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 OCaml (4.02.3) (judge verdict: Wrong Answer)

let dir = function "R" -> 0 | "U" -> 1 | "L" -> 2 | _ -> 3
let foi = float_of_int
let () =
  Scanf.scanf "%d" @@ fun n ->
  let xyd = Array.init n (fun _ -> Scanf.scanf " %d %d %s" @@ fun x y d -> x, y, dir d) in
  let calc1 d =
    let fld, v0, g =
      if d < 2 then max, min_int, (fun x -> x)
      else          min, max_int, (fun x -> -x)
    in
    let c = Array.fold_left (fun v (x,y,d') ->
      if (d-d'+4) mod 2 = 1 then fld v (if d mod 2 = 0 then y else x) else v
    ) v0 xyd
    in
    let i = Array.fold_left (fun v (x,y,d') ->
      if d = d' then fld v (if d mod 2 = 0 then x else y) else v
    ) v0 xyd in
    let r = Array.fold_left (fun v (x,y,d') ->
      if (d-d'+4) mod 4 = 2 then fld v (if d mod 2 = 0 then x else y) else v
    ) v0 xyd
    in
    [ 2 * g (r - c);
      2 * g (c - i);
      g (r - i); ]
  in
  let calc2 t =
    let xmin, xmax, ymin, ymax =
      Array.fold_left (fun (xmin, xmax, ymin, ymax) (x,y,d) ->
        let x, y = foi x, foi y in
        let x, y = match d with
          | 0 -> x +. t, y
          | 1 -> x,      y +. t
          | 2 -> x -. t, y
          | _ -> x,      y -. t
        in
        min xmin x, max xmax x, min ymin y, max ymax y) (1e30, -.1e30, 1e30, -.1e30) xyd
    in
    (xmax -. xmin) *. (ymax -. ymin)
  in
  List.concat [calc1 0; calc1 1; calc1 2; calc1 3; ]
  |> List.filter (fun x -> 0 <= x && x <= 100000)
  |> List.sort_uniq (-)
  |> List.fold_left (fun v t -> min v (calc2 (foi t /. 2.0))) max_float
  |> Printf.printf "%.10f\n"