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.

Examples

Input

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


Output

4


Input

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


Output

18


Input

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


Output

41
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using VI = vector<int>;
using VVI = vector<VI>;


int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n; cin >> n;
  VI iw(n), ib(n);
  for (int i = 0; i < 2 * n; i++) {
    char c; cin >> c;
    int a; cin >> a;
    a--;
    (c == 'W' ? iw[a] : ib[a]) = i;
  }

  VVI ww(n, VI(n + 1)), wb(n, VI(n + 1)), bw(n, VI(n + 1)), bb(n, VI(n + 1));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ww[i][j + 1] = ww[i][j] + (iw[i] < iw[j]);
      wb[i][j + 1] = wb[i][j] + (iw[i] < ib[j]);
      bw[i][j + 1] = bw[i][j] + (ib[i] < iw[j]);
      bb[i][j + 1] = bb[i][j] + (ib[i] < ib[j]);
    }
  }

  VVI dp(n + 1, VI(n + 1));
  for (int i = 0; i < n; i++) {
    dp[i + 1][0] = dp[i][0] + ww[i][i];
    dp[0][i + 1] = dp[0][i] + bb[i][i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      dp[i + 1][j + 1] = min(dp[i][j + 1] + ww[i][i] + wb[i][j + 1], dp[i + 1][j] + bw[j][i + 1] + bb[j][j]);
    }
  }

  cout << dp[n][n] << '\n';
  return 0;
}