#include<bits/stdc++.h> 
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define REP(i,a,b) for(int i=int(a);i<=int(b);i++)
#define MOD 1000000007
#define INF 2000000000
#define DESPACITO 1000000000000000000
#define PI acos(-1);
#define E 998244353
#define pb push_back
#define mp make_pair
#define ll long long
#define double long double

using namespace std;

double x[100001],y[100001];
char dir[100001];
int n;

double check(double t)
{
	double maxx = -INF, maxy = -INF, minx = INF, miny = INF;
	REP(i,1,n)
	{
		double valx = x[i];
		double valy = y[i];
		if(dir[i] == 'U')
			valy+=t;
		else if(dir[i]=='D')
			valy-=t;
		else if(dir[i]=='R')
			valx+=t;
		else
			valx-=t;
		minx=min(minx,valx);
		maxx=max(maxx,valx);
		miny=min(miny,valy);
		maxy=max(maxy,valy);
	}
	double ans=(maxx-minx)*(maxy-miny);
	return ans;
}

int main()
{
	IOS;
	cin >> n;
	REP(i,1,n)
		cin >> x[i] >>  y[i] >> dir[i];
	double l=0,r=INF;
	REP(i,1,500)
	{
		double mid1=(2*l+r)/3, mid2 = (l+2*r)/3;
		if(check(mid1)<=check(mid2))
			r=mid2;
		else
			l=mid1;
	}
	double mid1=(2*l+r)/3, mid2 = (l+2*r)/3;
	cout << std::fixed;
	cout << setprecision(9) << min(check(mid1),check(mid2));
}	

### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h> 
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define REP(i,a,b) for(int i=int(a);i<=int(b);i++)
#define MOD 1000000007
#define INF 2000000000
#define DESPACITO 1000000000000000000
#define PI acos(-1);
#define E 998244353
#define pb push_back
#define mp make_pair
#define ll long long
#define double long double

using namespace std;

double x[100001],y[100001];
char dir[100001];
int n;

double check(double t)
{
	double maxx = -INF, maxy = -INF, minx = INF, miny = INF;
	REP(i,1,n)
	{
		double valx = x[i];
		double valy = y[i];
		if(dir[i] == 'U')
			valy+=t;
		else if(dir[i]=='D')
			valy-=t;
		else if(dir[i]=='R')
			valx+=t;
		else
			valx-=t;
		minx=min(minx,valx);
		maxx=max(maxx,valx);
		miny=min(miny,valy);
		maxy=max(maxy,valy);
	}
	double ans=(maxx-minx)*(maxy-miny);
	return ans;
}

int main()
{
	IOS;
	cin >> n;
	REP(i,1,n)
		cin >> x[i] >>  y[i] >> dir[i];
	double l=0,r=INF;
	REP(i,1,500)
	{
		double mid1=(2*l+r)/3, mid2 = (l+2*r)/3;
		if(check(mid1)<=check(mid2))
			r=mid2;
		else
			l=mid1;
	}
	double mid1=(2*l+r)/3, mid2 = (l+2*r)/3;
	cout << std::fixed;
	cout << setprecision(9) << min(check(mid1),check(mid2));
}	

```