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: Accepted)

#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>
#include <math.h>

using namespace std;

using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

template<int Size>
class FenwickTree
{
public:

	i64 sum(int n)
	{
		i64 ans = 0;

		while (n > 0)
		{
			ans += data[n];
			n -= (n & -n);
		}

		return ans;
	}

	void add(int n, i64 num)
	{
		while (n < Size)
		{
			data[n] += num;
			n += (n & -n);
		}
	}

	int search(i64 k)
	{
		int lo = 0, hi = Size - 1;
		int ans = 0;

		while (lo <= hi)
		{
			int mid = (lo + hi) / 2;

			i64 q = sum(mid);

			if (q >= k)
			{
				ans = mid;
				hi = mid - 1;
			}
			else
			{
				lo = mid + 1;
			}
		}

		return ans;
	}

private:
	int data[Size];
};

int w[2005], b[2005];
int nextw[2005], nextb[2005];
int table[2005][2005];
int widx[2005], bidx[2005];
FenwickTree<2005> wtree[2005];
FenwickTree<2005> btree[2005];

int solve(int endw, int endb)
{
	if (endw == 0 && endb == 0)
		return 0;

	if (table[endw][endb] != -1)
		return table[endw][endb];

	int& res = table[endw][endb];
	res = 987654321;

	if (endb != 0)
	{
		int bi = bidx[endb];
		int bd = btree[bi].sum(endb - 1);
		int wi = nextw[bi];
		int wd = wtree[wi].sum(endw);

		res = min(res, solve(endw, endb - 1) + bd + wd);
	}

	if (endw != 0)
	{
		int wi = widx[endw];
		int wd = wtree[wi].sum(endw - 1);
		int bi = nextb[wi];
		int bd = btree[bi].sum(endb);

		res = min(res, solve(endw - 1, endb) + bd + wd);
	}

	return res;
}

int main()
{
	int n;
	scanf("%d", &n);

	int wi = 1, bi = 1;

	for (int i = 1; i <= n; i++)
		nextw[i] = nextb[i] = n + 1;

	for (int i = 0; i < 2 * n; i++)
	{
		char c[2];
		int k;

		scanf("%s %d", c, &k);

		if (c[0] == 'W')
		{
			w[wi] = k;
			widx[k] = wi;

			for (int j = 1; j < bi; j++)
				if (nextw[j] == n + 1)
					nextw[j] = wi;

			wi++;
		}
		else
		{
			b[bi] = k;
			bidx[k] = bi;

			for (int j = 1; j < wi; j++)
				if (nextb[j] == n + 1)
					nextb[j] = bi;

			bi++;
		}
	}

	for (int i = 1; i <= n; i++)
	{
		for (int j = i; j <= n; j++)
		{
			wtree[i].add(w[j], 1);
			btree[i].add(b[j], 1);
		}
	}

	memset(table, -1, sizeof(table));

	printf("%d\n", solve(n, n));

	return 0;
}