#include<bits/stdc++.h>
using namespace std;
#define MaxN 100003
#define inf INT_MAX

int N;
char opt;
struct point {
	long long x, y;
	double dtx, dty;
}a[MaxN];

double Chk(double tim) {
	double x, y, d = inf, u = -inf, l = inf, r = -inf;
	for(register int i = 1; i <= N; ++i) {
		x = a[i].x + a[i].dtx * tim, y = a[i].y + a[i].dty * tim;
		if(y < d)
			d = y;
		if(y > u)
			u = y;
		if(x < l)
			l = x;
		if(x > r)
			r = x;
	}
	return (r - l) * (u - d);
}

int main() {
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for(register int i = 1; i <= N; ++i) {
		cin >> a[i].x >> a[i].y >> opt;
		if(opt == 'R')
			a[i].dtx = 1.0;
		else if(opt == 'L')
			a[i].dtx = -1.0;
		else if(opt == 'U')
			a[i].dty = 1.0;
		else
			a[i].dty = -1.0;
	}
	double L = 0.0, R = 100000000.0, Mid1, Mid2, dlt, r1, r2, Ans;
	for (int i = 1; i <= 100; ++i) {
		dlt = (R - L) / 3.0;
		Mid1 = L + dlt, Mid2 = Mid1 + dlt;
		r1 = Chk(Mid1), r2 = Chk(Mid2);
		if(r1 < r2) {
			R = Mid2;
			Ans = r1;
		}
		else {
			L = Mid1;
			Ans = r2;
		}
	}
	cout << fixed << setprecision(10) << Ans << endl;
	return 0;
}
### Prompt

Please formulate a cpp solution to 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>
using namespace std;
#define MaxN 100003
#define inf INT_MAX

int N;
char opt;
struct point {
	long long x, y;
	double dtx, dty;
}a[MaxN];

double Chk(double tim) {
	double x, y, d = inf, u = -inf, l = inf, r = -inf;
	for(register int i = 1; i <= N; ++i) {
		x = a[i].x + a[i].dtx * tim, y = a[i].y + a[i].dty * tim;
		if(y < d)
			d = y;
		if(y > u)
			u = y;
		if(x < l)
			l = x;
		if(x > r)
			r = x;
	}
	return (r - l) * (u - d);
}

int main() {
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for(register int i = 1; i <= N; ++i) {
		cin >> a[i].x >> a[i].y >> opt;
		if(opt == 'R')
			a[i].dtx = 1.0;
		else if(opt == 'L')
			a[i].dtx = -1.0;
		else if(opt == 'U')
			a[i].dty = 1.0;
		else
			a[i].dty = -1.0;
	}
	double L = 0.0, R = 100000000.0, Mid1, Mid2, dlt, r1, r2, Ans;
	for (int i = 1; i <= 100; ++i) {
		dlt = (R - L) / 3.0;
		Mid1 = L + dlt, Mid2 = Mid1 + dlt;
		r1 = Chk(Mid1), r2 = Chk(Mid2);
		if(r1 < r2) {
			R = Mid2;
			Ans = r1;
		}
		else {
			L = Mid1;
			Ans = r2;
		}
	}
	cout << fixed << setprecision(10) << Ans << endl;
	return 0;
}
```