Problem:
Problem C: Find the Point

We understand that reading English is a great pain to many of you. So we’ll keep this problem statememt
simple. Write a program that reports the point equally distant from a set of lines given as the input. In
case of no solutions or multiple solutions, your program should report as such.

Input

The input consists of multiple datasets. Each dataset is given in the following format:

n

x1,1 y1,1 x1,2 y1,2

x2,1 y2,1 x2,2 y2,2

...

xn,1 yn,1 xn,2 yn,2

n is the number of lines (1 ≤ n ≤ 100); (xi,1, yi,1) and (xi,2, yi,2) denote the different points the i-th line passes through. The lines do not coincide each other. The coordinates are all integers between -10000
and 10000.

The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed.

Output

For each dataset, print a line as follows. If there is exactly one point equally distant from all the given lines, print the x- and y-coordinates in this order with a single space between them. If there is more than one such point,
just print "Many" (without quotes). If there is none, just print "None" (without quotes).

The coordinates may be printed with any number of digits after the decimal point, but should be accurate
to 10-4.

Sample Input

2
-35 -35 100 100
-49 49 2000 -2000
4
0 0 0 3
0 0 3 0
0 3 3 3
3 0 3 3
4
0 3 -4 6
3 0 6 -4
2 3 6 6
-1 2 -4 6
0

Output for the Sample Input

Many
1.5000 1.5000
1.000 1.000

Solution in C++ (judge verdict: Wrong Answer)

#include<iostream>
#include<complex>
#include<cmath>
#include<vector>
using namespace std;
#define REP(i,b,n) for(int i=b;i<n;i++)
#define rep(i,n)   REP(i,0,n)
#define mp         make_pair
#define pb         push_back
#define fir        first
#define sec        second

typedef complex<double> P;
typedef pair<P,P>       Line;

const double eps = 1e-10;
const double pi = acos(-1);
#define EQ(a,b) (abs(a-b)<eps)
#define EQV(a,b) ( EQ((a).real(),(b).real())&&EQ((a).imag(),(b).imag()))

double cross(P a,P b){
  return a.real()*b.imag()-a.imag()*b.real();
}

double distance_lp(P a,P b,P c){
  return fabs(cross(b-a,c-a))/abs(b-a);
}

bool is_intersected_ll(P a1,P a2,P b1,P b2){
  return !(EQ(cross(a1-a2,b1-b2),0.0));
}

P intersection_ll(P a1,P a2,P b1,P b2){
  P a=a2-a1,b=b2-b1;
  return a1+a*cross(b,b1-a1)/cross(b,a);
}

P rot(P a,P b){
  double theta=arg(a);
  double phi=arg(b);
  if (phi < theta)phi+=2*pi;
  phi-=theta;
  phi/=2;
  P ret;
  ret.real()=cos(phi)*a.real()-sin(phi)*a.imag();
  ret.imag()=sin(phi)*a.real()+cos(phi)*a.imag();
  return ret;
}

void makeall(vector<Line> &a,vector<Line> &b){
  rep(i,1&&a.size()){
    REP(j,i+1,a.size()){
      if (is_intersected_ll(a[i].fir,a[i].sec,a[j].fir,a[j].sec)){
	P tmp = intersection_ll(a[i].fir,a[i].sec,a[j].fir,a[j].sec);
	P at=EQV(a[i].fir,tmp)?a[i].sec:a[i].fir;
	P bt=EQV(a[j].fir,tmp)?a[j].sec:a[j].fir;
	b.pb(mp(tmp,tmp+rot(at-tmp,bt-tmp)));
	b.pb(mp(tmp,tmp+rot(bt-tmp,at-tmp)));
	
      }
    }
  }
}

bool check(vector<Line> &a,P &b){
  double tmp = distance_lp(a[0].fir,a[0].sec,b);
  REP(i,1,a.size()){
    if (!(fabs(tmp-distance_lp(a[i].fir,a[i].sec,b))<eps))return false;
  }
  return true;
}

void solve(vector<Line> &a){
  vector<Line> b;
  makeall(a,b);
  P ans;
  int cnt=0;
  vector<P> candy;
  rep(i,b.size()&&cnt<2){
    REP(j,i+1,b.size()&&cnt<2){
      if (is_intersected_ll(b[i].fir,b[i].sec,b[j].fir,b[j].sec)){
	P cand=intersection_ll(b[i].fir,b[i].sec,b[j].fir,b[j].sec);
	candy.pb(cand);
	if (cnt == 1 && EQV(ans,cand))continue;
	if (check(a,cand)){
	  ans=cand;
	  cnt++;
	  if (cnt ==2)break;
	}
      }
    }
  }

  if (cnt == 0){
    printf("None\n");
  }else if(cnt == 1){
    printf("%.10lf %.10lf\n",ans.real(),ans.imag());
  }else {
    printf("Many\n");
  }
}

main(){
  int n;
  while(cin>>n && n){
    vector<Line> a(n);
    rep(i,n){
      rep(j,2){
	cin>>a[i].fir.real()>>a[i].fir.imag();
	swap(a[i].fir,a[i].sec);
      }
    }

    rep(i,a.size()){
      for(int j=(int)a.size()-1;j>i;j--){
	if(distance_lp(a[i].fir,a[i].sec,a[j].sec)<eps &&
	   distance_lp(a[i].fir,a[i].sec,a[j].fir)<eps){
	  a.erase(a.begin()+j);
	}
      }
    }

    if (a.size() == 1||a.size() ==2){
      printf("Many\n");
    }else solve(a);
  }
  return false;
}