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: Runtime Error)

#include <iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;
    char c[2003];
    int a[2003];
    for(int i = 0; i < n * 2; i++) cin >> c[i] >> a[i];
    c[n * 2] = '?';
    a[n * 2] = 1000000;
    int ans = 0;
    int last = n * 2 - 1;
    while(last){
        bool f = false;
        for(int i = last - 1; i >= 0; i--){
            if(c[i] == c[last] && a[i] > a[last]){
                f = true;
                int t = a[i];
                int j = i;
                for(; j < last; j++){
                    c[j] = c[j + 1];
                    a[j] = a[j + 1];
                }
                while(c[j + 1] == c[last] && t > a[j + 1]){
                    a[j] = a[j + 1];
                    j++;
                }
                a[j] = t;
                ans += j - i;
                last--;
            }
        }
        if(!f) last--;
    }
    cout << ans << endl;
}