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<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
 
const int maxn=2e3+10,maxn2=2*2e3+10;
int n,a[maxn2],dp[maxn][maxn],rnkw[maxn],rnkb[maxn];
int bit[maxn2];
char c[maxn2][2];
 
inline void add(int pos,int val){
	while(pos<maxn2){
		bit[pos]+=val;
		pos+=pos&-pos;
	}
}
inline int sum(int pos){
	int res=0;
	while(pos){
		res+=bit[pos];
		pos-=pos&-pos;
	}
	return res;
}
inline void chkmin(int&a,int b){
	if(a>b)a=b;
}
 
int main(){
	cin>>n;
	for(int i=1;i<=2*n;++i){
		cin>>c[i]>>a[i];
		if(c[i][0]=='W')
			rnkw[a[i]]=i;
		else
			rnkb[a[i]]=i;
	}
	for(int i=0;i<=n;++i)
		for(int j=0;j<=n;++j)
			dp[i][j]=1e9;
	dp[0][0]=0;
	for(int i=1;i<=n;++i)
		add(rnkw[i],1);
	for(int i=0;i<=n;++i){
		if(i)
			add(rnkw[i],-1);
		for(int j=1;j<=n;++j)
			add(rnkb[j],1);
		for(int j=0;j<=n;++j){
			if(j)
				add(rnkb[j],-1);
			if(i<n)
				chkmin(dp[i+1][j],dp[i][j]+sum(rnkw[i+1])-1);
			if(j<n)
				chkmin(dp[i][j+1],dp[i][j]+sum(rnkb[j+1])-1);
		}
	}
	cout<<dp[n][n]<<endl;
	return 0;
}