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 4005
using namespace std;
int n,ans,f[N][N],a[N],w[N],b[N],sumw[N][N],sumb[N][N];
char c[N];
int main(){
	scanf("%d",&n);getchar();
	for (int i=1;i<=2*n;i++){
		c[i]=getchar();
		scanf("%d",&a[i]);getchar();
		if (c[i]=='W') w[a[i]]=i; else b[a[i]]=i;
	}
	for (int i=1;i<=2*n;i++){
		for (int j=1;j<=n;j++){
			if (c[i]=='W'){
				if (a[i]<=j) sumw[i][j]=sumw[i-1][j]+1; else sumw[i][j]=sumw[i-1][j];
				sumb[i][j]=sumb[i-1][j];
			} else
			if (c[i]=='B'){
				if (a[i]<=j) sumb[i][j]=sumb[i-1][j]+1; else sumb[i][j]=sumb[i-1][j];
				sumw[i][j]=sumw[i-1][j];
			}
		}
	}
	memset(f,0x3f3f3f3f,sizeof(f));
	f[0][0]=0;
	for (int i=0;i<=n;i++)
	for (int j=0;j<=n;j++){
		int t=w[i+1];
		f[i+1][j]=min(f[i+1][j],f[i][j]+t-sumw[t][i]-sumb[t][j]-1);
		t=b[j+1];
		f[i][j+1]=min(f[i][j+1],f[i][j]+t-sumw[t][i]-sumb[t][j]-1);
	}
	printf("%d\n",f[n][n]);
}