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;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i, x) for(int i=0;i<(int)(x);i++)
#define REPS(i, x) for(int i=1;i<=(int)(x);i++)
#define RREP(i, x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i, x) for(int i=((int)(x));i>0;i--)
#define FOR(i, c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i, c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a, b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define __builtin_popcount __builtin_popcountll

template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T>
ostream &operator<<(ostream &os, const vector<T> &t) {
    os << "[";
    FOR(it, t) {
        if (it != t.begin()) os << ",";
        os << *it;
    }
    os << "]";
    return os;
}

template<class T>
ostream &operator<<(ostream &os, const set<T> &t) {
    os << "{";
    FOR(it, t) {
        if (it != t.begin()) os << ",";
        os << *it;
    }
    os << "}";
    return os;
}

template<class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &t) { return os << "(" << t.first << "," << t.second << ")"; }

template<class S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first + t.first, s.second + t.second);
}

template<class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first - t.first, s.second - t.second);
}

const int INF = 1 << 28;
const double EPS = 1e-8;
const int MOD = 1000000007;

int n, m;

int main(int argc, char *argv[]) {
    cin >> n;
    vi w(n), b(n);
    vector<vi> wl(2 * n, vi(n + 1)), bl(2 * n, vi(n + 1));
    REP(i, 2 * n) {
        char c;
        int x;
        cin >> c >> x;
        x--;
        if (c == 'W') w[x] = i;
        else b[x] = i;
        REP(j, i + 1) {
            (c == 'W' ? wl : bl)[j][x + 1] ++;
        }
    }
    REP(i, 2 * n) {
        REP(j, n) {
            wl[i][j + 1] += wl[i][j];
            bl[i][j + 1] += bl[i][j];
        }
    }
    vector<vi> dp(n + 1, vi(n + 1, MOD));
    dp[0][0] = 0;

    REP(i, n + 1) {
        REP(j, n + 1) {
            if (i < n) {
                chmin(dp[i+1][j], dp[i][j] + wl[w[i]][i] + bl[w[i]][j]);
            }
            if (j < n) {
                chmin(dp[i][j+1], dp[i][j] + wl[b[j]][i] + bl[b[j]][j]);
            }
        }
    }
    cout << dp[n][n] << endl;
    return 0;
}