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

using namespace std;

const int maxn=1e5+10;
int n,x[maxn],y[maxn],d[maxn];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

double Min(double a,double b)
{
    return a-b>1e-9?b:a;
}
double Max(double a,double b)
{
    return a-b>1e-9?a:b;
}

double func(double s)
{
    double x1=1e9,x2=-1e9,y1=1e9,y2=-1e9,t1,t2;
    //cout<<setprecision(2)<<s<<' ';
    for(int i=0;i<n;i++)
    {
        t1=x[i]+s*dir[d[i]][0];
        t2=y[i]+s*dir[d[i]][1];
        x1=min(x1,t1);
        y1=min(y1,t2);
        x2=max(x2,t1);
        y2=max(y2,t2);
        //cout<<t[0]<<' '<<t[1]<<x[0]<<' '<<x[1]<<' '<<y[0]<<' '<<y[1]<<endl;
    }
    return (y2-y1)*(x2-x1);
}

int main()
{
    int i,j; char ch;
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>x[i]>>y[i]>>ch;
        switch(ch)
        {
        case 'R':
            d[i]=0; break;
        case 'L':
            d[i]=1; break;
        case 'U':
            d[i]=2; break;
        case 'D':
            d[i]=3;
        }
    }
    double res = 1e18,l=0,r=1e9,lm,rm,are1,are2;
    for(i=0;i<200;i++)
    {
        lm=(l*2+r)/3,rm=(l+r*2)/3;
        are1=func(lm),are2=func(rm);
        //cout<<setprecision(4)<<are1<<' '<<are2<<' '<<res<<endl;
        res=min(res,are1);
        res=min(res,are2);
        if(are1<are2)
        {
            r=rm;
        }
        else
            l=lm;
        //cout<<res<<endl;
        //printf("%.15lf\n",res);
        //cout<<res<<endl;
        //fflush(stdin);
    }
    printf("%.15lf\n",res);
    return 0;
}