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 <stdio.h>
#include <string.h>
#include <algorithm>

#define rep(i,st,ed) for (int i=st;i<=ed;++i)

const int INF=0x3f3f3f3f;
const int N=4005;

int f[N][N],a[N],w[N],b[N];
int cw[N][N],cb[N][N];

int main(void) {
	int n; scanf("%d",&n);
	rep(i,1,n*2) {
		char ch[2]; scanf("%s%d",ch,&a[i]);
		if (ch[0]=='W') w[a[i]]=i;
		else b[a[i]]=i;
	}
	rep(i,1,n) {
		rep(j,1,i-1) cw[i][0]+=(w[j]>w[i]);
		rep(j,1,n) cw[i][j]=cw[i][j-1]+(b[j]>w[i]);
	}
	rep(i,1,n) {
		rep(j,1,i-1) cb[0][i]+=(b[j]>b[i]);
		rep(j,1,n) cb[j][i]=cb[j-1][i]+(w[j]>b[i]);
	}
	rep(i,1,n) f[i][0]=f[i-1][0]+cw[i][0];
	rep(j,1,n) f[0][j]=f[0][j-1]+cb[0][j];
	rep(i,1,n) rep(j,1,n) {
		f[i][j]=std:: min(f[i-1][j]+cw[i][j],f[i][j-1]+cb[i][j]);
	}
	printf("%d\n", f[n][n]);
	return 0;
}