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 <iostream>
#include <algorithm>
using namespace std;

int N;
long double X[1 << 18], Y[1 << 18];
char C[1 << 18];

long double solve(long double v) {
	long double lx = 1e9, ly = 1e9, rx = -1e9, ry = -1e9;
	
	for (int i = 0; i < N; i++) {
		long double cx = X[i], cy = Y[i];
		if (C[i] == 'L') cx -= v;
		if (C[i] == 'R') cx += v;
		if (C[i] == 'D') cy -= v;
		if (C[i] == 'U') cy += v;
		lx = min(lx, cx); rx = max(rx, cx);
		ly = min(ly, cy); ry = max(ry, cy);
	}
	return (rx - lx) * (ry - ly);
}

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> X[i] >> Y[i] >> C[i];
	}

	long double cl = 0.0, cr = 1e9, c1, c2, minx = 1e18;
	for (int i = 0; i < 130; i++) {
		c1 = (cl + cl + cr) / 3.0;
		c2 = (cl + cr + cr) / 3.0;
		long double p1 = solve(c1);
		long double p2 = solve(c2);
		if (p1 < p2) { cr = c2; minx = min(minx, p1); }
		else { cl = c1; minx = min(minx, p2); }
	}
	printf("%.13Lf\n", minx);
	return 0;
}