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<iostream>
#include<cstdio>
#include<cstring>
#define N (4020)
#define INF (21000010)
using namespace std;
int p[2][N],A[2][N][2][N],n,dp[N][N],a[N],C[2][N][N];
//0 B/1 W			//A存每个球后面的球的情况 
char c[N];
int main(){
	scanf("%d",&n); int j,i,i1,j1; getchar();
	for (i=1;i<=2*n;i++){
		scanf("%c %d",&c[i],&a[i]); getchar();
		p[(c[i]=='W')][a[i]]=i; 
	}
	for (i=1;i<=2*n;i++)
		for (j=i+1;j<=2*n;j++)
			A[(c[i]=='W')][a[i]][(c[j]=='W')][a[j]]=1;
	for (i1=0;i1<2;i1++) for(i=1;i<=n;i++)
		for (j1=0;j1<2;j1++) for (j=1;j<=n;j++)
			A[i1][i][j1][j]+=A[i1][i][j1][j-1];
 	
	for (i=0;i<=n;i++) for (j=0;j<=n;j++){
 		C[0][i][j]=A[0][i+1][0][i]+A[0][i+1][1][j]+p[0][i+1]-(i+j+1);
 		C[1][i][j]=A[1][j+1][0][i]+A[1][j+1][1][j]+p[1][j+1]-(i+j+1);
 	}

 	for (i=0;i<=n;i++) for (j=0;j<=n;j++){
 		if (!i&&!j) continue;
 		dp[i][j]=INF; 
 		if (i>0) dp[i][j]=min(dp[i][j],dp[i-1][j]+C[0][i-1][j]);
 		if (j>0) dp[i][j]=min(dp[i][j],dp[i][j-1]+C[1][i][j-1]);
	}
	cout<<dp[n][n];
}