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: Runtime Error)

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MAXN 2010
int n, a[MAXN << 1], c[MAXN << 1], f[MAXN][MAXN], p[2][MAXN];
int g[2][MAXN][MAXN], h[2][MAXN][MAXN];
char buf[3];
inline void up(int &x, int y) {
	if (y < x) x = y;
}
int main() {
	scanf("%d", &n);
	n <<= 1;
	for (int i = 1; i <= n; i++) {
		scanf("%s%d", buf, a + i);
		p[c[i] = buf[0] == 'B'][a[i]] = i;
	}
	for (int j = 0; j < 2; j++) {
		for (int i = 1; i <= n; i++) {
			for (int k = 1; k <= n; k++) {
				g[j][i][k] = g[j][i - 1][k] + (a[i] == k && c[i] == j);
			}
		}
	}
	for (int j = 0; j < 2; j++) {
		for (int i = 1; i <= n; i++) {
			for (int k = n; k >= 1; k--) {
				g[j][i][k] += g[j][i][k + 1];
			}
		}
	}
	for (int j = 0; j < 2; j++) {
		for (int i = n; i >= 1; i--) {
			for (int k = 1; k <= n; k++) {
				h[j][i][k] = h[j][i][k - 1];
			}
		}
	}
	memset(f, 0x3F, sizeof(f));
	f[0][0] = 0;
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) {
			for (int k = 0; k < 2; k++) {
				int pos = p[k][(k ? j : i) + 1];
				int cost = 0;
				cost += g[0][pos - 1][i + 1];
				cost += g[1][pos - 1][j + 1];
				cost += h[0][pos + 1][i - 1];
				cost += h[1][pos + 1][j - 1];
				up(f[i + !k][j + k], f[i][j] + cost);
			}
		}
	}
	printf("%d\n", f[n][n]);
	return 0;
}