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>
#define f first
#define s second
#define MOD 1000000007
#define PMOD 998244353
#define pb(x) push_back(x)
using namespace std;

typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> plii;
typedef pair<int, pii> piii;
const int INF = 1e9+10;
const ll LINF = 1LL*INF*INF;
const int MAXN = 2e5+10;
const int MAXM = 2e3+10;

priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;

int A[MAXN][2];
int dp[MAXM][MAXM];
int id[MAXM][2];

int main()
{
    int n,m,k,a,b,x,y,q;
    int sum = 0;
    int cnt = 0;
    int mx = 0;
    int mn = INF;
    int cur = 0, idx = -1;
    int tc;
    char c;

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin>>n;

    for(int i=0;i<2*n;i++)
    {
        cin>>c>>x;
        if(c=='W'){
            dp[x][0] = i-cnt;
            A[cnt++][0] = x;
            id[x][0] = i;
        }
        else  {
            A[i-cnt][1] = x;
            id[x][1] = i;
        }
    }

    int res = 0;

    for(int i=0;i<n;i++)
        for(int j=0;j<i;j++){
            if(A[i][0]<A[j][0])res++;
            if(A[i][1]<A[j][1])res++;
        }

    int inv;

    for(int i=1;i<=n;i++){
        mn = dp[i-1][0];
        inv = dp[i][0];
        dp[i][0]+=dp[i-1][0];
        for(int j=1;j<=n;j++){
            mn = min(mn,dp[i-1][j]);
            if(id[j][1]<id[i][0])inv--;
            else inv++;
            dp[i][j] = inv+mn;
        }
    }

    mn = INF;

    for(int i=0;i<=n;i++)
        mn = min(mn,dp[n][i]);

    res+=mn;

    cout<<res<<"\n";

    return 0;
}