Problem: In the International City of Pipe Construction, it is planned to repair the water pipe at a certain point in the water pipe network. The network consists of water pipe segments, stop valves and source point. A water pipe is represented by a segment on a 2D-plane and intersected pair of water pipe segments are connected at the intersection point. A stop valve, which prevents from water flowing into the repairing point while repairing, is represented by a point on some water pipe segment. In the network, just one source point exists and water is supplied to the network from this point.

Of course, while repairing, we have to stop water supply in some areas, but, in order to reduce the risk of riots, the length of water pipes stopping water supply must be minimized. What you have to do is to write a program to minimize the length of water pipes needed to stop water supply when the coordinates of end points of water pipe segments, stop valves, source point and repairing point are given.



Input

A data set has the following format:

> N M
>  xs1 ys1 xd1 yd1
>  ...
>  xsN ysN xdN ydN
>  xv1 yv1
>  ...
>  xvM yvM
>  xb yb
>  xc yc
>

The first line of the input contains two integers, N (1 ≤ N ≤ 300) and M (0 ≤ M ≤ 1,000) that indicate the number of water pipe segments and stop valves. The following N lines describe the end points of water pipe segments. The i-th line contains four integers, xsi, ysi, xdi and ydi that indicate the pair of coordinates of end points of i-th water pipe segment. The following M lines describe the points of stop valves. The i-th line contains two integers, xvi and yvi that indicate the coordinate of end points of i-th stop valve. The following line contains two integers, xb and yb that indicate the coordinate of the source point. The last line contains two integers, xc and yc that indicate the coordinate of the repairing point.

You may assume that any absolute values of coordinate integers are less than 1,000 (inclusive.) You may also assume each of the stop valves, the source point and the repairing point is always on one of water pipe segments and that that each pair among the stop valves, the source point and the repairing point are different. And, there is not more than one intersection between each pair of water pipe segments. Finally, the water pipe network is connected, that is, all the water pipes are received water supply initially.

Output

Print the minimal length of water pipes needed to stop water supply in a line. The absolute or relative error should be less than or 10-6. When you cannot stop water supply to the repairing point even though you close all stop valves, print "`-1`" in a line.

Examples

Input

1 2
0 0 10 0
1 0
9 0
0 0
5 0


Output

9.0


Input

5 3
0 4 2 4
0 2 2 2
0 0 2 0
0 0 0 4
2 0 2 4
0 2
1 0
2 2
1 4
2 1


Output

3.0


Input

2 1
0 0 0 4
0 2 2 2
1 2
0 1
0 3


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
#define valid(y,x,h,w) (0<=y&&y<h&&0<=x&&x<w)
#define tpl(...) make_tuple(__VA_ARGS__)
const int INF = 0x3f3f3f3f;
const long double EPS = 1e-12;
#define double long double
const double PI = acos(-1);
typedef long long ll;
typedef pair<int,int> pii;
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 &o,const vector<T>&t){o<<'[';FOR(i,t){if(i!=t.begin())o<<',';o<<*i;}return o<<']';}
template<class S,class T>ostream&operator<<(ostream &o,const pair<S,T>&t){return o<<'('<<t.first<<','<<t.second<<')';}
template<int N,class Tp>void output(ostream&,const Tp&){}
template<int N,class Tp,class,class ...Ts>void output(ostream &o,const Tp&t){if(N)o<<',';o<<get<N>(t);output<N+1,Tp,Ts...>(o,t);}
template<class ...Ts>ostream&operator<<(ostream&o,const tuple<Ts...>&t){o<<'(';output<0,tuple<Ts...>,Ts...>(o,t);return o<<')';}
template<class T>void output(T t,char z=10){if(t<0)t=-t,putchar(45);int c[20];
int k=0;while(t)c[k++]=t%10,t/=10;for(k||(c[k++]=0);k;)putchar(c[--k]^48);putchar(z);}
template<class T>void outputs(T t){output(t);}
template<class S,class ...T>void outputs(S a,T...t){output(a,32);outputs(t...);}
template<class T>void output(T *a,int n){REP(i,n)cout<<a[i]<<(i!=n-1?',':'\n');}
template<class T>void output(T *a,int n,int m){REP(i,n)output(a[i],m);}
template<class T>bool input(T &t){int n=1,c;for(t=0;!isdigit(c=getchar())&&~c&&c-45;);
if(!~c)return 0;for(c-45&&(n=0,t=c^48);isdigit(c=getchar());)t=10*t+c-48;t=n?-t:t;return 1;}
template<class S,class ...T>bool input(S&a,T&...t){input(a);return input(t...);}

