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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n)  for(long long i=0;i<(long long)(n);i++)
#define REP(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define all(a) a.begin(),a.end()
#define pb push_back
#define eb emplace_back
#define lb(v,k) (lower_bound(all(v),k)-v.begin())
#define ub(v,k) (upper_bound(all(v),k)-v.begin())
#define fi first
#define se second
#define pi M_PI
#define PQ(T) priority_queue<T>
#define SPQ(T) priority_queue<T,vector<T>,greater<T>>
#define dame(a) {out(a);return 0;}
#define decimal cout<<fixed<<setprecision(15);
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef tuple<ll,ll,ll,ll> PPP;
typedef multiset<ll> S;
using vi=vector<ll>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;
using vp=vector<P>;
using vvp=vector<vp>;
using vb=vector<bool>;
using vvb=vector<vb>;
const ll inf=1001001001001001001;
const int INF=1001001001;
const int mod=998244353;
const double eps=1e-10;
template<class T> bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<class T> bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
template<class T> void out(T a){cout<<a<<'\n';}
template<class T> void outp(T a){cout<<'('<<a.fi<<','<<a.se<<')'<<'\n';}
template<class T> void outvp(T v){rep(i,v.size())cout<<'('<<v[i].fi<<','<<v[i].se<<')';cout<<'\n';}
template<class T> void outvvp(T v){rep(i,v.size())outvp(v[i]);}
template<class T> void outv(T v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<'\n';}
template<class T> void outvv(T v){rep(i,v.size())outv(v[i]);}
template<class T> bool isin(T x,T l,T r){return (l)<=(x)&&(x)<=(r);}
template<class T> void yesno(T b){if(b)out("yes");else out("no");}
template<class T> void YesNo(T b){if(b)out("Yes");else out("No");}
template<class T> void YESNO(T b){if(b)out("YES");else out("NO");}
template<class T> void noyes(T b){if(b)out("no");else out("yes");}
template<class T> void NoYes(T b){if(b)out("No");else out("Yes");}
template<class T> void NOYES(T b){if(b)out("NO");else out("YES");}
void outs(ll a,ll b){if(a>=inf-100)out(b);else out(a);}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll modpow(ll a,ll b){a%=mod;if(b==0)return 1;if(b&1)return a*modpow(a,b-1)%mod;ll k=modpow(a,b/2);return k*k%mod;}
vi seg,seg2;
ll N;
void init(ll n){
    N=1;
    while(N<n)N*=2;
    seg=vi(N*2-1);
    rep(i,n)seg[i+N-1]=1;
    for(int i=N-2;i>=0;i--)seg[i]=seg[i*2+1]+seg[i*2+2];
}
void init2(ll n){
    seg2=vi(N*2-1);
    rep(i,n)seg2[i+N-1]=1;
    for(int i=N-2;i>=0;i--)seg2[i]=seg2[i*2+1]+seg2[i*2+2];
}
void update(ll i,ll x){
    i=N+i-1;
    seg[i]=x;
    while(i>0){
        i=(i-1)/2;
        seg[i]=seg[i*2+1]+seg[i*2+2];
    }
}
ll getsum(ll a,ll b,ll k,ll l,ll r){
    if(r<=a||b<=l)return 0;
    if(a<=l&&r<=b)return seg[k];
    ll c1=getsum(a,b,k*2+1,l,(l+r)/2);
    ll c2=getsum(a,b,k*2+2,(l+r)/2,r);
    return c1+c2;
}
void update2(ll i,ll x){
    i=N+i-1;
    seg2[i]=x;
    while(i>0){
        i=(i-1)/2;
        seg2[i]=seg2[i*2+1]+seg2[i*2+2];
    }
}
ll getsum2(ll a,ll b,ll k,ll l,ll r){
    if(r<=a||b<=l)return 0;
    if(a<=l&&r<=b)return seg2[k];
    ll c1=getsum2(a,b,k*2+1,l,(l+r)/2);
    ll c2=getsum2(a,b,k*2+2,(l+r)/2,r);
    return c1+c2;
}
int main(){
    ll n;cin>>n;
    vi a(n),b(n);
    rep(i,n*2){
        char c;ll d;cin>>c>>d;
        if(c=='B')a[d-1]=i;
        else b[d-1]=i;
    }
    vvi dp(n+1,vi(n+1,inf));
    dp[0][0]=0;
    init(n*2);init2(n*2);
    rep(i,n+1){
        rep(j,n+1){
            if(i<n)chmin(dp[i+1][j],dp[i][j]+getsum(0,a[i],0,0,N));
            if(j<n)chmin(dp[i][j+1],dp[i][j]+getsum2(0,b[j],0,0,N));
            if(j<n)update2(b[j],0);
            if(j<n)update(b[j],0);
        }
        init(n*2);init2(n*2);
        rep(j,i+1){
            update(a[j],0);
            update2(a[j],0);
        }
    }
    out(dp[n][n]);
}