Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

4

The objective can be achieved in four operations, for example, as follows:

Swap the black 3 and white 1.

Swap the white 1 and white 2.

Swap the black 3 and white 3.

Swap the black 3 and black 2.

Sample Input 2

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

Sample Output 2

18

Sample Input 3

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

Sample Output 3

41

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

require 'pp'
# IS_DEBUG = true
IS_DEBUG = false
def dputs str
	if(IS_DEBUG)
		puts str
	end
end
def dpp str
	if(IS_DEBUG)
		pp str
	end
end

require 'pp'

N,d=gets.chomp.split(' ').map{|n| n.to_i}
A = Array.new(2*N){Array.new(2)}
W = Array.new(N)
B = Array.new(N)
# all 0 indexed
0.upto(2*N-1){|d|
	tmp=gets.chomp.split(' ')
	A[d]=[tmp[0],tmp[1].to_i-1]
	if(tmp[0]=="W")
		W[tmp[1].to_i-1] = d
	else
		B[tmp[1].to_i-1] = d
	end
}

# cost_w[i][j] : already i whites, j blacks. cost (inversion number) of i+1 white
cost_w = Array.new(N+1){Array.new(N+1)}
cost_b = Array.new(N+1){Array.new(N+1)}
(N-1).downto(0){|n|
	# wがn-1個, bがN個
	wfi = W[n]
	cost_w[n][N] = A.each_with_index.count{|tmp,wi|
		# パターンマッチしてくれない。|c,m,wi|だとダメ
		c = tmp[0]
		m = tmp[1]
		# puts "c,m,wi=#{c},#{m},#{wi}"
		# puts " n,fwi=  #{n},#{wfi}"
		c=="W" && wfi > wi && n < m
	}
	(N-1).downto(0){|m|
		cost_w[n][m] = cost_w[n][m+1] + (B[m] < W[n] ? 1 : 0)
	}
}
(N-1).downto(0){|n|
	bfi = B[n]
	cost_b[N][n] = A.each_with_index.count{|tmp,bi|
		# パターンマッチしてくれない。|c,m,wi|だとダメ
		c = tmp[0]
		m = tmp[1]
		# puts "c,m,wi=#{c},#{m},#{wi}"
		# puts " n,fwi=  #{n},#{wfi}"
		c=="B" && bfi > bi && n < m
	}
	(N-1).downto(0){|m|
		cost_b[m][n] = cost_b[m+1][n] + (W[m] < B[n] ? 1 : 0)
	}
}
dpp cost_w
dpp cost_b

dp = Array.new(N+1){Array.new(N+1)}
dp[0][0] = 0
1.upto(N){|i|
	dp[i][0] = dp[i-1][0] + cost_w[i-1][0]
	dp[0][i] = dp[0][i-1] + cost_b[0][i-1]
}
1.upto(N){|i|
	1.upto(N){|j|
		dp[i][j] = [dp[i-1][j] + cost_w[i-1][j],
					dp[i][j-1] + cost_b[i][j-1]].min
	}
}
puts dp[N][N]