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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdlib.h>
#include <stdio.h>

int N;
int XYD[10001][3];

double calc_area(double n) {
  const int tbl[][2] = {
    {1, 0}, {-1, 0}, {0, 1}, {0, -1},
  };
  double min_x, min_y, max_x, max_y;
  min_x = min_y = 1e9;
  max_x = max_y = -1e9;
  
  for (int i = 0; i < N; ++i) {
    // printf("%lf %lf %lf %lf\n", max_x, min_x, max_y, min_y);
    // printf("%d %d\n", tbl[XYD[i][2]][0], tbl[XYD[i][2]][1]);
    double x = XYD[i][0];
    double y = XYD[i][1];
    x += n * tbl[XYD[i][2]][0];
    y += n * tbl[XYD[i][2]][1];
    if (x < min_x)
      min_x = x;
    if (x > max_x)
      max_x = x;
    if (y < min_y)
      min_y = y;
    if (y > max_y)
      max_y = y;
  }
  return (max_x - min_x) * (max_y - min_y);
}

int main(void) {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    int x, y, z;
    char d;
    scanf("%d %d %c", &x, &y, &d);
    XYD[i][0] = x;
    XYD[i][1] = y;
    switch (d) {
    case 'R':
      XYD[i][2] = 0;
      break;
    case 'L':
      XYD[i][2] = 1;
      break;
    case 'U':
      XYD[i][2] = 2;
      break;
    default:
      XYD[i][2] = 3;
      break;
    }
  }

  double l = 0.0;
  double r = 1e9;

  for (int i = 0; i < 200; ++i) {
    if (calc_area((l * 2 + r) / 3) < calc_area((l + r * 2) / 3))
      r = (l + r * 2) / 3;
    else
      l = (l * 2 + r) / 3;
  }
  double c = (l + r) / 2;
  // printf("%lf\n", c);
  printf("%lf\n", calc_area(c));

  return 0;
}