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
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

using ld = long double;
using Point =  complex<ld>;
const ld eps = 1e-10;
const ld pi = acos(-1.0);

bool eq(ld a, ld b) {
	return (abs(a - b) < eps);
}

bool cmp(Point x,Point y){
	if(eq(x.real(),y.real()))return x.imag()<y.imag();
	return x.real()<y.real();
}

bool eqq(Point x,Point y){
	return eq(x.real(),y.real())&&eq(x.imag(),y.imag());
}
//内積
ld dot(Point a, Point b) {
	return real(conj(a)*b);
}
//外積
ld cross(Point a, Point b) {
	return imag(conj(a)*b);
}


//線分
//直線にするなら十分通い２点を端点とすればよい
class Line {
public:
	Point a, b;
};
//円
class Circle {
public:
	Point p;
	ld r;
};
//3点の位置関係
int ccw(Point a, Point b, Point c) {
	b -= a; c -= a;
	if (cross(b, c) > eps)return 1;//a,b,cが反時計回り
	if (cross(b, c) < -eps)return -1;//a,b,cが時計回り
	if (dot(b, c) < 0)return 2;//c,a,bの順に一直線
	if (norm(b) < norm(c))return -2;//a,b,cの順に一直線
	return 0;//a,c,bの順に一直線
}
//2直線の交差判定
bool isis_ll(Line l, Line m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}
//直線と線分の交差判定
bool isis_ls(Line l, Line s) {
	return (cross(l.b - l.a, s.a - l.a)*cross(l.b - l.a, s.b - l.a) < eps);
}
//点が直線上に存在するか
bool isis_lp(Line l, Point p) {
	return (abs(cross(l.b - p, l.a - p)) < eps);
}
//点が線分上に存在するか
bool isis_sp(Line s, Point p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}
//線分と線分の交差判定
bool isis_ss(Line s, Line t) {
	if (isis_sp(s, t.a) || isis_sp(s, t.b) || isis_sp(t, s.a) || isis_sp(t, s.b))return true;
	return(cross(s.b - s.a, t.a - s.a)*cross(s.b - s.a, t.b - s.a) < -eps && cross(t.b - t.a, s.a - t.a)*cross(t.b - t.a, s.b - t.a) < -eps);
}
//点から直線への垂線の足
Point proj(Line l, Point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}
//直線と直線の交点
//平行な２直線に対しては使うな！！！！
Point is_ll(Line s, Line t) {
	Point sv = s.b - s.a; Point tv = t.b - t.a;
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}
//直線と点の距離
ld dist_lp(Line l, Point p) {
	return abs(p - proj(l, p));
}
//直線と直線の距離
ld dist_ll(Line l, Line m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}
//線分と直線の距離
ld dist_ls(Line l, Line s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}
//線分と点の距離
ld dist_sp(Line s, Point p) {
	Point r = proj(s, p);
	return isis_sp(s, r) ? abs(p - r) : min(abs(p - s.a), abs(p - s.b));
}
//線分と線分の距離
ld dist_ss(Line s, Line t) {
	if (isis_ss(s, t))return 0;
	return min({ dist_sp(s,t.a),dist_sp(s,t.b),dist_sp(t,s.a),dist_sp(t,s.b) });
}

vector<Line> calc(Line l,Line r){
    if(abs(cross(l.b-l.a,r.b-r.a))<eps){
        Point mid=(l.a+r.a)/(ld)2.0;
        return {{mid, mid + l.b-l.a}};
    }
    Point mid=is_ll(l,r);
    Point pl = (l.b-l.a)/abs(l.b-l.a);
    Point pr = (r.b-r.a)/abs(r.b-r.a);
    vector<Line> ret;
    ret.push_back(Line{mid,mid+(pl+pr)/(ld)2.0});
    ret.push_back(Line{mid,mid+(pl-pr)/(ld)2.0});
    return ret;
}
void solve(int n){
    Line l[n];
    rep(i,n){
        int x,y,xx,yy;
        cin>>x>>y>>xx>>yy;
        l[i]=Line{Point(x,y),Point(xx,yy)};
    }
    if(n<=2){
        cout<<"Many"<<endl;
        return;
    }
    auto v1=calc(l[0],l[1]);
    auto v2=calc(l[0],l[2]);
    vector<Point> p;
    for(auto e : v1)for(auto f : v2){
        if(abs(cross(e.a-e.b,f.a-f.b))<eps)continue;
        p.push_back(is_ll(e,f));
    }
    vector<Point> ans;
    for(auto q : p){
        bool ok=true;
        ld dist = dist_lp(l[0],q);
        rep(i,n){
            if(!eq(dist_lp(l[i],q),dist))ok=false;
        }
        if(ok)ans.push_back(q);
    }
    sort(ans.begin(),ans.end(),cmp);
    ans.erase(unique(ans.begin(),ans.end(),eqq),ans.end());
    if(ans.size()>1){
        cout<<"Many"<<endl;
    }
    else if(ans.size()==0){
        cout<<"None"<<endl;
    }
    else cout<<ans[0].real()<<" "<<ans[0].imag()<<endl;
}
int main(){
    cout<<fixed<<setprecision(12);
    int n;
    while(cin>>n,n!=0)solve(n);
    return 0;
}
