Problem: Hide-and-seek is a children’s game. Players hide here and there, and one player called it tries to find all the other players.

Now you played it and found all the players, so it’s turn to hide from it. Since you have got tired of running around for finding players, you don’t want to play it again. So you are going to hide yourself at the place as far as possible from it. But where is that?

Your task is to find the place and calculate the maximum possible distance from it to the place to hide.



Input

The input contains a number of test cases.

The first line of each test case contains a positive integer N (N ≤ 1000). The following N lines give the map where hide-and-seek is played. The map consists of N corridors. Each line contains four real numbers x1, y1, x2, and y2, where (x1, y1 ) and (x2, y2 ) indicate the two end points of the corridor. All corridors are straight, and their widths are negligible. After these N lines, there is a line containing two real numbers sx and sy, indicating the position of it. You can hide at an arbitrary place of any corridor, and it always walks along corridors. Numbers in the same line are separated by a single space.

It is guaranteed that its starting position (sx, sy) is located on some corridor and linked to all corridors directly or indirectly.

The end of the input is indicated by a line containing a single zero.

Output

For each test case, output a line containing the distance along the corridors from ‘it”s starting position to the farthest position. The value may contain an error less than or equal to 0.001. You may print any number of digits below the decimal point.

Example

Input

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


Output

4.243

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <complex>
#include <vector>
#include <algorithm>
#include <cmath>
#include <array>
#include <map>
#include <utility>
#include <queue>
using namespace std;
const double EPS = 1e-8;
const double INF = 1e12;
const double PI = acos(-1);
#define EQ(n,m) (abs((n)-(m)) < EPS)
#define X real()
#define Y imag()

typedef complex<double> P;
typedef vector<P> VP;
struct L : array<P, 2>{
    L(const P& a, const P& b){ at(0)=a; at(1)=b; }
    L(){}
};

namespace std{
    bool operator < (const P& a, const P& b){
        return !EQ(a.X,b.X) ? a.X<b.X : a.Y+EPS<b.Y;
    }
    bool operator == (const P& a, const P& b){
        return abs(a-b) < EPS;
    }
}

double dot(P a, P b){
    return (conj(a)*b).X;
}
double cross(P a, P b){
    return (conj(a)*b).Y;
}
int ccw(P a, P b, P c){
    b -= a;
    c -= a;
    if(cross(b,c) > EPS) return +1; //ccw
    if(cross(b,c) < -EPS) return -1; //cw
    if(dot(b,c) < -EPS) return +2; //c-a-b
    if(abs(c)-abs(b) > EPS) return -2; //a-b-c
    return 0; //a-c-b
}

bool intersectSS(const L& a, const L& b){
    return ( ccw(a[0],a[1],b[0]) *ccw(a[0],a[1],b[1]) <= 0 ) &&
        ( ccw(b[0],b[1],a[0]) *ccw(b[0],b[1],a[1]) <= 0 );
}
bool intersectSP(const L& s, const P &p){
    return abs(cross(s[0]-p, s[1]-p))<EPS  && dot(s[0]-p, s[1]-p)<EPS;
}

bool isParallel(const P &a, const P &b){
    return abs(cross(a,b)) < EPS;
}
bool isParallel(const L &a, const L &b){
    return isParallel(a[1]-a[0], b[1]-b[0]);
}

P crosspointLL(const L &l, const L &m) {
  double A = cross(l[1]-l[0], m[1]-m[0]);
  double B = cross(l[1]-l[0], l[1]-m[0]);
  return m[0] + B/A *(m[1]-m[0]);
}

pair<vector<vector<pair<int, double> > >, VP> arrangementEX(const vector<L> &l, const P &sp){
    vector<VP> cp(l.size());
    VP plist;
    plist.push_back(sp);
    for(int i=0; i<(int)l.size(); i++){
        for(int j=i+1; j<(int)l.size(); j++){
            if(!isParallel(l[i], l[j]) && intersectSS(l[i], l[j])){
                P cpij = crosspointLL(l[i], l[j]);
                cp[i].push_back(cpij);
                cp[j].push_back(cpij);
                plist.push_back(cpij);
            }
        }
        if(intersectSP(l[i], sp)){
            cp[i].push_back(sp);
        }
        cp[i].push_back(l[i][0]);
        cp[i].push_back(l[i][1]);
        plist.push_back(l[i][0]);
        plist.push_back(l[i][1]);
        sort(cp[i].begin(), cp[i].end());
        cp[i].erase(unique(cp[i].begin(), cp[i].end()), cp[i].end());
        for(int j=cp[i].size()-2; j>=0; j--){
            P mid = (cp[i][j] +cp[i][j+1]) /2.0;
            cp[i].push_back(mid);
            plist.push_back(mid);
        }
        sort(cp[i].begin(), cp[i].end());
        cp[i].erase(unique(cp[i].begin(), cp[i].end()), cp[i].end());
    }
    sort(plist.begin(), plist.end());
    plist.erase(unique(plist.begin(), plist.end()), plist.end());

    int n = plist.size();
    map<P, int> conv;
    for(int i=0; i<n; i++){
        conv[plist[i]] = i;
    }
    vector<vector<pair<int, double> > > adj(n);
    for(int i=0; i<(int)cp.size(); i++){
        for(int j=0; j<(int)cp[i].size()-1; j++){
            int jidx = conv[cp[i][j]];
            int jp1idx = conv[cp[i][j+1]];
            double dist = abs(cp[i][j] -cp[i][j+1]);
            adj[jidx].emplace_back(jp1idx, dist);
            adj[jp1idx].emplace_back(jidx, dist);
        }
    }
    for(int i=0; i<n; i++){
        sort(adj[i].begin(), adj[i].end());
        adj[i].erase(unique(adj[i].begin(), adj[i].end()), adj[i].end());
    }
    return make_pair(adj, plist);
}

int main(){
    while(1){
        int n;
        cin >> n;
        if(n == 0) break;

        vector<L> l(n);
        for(int i=0; i<n; i++){
            double xs,ys,xt,yt;
            cin >> xs >> ys >> xt >> yt;
            l[i] = L(P(xs, ys), P(xt, yt));
        }
        double sx,sy;
        cin >> sx >> sy;
        P s(sx, sy);

        pair<vector<vector<pair<int, double> > >, VP> ret = arrangementEX(l, s);
        vector<vector<pair<int, double> > > &adj = ret.first;
        VP &plist = ret.second;
        int m = plist.size();
        int sidx = lower_bound(plist.begin(), plist.end(), s) -plist.begin();

        priority_queue<pair<double, int> > wait;
        wait.push(make_pair(0, sidx));
        vector<double> mincost(m, INF);
        mincost[sidx] = 0;
        while(!wait.empty()){
            double dist = -wait.top().first;
            int pos = wait.top().second;
            wait.pop();
            if(dist > mincost[pos]) continue;

            for(pair<int, double> next: adj[pos]){
                int npos = next.first;
                double ndist = dist +next.second;
                if(ndist +EPS < mincost[npos]){
                    mincost[npos] = ndist;
                    wait.push(make_pair(-ndist, npos));
                }
            }
        }

        cout << fixed << setprecision(10);
        cout << *max_element(mincost.begin(), mincost.end()) << endl;
    }
    return 0;
}