#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
typedef long long ll;
const int maxn=1e5+10;
int n;
struct node{
	int x,y;
	char c;
}cor[maxn];
double cal(double tim){
	double ml=1e10,mr=-1e10,mu=-1e10,md=1e10;
	for(int i=1;i<=n;i++){
		pair<double,double>tmp;
		tmp=make_pair(cor[i].x,cor[i].y);
		switch(cor[i].c){
			case 'U':tmp.second+=tim;break;
			case 'D':tmp.second-=tim;break;
			case 'L':tmp.first-=tim;break;
			case 'R':tmp.first+=tim;break;
		}
		ml=min(ml,tmp.first);
		mr=max(mr,tmp.first);
		md=min(md,tmp.second);
		mu=max(mu,tmp.second);
	}
	return (mu-md)*(mr-ml);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d %c",&cor[i].x,&cor[i].y,&cor[i].c);
	}
	double l=0,r=1e12,ans=1e18;
	for(int i=1;i<=300;i++){
		double m1=l+(r-l)/3,m2=l+(r-l)*2/3;
		double tmp1=cal(m1),tmp2=cal(m2);
		if(tmp1>tmp2)l=m1;
		else r=m2;
		ans=min(ans,tmp1);
		ans=min(ans,tmp2); 
	}
	printf("%.10f\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;
typedef unsigned long long ULL;
typedef long long ll;
const int maxn=1e5+10;
int n;
struct node{
	int x,y;
	char c;
}cor[maxn];
double cal(double tim){
	double ml=1e10,mr=-1e10,mu=-1e10,md=1e10;
	for(int i=1;i<=n;i++){
		pair<double,double>tmp;
		tmp=make_pair(cor[i].x,cor[i].y);
		switch(cor[i].c){
			case 'U':tmp.second+=tim;break;
			case 'D':tmp.second-=tim;break;
			case 'L':tmp.first-=tim;break;
			case 'R':tmp.first+=tim;break;
		}
		ml=min(ml,tmp.first);
		mr=max(mr,tmp.first);
		md=min(md,tmp.second);
		mu=max(mu,tmp.second);
	}
	return (mu-md)*(mr-ml);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d %c",&cor[i].x,&cor[i].y,&cor[i].c);
	}
	double l=0,r=1e12,ans=1e18;
	for(int i=1;i<=300;i++){
		double m1=l+(r-l)/3,m2=l+(r-l)*2/3;
		double tmp1=cal(m1),tmp2=cal(m2);
		if(tmp1>tmp2)l=m1;
		else r=m2;
		ans=min(ans,tmp1);
		ans=min(ans,tmp2); 
	}
	printf("%.10f\n",ans);
}


```