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
import sys
input = sys.stdin.readline

N = int(input())
c = [0] * (2 * N)
a = [0] * (2 * N)
for i in range(2 * N):
    c[i], a[i] = input().split()
    a[i] = int(a[i]) - 1

idx = [[0] * 2 for _ in range(N)]
for i in range(2 * N):
    idx[a[i]][c[i] == 'W'] = i

costB = [[0] * (N + 1) for _ in range(N)]
for i in range(N):
    j = idx[i][0]
    cnt = 0
    for k in range(i):
        if j < idx[k][0]:
            cnt += 1
    costB[i][0] = j + cnt - i
    cnt2 = 0
    for k in range(1, N + 1):
        if j < idx[k - 1][1]:
            cnt2 += 1
        costB[i][k] = j + cnt + cnt2 - (i + k)

costW = [[0] * (N + 1) for _ in range(N)]
for i in range(N):
    j = idx[i][1]
    cnt = 0
    for k in range(i):
        if j < idx[k][1]:
            cnt += 1
    costW[i][0] = j + cnt - i
    cnt2 = 0
    for k in range(1, N + 1):
        if j < idx[k - 1][0]:
            cnt2 += 1
        costW[i][k] = j + cnt + cnt2 - (i + k)

dp = [[10**18] * (N + 1) for _ in range(2 * N + 1)]
dp[0][0] = 0
for i in range(1, 2 * N + 1):
    for j in range(max(0, i - N), min(i, N) + 1):
        dp[i][j] = dp[i-1][j] + costW[i-j-1][j]
        if j > 0:
            dp[i][j] = min(dp[i][j], dp[i-1][j-1] + costB[j-1][i-j])
print(dp[2*N][N])