#include <bits/stdc++.h>

using namespace std;

int n, dx[100005], dy[100005];
pair <int, int> a[100005];
char s[5];

double solve(double t) {
	double xmx = -1e18;
	double xmn = 1e18;
	double ymx = -1e18;
	double ymn = 1e18;
	
	for (int i = 0; i < n; i++) {
		double x = t * dx[i] + a[i].first;
		double y = t * dy[i] + a[i].second;
		xmx = max(xmx, x);
		xmn = min(xmn, x);
		ymx = max(ymx, y);
		ymn = min(ymn, y);
	}
	
	return (xmx - xmn) * (ymx - ymn);
}

int main() {
	scanf("%d", &n);
	
	for (int i = 0; i < n; i++) {
		scanf("%d%d%s", &a[i].first, &a[i].second, s);
		if (s[0] == 'U') dy[i] = 1;
		else if (s[0] == 'D') dy[i] = -1;
		else if (s[0] == 'L') dx[i] = -1;
		else dx[i] = 1;
	}
	
	double lb = 0.0;
	double ub = 1e9;
	double res = 1e18;
	
	for (int i = 0; i < 200; i++) {
		double lmid = (lb * 2 + ub) / 3;
		double rmid = (lb + ub * 2) / 3;
		double lv = solve(lmid);
		double rv = solve(rmid);
		
		res = min(res, lv);
		res = min(res, rv);
		
		if (lv > rv) lb = lmid;
		else ub = rmid;
	}
	
	printf("%.12lf\n", res);
}

### 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>

using namespace std;

int n, dx[100005], dy[100005];
pair <int, int> a[100005];
char s[5];

double solve(double t) {
	double xmx = -1e18;
	double xmn = 1e18;
	double ymx = -1e18;
	double ymn = 1e18;
	
	for (int i = 0; i < n; i++) {
		double x = t * dx[i] + a[i].first;
		double y = t * dy[i] + a[i].second;
		xmx = max(xmx, x);
		xmn = min(xmn, x);
		ymx = max(ymx, y);
		ymn = min(ymn, y);
	}
	
	return (xmx - xmn) * (ymx - ymn);
}

int main() {
	scanf("%d", &n);
	
	for (int i = 0; i < n; i++) {
		scanf("%d%d%s", &a[i].first, &a[i].second, s);
		if (s[0] == 'U') dy[i] = 1;
		else if (s[0] == 'D') dy[i] = -1;
		else if (s[0] == 'L') dx[i] = -1;
		else dx[i] = 1;
	}
	
	double lb = 0.0;
	double ub = 1e9;
	double res = 1e18;
	
	for (int i = 0; i < 200; i++) {
		double lmid = (lb * 2 + ub) / 3;
		double rmid = (lb + ub * 2) / 3;
		double lv = solve(lmid);
		double rv = solve(rmid);
		
		res = min(res, lv);
		res = min(res, rv);
		
		if (lv > rv) lb = lmid;
		else ub = rmid;
	}
	
	printf("%.12lf\n", res);
}

```