Problem: Convex polygon pillar industrial city

The Industrial Convex Pillar City (ICPC) is a city of buildings in the shape of several convex polygonal columns. You are about to walk through this city from your current location S to your destination T. The sun is strong today, so I want to go to my destination without passing through the sun as much as possible. If the building is on a straight line connecting the point where you are standing and the sun, you are behind the building and will not be exposed to the sun. Also, since all the outer circumferences of the buildings in this city have eaves, while walking along the outer circumference of the buildings, even if you walk along the edge of the sun, you will not receive the sunlight. You are free to walk around the city anywhere except inside the building.

Create a program that outputs the walking distance of Hinata when walking from the current location to the destination so as not to receive the sun.

<image>

Figure E1: For the first input

<image>

Figure E2: For the second input

<image>

Figure E3: For the third input

Input

The input consists of multiple datasets. The maximum number of data sets does not exceed 30. Each dataset is represented in the following format.

> N
> NV1 H1 X1,1 Y1,1 X1,2 Y1,2 ... X1, NV1 Y1, NV1
> ...
> NVN HN XN, 1 YN, 1 XN, 2 YN, 2 ... XN, NVN YN, NVN
> θ φ
> Sx Sy Tx Ty
>

N in the first line represents the number of buildings. The following N lines specify the shape of each building. NVi represents the number of vertices of the polygon when the i-th building is viewed from above, and Hi represents the height of the i-th building. Xi, j and Yi, j represent the x-coordinate and y-coordinate of the j-th vertex of the polygon when the i-th building is viewed from above. The vertices are given in counterclockwise order. All buildings are convex polygons when viewed from above, and there are no other buildings inside the building, and the vertices and sides do not overlap with other polygons. The following lines are given θ and φ, which represent the direction of the sun, θ represents the direction of the sun as a counterclockwise angle from the positive direction of x, and φ is the elevation angle of the sun from the horizon, that is, looking up at the sun. It represents the angle between the direction of the line of sight and the ground surface. However, the sun is at infinity and does not change its position during movement. The following lines are given the coordinates of the current location and destination, (Sx, Sy) and (Tx, Ty).

> All the numerical values ​​given by the input are integers and satisfy the following conditions.

> 1 ≤ N ≤ 100
> 3 ≤ NVi ≤ 12
> 1 ≤ Hi ≤ 1,000
> 0 ≤ θ <360
> 0 <φ <90

All coordinates are -1,000 or more and 1,000 or less. The current location and the destination are different, and neither exists inside or outside the building.

> The end of the input is represented by a single zero line.

> ### Output

For each dataset, output the shortest walking distance in Hinata on one line. The output must not have an absolute error greater than 0.001.

> ### Sample Input


2
4 1 0 0 1 0 1 1 0 1
4 2 2 2 3 2 3 3 2 3
60 45
-1 -1 4 4
Four
4 1 0 0 3 1 1 2 0 1
3 2 10 7 8 2 12 4
6 8 7 12 8 13 9 15 10 19 11 24 10 25
5 4 16 2 16 4 12 8 14 2 15 0
167 38
3 3 15 21
12
4 3 -8 -3 -9 -3 -9 -5 -8 -6
4 5 -4 -5 -7 -5 -7 -6 -5 -6
4 2 -4 1 -5 1 -5 -4 -4 -4 -4
4 1 -1 1 -2 1 -2 -4 -1 -3
4 2 2 3 -1 3 -2 2 3 2
4 1 3 1 2 1 2 -3 3 -4
4 7 1 0 0 0 0 -1 1 -1
4 4 9 5 7 5 7 4 10 4
4 3 6 5 5 4 5 0 6 0
4 5 8 -1 5 -1 6 -2 8 -2
4 1 10 0 9 0 9 -2 10 -1
4 6 10 2 8 2 8 1 10 1
131 78
-10 10 10 -10
0


Output for Sample Input


1.93185
7.87174
20.86840





Example

Input

2
4 1 0 0 1 0 1 1 0 1
4 2 2 2 3 2 3 3 2 3
60 45
-1 -1 4 4
4
4 1 0 0 3 1 1 2 0 1
3 2 10 7 8 2 12 4
6 8 7 12 8 13 9 15 10 19 11 24 10 25
5 4 16 2 16 4 12 8 14 2 15 0
167 38
3 3 15 21
12
4 3 -8 -3 -9 -3 -9 -5 -8 -6
4 5 -4 -5 -7 -5 -7 -6 -5 -6
4 2 -4 1 -5 1 -5 -4 -4 -4
4 1 -1 1 -2 1 -2 -4 -1 -3
4 2 2 3 -1 3 -2 2 3 2
4 1 3 1 2 1 2 -3 3 -4
4 7 1 0 0 0 0 -1 1 -1
4 4 9 5 7 5 7 4 10 4
4 3 6 5 5 4 5 0 6 0
4 5 8 -1 5 -1 6 -2 8 -2
4 1 10 0 9 0 9 -2 10 -1
4 6 10 2 8 2 8 1 10 1
131 78
-10 10 10 -10
0


