#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mn=1e5+10;
const ll mod=1e9+7;
double x[mn],y[mn];
double eps=1e-13;
char d[mn];
int n;
double eval(long double t){
    long double x1=1e300,x2=-x1,y1=x1,y2=x2;
    for(int i=0;i<n;i++){
        if(d[i]=='U'){
            x1=min(x1,x[i]+0.L);
            x2=max(x2,x[i]+0.L);
            y1=min(y1,y[i]+t);
            y2=max(y2,y[i]+t);
        }
        else if(d[i]=='D'){
            x1=min(x1,x[i]+0.L);
            x2=max(x2,x[i]+0.L);
            y1=min(y1,y[i]-t);
            y2=max(y2,y[i]-t);
        }
        else if(d[i]=='R'){
            x1=min(x1,x[i]+t);
            x2=max(x2,x[i]+t);
            y1=min(y1,y[i]+0.L);
            y2=max(y2,y[i]+0.L);
        }
        else{
            x1=min(x1,x[i]-t);
            x2=max(x2,x[i]-t);
            y1=min(y1,y[i]+0.L);
            y2=max(y2,y[i]+0.L);
        }
    }
    return (x2-x1)*(y2-y1);
}
int main(){
    int i;
    scanf("%d",&n);
    for(i=0;i<n;i++)scanf("%lf %lf %c",x+i,y+i,d+i);
    long double l=0,r=1e9;
    for(int i=0;i<200;i++){
        long double a=(2*l+r)/3,b=(2*r+l)/3;
        if(eval(b)>eval(a))r=b;
        else l=a;
    }
    printf("%.20f",eval(l));
}

### 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 long long ll;
const int mn=1e5+10;
const ll mod=1e9+7;
double x[mn],y[mn];
double eps=1e-13;
char d[mn];
int n;
double eval(long double t){
    long double x1=1e300,x2=-x1,y1=x1,y2=x2;
    for(int i=0;i<n;i++){
        if(d[i]=='U'){
            x1=min(x1,x[i]+0.L);
            x2=max(x2,x[i]+0.L);
            y1=min(y1,y[i]+t);
            y2=max(y2,y[i]+t);
        }
        else if(d[i]=='D'){
            x1=min(x1,x[i]+0.L);
            x2=max(x2,x[i]+0.L);
            y1=min(y1,y[i]-t);
            y2=max(y2,y[i]-t);
        }
        else if(d[i]=='R'){
            x1=min(x1,x[i]+t);
            x2=max(x2,x[i]+t);
            y1=min(y1,y[i]+0.L);
            y2=max(y2,y[i]+0.L);
        }
        else{
            x1=min(x1,x[i]-t);
            x2=max(x2,x[i]-t);
            y1=min(y1,y[i]+0.L);
            y2=max(y2,y[i]+0.L);
        }
    }
    return (x2-x1)*(y2-y1);
}
int main(){
    int i;
    scanf("%d",&n);
    for(i=0;i<n;i++)scanf("%lf %lf %c",x+i,y+i,d+i);
    long double l=0,r=1e9;
    for(int i=0;i<200;i++){
        long double a=(2*l+r)/3,b=(2*r+l)/3;
        if(eval(b)>eval(a))r=b;
        else l=a;
    }
    printf("%.20f",eval(l));
}

```