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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x7fffffff,MAXN=2005;
int n,a[MAXN<<1],f[MAXN][MAXN],Sum[MAXN<<1][MAXN<<1];
int Get(int x,int B,int W){return (Sum[x][B]-Sum[x-1][B]+Sum[x][W])-(Sum[x][n]+Sum[x-1][W]-Sum[x-1][n]);}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=(n<<1);++i){
    	char ch[10];scanf("%s%d",ch,&a[i]),a[i]+=(ch[0]=='B')*n;
        for(int j=1;j<i;++j) Sum[a[j]][a[i]]=1;
    }
    memset(f,63,sizeof(f));f[0][0]=0;
    for(int i=1;i<=(n<<1);++i) for(int j=1;j<=(n<<1);++j) Sum[i][j]+=Sum[i-1][j]+Sum[i][j-1]-Sum[i-1][j-1];
    for(int i=0;i<=n;++i) for(int j=(i==0);j<=n;++j) f[i][j]=min(i?f[i-1][j]+Get(i,i,j+n):INF,j?f[i][j-1]+Get(j+n,i,j+n):INF);
    printf("%d\n",f[n][n]);
    return 0;
}