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>
using namespace std;
long long costb[5000][5000],costw[5000][5000],N;
long long memo[5000][5000],barr[2500],warr[2500];

int main(){
	ios::sync_with_stdio(0);
	cin>>N;
	for(long long i=0;i<2500;++i){
		barr[i]=-1;
		warr[i]=-1;
	}
	for(long long i=0;i<2*N;++i){
		char c;
		long long a;
		cin>>c>>a;
		if(c=='W')warr[a]= i;
		else barr[a]=i;
	}
	for(long long i=0;i<=2*N;++i){
		costb[i][N]=0;
		for(long long j=N-1;j>=0;--j){
			costb[i][j]=costb[i][j+1];
			costb[i][j]+=(barr[j+1]<i&&barr[j+1]>=0);
		}
	}
	for(long long i=0;i<=2*N;++i){
		costw[i][N]=0;
		for(long long j=N-1;j>=0;--j){
			costw[i][j]=costw[i][j+1];
			costw[i][j]+=(warr[j+1]<i&&warr[j+1]>=0);
		}
	}
	for(long long i=0;i<=2*N;++i){
		for(long long j=0;j<=2*N;++j){
			memo[i][j]=1e9;
		}
	}
	memo[0][0]=0;
	for(long long i=0;i<=N;++i){
		for(long long j=0;j<=N;++j){
			if(i!=N){
				int x=costb[barr[i+1]][i+1];
				x+=costw[barr[i+1]][j];
				memo[i+1][j]=min(memo[i+1][j],memo[i][j]+x);
			}
			if(j!=N){
				int x=costw[warr[j+1]][j+1];
				x+=costb[warr[j+1]][i];
				memo[i][j+1]=min(memo[i][j+1],memo[i][j]+x);
			}
		}
	}
	cout<<memo[N][N]<<"\n";
	return 0;
}