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 <memory.h>
#include <algorithm>
using namespace std;

#define CIDX(cchar)		((cchar == 'W') ? 0 : 1)

char org_c[4003];
int d[2002][2002], org_n[4003], idx[2][2003], cum[4003][2003][2];

int main()
{
	int N; scanf("%d", &N);
	for(int i = 1; i <= 2 * N; i++){
		scanf(" %c %d", &org_c[i], &org_n[i]);
		idx[CIDX(org_c[i])][org_n[i]] = i;
	}

	for(int i = 1; i <= 2 * N; i++)
		for(int j = 1; j <= N; j++){
			cum[i][j][0] += cum[i - 1][j][0] + cum[i][j - 1][0] - cum[i - 1][j - 1][0];
			cum[i][j][1] += cum[i - 1][j][1] + cum[i][j - 1][1] - cum[i - 1][j - 1][1];
			if(org_n[i] == j) cum[i][j][CIDX(org_c[i])]++;
		}

	memset(d, 0x3f, sizeof(d));
	d[0][0] = 0;
	for(int i = 1; i <= N; i++){
		d[i][0] = d[i - 1][0] + idx[0][i] - cum[idx[0][i]][i][0];
		d[0][i] = d[0][i - 1] + idx[1][i] - cum[idx[1][i]][i][1];
	}
	for(int i = 0; i <= N; i++)
		for(int j = 0; j <= N; j++){
			d[i + 1][j] = min(d[i + 1][j], d[i][j] + idx[0][i + 1] - cum[idx[0][i + 1]][i + 1][0] - cum[idx[0][i + 1]][j][1]);
			d[i][j + 1] = min(d[i][j + 1], d[i][j] + idx[1][j + 1] - cum[idx[1][j + 1]][i][0] - cum[idx[1][j + 1]][j + 1][1]);
		}

	printf("%d\n", d[N][N]);
	return 0;
}