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;
typedef long long ll;
typedef pair<char,int> P;
P a[4005];
bool b[2005],w[2005],B[2005],W[2005];
int n;
ll ans;
int main(void){
	cin>>n;
	for(int i=0;i<2004;i++){
		b[i]=false; w[i]=false; B[i]=false; W[i]=false; 
	}
	for(int i=0;i<2*n;i++){
		cin>>a[i].first>>a[i].second;
		bool aaa=true;
		if(a[i].first=='B'){
			B[a[i].second]=true;
			for(int j=1;j<a[i].second;j++){
				if(B[j]){
					
				}else{
					aaa=false;
				}
			}
			if(aaa){
				b[a[i].second]=true;
			}
		}else{
			W[a[i].second]=true;
			for(int j=1;j<a[i].second;j++){
				if(W[j]){
					
				}else{
					aaa=false;
				}
			}
			if(aaa){
				w[a[i].second]=true;
			}
		}
	}
	for(int i=2*n-1;i>=0;i--){
		if(a[i].first=='B'){
			if(b[a[i].second]){
				
			}else{
				int k=i;
				for(int j=i;j<2*n;j++){
					if(a[i].first==a[j].first&&a[i].second>a[j].second){
						k=j;
					}
				}
				//cout<<k<<endl;
				for(int j=i;j<k;j++){
					//cout<<j<<endl;
					swap(a[j],a[j+1]);
					ans++;
				}
			}
		}else{
			if(w[a[i].second]){
				
			}else{
				int k=i;
				for(int j=i;j<2*n;j++){
					if(a[i].first==a[j].first&&a[i].second>a[j].second){
						k=j;
					}
				}
				//cout<<k<<endl;
				for(int j=i;j<k;j++){
					//cout<<j<<endl;
					swap(a[j],a[j+1]);
					ans++;
				}
			}
		}
	}
	cout<<ans<<endl;
	
	
}