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: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

const int maxN = 2 * 2018;

int n, i, j;
char c[maxN];
int v[maxN];
int wh[2][maxN];

int data[maxN];
int dp[maxN][maxN];

int zrs(int x) {
    return (x ^ (x - 1)) & x;
}

void add(int pos, int val) {
    while (pos <= 2 * n) {
        data[pos] += val;
        pos += zrs(pos);
    }
}

int sum(int pos) {
    int ans = 0;
    while (pos > 0) {
        ans += data[pos];
        pos -= zrs(pos);
    }
    return ans;
}

int main()
{
  //  freopen("test.in","r",stdin);
    
    scanf("%d\n", &n);
    for (i = 1; i <= 2 * n; i++) {
        scanf("%c %d\n", &c[i], &v[i]);

        if (c[i] == 'W') {
            wh[0][v[i]] = i;
        } else {
            wh[1][v[i]] = i;
        }
    }
    
    for (i = 0; i <= n; i++)
        for (j = 0; j <= n; j++)
            dp[i][j] = 10 * n * n;
    dp[0][0] = 0;

    for (i = 0; i <= n; i++) {
        if (i > 0) add(wh[0][i], +1);
        for (j = 0; j <= n; j++) {
            
            if (j > 0)
                add(wh[1][j], +1);

            if (i < n)
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + wh[0][i + 1] - 1 -
                sum(wh[0][i + 1]));

            if (j < n)
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + wh[1][j + 1] - 1 -
                sum(wh[1][j + 1]));
        }

        for (j = 1; j <= n; j++) 
            add(wh[1][j], -1);
    }

    /*for (i = 0; i <= n; i++, cerr << '\n') {
        for (j = 0; j <= n; j++)
            cerr << dp[i][j] << ' ';
    }*/

    printf("%d", dp[n][n]);

    return 0;
}