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;
const int N = 2005;

int n, a[2*N], c[2*N], id[2][N], sum[2][N][2*N], dt[N][N];

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=2*n;i++) {
		char T[2];
		scanf("%s%d",T,&a[i]);
		c[i] = (T[0] == 'B');
		id[c[i]][a[i]] = i;
		for(int j=a[i];j<=n;j++) {
			sum[c[i]][j][i]++;
		}
	}
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=2*n;j++) {
			sum[0][i][j] += sum[0][i][j-1];
			sum[1][i][j] += sum[1][i][j-1];
		}
	}
	for(int i=0;i<=n;i++) {
		for(int j=0;j<=n;j++) {
			int A = id[0][i], B = id[1][j];
			if(i == 0 && j == 0) continue;
			else if(i == 0) {
				dt[i][j] = dt[i][j-1] + B-1 - sum[0][i][B] - sum[1][j-1][B];
			}
			else if(j == 0) {
				dt[i][j] = dt[i-1][j] + A-1 - sum[0][i-1][A] - sum[1][j][A];
			}
			else {
				dt[i][j] = min(dt[i-1][j] + A-1 - sum[0][i-1][A] - sum[1][j][A], dt[i][j-1] + B-1 - sum[0][i][B] - sum[1][j-1][B]);
			}
		}
	}
	printf("%d\n",dt[n][n]);
}
