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 n,c[4010],a[4010],
    pos[2010],pos1[2010],
    dp[2010][2010],
    cost[4010][2010],cost1[4010][2010];
    
    
    
int main()
{
    
    
    
    //color
    
    scanf("%d",&n);
    for (int i=1;i<=2*n;i++)
    {
        char s;
        cin>>s>>a[i];
        c[i]=(s=='W'?0:1);
        if(c[i]==0) pos[a[i]]=i; else pos1[a[i]]=i;
    }
    
    
    
    

    
    for (int i=1;i<=2*n;i++) for (int j=0;j<n;j++)
    {
        cost[i][j]=cost[i-1][j]+(a[i-1]>j&&c[i-1]==0);
        cost1[i][j]=cost1[i-1][j]+(a[i-1]>j&&c[i-1]==1);
    }




    for (int i=0;i<=n;i++) for (int j=0;j<=n;j++)
    {
        if (i==0 &&j==0) continue;
        dp[i][j]=0x3f3f3f3f;
        
        int p=pos[i];
        if (i) dp[i][j] = min(dp[i][j],dp[i-1][j]+cost[p][i-1]+cost1[p][j]);



        int p1=pos1[j];
        if (j) dp[i][j] = min(dp[i][j],dp[i][j-1]+cost1[p1][j-1]+cost[p1][i]);
    }


    cout<<dp[n][n]<<endl;


    return 0;
}