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 <iostream>
#include <cstring>
#define N 2005
using namespace std;

int n, x, w[N], b[N], dw[2*N][N], db[2*N][N], dp[N][N];
char c;
int main() {
	int i, j;
	cin>>n;
	for(i=1; i<=2*n; i++) {
		for(j=1; j<=n; j++) dw[i][j] = dw[i-1][j], db[i][j] = db[i-1][j];
		cin>>c>>x;
		if(c=='W') {
			w[x] = i;
			for(j=x; j; j--) dw[i][j]++;
		} else {
			b[x] = i;
			for(j=x; j; j--) db[i][j]++;
		}
	}
	memset(dp, 0x3f, sizeof(dp));
	dp[0][0] = 0;
	for(i=0; i<=n; i++) for(j=0; j<=n; j++) {
		if(i) dp[i][j] = min(dp[i][j], dp[i-1][j] + dw[w[i]][i+1] + db[w[i]][j+1]);
		if(j) dp[i][j] = min(dp[i][j], dp[i][j-1] + dw[b[j]][i+1] + db[b[j]][j+1]);
	}
	cout<<dp[n][n];
	return 0;
}