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;

int b[2001], w[2001];
int cb[2001][2001], cw[2001][2001], f[2001][2001];

int main(){
  memset(cb, 0, sizeof(cb));
  memset(cw, 0, sizeof(cb));
  memset(f, 0, sizeof(f));
  
  int N;cin>>N;
  for (int i=1;i<=2*N;++i){
    char c;cin>>c;int a;cin>>a;
    if(c=='W') w[a]=i;
    else b[a]=i;
  }

  for (int i=1;i<=N;++i){
    for (int j=1;j<i;++j){
      cb[i][0]+=(b[j]>b[i]);
    }

    for (int j=1;j<=N;++j){
      cb[i][j]=cb[i][j-1]+(b[i]<w[j]);
    }
  }

  for (int i=1;i<=N;++i){
    for (int j=1;j<i;++j){
      cw[0][i]+=(w[j]>w[i]);
    }

    for (int j=1;j<=N;++j){
      cw[j][i]=cw[j-1][i]+(w[i]<b[j]);
    }
  }

  for (int i=1;i<=N;++i){
    f[0][i]=f[0][i-1]+cw[0][i];
    f[i][0]=f[i-1][0]+cb[i][0];
  }

  for (int i=1;i<=N;++i){
    for (int j=1;j<=N;++j){
      f[i][j]=min(f[i-1][j]+cb[i][j],f[i][j-1]+cw[i][j]);
    }
  }
  cout << f[N][N] << endl;
  return 0;
}