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())
MB = {}; MW = {}
for i in range(2*N):
    c, a = input().split()
    c = "BW".index(c); a = int(a)
    if c:
        MW[a-1] = i
    else:
        MB[a-1] = i

P = [[0]*(N+1) for i in range(N+1)]
Q = [[0]*(N+1) for i in range(N+1)]

for p in range(N):
    cnt = MW[p] - sum(MW[p0] < MW[p] for p0 in range(p))

    P[p][0] = cnt
    for q in range(N):
        if MB[q] < MW[p]:
            cnt -= 1
        P[p][q+1] = cnt

for q in range(N):
    cnt = MB[q] - sum(MB[q0] < MB[q] for q0 in range(q))

    Q[q][0] = cnt
    for p in range(N):
        if MW[p] < MB[q]:
            cnt -= 1
        Q[q][p+1] = cnt

dp = [[10**9]*(N+1) for i in range(N+1)]
dp[0][0] = 0
for i in range(N):
    dp[i+1][0] = dp[i][0] + P[i][0]
for i in range(N):
    dp[0][i+1] = dp[0][i] + Q[i][0]
for i in range(N):
    for j in range(N):
        dp[i+1][j+1] = min(dp[i][j+1] + P[i][j+1], dp[i+1][j] + Q[j][i+1])
print(dp[N][N])