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 <bits/stdc++.h>
template<typename T1, typename T2>
inline bool cmn (T1& a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  std::cin.tie(0); std::cin.sync_with_stdio(false);
  int n; std::cin >> n;
  int N = 2 * n;
  std::vector<char> c(N);
  std::vector<int> a(N);
  for (int i = 0; i < N; i++){
    std::cin >> c[i] >> a[i];
    a[i]--;
  }
  auto cum = [&](char d){
    std::vector<std::vector<int>> cum(n + 1, std::vector<int>(N + 1, 0));
    for (int i = 0; i <= n; i++) {
      for (int j = 1; j <= N; j++) {
        cum[i][j] = cum[i][j - 1] + (c[j - 1] == d && a[j - 1] >= i);
      }
    }
    return cum;
  };
  auto cnt_w = cum('W');
  auto cnt_b = cum('B');
  auto count = [&] (int i, int j, int pos) {
    assert(0 <= i && i <= n);
    assert(0 <= j && j <= n);
    assert(0 <= pos && pos < N);
    return cnt_w[i][pos] + cnt_b[j][pos];
  };
  std::vector<int> w(n), b(n);
  for (int i = 0; i < N; i++) {
    (c[i] == 'W' ? w : b).at(a[i]) = i;
  }
  constexpr int inf = 1 << 30;
  std::vector<std::vector<int>> dp(n + 1, std::vector<int>(n + 1, inf));
  dp[0][0] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (i < n) cmn(dp[i + 1][j], dp[i][j] + count(i, j, w[i]));
      if (j < n) cmn(dp[i][j + 1], dp[i][j] + count(i, j, b[j]));
    }
  }
  std::cout << dp[n][n] << std::endl;
  return 0;
}