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>
#define LL long long
#define Accel ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define Re(a,b) memset(a,b,sizeof a)
#define F first
#define S second
#define MP make_pair
#define PB push_back
#define ALL(X) (X).begin(), (X).end()
#define FOR(I, A, B) for (int I = (A); I <= (B); I++)
#define REP(I, N) for (int I = 0; I < (N); I++)
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define SZ(a) ((int)a.size())
using namespace std;
const int N=4e3+10;
int n;
int p[2][N];//B/W
int f[N][N];
int c[2][N][N];//B,W
int a[N];
char b[N];
int main(){
	cin>>n;n*=2;
	FOR(i,1,n){
		cin>>b[i]>>a[i];
		if(b[i]=='B')
			p[0][a[i]]=i;
		else p[1][a[i]]=i;
	}
	FOR(i,1,n/2){
		int s=0;
		//B
		for(int j=p[0][i]+1;j<=n;j++){
			if(b[j]=='B'&&a[j]<i)
				s++;
		}
		for(int j=0;j<=n/2;j++){
			if(p[1][j]>p[0][i])s++;
			c[0][i-1][j]=s;
		}
		s=0;
		for(int j=p[1][i]+1;j<=n;j++){
			if(b[j]=='W'&&a[j]<i)s++;
		}
		for(int j=0;j<=n/2;j++){
			if(p[0][j]>p[1][i])s++;
			c[1][j][i-1]=s;
		}
		
	}
	FOR(i,1,n/2){
		f[i][0]=f[i-1][0]+c[0][i-1][0];
		f[0][i]=f[0][i-1]+c[1][0][i-1];
	}
	FOR(i,1,n/2)
		FOR(j,1,n/2)
			f[i][j]=min(f[i-1][j]+c[0][i-1][j],f[i][j-1]+c[1][i][j-1]);
	cout<<f[n/2][n/2]<<endl;
		
	
		
}