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
//Zory-2018
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
#include<set>
#include<queue>
#include<deque>
#include<bitset>
#include<vector>
#include<algorithm>
#include<iostream>
#include<stack>
using namespace std;

namespace mine
{
	const int MAX_N=2100;
	const int INF=0x3f3f3f3f;
	typedef long long ll;
	int mymax(int x,int y) {return x>y?x:y;}
	int mymin(int x,int y) {return x<y?x:y;}

	int p1[MAX_N*2];
	int f[MAX_N*2][MAX_N];
	int nw[MAX_N*2][MAX_N*2];
	void main()
	{
		int n;scanf("%d",&n);
		for(int i=1;i<=2*n;i++)
		{
			char s[5];int id;scanf("%s%d",s,&id);
			p1[id+n*(s[0]=='B')]=i;
		}

		for(int i=1;i<=2*n;i++)
			for(int t=1;t<=2*n;t++)
				nw[i][t]=nw[i][t-1]+(p1[t]>p1[i]);

		memset(f,63,sizeof f);f[0][0]=0;
		for(int x=1;x<=2*n;x++)
		{
			for(int a=0;a<=x and a<=n;a++)
			{
				int b=x-a;
				if(a>0) f[x][a]=mymin(f[x][a],
					f[x-1][a-1]+nw[a][a-1]+nw[a][n+b]-nw[a][n]
				);
				if(b>0) f[x][a]=mymin(f[x][a],
					f[x-1][a]+nw[n+b][a]+nw[n+b][n+b-1]-nw[n+b][n]
				);
			}
		}
		printf("%d",f[2*n][n]);
	}
};

int main()
{
	mine::main();
}
