#include <bits/stdc++.h>
typedef long long LL;
using namespace std;

int n;
vector<pair<int,int>> x, y;

double find_size(double mid)  {
	double mn = 2e18, mx = -2e18;
	for (int i = 0; i < n; ++i) {
		mn = min(mn, x[i].first + x[i].second * mid);
		mx = max(mx, x[i].first + x[i].second * mid);
	}
	double sz = (mx - mn);
	mn = 2e18, mx = -2e18;
	for (int i = 0; i < n; ++i) {
		mn = min(mn, y[i].first + y[i].second * mid);
		mx = max(mx, y[i].first + y[i].second * mid);
	}
	sz *= (mx - mn);
	return sz;
}

int main(int argc, char **argv) {
	cin >> n;
	x.resize(n);
	y.resize(n);
	for (int i = 0; i < n; ++i) {
		char d;
		int a, b;
		cin >> a >> b >> d;
		switch (d) {
			case 'L':
				x[i] = {a, -1};
				y[i] = {b, 0};
				break;
			case 'R':
				x[i] = {a, 1};
				y[i] = {b, 0};
				break;
			case 'D':
				x[i] = {a, 0};
				y[i] = {b, -1};
				break;
			case 'U':
				x[i] = {a, 0};
				y[i] = {b, 1};
				break;
		}
	}

	double lo = 0., hi = 1e12;
	for (int i = 0; i < 500; ++i) {
		double t1 = lo + (hi - lo) / 3.;
		double t2 = lo + (hi - lo) / 3. * 2;
		if (find_size(t1) < find_size(t2))
			hi = t2;
		else
			lo = t1;
	}

	cout << fixed;
	cout.precision(9);
	cout << find_size(lo) << '\n';

	return 0;
}


### Prompt

Please create a solution in CPP 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>
typedef long long LL;
using namespace std;

int n;
vector<pair<int,int>> x, y;

double find_size(double mid)  {
	double mn = 2e18, mx = -2e18;
	for (int i = 0; i < n; ++i) {
		mn = min(mn, x[i].first + x[i].second * mid);
		mx = max(mx, x[i].first + x[i].second * mid);
	}
	double sz = (mx - mn);
	mn = 2e18, mx = -2e18;
	for (int i = 0; i < n; ++i) {
		mn = min(mn, y[i].first + y[i].second * mid);
		mx = max(mx, y[i].first + y[i].second * mid);
	}
	sz *= (mx - mn);
	return sz;
}

int main(int argc, char **argv) {
	cin >> n;
	x.resize(n);
	y.resize(n);
	for (int i = 0; i < n; ++i) {
		char d;
		int a, b;
		cin >> a >> b >> d;
		switch (d) {
			case 'L':
				x[i] = {a, -1};
				y[i] = {b, 0};
				break;
			case 'R':
				x[i] = {a, 1};
				y[i] = {b, 0};
				break;
			case 'D':
				x[i] = {a, 0};
				y[i] = {b, -1};
				break;
			case 'U':
				x[i] = {a, 0};
				y[i] = {b, 1};
				break;
		}
	}

	double lo = 0., hi = 1e12;
	for (int i = 0; i < 500; ++i) {
		double t1 = lo + (hi - lo) / 3.;
		double t2 = lo + (hi - lo) / 3. * 2;
		if (find_size(t1) < find_size(t2))
			hi = t2;
		else
			lo = t1;
	}

	cout << fixed;
	cout.precision(9);
	cout << find_size(lo) << '\n';

	return 0;
}


```