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 maxn=2e3+45;
const int INF=1e9+7;
#define charmax(x,y) x=max(x,y)
#define charmin(x,y) x=min(x,y)
int pos[2][maxn];int dp[maxn][maxn];int num[2*maxn];
int w;
void add(int x,int p){for(int i=x;i<=w;i+=i&-i) num[i]+=p;}
int sum(int x){int ans=0;for(int i=x;i>=1;i-=i&-i) ans+=num[i];return ans;}
int main()
{
    int n;
    scanf("%d",&n);
    w=n*2;
    memset(dp,0,sizeof dp);
    //memset(dp,INF,sizeof dp);
    for(int i=1;i<=w;i++){
        char s;int x;
        cin>>s>>x;
        pos[s=='W'][x]=i;
        add(i,1);
    }
    pos[0][0]=pos[1][0]=w+1;
    for(int i=0;i<=n;i++){
        add(pos[0][i],-1);
        for(int j=0;j<=n;j++){
            add(pos[1][j],-1);
            if(i||j){
                dp[i][j]=INF;
                if(i){
                    charmin(dp[i][j],dp[i-1][j]+sum(pos[0][i]-1));
                }
                if(j){
                    charmin(dp[i][j],dp[i][j-1]+sum(pos[1][j]-1));
                }
            }
        }
        for(int j=0;j<=n;j++){
            add(pos[1][j],1);
        }
    }
    printf("%d\n",dp[n][n]);
    return 0;
}
