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<algorithm>
#include<cstring>
#define MN 2005
using namespace std;
int n,T[1<<12],f[MN][MN],col[2][MN<<1];
void add(int x,int y){while(x<=2*n)T[x]+=y,x+=(x&(-x));}
int que(int x){int res=0;while(x)res+=T[x],x-=(x&(-x));return res;}
int main(){
	scanf("%d",&n);char ch[5];int x;
	for(int i=1;i<=2*n;i++)scanf("%s%d",ch,&x),col[ch[0]=='B'][x]=i;memset(f,0x3f,sizeof(f));f[0][0]=0;
	for(int i=0;i<=n;i++){
		memset(T,0,sizeof(T));
		for(int j=1;j<=2*n;j++)add(j,1);
		for(int j=1;j<=i;j++)add(col[0][j],-1);
		for(int j=0;j<=n;j++){
			if(j)add(col[1][j],-1);
			if(j)f[i][j]=min(f[i][j],f[i][j-1]+que(col[1][j]));
			if(i)f[i][j]=min(f[i][j],f[i-1][j]+que(col[0][i]));
		}
	}printf("%d\n",f[n][n]);
}