template<class T>
struct Pt {
  T x,y;
  Pt(T x=T(), T y=T()) :x(x),y(y) {}
  bool operator<(const Pt &p) const { return x!=p.x?x<p.x:y<p.y; }
  bool operator==(const Pt &p) const { return x==p.x&&y==p.y; }
  Pt operator+=(const Pt &p) { x+=p.x, y+=p.y; return *this; }
  Pt operator-=(const Pt &p) { x-=p.x, y-=p.y; return *this; }
  Pt operator*=(const Pt &p) { T t=x;x=x*p.x-y*p.y,y=t*p.y+y*p.x; return *this;}
  Pt operator*=(const T &v) { x*=v; y*=v; return *this;}
  friend ostream& operator<<(ostream& os, const Pt &p) {return os<<'('<<p.x<<','<<p.y<<')';}
  friend istream& operator>>(istream& is, Pt &p) {return is>>p.x>>p.y;}

  Pt operator+(const Pt &p) const { return Pt(*this)+=p; }
  Pt operator-(const Pt &p) const { return Pt(*this)-=p; }
  Pt operator*(const Pt &p) const { return Pt(*this)*=p; }
  Pt operator*(const T &v) const { return Pt(*this)*=v; }
  friend Pt operator*(const T &v, const Pt &p) { return p*v; }
  bool operator!=(const Pt &p) const { return !(*this == p); }

  T abs() const { return hypot(x,y); }
  T norm() const { return x*x+y*y; }
  T arg() const { return atan2(y,x); }
  T dot(const Pt &p) const { return x*p.x+y*p.y; }
  T det(const Pt &p) const { return x*p.y-y*p.x; }
  friend T abs(const Pt &p) { return p.abs(); }
  friend T norm(const Pt &p) { return p.norm(); }
  friend T arg(const Pt &p) { return p.arg(); }
  friend T dot(const Pt &p, const Pt &q) { return p.dot(q); }
  friend T det(const Pt &p, const Pt &q) { return p.det(q); }
};

int sig(double r) {return r<-EPS?-1:r>EPS?1:0;}
int sig(ll r) {return r<0?-1:r>0?1:0;}
bool eq(double a, double b) {return sig(a-b)==0;}
bool eq(ll a, ll b) {return a==b;}

template<class T>int ccw(Pt<T> a, Pt<T> b, Pt<T> c); // 2(c--a--b), -2(a--b--c), 0(a=-c-=b)

template<> int ccw(Pt<ll> a, Pt<ll> b, Pt<ll> c) {
  b -= a; c -= a;
  return det(b,c) ? sig(det(b,c)) : dot(b,c)<0 ? 2 : (0<b.norm()&&b.norm()<c.norm()) ? -2 : 0;
}
template<> int ccw(Pt<double> a, Pt<double> b, Pt<double> c) {
  b -= a; c -= a;
  if (det(b,c)) return sig(det(b,c));
  if (sig(dot(b, c)) == -1) return +2; // c--a--b
  if (EPS < b.norm() && b.norm() < c.norm()-EPS) return -2; // a--b--c
  return 0;
}

typedef Pt<double> P;

class L {
  vector<P> v;
public:
  L(const P &a, const P &b) : v({a,b,b-a}) {}
  const P operator[](int i) const { return v[i]; }
  void sub(const P&p) { v[0] -= p; v[1] -= p; }
  friend ostream &operator<<(ostream &os, const L &l) {
    return os << l[0] << "->" << l[1];
  }
};

bool iSS(const L &s, const L &t) {
  return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
         ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
bool iSP(const L &s, const P &p) {

  return sig(abs(s[0]-p)+abs(s[1]-p)-abs(s[2])) <= 0;
};

