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 MAX_N = 2005, INF = 1e9 + 7;
int B[MAX_N], W[MAX_N], dp[MAX_N][MAX_N], BB[2*MAX_N][MAX_N], WW[2*MAX_N][MAX_N];
int costB[MAX_N][MAX_N], costW[MAX_N][MAX_N];

int main(){
    int N; cin >> N;
    for(int i=1;i<=2*N;i++){
        char c; int a; cin >> c >> a;
        if(c=='B'){B[a] = i; BB[i+1][a+1]++;}
        else{W[a] = i; WW[i+1][a+1]++;}
    }

    for(int i=1;i<=2*N+1;i++){
        for(int j=1;j<=N;j++){BB[i][j+1] += BB[i][j]; WW[i][j+1] += WW[i][j];}
    }

    for(int i=1;i<=2*N;i++){
        for(int j=1;j<=N+1;j++){BB[i+1][j] += BB[i][j]; WW[i+1][j] += WW[i][j];}
    }

    for(int j=0;j<=N;j++)
        for(int i=1;i<=N;i++) costB[i][j] = BB[B[i]][N+1] - BB[B[i]][i] + WW[B[i]][N+1] - WW[B[i]][j+1];
    for(int i=0;i<=N;i++)
        for(int j=1;j<=N;j++) costW[i][j] = BB[W[j]][N+1] - BB[W[j]][i+1] + WW[W[j]][N+1] - WW[W[j]][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(0<i) dp[i][j] = dp[i-1][j] + costB[i][j];
            if(0<j) dp[i][j] = min(dp[i][j], dp[i][j-1] + costW[i][j]);
        }
    }

    cout << dp[N][N] << endl;
    return 0;
}