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>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <iostream>
#include <bitset>
#include <cassert>
#include <queue>
#include <random>
#include <stack>
#include <iomanip>
#include <math.h>
using namespace std;
#define rep(i, n) for (ll i = 0; i < (ll)n; i++)
#define repf(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define repr(i, a, b) for (ll i = (ll)a; i > (ll)b; i--)
#define repv(x, arr) for (auto &x : arr)
#define all(v) (v).begin(), (v).end()
#define vec(name, num) vector<ll> name((num), 0);
#define mp(a, b) make_pair(a, b)
#define pb(x) push_back(x)
#define F first
#define S second
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vii;
typedef vector<vii> vvii;
const ll mod = 1e9 + 7;
const int infi = 2147483600;
const ll infl = 1e17;

int main()
{
    int n;
    cin>>n;
    ll b = 0, w = 0;
    vll ball(2 * n);
    map<ll,ll> idx;
    rep(i,2*n){
        char ci;
        cin >> ci;
        ll ai;
        cin >> ai;
        ai--;
        if (ci=='B'){
            ball[i] = ai;
            idx[ai] = i;
        }
        else{
            ball[i] = ai + n;
            idx[ai + n] = i;
        }
    }
    vvll sb(2 * n + 1, vll(n + 1, 0));
    vvll sw(2 * n + 1, vll(n + 1, 0));
    vll now_b(n, 0), now_w(n, 0);
    repr(i,2*n-1,-1){
        rep(j,n){
            sb[i][j + 1] = sb[i][j] + now_b[j];
            sw[i][j + 1] = sw[i][j] + now_w[j];
        }
        if (ball[i]<n){
            now_b[ball[i]]++;
        }
        else{
            now_w[ball[i] - n]++;
        }
    }
    vvll dp(2 * n + 1, vll(n + 1, infl));
    dp[0][0] = 0;
    rep(i,2*n){
        rep(j,n+1){
            if (i-j<0)
                continue;
            if (j<n){
                dp[i + 1][j + 1] 
                = min(dp[i + 1][j + 1], dp[i][j] + sb[idx[j]][j] + sw[idx[j]][i - j]);
            }
            if (i-j<n){
                dp[i + 1][j] 
                = min(dp[i + 1][j], dp[i][j] + sw[idx[n + i - j]][i - j] + sb[idx[n + i - j]][j]);
            }
        }
    }
    cout << dp[2 * n][n] << endl;
}