Problem: 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.

Example

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

Many
1.5000 1.5000
1.000 1.000

Solution (in CPP):

Solution:
#include<iostream>
#include<sstream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
#define pb push_back
#define each(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define dbg(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<endl

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
const int inf = (int)1e9;
const double INF = 1e12, EPS = 5e-7;

typedef complex<double> P;
namespace std{
	bool operator<(const P& a, const P& b){
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
	}
}
inline double cross(const P& a, const P& b){ return imag(conj(a)*b); }
inline double dot(const P& a, const P& b){ return real(conj(a)*b); }
struct L : public vector<P>{
	L(const P &a, const P &b) {
		push_back(a); push_back(b);
	}
};
typedef vector<P> G;

inline int ccw(P a, P b, P c) {
	b -= a; c -= a;
	if(cross(b, c) > 0)   return +1;		// counter clockwise
	if(cross(b, c) < 0)   return -1;		// clockwise
	if(dot(b, c) < 0)     return +2;		// c--a--b on line
	if(norm(b) < norm(c)) return -2;		// a--b--c on line
	return 0;
}
P crosspoint(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]);
  if (abs(A) < EPS && abs(B) < EPS) return P(rand(), rand()); // same line
  if (abs(A) < EPS) return P(rand(), rand());// !!!PRECONDITION NOT SATISFIED!!!
  return m[0] + B / A * (m[1] - m[0]);
}
P projection(const L &l, const P &p) {
  double t = dot(p-l[0], l[0]-l[1]) / norm(l[0]-l[1]);
  return l[0] + t*(l[0]-l[1]);
}
double distanceLP(const L &l, const P &p) {
  return abs(p - projection(l, p));
}
P incenter(const P &x, const P &y, const P &z){
	P a = y - x, b = z - x;
	P d1 = a / abs(a) + b / abs(b);
	P d2 = -a / abs(a) + (b - a) / abs(b - a);
	return crosspoint(L(P(0, 0), d1), L(a, a + d2)) + x;
}
G excenter(const G &t){
	G res;
	rep(i, 3){
		P a = t[i], b = t[(i + 1) % 3], c = t[(i + 2) % 3];
		res.pb(crosspoint(L(a, a + (a - c) / abs(a - c) + (b - a) / abs(b - a)),
		L(b, b + (b - c) / abs(b - c) + (a - b) / abs(a - b))));
	}
	return res;
}
bool parallel(const L &l, const L &m){
	return abs(cross(l[0] - l[1], m[0] - m[1])) < EPS;
}

int n;
int main(){
	while(cin >> n && n){
		vector<L> ls;
		rep(i, n){
			int a, b, c, d;
			cin >> a >> b >> c >> d;
			L l(P(a, b), P(c, d));
			rep(j, i) if(parallel(ls[j], l) && abs(cross(ls[j][0] - l[1], l[0] - l[1])) < EPS) goto SKIP;
			
			ls.pb(l);
			SKIP:;
		}
		n = ls.size();
		if(n < 3){
			cout << "Many" << endl;
			continue;
		}
		
		P p(INF, INF);
		G cand;
		bool pa[100][100] = {};
		rep(i, n) rep(j, i) pa[i][j] = pa[j][i] = parallel(ls[i], ls[j]);
		rep(i, n) rep(j, i) if(pa[i][j]) rep(k, j) if(pa[j][k]){
			cout << "None" << endl;
			goto END;
		}
		
		rep(i, n) rep(j, i) rep(k, j){
			if(pa[i][j] || pa[j][k] || pa[k][i]) continue;
			P x = crosspoint(ls[i], ls[j]), y = crosspoint(ls[j], ls[k]), z = crosspoint(ls[k], ls[i]);
			if(abs(x - y) < EPS && abs(y - z) < EPS){
				cand.pb(x);
				goto OUT;
			}
			
			cand.pb(incenter(x, y, z));
			G g; g.pb(x); g.pb(y); g.pb(z);
			g = excenter(g);
			cand.insert(cand.end(), all(g));
			goto OUT;
		}
		OUT:
		//each(i, cand) dbg(*i);
		
		
		if(cand.size()){
			rep(it, cand.size()){
				bool ok = 1;
				double dist = distanceLP(ls[0], cand[it]);
				rep(i, n) if(abs(dist - distanceLP(ls[i], cand[it])) > EPS){
					ok = 0;
					break;
				}
				if(ok){
					if(p.real() != INF){
						cout << "Many" << endl;
						goto END;
					}
					p = cand[it];
				}
			}
			if(p.real() == INF){
				cout << "None" << endl;
				goto END;
			}
		}
		else{
			if(n < 4){
				cout << "Many" << endl;
				continue;
			}
			sort(all(ls));
			do{
				if(parallel(ls[0], ls[1]) && parallel(ls[2], ls[3])) break;
			}while(next_permutation(all(ls)));
			
			vector<L> t;
			rep(i, 2){
				P m = (ls[i * 2][0] + ls[i * 2 + 1][0]) * 0.5;
				t.pb(L(m, m + ls[i * 2][1] - ls[i * 2][0]));
			}
			p = crosspoint(t[0], t[1]);
			double dist = distanceLP(ls[0], p);
			rep(i, 4) if(abs(distanceLP(ls[i], p) - dist) > EPS){
				cout << "None" << endl;
				goto END;
			}
		}
		
		printf("%.9f %.9f\n", p.real(), p.imag());
		END:;
	}
	return 0;
}