Problem:
Score : 600 points

Problem Statement

There are 2N balls, N white and N black, arranged in a row. The integers from 1 through N are written on the white balls, one on each ball, and they are also written on the black balls, one on each ball.
The integer written on the i-th ball from the left (1 ≤ i ≤ 2N) is a_i, and the color of this ball is represented by a letter c_i.
c_i = W represents the ball is white; c_i = B represents the ball is black.

Takahashi the human wants to achieve the following objective:

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the white ball with i written on it is to the left of the white ball with j written on it.

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the black ball with i written on it is to the left of the black ball with j written on it.

In order to achieve this, he can perform the following operation:

Swap two adjacent balls.

Find the minimum number of operations required to achieve the objective.

Constraints

1 ≤ N ≤ 2000

1 ≤ a_i ≤ N

c_i = W or c_i = B.

If i ≠ j, (a_i,c_i) ≠ (a_j,c_j).

Input

Input is given from Standard Input in the following format:

N
c_1 a_1
c_2 a_2
:
c_{2N} a_{2N}

Output

Print the minimum number of operations required to achieve the objective.

Sample Input 1

3
B 1
W 2
B 3
W 1
W 3
B 2

Sample Output 1

4

The objective can be achieved in four operations, for example, as follows:

Swap the black 3 and white 1.

Swap the white 1 and white 2.

Swap the black 3 and white 3.

Swap the black 3 and black 2.

Sample Input 2

4
B 4
W 4
B 3
W 3
B 2
W 2
B 1
W 1

Sample Output 2

18

Sample Input 3

9
W 3
B 1
B 4
W 1
B 5
W 9
W 2
B 6
W 5
B 3
W 8
B 9
W 7
B 2
B 8
W 4
W 6
B 7

Sample Output 3

41

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let () = Scanf.scanf "%d\n" @@ fun n ->
  let cas = Array.init (2 * n) @@ fun _ -> Scanf.scanf "%c %d\n" @@ fun c a -> c, a in
  let inv_ws = Array.make n 0 in
  let inv_bs = Array.make n 0 in
  let acc_ws = Array.make_matrix (2 * n + 1) (n + 1) 0 in
  let acc_bs = Array.make_matrix (2 * n + 1) (n + 1) 0 in
  for i = 2 * n - 1 downto 0 do
    begin match cas.(i) with
    | ('W', a) ->
        inv_ws.(a - 1) <- i;
        acc_ws.(i).(a) <- 1 + acc_ws.(i).(a)
    | ('B', a) ->
        inv_bs.(a - 1) <- i;
        acc_bs.(i).(a) <- 1 + acc_bs.(i).(a)
    end;
    for j = 1 to n do
      acc_ws.(i).(j) <- acc_ws.(i).(j) + acc_ws.(i).(j - 1);
      acc_bs.(i).(j) <- acc_bs.(i).(j) + acc_bs.(i).(j - 1)
    done;
    for j = 0 to n do
      acc_ws.(i).(j) <- acc_ws.(i).(j) + acc_ws.(i + 1).(j);
      acc_bs.(i).(j) <- acc_bs.(i).(j) + acc_bs.(i + 1).(j)
    done
  done;
  let dp = Array.make_matrix (n + 1) (n + 1) max_int in
  dp.(0).(0) <- 0;
  for w = 1 to n do
    dp.(w).(0) <- dp.(w - 1).(0) - w + 1
      + inv_ws.(w - 1)
      + acc_ws.(inv_ws.(w - 1) + 1).(w)
  done;
  for b = 1 to n do
    dp.(0).(b) <- dp.(0).(b - 1) - b + 1
      + inv_bs.(b - 1)
      + acc_bs.(inv_bs.(b - 1) + 1).(b)
  done;
  for w = 1 to n do
    for b = 1 to n do
      dp.(w).(b) <- min
        ( dp.(w - 1).(b) + inv_ws.(w - 1)
          + acc_ws.(inv_ws.(w - 1) + 1).(w)
          + acc_bs.(inv_ws.(w - 1) + 1).(b) )
        ( dp.(w).(b - 1) + inv_bs.(b - 1)
          + acc_ws.(inv_bs.(b - 1) + 1).(w)
          + acc_bs.(inv_bs.(b - 1) + 1).(b) ) - w - b + 1
    done
  done;
  Printf.printf "%d\n" @@ dp.(n).(n)