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<stdio.h>
#include<math.h>
#define N 100000000
#define M 100005
int main()
{
	int n;
	double x[M],y[M];
	char c[M];
	double minS=-1,S1,S2,l=0,r=N;
	double minl1=N,maxr1=-N,maxu1=-N,mind1=N,xx1,yy1,move1=0;
	double minl2=N,maxr2=-N,maxu2=-N,mind2=N,xx2,yy2,move2=N;
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%lf%lf %c",&x[i],&y[i],&c[i]);
    for(int k=0;k<=10000&&(move2-move1>0.00000000001);k++)
	{
		minl1=N,maxr1=-N,maxu1=-N,mind1=N;
		minl2=N,maxr2=-N,maxu2=-N,mind2=N;
        move1=(r/3+2*l/3),move2=(2*r/3+l/3);
        for(int i=0;i<n;i++)
		{
	        xx1=xx2=x[i];
			yy1=yy2=y[i];
			if(c[i]=='U')
			{
				yy1+=move1;
				yy2+=move2;
			}
			if(c[i]=='D')
			{
				yy1-=move1;
				yy2-=move2;
			}
			if(c[i]=='L')
			{
				xx1-=move1;
				xx2-=move2;
			}
			if(c[i]=='R')
			{
				xx1+=move1;
				xx2+=move2;
			}
	        if(xx1<minl1)minl1=xx1;
	        if(xx1>maxr1)maxr1=xx1;
	        if(yy1<mind1)mind1=yy1;
	        if(yy1>maxu1)maxu1=yy1;
	        if(xx2<minl2)minl2=xx2;
	        if(xx2>maxr2)maxr2=xx2;
	        if(yy2<mind2)mind2=yy2;
	        if(yy2>maxu2)maxu2=yy2;
	    }
	    S1=(maxr1-minl1)*(maxu1-mind1);
	    S2=(maxr2-minl2)*(maxu2-mind2);
        if(S1<minS||minS==-1)minS=S1;
        if(S2<minS||minS==-1)minS=S2;
        if(S1>S2)l=move1;
        else r=move2;
    }
    printf("%.10f\n",minS);
}