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;

int n, a[4003][2], b[2003][2], c[4003][2003][2], dp[2003][2003];
char ch;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for (int i = 1; i <= 2 * n; ++i) {
		cin >> ch >> a[i][1];
		a[i][0] = ch == 'B' ? 0 : 1;
		b[a[i][1]][a[i][0]] = i;
	}
	for(int i = 2 * n; i >= 1; --i) {
		for(int j = 1; j <= n; ++j) {
			c[i][j][0] = c[i + 1][j][0];
			c[i][j][1] = c[i + 1][j][1];
		}
		++c[i][a[i][1]][a[i][0]];
	}
	for(int i = 1; i <= 2 * n; ++i) {
		for(int j = 2; j <= n; ++j) {
			c[i][j][0] += c[i][j - 1][0];
			c[i][j][1] += c[i][j - 1][1];
		}
	}
	for(int i = 0; i <= n; ++i) {
		for(int j = 0; j <= n; ++j) {
			if(i + j) dp[i][j] = 1e9;
			if(i) {
				dp[i][j] = dp[i - 1][j] + (b[i][0] - (i + j) + (c[b[i][0]][i - 1][0] + c[b[i][0]][j][1]));
			}
			if(j) {
				dp[i][j] = min(dp[i][j], dp[i][j - 1] + (b[j][1] - (i + j) + (c[b[j][1]][i][0] + c[b[j][1]][j - 1][1])));
			}
		}
	}
	cout << dp[n][n] << endl;
	return 0;
}
