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:
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()
#define pb push_back

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 1e5 + 20 , inf = 1e9 + 20 ;
int n , x[N] , y[N] ;
char d[N] ;

inline ld get (ld k) {
	ld xmn = inf , xmx = -inf , ymn = inf , ymx = -inf ;

	for (int i = 0 ; i < n ; i ++) {
		ld nx = x[i] + ((d[i] == 'R') - (d[i] == 'L')) * k ;
		ld ny = y[i] + ((d[i] == 'U') - (d[i] == 'D')) * k ;

		xmn = min(xmn , nx) , xmx = max(xmx , nx) ;
		ymn = min(ymn , ny) , ymx = max(ymx , ny) ;
	}

	return (xmx - xmn) * (ymx - ymn) ;
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n ;
	
	for (int i = 0 ; i < n ; i ++) cin >> x[i] >> y[i] >> d[i] ;

	ld low = 0 , high = 2e8 + 20 ;

	for (int _ = 0 ; _ < 200 ; _ ++) {
		ld f1 = (2 * low + high) / 3 , f2 = (low + 2 * high) / 3 ;

		if (get(f1) < get(f2)) high = f2 ;
		else low = f1 ;
	}

	cout << fixed << setprecision(10) << get(low) << '\n' ;
}