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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
 
using namespace std;
 
struct st {
	double x, y;
	char d;
	st() {}
	st(double a, double b, char c) {
		x = a, y = b, d = c;
	}
}take[(int)1e5+5];
 
int n;
 
double f(double t) {
	double x_max = -1e10, x_min = 1e10, y_max = -1e10, y_min = 1e10;
	for(int i=0; i<n; i++) {
		double x = take[i].x, y = take[i].y;
		if(take[i].d == 'R') {
			x += t;	
		}
		else if(take[i].d == 'L') {
			x -= t;
		}
		else if(take[i].d == 'U') {
			y += t;
		}
		else if(take[i].d == 'D') {
			y -= t;
		}
		x_max = max(x_max, x);
		x_min = min(x_min, x);
		y_max = max(y_max, y);
		y_min = min(y_min, y);
	}
	return (x_max - x_min) * (y_max - y_min);
}
 
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for(int i=0; i<n; i++) {
		double x, y;
		char d;
		cin >> x >> y >> d;
		take[i] = st(x, y, d);
	}
	double l, r, m1, m2;
	l = 0, r = 1e12;
	for(int i=0; i<300; i++){
		m1 = l + (r - l) / 3.0;
		m2 = l + 2.0 * (r - l) / 3.0;
		if (f(m1) < f(m2)){
			r = m2;
		}
		else{
			l = m1;
		}
	}
	cout << fixed;
	cout << f(l) << endl;
	return 0;
}