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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <queue>
#include <stack>
#include <cstdlib>
#include <map>
#include <iomanip>
#include <set>
#include <stdio.h>
#include <ctype.h>
#include <random>
#include <string.h>
#include <cmath>
#include <unordered_map>
#include <cstdio>
using namespace std;
#define pie 3.141592653589793238462643383279
#define int long long
#define P pair<int,int>
#define all(vec) vec.begin(),vec.end()

int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
int lcm(int x, int y) {
	return x / gcd(x, y)*y;
}
int kai(int x, int y, int m) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= i; res %= m;
	}
	return res;
}
int mod_pow(int x, int y, int m) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}

int comb(int x, int y, int m) {
	if (y > x)return 0;
	return kai(x, y, m) * mod_pow(kai(y, y, m), m - 2, m) % m;
}

int n;
P x[4010];
int a[2010], b[2010];
int cnt1[2010][2010], cnt2[2010][2010];
bool bl1[2010], bl2[2010];
int d[4010][2010];

signed main() {
	cin >> n;
	for (int i = 1; i <= 2 * n; i++) {
		char c; int p; cin >> c >> p;
		if (c == 'W')a[p] = i, x[i] = make_pair(p, 0);
		else b[p] = i, x[i] = make_pair(p, 1);
	}
	for (int i = 2 * n; i >= 1; i--) {
		if (x[i].second == 0) {
			int cnt = 0;
			for (int j = 1; j < x[i].first; j++)if (bl1[j])cnt++;
			for (int j = 0; j <= n; j++) {
				if (bl2[j])cnt++;
				cnt1[x[i].first][j] = cnt;
			}
			bl1[x[i].first] = true;
		}
		else {
			int cnt = 0;
			for (int j = 1; j < x[i].first; j++)if (bl2[j])cnt++;
			for (int j = 0; j <= n; j++) {
				if (bl1[j])cnt++;
				cnt2[x[i].first][j] = cnt;
			}
			bl2[x[i].first] = true;
		}
	}
	for (int i = 0; i <= 2 * n; i++)for (int j = 0; j <= n; j++)d[i][j] = 1000000000;
	d[0][0] = 0;
	for (int i = 1; i <= 2 * n; i++) {
		for (int j = 0; j <= min(i - 1, n); j++)d[i][j] = min(d[i][j], d[i - 1][j] + cnt2[i - j][j]);
		for (int j = 1; j <= min(i, n); j++)d[i][j] = min(d[i][j], d[i - 1][j - 1] + cnt1[j][i - j]);
	}
	cout << d[2 * n][n] << endl;
}