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>
#define F first
#define S second
#define LL long long
#define PB push_back
#define MP make_pair
#define I insert
 
const LL N = 2e3+69, mod = 1000000007;
 
using namespace std;
 
LL n, m, i, j, k, l, r, t, bl, wh, ans;
LL dp[N][N], xe[N*2], a[2][N];
char c;
 
LL get ( LL x ){
	LL sum = 0;
	while ( x > 0 ){
		sum += xe[x];
		x -= (x & -x);
	}
	return sum;
}
 
void upd ( LL x, LL val ){
	if ( !x ) return;
	while ( x < 2*N ){
		//if ( val == -1 ) cout << x << endl;
		xe[x] += val;
		x += (x & -x);
	}
}
 
int main(){
	// B - 66;  W - 87
	// B - 0    W - 1
	cin >> n;
	for ( i = 1; i <= 2 * n; ++i ){
		cin >> c >> k;
		a[c%2][k] = i;
	}
	
	for ( i = 0; i < N; ++i ){
		for ( j = 0; j < N; ++j ){
			dp[i][j] = N*N;
		}
	}
	
	dp[0][0] = 0;
	
	for ( bl = 0; bl <= n; bl++ ){
		for ( i = 0; i < 2*N; ++i ) xe[i] = 0;
		for ( i = 1; i <= n; ++i ){
			upd(a[1][i],1);
		}
		for ( i = bl + 1; i <= n; ++i ){
			upd(a[0][i],1);
		}
		for ( wh = 0; wh <= n; ++wh ){
			if ( wh ) upd(a[1][wh],-1);
			if ( bl < n ){
				t = get ( a[0][bl+1] - 1 );
				dp[bl+1][wh] = min ( dp[bl+1][wh], dp[bl][wh] + t );
			}
			if ( wh < n ){
				t = get ( a[1][wh+1] - 1 );
				dp[bl][wh+1] = min ( dp[bl][wh+1], dp[bl][wh] + t );
			}
		}
	}/*/
	for ( i = 0; i <= n; ++i ){
		for ( j = 0; j <= n; ++j ){
			cout << dp[i][j] << " ";
		}
		cout << endl;
	}/**/
	cout << dp[n][n] << endl;
	/*/ans = dp[n][n];
	for ( i = 1; i <= n; ++i ){
		for ( j = 1; j < i; ++j ){
			if ( a[0][i] < a[0][j] ) ans++;
			if ( a[1][i] < a[1][j] ) ans++;
		}
	}
	cout << ans << endl;/**/
	return 0;
}
 
/**/
//      IIIIIIIII      OOOOO             A          NN        N    EEEEEEEEEE
//          I         O     O           A A         N N       N    E
//          I        O       O         A   A        N  N      N    E
//          I        O       O        A     A       N   N     N    E
//          I        O       O       AAAAAAAAA      N    N    N    EEEEEEEE
//          I        O       O      A         A     N     N   N    E
//          I        O       O     A           A    N      N  N    E
//          I         O     O     A             A   N       N N    E
//      IIIIIIIII      OOOOO     A               A  N        NN    EEEEEEEEEE ___ KAPANADZE