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;

void chmin(int& a, int b){
    a = min(a, b);
}

int main(){
    int N;
    cin >> N;
    vector<int> Bi(N), Wi(N);
    vector<vector<int>> Bsum(2*N+1, vector<int>(N+1)), Wsum(2*N+1, vector<int>(N+1));
    for(int i=0; i<2*N; i++){
        char c;
        int a;
        cin >> c >> a;
        a--;
        Bsum[i+1] = Bsum[i];
        Wsum[i+1] = Wsum[i];
        if(c == 'B'){
            Bi[a] = i;
            for(int j=0; j<=a; j++) Bsum[i+1][j]++;
        }else{
            Wi[a] = i;
            for(int j=0; j<=a; j++) Wsum[i+1][j]++;
        }
    }
    
    vector<vector<int>> dp(N+1, vector<int>(N+1, 1e9));
    dp[0][0] = 0;
    for(int i=0; i<=N; i++) for(int j=0; j<=N; j++){
        if(i<N) chmin(dp[i+1][j], dp[i][j] + Wsum[Wi[i]][i] + Bsum[Wi[i]][j]);
        if(j<N) chmin(dp[i][j+1], dp[i][j] + Wsum[Bi[j]][i] + Bsum[Bi[j]][j]);
    }
    cout << dp[N][N] << endl;
}
