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;

#define pb push_back

const int MAXN=2005;

int n,a[MAXN*2],pos[2][MAXN],dp[MAXN][MAXN];
bool c[MAXN*2];
vector <int> q[2][MAXN];

void add(vector <int> q[],int x,int p)
{
	for (;x<=n;x+=x&-x)
		q[x].pb(p);
}

int query(vector <int> q[],int x,int p)
{
	int ret=0;
	for (;x;x-=x&-x)
	{
		ret+=q[x].end()-upper_bound(q[x].begin(),q[x].end(),p);
	}
	return ret;
}

int cal(int p,int a,int b)
{
	int now=p+query(q[0],a,p)+query(q[1],b,p);
	return now-(a+b+1);
}

void update(int &a,int b)
{
	if (a>b)
		a=b;
}

int main()
{
	//freopen("read.txt","r",stdin);
	scanf("%d",&n);
	char s[5];
	for (int i=1;i<=2*n;i++)
	{
		scanf("%s%d",s,&a[i]);
		if (s[0]=='W')
		{
			c[i]=true;
			pos[1][a[i]]=i;
			add(q[1],a[i],i);
		}
		else
		{
			pos[0][a[i]]=i;
			add(q[0],a[i],i);
		}
	}
	for (int j=0;j<2;j++)
		for (int i=1;i<=n;i++)
			sort(q[j][i].begin(),q[j][i].end());
	memset(dp,0x3f,sizeof(dp));
	dp[0][0]=0;
	for (int i=0;i<=n;i++)
		for (int j=0;j<=n;j++)
		{
			if (i<n)
			{
				update(dp[i+1][j],dp[i][j]+cal(pos[0][i+1],i,j));
			}
			if (j<n)
			{
				update(dp[i][j+1],dp[i][j]+cal(pos[1][j+1],i,j));
			}
		}
	printf("%d",dp[n][n]);
	return 0;
}