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;
const int N=2005;
int a[N<<1],b[N<<1];
int s0[N<<1][N],s1[N<<1][N],f[N][N],pos1[N],pos0[N];
char s[555];
int main(){
	int n;
	scanf("%d",&n);
	for (int i=1;i<=2*n;i++){
		scanf("%s",s+1);
		scanf("%d",&a[i]);
		if (s[1]=='W') b[i]=0,pos0[a[i]]=i;
		else b[i]=1,pos1[a[i]]=i;
	}
	for (int i=1;i<=2*n;i++){
		for (int j=0;j<=n;j++){
			s0[i][j]=s0[i-1][j];
			s1[i][j]=s1[i-1][j];
			if (a[i-1]>j&&b[i-1]==0) s0[i][j]++;
			if (a[i-1]>j&&b[i-1]==1) s1[i][j]++;
		}
	}
	for (int i=0;i<=n;i++){
		for (int j=0;j<=n;j++){
			if (i==0&&j==0) continue;
			int p0=pos0[i],p1=pos1[j];
			f[i][j]=1e9;
			if (i) f[i][j]=min(f[i][j],f[i-1][j]+s0[p0][i-1]+s1[p0][j]);
			if (j) f[i][j]=min(f[i][j],f[i][j-1]+s0[p1][i]+s1[p1][j-1]);
		}
	}
	printf("%d\n",f[n][n]);
	return 0;
}