Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
B 1
W 2
B 3
W 1
W 3
B 2

Sample Output 1

4

The objective can be achieved in four operations, for example, as follows:

Swap the black 3 and white 1.

Swap the white 1 and white 2.

Swap the black 3 and white 3.

Swap the black 3 and black 2.

Sample Input 2

4
B 4
W 4
B 3
W 3
B 2
W 2
B 1
W 1

Sample Output 2

18

Sample Input 3

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

Sample Output 3

41

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
using namespace std;

int N, a[4000];
char c[4000];
int b_idx[2000], w_idx[2000];


int main(){
    cin >> N;
    for(int i = 0; i <2*N; i++){
        cin >> c[i] >> a[i];
        if(c[i] == 'W'){
            w_idx[a[i]-1] = i;
        }else{
            b_idx[a[i]-1] = i;
        }
    }
    for(int i = 0; i < N; i++){
        //printf("black %d = %d\n", i, b_idx[i]);
    }

    long long ans = 0;
    int w_now = 0, b_now = 0;
    for(int i = 0; i < 2*N; i++){
        if(b_now >= N || (w_now < N && w_idx[w_now] < b_idx[b_now])){
            ans += w_idx[w_now] - i;
            for(int j = 0; j < N; j++){
                if(w_idx[j] < w_idx[w_now]){
                    w_idx[j]++;
                }
                if(b_idx[j] < w_idx[w_now]){
                    b_idx[j]++;
                }
            }
            w_now++;
            //printf("i = %d, white\n", i);
        }else{
            ans += b_idx[b_now] - i;
            for(int j = 0; j < N; j++){
                if(w_idx[j] < b_idx[b_now]){
                    w_idx[j]++;
                }
                if(b_idx[j] < b_idx[b_now]){
                    b_idx[j]++;
                }
            }
            b_now++;
            //printf("i = %d, black\n", i);
        }
        //printf("i = %d, ans = %lld\n", i, ans);
    }
    cout << ans << endl;

    
}