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;
#define all(vec) vec.begin(),vec.end()
typedef long long int ll;
typedef pair<int,int> P;
const ll MOD=1000000007;
const ll INF=1000000010;
const ll LINF=4000000000000000010LL;
const int MAX=310;
const double EPS=1e-9;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int c[2][4010][2010];
int p[2][2010];
int dp[2010][2010];
int main(){
    int n;cin>>n;
    vector<P> v(n*2+2);
    for(int i=1;i<=2*n;i++){
        char cc;cin>>cc>>v[i].second;
        if(cc=='B'){
            v[i].first=1;
        }else{
            v[i].first=0;
        }
        p[v[i].first][v[i].second]=i;
    }
    for(int i=1;i<=2*n;i++){
        for(int j=1;j<=n;j++){
            c[0][i][j]=c[0][i-1][j];
            c[1][i][j]=c[1][i-1][j];
            if(v[i].second<=j)c[v[i].first][i][j]++;
        }
    }
    for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)dp[i][j]=INF;
    dp[0][0]=0;
    for(int i=0;i<=n;i++){
        for(int j=0;j<=n;j++){
            if(i)dp[i][j]=min(dp[i][j],dp[i-1][j]+p[0][i]-c[0][p[0][i]][i-1]-c[1][p[0][i]][j]-1);
            if(j)dp[i][j]=min(dp[i][j],dp[i][j-1]+p[1][j]-c[0][p[1][j]][i]-c[1][p[1][j]][j-1]-1);
        }
    }
    cout<<dp[n][n]<<endl;
    return 0;
}
