2)/3;
		L=f(lmid),R=f(rmid);
		res=min(res,min(L,R));
		if (L>R) l=lmid;
		else r=rmid;
	}
	printf("%.15f\n", res);
	return 0;
}
```