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 <bits/stdc++.h>

using namespace std;

using ll = long long;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> a(2 * n), b(n), w(n);
    vector<char> c(2 * n);
    for (int i = 0; i < 2 * n; i++) {
        cin >> c[i] >> a[i];
        a[i]--;
        if (c[i] == 'B') b[a[i]] = i;
        else w[a[i]] = i;
    }

    ll ans = 0;
    for (int i = 0, j = 0, k = 0; k < 2 * n; k++) {
        int tar = -1;
        if (i == n) {
            tar = w[j];
            j++;
        } else if (j == n) {
            tar = b[i];
            i++;
        } else {
            if (abs(b[i] - k) < abs(w[j] - k)) {
                tar = b[i];
                i++;
            } else {
                tar = w[j];
                j++;
            }
        }

        for (int x = tar; x > k; x--) {
            if (c[x - 1] == 'B') b[a[x - 1]] = x;
            else w[a[x - 1]] = x;
            if (c[x] == 'B') b[a[x]] = x - 1;
            else w[a[x]] = x - 1; 
            swap(c[x - 1], c[x]);
            swap(a[x - 1], a[x]);    
            ans++;
        }
    }

    cout << ans << endl;
    return 0;
}