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 <vector>
#include <algorithm>

using namespace std;

int main(int argc, char *argv[])
{
	int N;
	double inf = 1e9;
	double lx[3] = {inf, inf, inf};
	double rx[3] = {-inf, -inf, -inf};
	double ly[3] = {inf, inf ,inf};
	double ry[3] = {-inf, -inf, -inf};

	cin >> N;

	for (int i = 0; i < N; i++) {
		double x, y;
		char d;
		cin >> x >> y >> d;
		if (d == 'R') {
			lx[2] = min(lx[2], x);
			rx[2] = max(rx[2], x);
			ly[1] = min(ly[1], y);
			ry[1] = max(ry[1], y);
		}
		if (d == 'L') {
			lx[0] = min(lx[0], x);
			rx[0] = max(rx[0], x);
			ly[1] = min(ly[1], y);
			ry[1] = max(ry[1], y);
		}
		if (d == 'U') {
			lx[1] = min(lx[1], x);
			rx[1] = max(rx[1], x);
			ly[2] = min(ly[2], y);
			ry[2] = max(ry[2], y);
		}
		if (d == 'D') {
			lx[1] = min(lx[1], x);
			rx[1] = max(rx[1], x);
			ly[0] = min(ly[0], y);
			ry[0] = max(ry[0], y);
		}
	}

	vector<double> ts;
	ts.push_back(0);
	ts.push_back(inf);
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < i; j++) {
			double t = (lx[j] - lx[i]) / (i - j);
			if (t > 0) {
				ts.push_back(t);
			}
			t = (rx[j] - rx[i]) / (i - j);
			if (t > 0) {
				ts.push_back(t);
			}
			t = (ly[j] - ly[i]) / (i - j);
			if (t > 0) {
				ts.push_back(t);
			}
			t = (ry[j] - ry[i]) / (i - j);
			if (t > 0) {
				ts.push_back(t);
			}
		}
	}
	sort(ts.begin(), ts.end());

	double ans = 1e18;
	for (double t : ts) {
		double dx, dy;
		double nl = inf, nr = -inf;
		for (int i = 0; i < 3; i++) {
			nl = min(nl, lx[i] + i * t);
			nr = max(nr, rx[i] + i * t);
		}
		dx = nr - nl;
		nl = inf;
		nr = -inf;
		for (int i = 0; i < 3; i++) {
			nl = min(nl, ly[i] + i * t);
			nr = max(nr, ry[i] + i * t);
		}
		dy = nr - nl;
		ans = min(ans, dx * dy);
	}

	printf("%.9lf\n", ans);

	return 0;
}