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)

//besmellah
#include <bits/stdc++.h>
using namespace std;
// __builtin_popcount

const long double inf = 1e18;
long double xmax[4] = {-inf, -inf, -inf, -inf}, xmin[4] = {inf, inf, inf, inf}, ymax[4] = {-inf, -inf, -inf, -inf}, ymin[4] = {inf, inf, inf, inf};

long double f(long double t){
	return (max(max(xmax[0], xmax[1]), max(xmax[2] - t, xmax[3] + t)) - min(min(xmin[0], xmin[1]), min(xmin[2] - t, xmin[3] + t))) *
		(max(max(ymax[0] - t, ymax[1] + t), max(ymax[2], ymax[3])) - min(min(ymin[0] - t, ymin[1] + t), min(ymin[2], ymin[3])));
}

int main(){
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	for (int i = 0; i < n; i++){
		long double x, y;
		char d;
		cin >> x >> y >> d;
		if (d == 'D'){
			xmax[0] = max(xmax[0], x);
			xmin[0] = min(xmin[0], x);
			ymax[0] = max(ymax[0], y);
			ymin[0] = min(ymin[0], y);
		}
		else if (d == 'U'){
			xmax[1] = max(xmax[1], x);
			xmin[1] = min(xmin[1], x);
			ymax[1] = max(ymax[1], y);
			ymin[1] = min(ymin[1], y);
		}
		else if (d == 'L'){
			xmax[2] = max(xmax[2], x);
			xmin[2] = min(xmin[2], x);
			ymax[2] = max(ymax[2], y);
			ymin[2] = min(ymin[2], y);
		}
		else{
			xmax[3] = max(xmax[3], x);
			xmin[3] = min(xmin[3], x);
			ymax[3] = max(ymax[3], y);
			ymin[3] = min(ymin[3], y);
		}
	}
	long double l = 0, r = 2e8 + 1, m;
	while (true){
		if (!(f(l) && f(r))){
			cout << 0;
			return 0;
		}
		if (r - l <= 1e-10){
			cout << setprecision(9) << min(f(round(l * 1000) / 1000), f(round(r * 1000) / 1000));
			// cout << ' ' << l / 2;
			return 0;
		}
		m = (l + r) / 2;
		// cout << l << ' ' << r << ' ' << m << ' ' << f(l) << ' ' << f(m) << ' ' << f(m + 1e-6) << ' ' << f(r) << endl;
		long double x = f(m), y = f(m + 1e-10);
		if (x > y){
			l = m;
		}
		else{
			r = m;
		}
	}
}