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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;


// 0-origin BIT
template <typename T>
class BIT {
public:
	vector<T> data;
	int n;
	BIT() {}
	BIT(int n_) {
		n = n_;
		data.assign(n, T(0));
	}
	// v[a] += w
	void add(int a, T w) {
		for (int x = a; x < n; x |= x + 1) data[x] += w;
	}
	// sum of v[i] for i in [0, a)
	T sum(int a) {
		T ret = 0;
		for (int x = a - 1; x >= 0; x = (x & (x + 1)) - 1) {
			ret += data[x];
		}
		return ret;
	}
};


int main() {
	int N;
	cin >> N;

	map<pair<char, int>, int> ca_to_pos;
	for(int i=0;i<2*N;i++) {
		char c;
		int a;
		cin >> c >> a;
		a--;
		ca_to_pos[make_pair(c, a)] = i;
	}

	int Wi = 0, Bi = 0;
	vector<int> dest_pos(2 * N);
	for(int i=0;i<2*N;i++) {
		if(Wi == N || (Bi < N && ca_to_pos[make_pair('B', Bi)] < ca_to_pos[make_pair('W', Wi)])) {
			dest_pos[ca_to_pos[make_pair('B', Bi++)]] = i;
		}else {
			dest_pos[ca_to_pos[make_pair('W', Wi++)]] = i;
		}
	}

	BIT<int> bit(2 * N);
	ll ans = 0;
	for(int i=0;i<2*N;i++) {
		ans += i - bit.sum(dest_pos[i]);
		bit.add(dest_pos[i], 1);
	}
	cout << ans << endl;

	return 0;
}