#include <bits/stdc++.h>
using namespace std;

#define INF 1e18

int N;
vector<long long> x;
vector<long long> y;
vector<char> d;

double f( double t )
{
	double xmax = -INF;
	double xmin = INF;
	double ymax = -INF;
	double ymin = INF;
	for( int i = 0; i < N; i++) {
		double xx = x[i];
		double yy = y[i];
		char dd = d[i];
		if( dd == 'R' ) {
			xx += t;
		}
		else if( dd == 'L' ) {
			xx -= t;
		}
		else if( dd == 'U' ) {
			yy += t;
		}
		else if( dd == 'D' ) {
			yy -= t;
		}
		xmax = max( xmax, xx );
		xmin = min( xmin, xx );
		ymax = max( ymax, yy );
		ymin = min( ymin, yy );
	}
	return ( xmax - xmin ) * ( ymax - ymin );
}

int main() {

	cin >> N;
	x = vector<long long>( N );
	y = vector<long long>( N );
	d = vector<char>( N );
	for( int i = 0; i < N; i++ ) {
		cin >> x[i] >> y[i] >> d[i];
	}

	double l = 0;
	double r = 1e9;
	double ans = INF;
	for( int i = 0; i < 200; i++ ) {
		double t1 = ( l * 2 + r ) / 3;
		double t2 = ( l + r * 2 ) / 3;
		double res1 = f( t1 );
		double res2 = f( t2 );
		if( res1 > res2 ) {
			l = t1;
		}
		else {
			r = t2;
		}
		ans = min( ans, min( res1, res2 ) );
	}

	cout << fixed << setprecision(12) << ans << endl;

	return 0;
}
### 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>
using namespace std;

#define INF 1e18

int N;
vector<long long> x;
vector<long long> y;
vector<char> d;

double f( double t )
{
	double xmax = -INF;
	double xmin = INF;
	double ymax = -INF;
	double ymin = INF;
	for( int i = 0; i < N; i++) {
		double xx = x[i];
		double yy = y[i];
		char dd = d[i];
		if( dd == 'R' ) {
			xx += t;
		}
		else if( dd == 'L' ) {
			xx -= t;
		}
		else if( dd == 'U' ) {
			yy += t;
		}
		else if( dd == 'D' ) {
			yy -= t;
		}
		xmax = max( xmax, xx );
		xmin = min( xmin, xx );
		ymax = max( ymax, yy );
		ymin = min( ymin, yy );
	}
	return ( xmax - xmin ) * ( ymax - ymin );
}

int main() {

	cin >> N;
	x = vector<long long>( N );
	y = vector<long long>( N );
	d = vector<char>( N );
	for( int i = 0; i < N; i++ ) {
		cin >> x[i] >> y[i] >> d[i];
	}

	double l = 0;
	double r = 1e9;
	double ans = INF;
	for( int i = 0; i < 200; i++ ) {
		double t1 = ( l * 2 + r ) / 3;
		double t2 = ( l + r * 2 ) / 3;
		double res1 = f( t1 );
		double res2 = f( t2 );
		if( res1 > res2 ) {
			l = t1;
		}
		else {
			r = t2;
		}
		ans = min( ans, min( res1, res2 ) );
	}

	cout << fixed << setprecision(12) << ans << endl;

	return 0;
}
```