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 <bits/stdc++.h>

#define mp make_pair
#define pb push_back

using namespace std;

typedef int32_t i32;
typedef uint32_t u32;
typedef int64_t i64;
typedef uint64_t u64;
typedef pair<int,int> pii;
typedef vector<int> vi;

const int MAX_N = 2000;
int n;
int x[2*MAX_N];
char col[2*MAX_N];
int bc[2*MAX_N][MAX_N+1];
int wc[2*MAX_N][MAX_N+1];
int dp[MAX_N+1][MAX_N+1];
int rb[MAX_N+1];
int rw[MAX_N+1];

int main()
{
  cin >> n;
  for (int i = 0; i < 2*n; ++i) {
    cin >> col[i] >> x[i];
    if (col[i] == 'B') {
      rb[x[i]] = i;
    } else {
      rw[x[i]] = i;
    }
  }

  for (int i = 0; i <= n; i++) {
    int w = 0, b = 0;
    for (int j = 0; j < 2*n; j++) {
      if (x[j] > i) {
        if (col[j] == 'B') b++;
        else w++;
      }
      bc[j][i] = b;
      wc[j][i] = w;
    }
  }

  dp[0][0] = 0;
  for (int j = 1; j <= n; j++)
    dp[0][j] = dp[0][j-1] + wc[rw[j]][j] + bc[rw[j]][0];
  for (int i = 1; i <= n; i++)
    dp[i][0] = dp[i-1][0] + bc[rb[i]][i] + wc[rb[i]][0];

  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = min(dp[i-1][j] + (bc[rb[i]][i] + wc[rb[i]][j]),
                     dp[i][j-1] + (wc[rw[j]][j] + bc[rw[j]][i]));
    }
  }

  cout << dp[n][n] << endl;

  return 0;
}
