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>	
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define ld long double
#define pii pair <int, int>
#define sz(a) (int)(a.size()) 
#define resize(v) v.resize(unique(all(v)) - v.begin()); 
#define all(a) a.begin(), a.end()
#define forit(it, s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)
 
// random_device rd; mt19937 g(rd());
 
using namespace std;
 
void Fast_Read_Out()
{
    ios_base::sync_with_stdio(0);
    cin.tie(), cout.tie();
}
 
void Random()
{
    unsigned int seed;                                                                        
    asm("rdtsc" : "=A" (seed));
    srand(seed);        
}
 
unsigned int Time()
{
     unsigned int time = clock() / 1000.00;                           
     return time;
}
 
const int mod = (int)1e9 + 7;
const int N = (int)1e5 + 123;
const ld eps = 1e-10;
 
int n;
ld x[N], y[N];
char d[N];
 
ld f(ld k)
{
	ld xmax = -1e9, xmin = 1e9;
	ld ymax = -1e9, ymin = 1e9;
	for(int i = 1; i <= n; i++)
	{
		ld curx = x[i], cury = y[i];
		if(d[i] == 'U') cury += k;
		if(d[i] == 'D') cury -= k;
		if(d[i] == 'R') curx += k;
		if(d[i] == 'L') curx -= k;
		xmax = max(xmax, curx);
		xmin = min(xmin, curx);
		ymax = max(ymax, cury);
		ymin = min(ymin, cury);
	}
	xmax -= xmin;
	ymax -= ymin;
	xmax *= ymax;
	return (ld)xmax;
}
 
int main ()
{                                  	
	#ifdef JUDGE
		freopen("input.txt", "r", stdin);
	#endif		
	Fast_Read_Out();
	Random();
	cin >> n;  
	for(int i = 1; i <= n; i++) cin >> x[i] >> y[i] >> d[i];
	ld l = 0.0, r = 1e9;
	for(int i = 1; i <= 500; i++)
	{
		ld m1 = l + (r - l) / 3;
		ld m2 = r - (r - l) / 3;
		if(f(m1) > f(m2)) l = m1;
		else r = m2;
	}
	cout << fixed << min(f(r), f(l));
}