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 N 4010
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int n,ps[2][N],F[N][N],tree[N<<2];
int S1[N][N],S2[N][N];
inline void Add(int x)
{
	for(;x<=2*n;x+=x&-x)
		tree[x]++;
}
inline int Ask(int x)
{
	if(!x) return 0;
	int ret=0; for(;x;x-=x&-x)
		ret+=tree[x]; return ret;
}
int main()
{
	n=read();
	for(int i=1;i<=2*n;i++)
	{
		char s[3]; scanf("%s",s);
		int x=read();
		if(s[0]=='B') ps[0][x]=i;
		else ps[1][x]=i;
	}
	memset(tree,0,sizeof tree);
	for(int i=1;i<=n;i++)
	{
		Add(ps[0][i]);
		for(int j=1;j<=2*n;j++)
			S1[i][j]=Ask(j-1);
	}
	memset(tree,0,sizeof tree);
	for(int i=1;i<=n;i++)
	{
		Add(ps[1][i]);
		for(int j=1;j<=2*n;j++)
			S2[i][j]=Ask(j-1);
	}
	memset(F,0x3f,sizeof F); F[0][0]=0;
	for(int i=0;i<=n;i++) for(int j=0;j<=n;j++)
	{
		if(i<n)
		{
			int xm=ps[0][i+1];
			F[i+1][j]=min(F[i+1][j],F[i][j]+(xm-1-S1[i][xm]-S2[j][xm]));
		}
		if(j<n)
		{
			int xm=ps[1][j+1];
			F[i][j+1]=min(F[i][j+1],F[i][j]+(xm-1-S1[i][xm]-S2[j][xm]));
		}
	}
	cout << F[n][n] << endl;
	return 0;
}