Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

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

Sample Output 3

273

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
double MAX(double a,double b){return a<b?b:a;}
double MIN(double a,double b){return a<b?a:b;}
int r=0;
double t[100],y[3][2],x[3][2];
void f(int a,int b,int c){
  t[r++]=MAX(0,b-a);
  t[r++]=MAX(0,c-b);
  t[r++]=MAX(0,(c-a)/2.0);
}
double g(double t){
  double l=MIN(MIN(x[0][0]+t,x[1][0]),x[2][0]-t);
  double r=MAX(MAX(x[0][1]+t,x[1][1]),x[2][1]-t);
  double u=MIN(MIN(y[0][0]+t,y[1][0]),y[2][0]-t);
  double d=MAX(MAX(y[0][1]+t,y[1][1]),y[2][1]-t);
  //printf("%lf %lf %lf %lf %lf\n",t,l,r,u,d);
  return (r-l)*(d-u);
}
int main(){
  int n,s,p,i,j;
  char c,cc[]={"URDL"};
  int a[]={1,2,1,0};
  int b[]={0,1,2,1};
  double ans=1e18;
  scanf("%d",&n);
  for(i=0;i<3;i++)x[i][0]=y[i][0]= 1e9;
  for(i=0;i<3;i++)x[i][1]=y[i][1]=-1e9;
  for(i=0;i<n;i++){
    scanf("%d %d %c",&s,&p,&c);
    for(j=0;c-cc[j];j++);
    x[a[j]][0]=MIN(x[a[j]][0],s);
    x[a[j]][1]=MAX(x[a[j]][1],s);
    y[b[j]][0]=MIN(y[b[j]][0],p);
    y[b[j]][1]=MAX(y[b[j]][1],p);
  }
  //for(i=0;i<3;i++)printf("x%lf %lf\n",x[i][0],x[i][1]);
  //for(i=0;i<3;i++)printf("y%lf %lf\n",y[i][0],y[i][1]);
  for(i=0;i<2;i++)f(x[0][i],x[1][i],x[2][i]);
  for(i=0;i<2;i++)f(y[0][i],y[1][i],y[2][i]);
  //for(i=0;i<r;i++)printf("%lf %lf\n",t[i],g(t[i]));
  for(i=0;i<r;i++)ans=MIN(g(t[i]),ans);
  printf("%lf\n",ans);
  return 0;
}