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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int MAX_N=2e3,INF=1e8;
int N,B[MAX_N],W[MAX_N];

int bit[MAX_N*2+1],dp[MAX_N+1][MAX_N+1];

void init(){
	memset(bit,0,sizeof(bit));
}
void add(int x,int y){
	while(x<=(N<<1)){
		bit[x]+=y;
		x+=x&(-x);
	}
}
int que(int x){
	int ret=0;
	while(x>0){
		ret+=bit[x];
		x-=x&(-x);
	}
	return ret;
}

int main(){
	cin>>N;
	REP(i,N<<1){
		char c;
		int a;
		cin>>c>>a;
		a--;
		if (c=='B'){
			B[a]=i+1;
		}else{
			W[a]=i+1;
		}
	}
	REP(i,N+1)REP(j,N+1)dp[i][j]=INF;
	dp[0][0]=0;
	REP(sm,N<<1){
		int b,w;
		b=min(sm,N);
		w=sm-b;
		init();
		REP(i,b){
			add(B[i],1);
		}
		REP(j,w){
			add(W[j],1);
		}
		int lop=(sm<=N ? sm+1 : (N<<1)+1-sm);
		REP(i,lop){
			if (b<N){
				dp[b+1][w]=min(dp[b+1][w],dp[b][w]+sm-que(B[b]));
			}
			if (w<N){
				dp[b][w+1]=min(dp[b][w+1],dp[b][w]+sm-que(W[w]));
			}
			if (b>0 && w<N){
				add(B[b-1],-1);
				b--;
				add(W[w],1);
				w++;
			}
		}
	}
	cout<<dp[N][N]<<endl;
	return 0;
}