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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

template<class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
using namespace std;
/***********************************************/
/* Dear online judge:
 * I've read the problem, and tried to solve it.
 * Even if you don't accept my solution, you should respect my effort.
 * I hope my code compiles and gets accepted.
 *  ___  __     _______    _______ 
 * |\  \|\  \  |\  ___ \  |\  ___ \
 * \ \  \/  /|_\ \   __/| \ \   __/|
 *  \ \   ___  \\ \  \_|/__\ \  \_|/__  
 *   \ \  \\ \  \\ \  \_|\ \\ \  \_|\ \ 
 *    \ \__\\ \__\\ \_______\\ \_______\
 *     \|__| \|__| \|_______| \|_______|
 */
const long long mod = 1000000007;

const int mxN = 2010;
const int inf = 10000010;
int BIT[mxN];
void upd(int ind, int val) {
	while (ind < mxN) {
		BIT[ind] += val;
		ind |= ind + 1;
	}
}

int get(int ind) {
	int res = 0;
	while (ind >= 0) {
		res += BIT[ind];
		ind = (ind & (ind + 1)) - 1;
	}
	return res;
}

int W[mxN],B[mxN];
int dp[mxN][mxN];
int main(int argc, char** argv) {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	int N,x;
	string s;
	cin >> N;
	for(int i = 0;i < (N<<1);i++) {
		cin>>s>>x;
		x--;
		if(s[0] == 'W') {
			W[x] = i;
		} else {
			B[x] = i;
		}
	}
	dp[N][N] = 0;
	for(int i = N;i >= 0;i--) {
		if(i < N) upd(B[i],1);
		for(int j = N - (i == N);j >= 0;j--) {
			dp[i][j] = inf;
			if(j < N) {
				upd(W[j],1);
				dp[i][j] = min(dp[i][j],dp[i][j+1] + get(W[j]-1));
			}
			if(i < N) {
				dp[i][j] = min(dp[i][j],dp[i+1][j] + get(B[i]-1));
			}
//			cerr<<dp[i][j]<<" \n"[j == 0];
		}
		for(int j = 0;j < N;j++) upd(W[j],-1);
	}
	cout<<dp[0][0]<<'\n';
	return 0;
}