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<bits/stdc++.h>
using namespace std;
using ll = long long;
template<class T> using vt = vector<T>;
template<class T> using vvt = vector<vt<T>>;
template<class T> using ttt = tuple<T,T>;
using tii = tuple<int,int>;
using vi = vector<int>;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define pb push_back
#define mt make_tuple
#define ALL(a) (a).begin(),(a).end()
#define FST first
#define SEC second
#define DEB cerr<<"!"<<endl
#define SHOW(a,b) cerr<<(a)<<" "<<(b)<<endl
#define DIV int(1e9+7)
const int INF = (INT_MAX/2);
const ll LLINF = (LLONG_MAX/2);
const double eps = 1e-8;
//const double PI = M_PI;  
inline ll pow(ll x,ll n,ll m){ll r=1;while(n>0){if((n&1)==1)r=r*x%m;x=x*x%m;n>>=1;}return r%m;}
inline ll lcm(ll d1, ll d2){return d1 / __gcd(d1, d2) * d2;}

/*Coding Space*/
template<class T>
struct FT{
    vector<T> d;
    FT(size_t n):d(n+1){}
    T sum(size_t i){T s{};for(++i;i>0;i-=i&-i)s+=d[i];return s;}
    T sum(size_t i,size_t j){return i?sum(j)-sum(i-1):sum(j);}
    void add(size_t i,T x){for(++i;i<d.size();i+=i&-i)d[i]+=x;}
};
int dp[2020][2020] = {};
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n; cin >> n;
  map<tii, int> m;
  rep(i,2*n){
    char c; int a; cin >> c >> a;
    m[tii{c,a}] = i;
  }
  fill_n(dp[0], 2020*2020, INF);
  dp[0][0] = 0;
  FT<int> memo(2*n);
  rep(i,n+1){rep(j,n+1){ // i : B <-> j : W
    int db = m[tii{'B', i+1}] + memo.sum( m[tii{'B', i+1}], 2*n) - (i+j);
    if(db < 0) db = 0;
    if(i != n)
      dp[i+1][j] = min(dp[i+1][j], dp[i][j] + db);
    int dw = m[tii{'W', j+1}] + memo.sum( m[tii{'W', j+1}], 2*n) - (i+j);
    if(dw < 0) dw = 0;
    if(j != n)
      dp[i][j+1] = min(dp[i][j+1], dp[i][j] + dw);
    memo.add(m[tii{'W',j+1}],1);
    //cerr << i << " " << j << " " << dp[i][j] << " " << memo.sum( m[tii{'B', i+1}], 2*n) << endl;
  }
  rep(j,n+1)  memo.add(m[tii{'W', j+1}], -1);
  memo.add(m[tii{'B',i+1}],1);
  }
  cout << dp[n][n] << endl;
}