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 N=4010;
int n,a[N],posa[N],b[N],posb[N],p1[N][N],p2[N][N],f[N][N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n+n;i++)
	{
		char s[2];int x;
		scanf("%s%d",s,&x);
		if (s[0]=='B') a[i]=x,posa[x]=i;
			else b[i]=x,posb[x]=i;
	}
	for (int i=1;i<=n+n;i++)
		for (int j=0;j<=n;j++)
		{
			p1[i][j]=p1[i-1][j]+(a[i]>j);
			p2[i][j]=p2[i-1][j]+(b[i]>j);
		}
	for (int i=0;i<=n;i++)
		for (int j=0;j<=n;j++)
		if (i||j)
		{
			int x=posa[i],y=posb[j];
			if (!i) { f[i][j]=f[i][j-1]+p1[y][i]+p2[y][j];continue;}
			if (!j) { f[i][j]=f[i-1][j]+p1[x][i]+p2[x][j];continue;}
			f[i][j]=min(f[i-1][j]+p1[x][i]+p2[x][j],f[i][j-1]+p1[y][i]+p2[y][j]);
		}
	printf("%d\n",f[n][n]);
	return 0;
}
