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.

Examples

Input

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


Output

4


Input

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


Output

18


Input

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


Output

41
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <functional>

using namespace std;

int sw[5000][5000];
int sb[5000][5000];
int dp[5000][5000];

int main() {
  int n;
  cin >> n;

  vector<bool> color(n * 2);
  vector<int> a(n * 2);
  vector<int> pw(n);
  vector<int> pb(n);
  for (int i = 0; i < n * 2; i++) {
    char c;
    cin >> c >> a[i];
    a[i]--;
    color[i] = c == 'B';
    if (color[i] == false) {
      pw[a[i]] = i;
    } else {
      pb[a[i]] = i;
    }
  }

  for (int i = 0; i < n * 2; i++) {
    for (int j = 0; j < n * 2; j++) {
      sw[i + 1][j] = sw[i][j];
      sb[i + 1][j] = sb[i][j];
      if (color[i] == false && a[i] >= j) {
        sw[i + 1][j]++;
      }
      if (color[i] == true && a[i] >= j) {
        sb[i + 1][j]++;
      }
    }
  }

  fill_n(*dp, 5000 * 5000, 1e9);
  dp[0][0] = 0;

  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (i + 1 <= n) {
        int k = pw[i];
        dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + sw[k][i] + sb[k][j]); 
      }
      if (j + 1 <= n) {
        int k = pb[j];
        dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + sw[k][i] + sb[k][j]);
      }
    }
  }
  cout << dp[n][n] << endl;
}