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)

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

double calc_diff(vector<double> &x) {
  return *max_element(x.begin(), x.end()) - *min_element(x.begin(), x.end());
}

double calc_s(vector<double> &x, vector<double> &y) {
  return calc_diff(x) * calc_diff(y);
}

int main()
{
  int n;
  cin >> n;
  vector<double> x(n), y(n), vx(n), vy(n);
  for (int i = 0; i < n; i++) {
    char d;
    cin >> x.at(i) >> y.at(i) >> d;
    switch(d) {
      case 'R':
        vx.at(i) = 1;
        vy.at(i) = 0;
        break;
      case 'L':
        vx.at(i) = -1;
        vy.at(i) = 0;
        break;
      case 'U':
        vx.at(i) = 0;
        vy.at(i) = 1;
        break;
      case 'D':
        vx.at(i) = 0;
        vy.at(i) = -1;
        break;
    }
  }
  vector<double> s;
  s.push_back(calc_s(x, y));
  cout << *min_element(s.begin(), s.end()) << endl;
}