#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pi pair<int,int>
#define ff first
#define ss second
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#include "string"
const int N = 1e5 + 5;
const int inf = 1e18;
pi a[N],d[N];
int n;
double okk(double m){
	double xmax = -inf,ymax = -inf;
	double xmin = inf,ymin = inf;

	for(int i=0;i<n;i++){
		double x = a[i].ff + m*d[i].ff;
		double y = a[i].ss + m*d[i].ss;
		xmax = max( xmax , x );
		xmin = min( xmin , x );
		ymax = max( ymax , y );
		ymin = min( ymin , y );
	}
	double ans = (xmax - xmin) * (ymax - ymin);
	return ans;
}


int32_t main(){
	boost;
	cout<<setprecision(20)<<fixed;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i].ff>>a[i].ss;
		char c;cin>>c;
		if(c == 'R')
			d[i] = {1,0};
		else if(c == 'L')
			d[i] = {-1,0};
		else if(c == 'U')
			d[i] = {0,1};
		else
			d[i] = {0,-1};
	}


	double l = 0,r = 2e8;
	for(int i=0;i<500;i++){
		double m1 = (2*l + r)/3;
		double m2 = (l + 2*r)/3;
		double t1 = okk(m1);
		double t2 = okk(m2);
		if(t1 <=t2)
			r = m2;
		else
			l = m1;
	}

	double ans = min(okk(l),okk(r));
	cout<<ans<<endl;

}


### Prompt

Your challenge is to write 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>
using namespace std;
#define int long long
#define pi pair<int,int>
#define ff first
#define ss second
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#include "string"
const int N = 1e5 + 5;
const int inf = 1e18;
pi a[N],d[N];
int n;
double okk(double m){
	double xmax = -inf,ymax = -inf;
	double xmin = inf,ymin = inf;

	for(int i=0;i<n;i++){
		double x = a[i].ff + m*d[i].ff;
		double y = a[i].ss + m*d[i].ss;
		xmax = max( xmax , x );
		xmin = min( xmin , x );
		ymax = max( ymax , y );
		ymin = min( ymin , y );
	}
	double ans = (xmax - xmin) * (ymax - ymin);
	return ans;
}


int32_t main(){
	boost;
	cout<<setprecision(20)<<fixed;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i].ff>>a[i].ss;
		char c;cin>>c;
		if(c == 'R')
			d[i] = {1,0};
		else if(c == 'L')
			d[i] = {-1,0};
		else if(c == 'U')
			d[i] = {0,1};
		else
			d[i] = {0,-1};
	}


	double l = 0,r = 2e8;
	for(int i=0;i<500;i++){
		double m1 = (2*l + r)/3;
		double m2 = (l + 2*r)/3;
		double t1 = okk(m1);
		double t2 = okk(m2);
		if(t1 <=t2)
			r = m2;
		else
			l = m1;
	}

	double ans = min(okk(l),okk(r));
	cout<<ans<<endl;

}


```