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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
int N;
char C[4040];
int A[4040];
int dp[2020][2020]; // 白1~i、黒1~jまで目標を達成するのに必要な最小のスワップ回数
int pos[2][2020]; // W==1,B==0の数jの位置

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin>>N;
  if(N>500) return 0;
  FOR(i,0,2*N){
    cin>>C[i]>>A[i];
    pos[C[i]=='W'][A[i]] = i;
  }
  FOR(i,0,2020) FOR(j,0,2020)dp[i][j] = 1e9;
  dp[0][0] = 0;
  FOR(i,0,N+1){
    FOR(j,0,N+1){
      // 白のi+1を持ってくる
      int cross = 0;
      FOR(k,0,pos[1][i+1]){
        if(C[k]=='W' && A[k] > i)cross++;
        if(C[k]=='B' && A[k] > j)cross++;
      }
      dp[i+1][j] = min(dp[i+1][j], dp[i][j] + cross);
      // 黒のj+1を持ってくる
      cross = 0;
      FOR(k,0,pos[0][j+1]){
        if(C[k]=='W' && A[k] > i)cross++;
        if(C[k]=='B' && A[k] > j)cross++;
      }
      dp[i][j+1] = min(dp[i][j+1], dp[i][j] + cross);
    }
  }
  cout<<dp[N][N]<<endl;
}