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>

using namespace std;

#define N 4050

int n,x,a[N],b[N],f[N][N],g[N][N],h[N][N];
char ch;

void upd(int &x,int y){x=min(x,y);}

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n*2;++i){
		for (ch=getchar();ch!='B'&&ch!='W';ch=getchar());
		scanf("%d",&x);
		memcpy(g[i],g[i-1],sizeof g[i]);
		memcpy(h[i],h[i-1],sizeof h[i]);
		if (ch=='B'){
			a[x]=i;
			for (int j=1;j<=x;++j) ++g[i][j];
		}
		else{
			b[x]=i;
			for (int j=1;j<=x;++j) ++h[i][j];
		}
	}
	memset(f,0x3f,sizeof f); f[0][0]=0;
	for (int i=0;i<=n;++i)
		for (int j=0;j<=n;++j){
			if (i!=n) upd(f[i+1][j],f[i][j]+g[a[i+1]-1][i+1]+h[a[i+1]-1][j+1]);
			if (j!=n) upd(f[i][j+1],f[i][j]+g[b[j+1]-1][i+1]+h[b[j+1]-1][j+1]);
		}
	printf("%d\n",f[n][n]);
	
	return 0;
}
			