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;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long
#define double long double

const int N = 1e5 + 5;

int n;
int x[N], y[N];
char d[N];

double calc(double t)
{
	double xmin = 1e12, xmax = -1e12, ymin = 1e12, ymax = -1e12;
	for(int i = 1; i <= n; i++)
	{	
		double cx = x[i], cy = y[i];
		if(d[i] == 'R')
			cx += t;
		else if(d[i] == 'L')
			cx -= t;
		else if(d[i] == 'U')
			cy += t;
		else
			cy -= t;
		xmin = min(xmin, cx);
		xmax = max(xmax, cx);
		ymin = min(ymin, cy);
		ymax = max(ymax, cy);
	}
	double ans = (xmax - xmin) * (ymax - ymin);
	return ans;
}

double ternary_search(double lo, double hi)
{
	for(int i = 1; i <= 500; i++)
	{
		double m1 = (lo * 2 + hi) / 3;
		double m2 = (lo + hi * 2) / 3;
		if(calc(m1) < calc(m2))
			hi = m2;
		else
			lo = m1;
	}
	return calc(lo);
}

int32_t main()
{
	IOS;
	cin >> n;
	for(int i = 1; i <= n; i++)
		cin >> x[i] >> y[i] >> d[i];
	double ans = ternary_search(0, 2e8);
	cout << fixed << setprecision(12) << ans;
	return 0;
}