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 ;

int n ;

pair < int , int > a[2010] ;

long long dp[2010][2010];
int foo[2010][2010][2][2] ;

int main(){
	#define NAME "arc097c"
	//freopen( NAME".inp" , "r" , stdin ) ; freopen( NAME".out" , "w" , stdout ) ;

	ios_base :: sync_with_stdio(false) ;

	cin >> n ;

	for (int i = 1; i <= 2 * n; ++ i){
		char c ; int id, type ; cin >> c >> id ;
		if(c == 'B') type = 0 ;
		else type = 1 ;
		a[i] = make_pair(type, id) ;
	}

	for (int i = 1; i <= 2 * n; ++ i){
		for (int j = i + 1; j <= 2 * n; ++ j){
			foo[a[i].second][a[j].second][a[i].first][a[j].first] ++ ;
		}
	}

	for (int i = 0; i <= n; ++ i){
		for (int j = 0; j <= n; ++ j){
			for (int x = 0; x <= 1; ++ x){
				for (int y = 0; y <= 1; ++ y){
					foo[i][j][x][y] += foo[i][j - 1][x][y] ;
				}
			}
		}
	}

	memset(dp, 0x3f, sizeof(dp)) ;

	dp[0][0] = 0 ;

	for (int i = 0; i <= n; ++ i){
		for (int j = 0; j <= n; ++ j){
			if(i) dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1ll * foo[i][i][0][0] + 1ll * foo[i][j][0][1]) ;
			if(j) dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1ll * foo[j][i][1][0] + 1ll * foo[j][j][1][1]) ; 
		}
	}

	cout << dp[n][n] << endl ;

	return 0 ;
}