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>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<map> 
using namespace std;
 
const int N=4005;
 
int n,x,y,ans,inf;
int a[N],dp[N][N],f[N][N],g[N][N],h[N];
char c[N],s[N];

void add(int x,int w){
	for (;x<=n;x+=x&-x) h[x]+=w;
}

int query(int x){
	int s=0;
	for (;x;x-=x&-x) s+=h[x];
	return s;
}

void init(){
	int i,j,k;
	for (i=0;i<=n;i++){
		k=0;
		memset(h,0,sizeof(h));
		for (j=1;j<=n+n;j++){
			if (c[j]=='W'){
				f[i][a[j]]=k-query(a[j]);
				add(a[j],1);
			}
			k++;
			if (c[j]=='B'&&a[j]<=i) k--;
		}
	}
	for (i=0;i<=n;i++){
		k=0;
		memset(h,0,sizeof(h));
		for (j=1;j<=n+n;j++){
			if (c[j]=='B'){
				g[i][a[j]]=k-query(a[j]);
				add(a[j],1);
			}
			k++;
			if (c[j]=='W'&&a[j]<=i) k--;
		}
	}
}

int main(){
	int i,j;
	scanf("%d",&n);
	memset(dp,0x3f,sizeof(dp));
	inf=dp[0][0];
	for (i=1;i<=n+n;i++){
		scanf("%s",s);
		scanf("%d",&a[i]);
		c[i]=s[0];
	}
	x=y=0; dp[0][0]=0;
	init();
	for (i=0;i<=n;i++)
		for (j=0;j<=n;j++) if (dp[i][j]<inf){
			if (i<n) dp[i+1][j]=min(dp[i+1][j],dp[i][j]+g[j][i+1]);
			if (j<n) dp[i][j+1]=min(dp[i][j+1],dp[i][j]+f[i][j+1]);
		} 
	ans=dp[n][n];
	printf("%d\n",ans);
	return 0;
}