P crosspoint(const L &l, const L &m) {
  auto A = det(l[2],m[2]);
  auto B = det(l[2],l[1]-m[0]);
  if (!sig(A) && !sig(B)) return m[0];// same line
  if (!sig(A)) assert(0); // !!!PRECONDITION NOT SATISFIED!!!
  return m[0] + B/A*m[2];
}

typedef vector<int> Edges;
typedef vector<Edges> Graph;

Graph segmentArrangement(const vector<L> &ss, vector<P> &ps) {
  int n = ss.size();
  REP(i,n) {
    ps.push_back( ss[i][0] );
    ps.push_back( ss[i][1] );
    for (int j = i+1; j < n; ++j)
      if (iSS(ss[i], ss[j]))
        ps.push_back( crosspoint(ss[i], ss[j]) );
  }
  sort(ALL(ps)); ps.erase(unique(ALL(ps)), ps.end());
  Graph g(ps.size());
  REP(i,ss.size()) {
    vector<pair<double,int> > v;
    REP(j,ps.size()) if (iSP(ss[i], ps[j])) {
      v.emplace_back(norm(ss[i][0]-ps[j]), j);
    }
    sort(ALL(v));
    REP(j,(int)v.size()-1) {
      int a = v[j].second, b = v[j+1].second;
      g[a].push_back(b); g[b].push_back(a);
    }
  }
  return g;
}


int main() {
  int n,m;
  while(input(n,m)) {
    vector<L> ss;
    double sum = 0;
    REP(i,n) {
      P a,b; cin >> a >> b;
      ss.emplace_back(a,b);
      sum += (a-b).abs();
    }
    vector<P> ps(m+2);
    REP(i,m+2) cin >> ps[i];
    vector<P> p(ps);
    Graph g = segmentArrangement(ss,ps);
    vector<int> p2ps(m+2,-1);
    vector<int> ps2p(ps.size(),-1);
    REP(i,m+2) {
      REP(j,ps.size())
        if (norm(p[i]-ps[j]) < EPS) {
          assert(p2ps[i] == -1);
          assert(ps2p[j] == -1);
          p2ps[i] = j;
          ps2p[j] = i;
        }
    }

    vector<bool> fg(ps.size()), fs(ps.size());
    {
      queue<int> Q;
      Q.push(p2ps[m+1]);
      bool ng = 0;
      fg[p2ps[m+1]] = 1;
      while(!Q.empty()) {
        int v = Q.front(); Q.pop();
        if (ps2p[v] != -1 && ps2p[v] < m) continue;
        if (ps2p[v] == m) {
          ng = 1;
          break;
        }
        FOR(it, g[v]) if (!fg[*it]) {
          fg[*it] = 1;
          Q.push(*it);
        }
      }
      if (ng) {
        puts("-1");
        continue;
      }
    }
    double ans = 0;
    {
      queue<int> Q;
      Q.push(p2ps[m]);
      fs[p2ps[m]] = 1;
      while(!Q.empty()) {
        int v = Q.front(); Q.pop();
        // cout << ps2p[v] << " " << ps[v] << " " << g[v].size() << endl;

        FOR(it, g[v]) {
          if (!fg[v] || !fg[*it])
            ans += abs(ps[v] - ps[*it]);
          if (!fs[*it] && !fg[v]) {
            fs[*it] = 1;
            Q.push(*it);
          }
        }
      }
    }
    ans = sum - ans / 2;
    printf("%.10Lf\n", ans);
  }
}