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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define LL long long
#define P 1000000007
#define M 100005
using namespace std;
int rx[]={1,-1,0,0};
int ry[]={0,0,1,-1};
//RLUD
struct node{
	int x,y,d;
}s[M];
int n;
double calc(double t){
	double x1=1e18,x2=-1e18;
	double y1=x1,y2=x2;
	for(int i=1;i<=n;i++){
		double x=s[i].x+t*rx[s[i].d];
		double y=s[i].y+t*ry[s[i].d];
		x1=min(x1,x);x2=max(x2,x);
		y1=min(y1,y);y2=max(y2,y);
	}
	return (x2-x1)*(y2-y1);
}
int main() { 
	char str[10];
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%d %d %s",&s[i].x,&s[i].y,str);
		if(str[0]=='R')s[i].d=0;
		if(str[0]=='L')s[i].d=1;
		if(str[0]=='U')s[i].d=2;
		if(str[0]=='D')s[i].d=3;
	}
	double l=0,r=1e8,res;
	for(int i=0;i<100;i++){
		double l1=l+(r-l)/3;
		double r1=l+(r-l)*2/3;
		double c1=calc(l1);
		double c2=calc(r1);
		if(c1<=c2){
			res=c1;r=r1;
		}else{
			res=c2;l=l1;
		}
	}
	printf("%.15f\n",res);
	return 0;
}