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
N = int(input())
L = []
C = []
A = []
b = w = 0
b_d = {}
w_d = {}
for i in range(2*N):
    c, a = input().split()
    a = int(a)
    n = 0
    for cc, aa in zip(C, A):
        if cc == c and aa > a:
            n += 1
    L.append([0] * (N + 1))
    cnt = 0
    for j in range(N, -1, -1):
        L[-1][j] = cnt+n
        if j in (b_d if c=="W" else w_d):
            cnt+=1

    if c=="B":
        b_d[a] = i
        b+=1
    else:
        w_d[a] = i
        w+=1
    C.append(c)
    A.append(a)
#print(L, b_d, w_d)

dp = [[0]*(N+2) for i in range(N+2)]
for i in range(N+1):
    for j in range(N+1):
        if i==j==0:
            continue
        if i==0:
            dp[i][j] = dp[i][j-1] + L[b_d[j]][i]
        elif j==0:
            dp[i][j] = dp[i-1][j] + L[w_d[i]][j]
        else:
            dp[i][j] = min(dp[i-1][j] + L[w_d[i]][j],
                           dp[i][j-1] + L[b_d[j]][i])
print(dp[-2][-2])
