Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

2
0 3 D
3 0 L

Sample Output 1

0

After three seconds, the two points will meet at the origin. The value in question will be 0 at that moment.

Sample Input 2

5
-7 -10 U
7 -6 U
-8 7 D
-3 3 D
0 -6 R

Sample Output 2

97.5

The answer may not be an integer.

Sample Input 3

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

Sample Output 3

273

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// F - Minimum Bounding Box
#include <bits/stdc++.h>

using namespace std;

#define	EPS	1e-9

int n;
vector<pair<long long, long long> > p, m;

double func(double t)
{
	double minx = 1e18, miny = 1e18, maxx = -1e18, maxy = -1e18;
	for (int i = 0; i < n; i++) {
		double x = p[i].first + t * m[i].first, y = p[i].second + t * m[i].second;
		minx = min(minx, x); maxx = max(maxx, x);
		miny = min(miny, y); maxy = max(maxy, y);
	}
	return (maxx - minx) * (maxy - miny);
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	cin >> n;
	for (int i = 0; i < n; i++) {
		long long x, y;
		string s;
		cin >> x >> y >> s;
		p.resize(n), m.resize(n);
		p[i] = make_pair(x, y);
		if (s[0] == 'R')      m[i] = make_pair(1, 0);
		else if (s[0] == 'L') m[i] = make_pair(-1, 0);
		else if (s[0] == 'U') m[i] = make_pair(0, 1);
		else if (s[0] == 'D') m[i] = make_pair(0, -1);
	}

	double l = 0, r = 1e9;
	for (int k = 0; k < 80; k++) {
		double m1 = l + (r - l) / 3, m2 = r - (r - l) / 3;
		if (func(m1) > func(m2)) l = m1;
		else r = m2;
	}
	double ans = func((l + r) / 2);

	cout << fixed << setprecision(15) << ans << endl;

	return 0;
}