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=4e3+10;
const int inf=0x3f3f3f3f;

char c[N];
int a[N], n, dp[N][N], b[N][N], w[N][N], sumb[N], sumw[N];

int main() {
	scanf("%d", &n);
	for(int i=1; i<=n*2; i++) {
		cin>>c[i];
		scanf("%d", &a[i]);
	}
	for(int i=1; i<=n*2; i++) {
		if (c[i]=='B') {
			for(int j=0; j<=n; j++) {
				b[a[i]][j]=sumb[a[i]+1]+sumw[j+1];
			}
			for(int j=0; j<=a[i]; j++) {
				sumb[j]+=1;
			}
		}
		else {
			for(int j=0; j<=n; j++) {
				w[j][a[i]]=sumb[j+1]+sumw[a[i]+1];
			}
			for(int j=0; j<=a[i]; j++) {
				sumw[j]+=1;
			}
		}
	}
	for(int i=0; i<=n; i++) {
		for(int j=0; j<=n; j++) {
			if (!i&&!j) dp[i][i]=0;else dp[i][j]=inf;
			if (i) dp[i][j]=min(dp[i][j], dp[i-1][j]+b[i][j]);
			if (j) dp[i][j]=min(dp[i][j], dp[i][j-1]+w[i][j]);
		}
	}
	cout<<dp[n][n]<<endl;
	return 0;
}
