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 <algorithm>
#include <cstring>
#define MN 4010

int dp[2010][2010];
int t[MN]; int a[MN];
int wz[MN][2];
int n;
int c[MN];

void add(int x, int v) {for(int i = x; i <= n * 2; i += i & -i) c[i] += v;}
int query(int x) {int ans = 0; for(int i = x; i; i -= i & -i) ans += c[i]; return ans;}

int main()
{
	memset(dp, 0x3f, sizeof(dp));
	scanf("%d", &n);
	for(int i = 1; i <= n * 2; i++)
	{
		char c[2];
		scanf("%s%d", c, &a[i]);
		if(c[0] == 'B') t[i] = 0;
		else t[i] = 1; 
		wz[a[i]][t[i]] = i;
	}
	dp[0][0] = 0;
	for(int i = 0; i <= n; i++)
	{
		memset(c, 0, sizeof(c));
		for(int j = 1; j <= 2 * n; j++) add(j, 1);
		for(int j = 1; j <= i; j++) add(wz[j][0], -1);
		for(int j = 0; j <= n; j++)
		{
			if(i == n && j == n) continue;
			if(j) add(wz[j][1], -1);
			if(i + 1 <= n) dp[i + 1][j] = std::min(dp[i + 1][j], dp[i][j] + query(wz[i + 1][0]) - 1);
			if(j + 1 <= n) dp[i][j + 1] = std::min(dp[i][j + 1], dp[i][j] + query(wz[j + 1][1]) - 1);
		}
	}
	printf("%d\n", dp[n][n]);
	return 0;
}