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 <stdio.h>
#include <algorithm>
#include <complex>
#include <vector>
#include <assert.h>
#include <string.h>

using namespace std;

typedef complex<double> Complex;
typedef Complex P;

static const double EPS = 1e-6;
inline bool EQ(double a, double b) { return abs(a - b) < EPS; }
inline bool EQP(Complex a, Complex b) { return norm(a - b) < EPS; }

struct Seg;
struct Line;
struct Circle;

double X(P p) { return p.real(); }
double Y(P p) { return p.imag(); }
double dot(P a, P b) { return X(conj(a) * b); }
double cross(P a, P b) { return Y(conj(a) * b); }
double ccw(P p1, P p2, P p3) { return cross(p2 - p1, p3 - p1); }
P proj(P p, P q) { return q * dot(p, q) / norm(q); }
P rot(P p) { return p * P(0, 1); }


struct Line {
  P p, v;
  double distance(P q) {
    return abs(cross(v, q - p)) / abs(v);
  }
  bool is_contain(P q) {
    return EQ(cross(v, q - p), 0.0);
  }
  bool is_parallel(Line rhs) {
    return EQ(cross(v, rhs.v), 0.0);
  }
  P intersection(Line rhs) {
    assert(!is_parallel(rhs));
    return p + v * cross(rhs.v, rhs.p - p) / cross(rhs.v, v);
  }
};


struct Triangle {
  P p1, p2, p3;
  Triangle(P p1, P p2, P p3) : p1(p1), p2(p2), p3(p3) {;}
  P incenter() {
    double le1 = abs(p2 - p3), le2 = abs(p3 - p1), le3 = abs(p1 - p2);
    return (p1 * le1 + p2 * le2 + p3 * le3) / (le1 + le2 + le3);
  }

  vector<P> excenter() {
    vector<P> ret;
    double le1 = abs(p2 - p3), le2 = abs(p3 - p1), le3 = abs(p1 - p2);
    ret.push_back((-p1 * le1 + p2 * le2 + p3 * le3) / (-le1 + le2 + le3));
    ret.push_back(( p1 * le1 - p2 * le2 + p3 * le3) / ( le1 - le2 + le3));
    ret.push_back(( p1 * le1 + p2 * le2 - p3 * le3) / ( le1 + le2 - le3));
    return ret;
  }
};

Line line[300];

int main() {
  int n;
  while (scanf("%d", &n), n) {
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    line[i].p = Complex(x1, y1);
    line[i].v = Complex(x2 - x1, y2 - y1);
    line[i].v /= abs(line[i].v);
  }
  if (n < 3) {
    puts("Many");
continue;
  }
  if (line[0].is_parallel(line[1]) && line[0].is_parallel(line[2])) {
    puts("None");
continue;
  }
  vector<P> point;
  vector<double> dist;
  bool fail[4];
  memset(fail, false, sizeof(fail));
  if (line[0].is_parallel(line[1]) || line[0].is_parallel(line[2]) || line[1].is_parallel(line[2])) {
    if (line[0].is_parallel(line[1])) {
    } else if (line[0].is_parallel(line[2])) {
      swap(line[1], line[2]);
    } else if (line[2].is_parallel(line[2])) {
      swap(line[0], line[2]);
    }
    P p1 = line[2].intersection(line[0]);
    P p2 = line[2].intersection(line[1]);
    P c = (p1 + p2) / 2.0;
    point.push_back(c + line[0].v * abs(p1 - c));
    point.push_back(c - line[0].v * abs(p1 - c));
    dist.push_back(line[0].distance(point[0]));
    dist.push_back(line[0].distance(point[1]));
  } else {
    P p1 = line[0].intersection(line[1]);
    P p2 = line[1].intersection(line[2]);
    P p3 = line[2].intersection(line[0]);
    if (EQP(p1, p2) && EQP(p1, p3)) {
      point.push_back(p1);
      dist.push_back(0.0);
    } else {
      Triangle tri(p1, p2, p3);
      point = tri.excenter();
      point.push_back(tri.incenter());
      dist.push_back(line[0].distance(point[0]));
      dist.push_back(line[0].distance(point[1]));
      dist.push_back(line[0].distance(point[2]));
      dist.push_back(line[0].distance(point[3]));
    }
  }
  int rest = (int)point.size();
  for (int i = 3; i < n; i++) {
    for (int j = 0; j < (int)point.size(); j++) {
      if (fail[j]) { continue; }
      double d = line[i].distance(point[j]);
      if (!EQ(d, dist[j])) {
        fail[j] = true;
        rest--;
      }
    }
  }
  P ans;
  for (int i = 0; i < (int)point.size(); i++) {
    if (!fail[i]) {
      ans = point[i];
    }
  }
  if (rest == 0) {
    puts("None");
  } else if (rest > 1) {
    puts("Many");
  } else {
    printf("%.4lf %.4lf\n", ans.real(), ans.imag());
  }
  }
}