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;
#define INF 1e9
int ww[4010][4010];
int wb[4010][4010];
int bb[4010][4010];
int bw[4010][4010];
int dp[4010][4010];

int main(){
  int n,a[4010];
  char c[4010];
  int wp[4010];
  int bp[4010];
  cin>>n;
  for(int i=0;i<2*n;i++){
    cin>>c[i]>>a[i];
    a[i]--;
    (c[i]=='B'?bp[a[i]]:wp[a[i]])=i;
  }
  
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      ww[i][j+1]=ww[i][j]+(wp[i]<wp[j]);
      wb[i][j+1]=wb[i][j]+(wp[i]<bp[j]);
      bb[i][j+1]=bb[i][j]+(bp[i]<bp[j]);
      bw[i][j+1]=bw[i][j]+(bp[i]<wp[j]);
    }
  }
  
  for(int i=0;i<4010;i++)
    for(int j=0;j<4010;j++)
      dp[i][j]=INF;
  
  dp[0][0]=0;
  
  for(int i=0;i<n+1;i++){
    for(int j=0;j<n+1;j++){
      if(i<n)dp[i+1][j]=min(dp[i+1][j],dp[i][j]+ww[i][i]+wb[i][j]);
      if(j<n)dp[i][j+1]=min(dp[i][j+1],dp[i][j]+bb[j][j]+bw[j][i]);
    }
  }
  cout<<dp[n][n]<<endl;
  return 0;
}
