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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

void initBIT(int *b){
  const int n=b[0];
  for(int i=1;i<=n;i++) b[i]=0;
}

void add(int *bit,int x,int v){
  const int n=bit[0];
  for(int i=x;i<=n;i+=i&-i) bit[i]+=v;
}

int sum(int *bit,int x){
  int res=0;
  for(int i=x;i>0;i-=i&-i) res+=bit[i];
  return res;
}

void run(void){
  int n;
  scanf("%d",&n);
  int *b=(int *)calloc(n+1,sizeof(int));
  int *w=(int *)calloc(n+1,sizeof(int));
  int i,j;
  for(i=0;i<2*n;i++){
    char s[10];
    int k;
    scanf("%s%d",s,&k);
    if(s[0]=='B'){
      b[k]=i+1;
    } else {
      w[k]=i+1;
    }
  }
  int *dp=(int *)calloc(2*(n+1),sizeof(int));
  int *bit=(int *)calloc(2*n+1,sizeof(int));
  bit[0]=2*n;
  dp[0]=0;
  for(i=1;i<=n;i++){
    add(bit,w[i],1);
    dp[i]=dp[i-1]+(i-sum(bit,w[i]));
  }
  int now=0;
  for(i=1;i<=n;i++){
    initBIT(bit);
    int inv=0;
    for(j=1;j<=i;j++){
      add(bit,b[j],1);
      inv+=j-sum(bit,b[j]);
    }
    int next=now^(n+1);
    dp[next]=inv;
    for(j=1;j<=n;j++){
      add(bit,w[j],1);
      int p=dp[next+j-1]+(i+j-sum(bit,w[j]));
      int q=dp[now+j]+(i+j-sum(bit,b[i]));
      dp[next+j]=p<q?p:q;
    }
    now=next;
  }
  printf("%d\n",dp[now+n]);
  return;
}

int main(void){
  run();
  return 0;
}