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>
#define inf 0x3f3f3f3f
using namespace std;
const int N=4005;
int n,dp[N][N],f[N][N],g[N][N],fw[N],fb[N];
struct node
{
    char c;
    int x;
}a[N];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=2*n;i++)
    {
        scanf(" %c %d",&a[i].c,&a[i].x);
        if(a[i].c=='W') fw[a[i].x]=i;
        else fb[a[i].x]=i;
    }
    for(int i=1;i<=2*n;i++)
    {
        for(int j=0;j<=n;j++)
            f[i][j]=f[i-1][j],g[i][j]=g[i-1][j];
        for(int j=a[i].x-1;j>=0;j--)
            if(a[i].c=='W') f[i][j]++;
        else g[i][j]++;
    }
    memset(dp,inf,sizeof(dp));
    dp[0][0]=0;
    for(int i=0;i<=n;i++)
        for(int j=0;j<=n;j++)
    {
        if(i+1<=n)
            dp[i+1][j]=min(dp[i+1][j],dp[i][j]+f[fw[i+1]][i]+g[fw[i+1]][j]-1);
        if(j+1<=n)
            dp[i][j+1]=min(dp[i][j+1],dp[i][j]+f[fb[j+1]][i]+g[fb[j+1]][j]-1);
    }
    printf("%d\n",dp[n][n]);
}
