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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N;
double x[100000], y[100000];
char d[100000];

double f(double t) {
	double XMIN = 10000000000;
	double XMAX = -10000000000;
	double YMIN = 10000000000;
	double YMAX = -10000000000;
	for (int i = 0; i < N; ++i) {
		if (d[i] == 'U') {
			XMAX = max(XMAX, x[i]);
			XMIN = min(XMIN, x[i]);
			YMIN = min(YMIN, y[i] + t);
			YMAX = max(YMAX, y[i] + t);
		}
		if (d[i] == 'D') {
			XMAX = max(XMAX, x[i]);
			XMIN = min(XMIN, x[i]);
			YMIN = min(YMIN, y[i] - t);
			YMAX = max(YMAX, y[i] - t);
		}
		if (d[i] == 'R') {
			XMAX = max(XMAX, x[i] + t);
			XMIN = min(XMIN, x[i] + t);
			YMIN = min(YMIN, y[i]);
			YMAX = max(YMAX, y[i]);
		}
		if(d[i]=='L'){
			XMAX = max(XMAX, x[i] - t);
			XMIN = min(XMIN, x[i] - t);
			YMIN = min(YMIN, y[i]);
			YMAX = max(YMAX, y[i]);
		}
	}
	return (XMAX - XMIN)*(YMAX - YMIN);
}

int main() {
	cin >> N;
	for (int i = 0; i < N; ++i) {
		cin >> x[i] >> y[i] >> d[i];
	}
	double LL, L, R, RR;
	LL = 0;
	RR = 10000000000;
	for (int i = 0; i < 1000; ++i) {
		L = (LL * 2 + RR) / 3;
		R = (LL + RR * 2) / 3;
		if (f(L) < f(R)) {
			RR = R;
		}
		else {
			LL = L;
		}
	}
	printf("%.11f\n", f(LL));
}