#include <bits/stdc++.h>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define double long double

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7;

string s = "RLUD";
int dir[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}, n;
int x[100005], y[100005];
char d[100005];

double f(double t) {
	double xM = -1e18, xm = 1e18, yM = -1e18, ym = 1e18;
	for(int i=0;i<n;i++) {
		int j = s.find(d[i]);
		double X = x[i] + t * dir[j][0],
			   Y = y[i] + t * dir[j][1];
		xM = max(xM, X);
		xm = min(xm, X);
		yM = max(yM, Y);
		ym = min(ym, Y);
	}
	return (xM - xm) * (yM - ym);
}

int main()
{
	IO_OP;
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> x[i] >> y[i] >> d[i];	
	double l = 0, r = 2e8+87;
	for(int i=0;i<300;i++) {
		double m1 = (l*2+r)/3,
			   m2 = (l+r*2)/3;
		double f1 = f(m1), f2 = f(m2);
		if(f1 >= f2)
			l = m1;
		else 
			r = m2; 
	}
	cout << fixed << setprecision(16) << f(l) << endl;
}


### Prompt

Create a solution in CPP for 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>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define double long double

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7;

string s = "RLUD";
int dir[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}, n;
int x[100005], y[100005];
char d[100005];

double f(double t) {
	double xM = -1e18, xm = 1e18, yM = -1e18, ym = 1e18;
	for(int i=0;i<n;i++) {
		int j = s.find(d[i]);
		double X = x[i] + t * dir[j][0],
			   Y = y[i] + t * dir[j][1];
		xM = max(xM, X);
		xm = min(xm, X);
		yM = max(yM, Y);
		ym = min(ym, Y);
	}
	return (xM - xm) * (yM - ym);
}

int main()
{
	IO_OP;
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> x[i] >> y[i] >> d[i];	
	double l = 0, r = 2e8+87;
	for(int i=0;i<300;i++) {
		double m1 = (l*2+r)/3,
			   m2 = (l+r*2)/3;
		double f1 = f(m1), f2 = f(m2);
		if(f1 >= f2)
			l = m1;
		else 
			r = m2; 
	}
	cout << fixed << setprecision(16) << f(l) << endl;
}


```