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<cmath>
#include<algorithm>
#include<iostream>
#include<utility>
#include<vector>
using namespace std;
const int maxn = 4008;
int ch[maxn],a[maxn],b[maxn];
int f[2012][2012];int n;
void add(int x)
{
	for(int i=x;i<=2*n;i+=i&(-i))
		ch[i]++;
	return ;
}
int query(int x)
{
	int sum = 0;
	for(int i=x;i;i-=i&(-i))
		sum += ch[i];
	return sum;
}
int cost(int x,int p)
{
	x += p-query(x);
	return x-p-1;
}
int main()
{
	cin>>n;getchar();
	for(int i=1;i<=2*n;i++)
	{
		char c = getchar();
		int x;scanf("%d",&x);
		if(c=='W')a[x] = i;
		else b[x] = i;
		getchar();
	}
/*	for(int i=1;i<=n;i++)
	cout<<a[i]<<" ";
	cout<<endl;
	for(int i=1;i<=n;i++)
	cout<<b[i]<<" ";
	cout<<endl;*/
	for(int i=0;i<=n;i++)
	for(int j=0;j<=n;j++)
	f[i][j] = 1<<30;
	f[0][0] = 0;
	for(int i=0;i<=n;i++)
	{
		memset(ch,0,sizeof(ch));
		for(int j=1;j<=i;j++)add(a[j]);
		for(int j=0;j<=n;j++)
		{
			if(i!=n)f[i+1][j] = min(f[i+1][j],f[i][j]+cost(a[i+1],i+j));
			if(j!=n)f[i][j+1] = min(f[i][j+1],f[i][j]+cost(b[j+1],i+j));
			if(j!=n)add(b[j+1]);
		}
	}
	cout << f[n][n] <<endl;
	return 0;
}