#include <iostream>
#include <vector>
#include <ios>
#include <iomanip>

constexpr long double INF = 1e20;

template<class T>
void chmin (T &a, T b) {
	if (a > b) a = b;
}

template<class T>
void chmax (T &a, T b) {
	if (a < b) a = b;
}

int main() {
	std::size_t n;
	std::cin >> n;
	std::vector<int> x(n), y(n), dx(n, 0), dy(n, 0);
	
	for (std::size_t i = 0; i < n; i++) {
		char c;
		std::cin >> x[i] >> y[i] >> c;
		if (c == 'U') dy[i] = 1;
		else if (c == 'D') dy[i] = -1;
		else if (c == 'L') dx[i] = -1;
		else dx[i] = 1;
	}
	
	auto f = [&](long double t) -> long double {
		long double x_max = -INF, x_min = INF;
		long double y_max = -INF, y_min = INF;
		
		for (std::size_t i = 0; i < n; i++) {
			long double cur_x = (0.0 + (t * dx[i]) + x[i]);
			long double cur_y = (0.0 + (t * dy[i]) + y[i]);
			chmax(x_max, cur_x); chmin(x_min, cur_x);
			chmax(y_max, cur_y); chmin(y_min, cur_y);
		}
		
		long double h = x_max - x_min, w = y_max - y_min;
		return (h * w);
	};
	
	int loop = 1000;
	long double l = 0.0, r = 1e18;
	while (loop--) {
		long double c1 = (l + l + r) / 3.0;
		long double c2 = (l + r + r) / 3.0;
		if (f(c1) >= f(c2)) l = c1;
		else r = c2;
	}
	
	std::cout << std::fixed << std::setprecision(12) << f(l) << '\n';
	
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <iostream>
#include <vector>
#include <ios>
#include <iomanip>

constexpr long double INF = 1e20;

template<class T>
void chmin (T &a, T b) {
	if (a > b) a = b;
}

template<class T>
void chmax (T &a, T b) {
	if (a < b) a = b;
}

int main() {
	std::size_t n;
	std::cin >> n;
	std::vector<int> x(n), y(n), dx(n, 0), dy(n, 0);
	
	for (std::size_t i = 0; i < n; i++) {
		char c;
		std::cin >> x[i] >> y[i] >> c;
		if (c == 'U') dy[i] = 1;
		else if (c == 'D') dy[i] = -1;
		else if (c == 'L') dx[i] = -1;
		else dx[i] = 1;
	}
	
	auto f = [&](long double t) -> long double {
		long double x_max = -INF, x_min = INF;
		long double y_max = -INF, y_min = INF;
		
		for (std::size_t i = 0; i < n; i++) {
			long double cur_x = (0.0 + (t * dx[i]) + x[i]);
			long double cur_y = (0.0 + (t * dy[i]) + y[i]);
			chmax(x_max, cur_x); chmin(x_min, cur_x);
			chmax(y_max, cur_y); chmin(y_min, cur_y);
		}
		
		long double h = x_max - x_min, w = y_max - y_min;
		return (h * w);
	};
	
	int loop = 1000;
	long double l = 0.0, r = 1e18;
	while (loop--) {
		long double c1 = (l + l + r) / 3.0;
		long double c2 = (l + r + r) / 3.0;
		if (f(c1) >= f(c2)) l = c1;
		else r = c2;
	}
	
	std::cout << std::fixed << std::setprecision(12) << f(l) << '\n';
	
	return 0;
}
```