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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 2050
int f[N][N],opt[N<<1],a[N<<1],cw[N<<1][N],cb[N<<1][N],pos[2][N],n;
char str[10];
int main() {
	scanf("%d",&n);
	int i,j;
	for(i=1;i<=(n<<1);i++) {
		scanf("%s%d",str,&a[i]);
		opt[i]=(str[0]=='B'); pos[opt[i]][a[i]]=i;
	}
	for(i=1;i<=(n<<1);i++) {
		for(j=0;j<=n;j++) {
			cw[i][j]=cw[i-1][j];
			cb[i][j]=cb[i-1][j];
			if(a[i-1]>j) opt[i-1]?cb[i][j]++:cw[i][j]++;
		}
	}
	memset(f,0x3f,sizeof(f));
	f[0][0]=0;
	for(i=0;i<=n;i++) {
		for(j=0;j<=n;j++) {
			int u=pos[0][i+1],v=pos[1][j+1];
			f[i+1][j]=min(f[i+1][j],f[i][j]+cw[u][i+1]+cb[u][j]);
			f[i][j+1]=min(f[i][j+1],f[i][j]+cw[v][i]+cb[v][j+1]);
		}
	}
	printf("%d\n",f[n][n]);
}