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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N = gets.to_i
P = N.times.map {
  x, y, d = gets.chomp.split

  [x.to_i, y.to_i, d]
}

def f(t)
  min_y = Float::INFINITY
  max_y = -Float::INFINITY
  min_x = Float::INFINITY
  max_x = -Float::INFINITY

  P.each do |x, y, d|
    case d
    when 'R'
      x += t
    when 'L'
      x -= t
    when 'U'
      y += t
    when 'D'
      y -= t
    end

    min_y = y if min_y > y
    min_x = x if min_x > x
    max_y = y if max_y < y
    max_x = x if max_x < x
  end

  # p [max_y, min_y, max_x, min_x]
  (max_y - min_y) * (max_x - min_x)
end

def ternary_search_tree(left, right)
  loop_cnt = 100

  loop_cnt.times do
    if yield((left * 2 + right) / 3.0) < yield((left + right * 2) / 3.0)
      right = (left + right * 2) / 3.0
    else
      left = (left * 2 + right) / 3.0
    end
  end

  (left + right) / 2.0
end

left = 0
right = 10 ** 8

time = ternary_search_tree(left, right) { |x| f(x) }

puts "%10.9f" % f(time)