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: Wrong Answer)

N = gets.to_i
R = []
L = []
U = []
D = []
N.times{
  sx, sy, d = gets.chomp.split(" ")
  x = sx.to_i
  y = sy.to_i
  case d
  when 'R' then
    R << [x, y]
  when 'L' then
    L << [x, y]
  when 'U' then
    U << [x, y]
  else
    D << [x, y]
  end
}

XMIN = []
XMAX = []
YMIN = []
YMAX = []
if !((U + D).empty?) then
  w1 = (U + D).min_by{|e| e[0]}[0]
  XMIN << Proc.new{|t| w1}
  w2 = (U + D).max_by{|e| e[0]}[0]
  XMAX << Proc.new{|t| w2}
end
if !((R + L).empty?) then
  w3 = (R + L).min_by{|e| e[1]}[1]
  YMIN << Proc.new{|t| w3}
  w4 = (R + L).max_by{|e| e[1]}[1]
  YMAX << Proc.new{|t| w4}
end

if !(R.empty?) then
  w5 = R.min_by{|e| e[0]}[0]
  XMIN << Proc.new{|t| w5 + t}
  w6 = R.max_by{|e| e[0]}[0]
  XMAX << Proc.new{|t| w6 + t}
end
if !(L.empty?) then
  w7 = L.min_by{|e| e[0]}[0]
  XMIN << Proc.new{|t| w7 - t}
  w8 = L.max_by{|e| e[0]}[0]
  XMAX << Proc.new{|t| w8 - t}
end
if !(U.empty?) then
  w9 = U.min_by{|e| e[1]}[1]
  YMIN << Proc.new{|t| w9 + t}
  wa = U.max_by{|e| e[1]}[1]
  YMAX << Proc.new{|t| wa + t}
end
if !(D.empty?) then
  wb = D.min_by{|e| e[1]}[1]
  YMIN << Proc.new{|t| wb - t}
  wc = D.max_by{|e| e[1]}[1]
  YMAX << Proc.new{|t| wc - t}
end

def calc(t)
  xmin = XMIN.map{|e| e.call(t)}.min
  xmax = XMAX.map{|e| e.call(t)}.max
  ymin = YMIN.map{|e| e.call(t)}.min
  ymax = YMAX.map{|e| e.call(t)}.max
  (xmax - xmin) * (ymax - ymin)
end

ans = 10 ** 18
wk = 10 ** 18
t = 0
l1, l2 = [0, 0]
while(true) do
  n = calc(t)
  break if n > wk
  wk = n
  t += 100000
  l2 = l1
  l1 = t
end

t = l2
while(true) do
  n = calc(t)
  break if n > ans
  ans = n
  t += 0.5
end

puts ans