#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
double x[100005],y[100005],lx[100005],ly[100005],maxi=1e9;
int n;
double check(double p){
	double lowx=maxi*maxi,lowy=maxi*maxi,highx=-maxi*maxi,highy=-maxi*maxi,tx,ty;
	FOR(i,1,n){
		tx=x[i]+lx[i]*p;
		ty=y[i]+ly[i]*p;
		lowx=min(lowx,tx);
		highx=max(highx,tx);
		lowy=min(lowy,ty);
		highy=max(highy,ty);
	}
	RE (highx-lowx)*(highy-lowy);
}
int main(){
	double t1,t2,ans=maxi*maxi,t3,t4;
	scanf("%d",&n);
	char c[1];
	FOR(i,1,n){
		scanf("%lf%lf",&x[i],&y[i]);
		scanf("%s",c);
		if(c[0]=='R')lx[i]=1;
		else if(c[0]=='L')lx[i]=-1;
		else if(c[0]=='U')ly[i]=1;
		else ly[i]=-1;
	}
	double l=0,r=maxi;
	FOR(i,1,200){
		t1=(l+r+r)/3.0;
		t2=(l+r+l)/3.0;
		t3=check(t1);
		t4=check(t2);
		ans=min(ans,t3);
		ans=min(ans,t4);
		if(t3>t4){
			r=t1;
		}else l=t2;
	}
	printf("%.12f",ans);
	RE 0;
}


### Prompt

Create a solution in Cpp for 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
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
double x[100005],y[100005],lx[100005],ly[100005],maxi=1e9;
int n;
double check(double p){
	double lowx=maxi*maxi,lowy=maxi*maxi,highx=-maxi*maxi,highy=-maxi*maxi,tx,ty;
	FOR(i,1,n){
		tx=x[i]+lx[i]*p;
		ty=y[i]+ly[i]*p;
		lowx=min(lowx,tx);
		highx=max(highx,tx);
		lowy=min(lowy,ty);
		highy=max(highy,ty);
	}
	RE (highx-lowx)*(highy-lowy);
}
int main(){
	double t1,t2,ans=maxi*maxi,t3,t4;
	scanf("%d",&n);
	char c[1];
	FOR(i,1,n){
		scanf("%lf%lf",&x[i],&y[i]);
		scanf("%s",c);
		if(c[0]=='R')lx[i]=1;
		else if(c[0]=='L')lx[i]=-1;
		else if(c[0]=='U')ly[i]=1;
		else ly[i]=-1;
	}
	double l=0,r=maxi;
	FOR(i,1,200){
		t1=(l+r+r)/3.0;
		t2=(l+r+l)/3.0;
		t3=check(t1);
		t4=check(t2);
		ans=min(ans,t3);
		ans=min(ans,t4);
		if(t3>t4){
			r=t1;
		}else l=t2;
	}
	printf("%.12f",ans);
	RE 0;
}


```