#include <bits/stdc++.h>
using namespace std;
int Dx[100005],Dy[100005];
int n;
int X[100005],Y[100005];
char dd;

double solve (double a)
{

	double mxx = -1e18;
	double mnx = 1e18;
	double mxy = -1e18;
	double mny = 1e18;
	for(int i = 1;i <= n;i++)
    {
		double x = X[i]+a*Dx[i];
		double y = Y[i]+a*Dy[i];
		mxx = max(mxx,x);
		mnx = min(mnx,x);
		mxy = max(mxy,y);
		mny = min(mny,y);
	}
	return (mxx-mnx)*(mxy-mny);
}

int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        scanf("%d %d %c", &X[i], &Y[i], &dd);
        switch(dd)
        {
            case 'U' :
            {
                Dy[i] = 1;
                break;
            }
            case 'D' :
            {
                Dy[i] = -1;
                break;
            }
            case 'L' :
            {
                Dx[i] = -1;
                break;
            }
            case 'R' :
            {
                Dx[i] = 1;
                break;
            }
        }
    }
    double ans = 1e18;
    double l = 0;
    double r = 1e9;
    for(int t = 0;t < 200;t++)
    {
        double uu = (l*2+r)/3;
        double dd = (l+r*2)/3;
        double uv = solve(uu);
        double dv = solve(dd);
        ans = min(ans,min(uv,dv));
        if(dv < uv)
        {
            l = uu;
        }
        else
        {
            r = dd;
        }
    }
    printf("%.15f",ans);
}
### Prompt

Develop a solution in cpp to the problem described below:

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;
int Dx[100005],Dy[100005];
int n;
int X[100005],Y[100005];
char dd;

double solve (double a)
{

	double mxx = -1e18;
	double mnx = 1e18;
	double mxy = -1e18;
	double mny = 1e18;
	for(int i = 1;i <= n;i++)
    {
		double x = X[i]+a*Dx[i];
		double y = Y[i]+a*Dy[i];
		mxx = max(mxx,x);
		mnx = min(mnx,x);
		mxy = max(mxy,y);
		mny = min(mny,y);
	}
	return (mxx-mnx)*(mxy-mny);
}

int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        scanf("%d %d %c", &X[i], &Y[i], &dd);
        switch(dd)
        {
            case 'U' :
            {
                Dy[i] = 1;
                break;
            }
            case 'D' :
            {
                Dy[i] = -1;
                break;
            }
            case 'L' :
            {
                Dx[i] = -1;
                break;
            }
            case 'R' :
            {
                Dx[i] = 1;
                break;
            }
        }
    }
    double ans = 1e18;
    double l = 0;
    double r = 1e9;
    for(int t = 0;t < 200;t++)
    {
        double uu = (l*2+r)/3;
        double dd = (l+r*2)/3;
        double uv = solve(uu);
        double dv = solve(dd);
        ans = min(ans,min(uv,dv));
        if(dv < uv)
        {
            l = uu;
        }
        else
        {
            r = dd;
        }
    }
    printf("%.15f",ans);
}
```