Output

1.93185
7.87174
20.86840

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
 
using namespace std;
 
const double EPS = 1e-8, PI = acos(-1);
 
inline bool eq(double a, double b) { return abs(b - a) < EPS; }
 
#define curr(P, i) P[i]
#define next(P, i) P[(i+1)%P.size()]
#define prev(P, i) P[(i+P.size()-1) % P.size()]
enum
{
  OUT, ON, IN
};
namespace Geometry
{
  struct Point
  {
    double x, y;
 
    Point() {};
 
    Point(double x, double y) : x(x), y(y) {};
 
    Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
 
    Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
 
    Point operator*(const double b) const { return Point(x * b, y * b); }
 
    Point operator*(const Point &b) const { return Point(x * b.x - y * b.y, x * b.y + y * b.x); }
 
    Point operator/(const double b) const { return Point(x / b, y / b); }
 
    bool operator<(const Point &b) const { return x != b.x ? x < b.x : y < b.y; }
 
    bool operator==(const Point &b) const { return eq(x, b.x) && eq(y, b.y); }
 
    double norm() { return x * x + y * y; }
 
    double arg() { return atan2(x, y); }
 
    double abs() { return sqrt(norm()); }
 
    Point rotate(double theta) { return Point(cos(theta) * x - sin(theta) * y, sin(theta) * x + cos(theta) * y); }
 
    Point rotate90() { return Point(-y, x); }
 
    friend ostream &operator<<(ostream &os, Point &p) { return os << "(" << p.x << "," << p.y << ")"; }
 
    friend istream &operator>>(istream &is, Point &a) { return is >> a.x >> a.y; }
  };
 
  struct Line
  {
    Point a, b;
 
    Line() {};
 
    Line(Point a, Point b) : a(a), b(b) {};
 
    friend ostream &operator<<(ostream &os, Line &p) { return os << "(" << p.a.x << "," << p.a.y << ") to (" << p.b.x << "," << p.b.y << ")"; }
 
    friend istream &operator>>(istream &is, Line &a) { return is >> a.a.x >> a.a.y >> a.b.x >> a.b.y; }
  };
 
  struct Segment
  {
    Point a, b;
 
    Segment() {};
 
    Segment(Point a, Point b) : a(a), b(b) {};
 
    friend ostream &operator<<(ostream &os, Segment &p) { return os << "(" << p.a.x << "," << p.a.y << ") to (" << p.b.x << "," << p.b.y << ")"; }
 
    friend istream &operator>>(istream &is, Segment &a) { return is >> a.a.x >> a.a.y >> a.b.x >> a.b.y; }
  };
 
  struct Circle
  {
    Point p;
    double r;
 
    Circle() {};
 
    Circle(Point p, double r) : p(p), r(r) {};
  };
 
  typedef vector< Point > Polygon;
  typedef vector< Segment > Segments;
  typedef vector< Line > Lines;
  typedef vector< Circle > Circles;
  typedef pair< Point, Point > PointPoint;
 
  double cross(const Point &a, const Point &b)
  {
    return a.x * b.y - a.y * b.x;
  }
 
  double dot(const Point &a, const Point &b)
  {
    return a.x * b.x + a.y * b.y;
  }
 
  int ccw(const Point &a, Point b, Point c)
  {
    b = b - a, c = c - a;
    if(cross(b, c) > EPS) return +1;  // a ??? b ??§ ???????¨??????????????????? c
    if(cross(b, c) < -EPS) return -1; // a ??? b ??§ ????¨??????????????????? c
    if(dot(b, c) < 0) return +2;  // c -- a -- b??§?????´??????
    if(b.norm() < c.norm()) return -2; // a -- b -- c??§?????´??????
    return 0;  // a -- c -- b??§?????´??????
  }
 
  Point Projection(const Line &l, const Point &p)
  {
    double t = dot(p - l.a, l.a - l.b) / (l.a - l.b).norm();
    return l.a + (l.a - l.b) * t;
  }
 
  Point Projection(const Segment &l, const Point &p)
  {
    double t = dot(p - l.a, l.a - l.b) / (l.a - l.b).norm();
    return l.a + (l.a - l.b) * t;
  }
 
  Point Reflection(const Line &l, const Point &p)
  {
    return p + (Projection(l, p) - p) * 2.0;
  }
 
  double Distance(const Line &l, const Point &p)
  { //OK
    return (p - Projection(l, p)).abs();
  }
 
  bool Intersect(const Line &l, const Line &m)
  {
    return abs(cross(l.b - l.a, m.b - m.a)) > EPS || abs(cross(l.b - l.a, m.b - l.a)) < EPS;
  }
 
  bool Intersect(const Line &l, const Segment &s)
  {
    return cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < EPS;
  }
 
  bool Intersect(const Line &l, const Point &p)
  {
    return abs(ccw(l.a, l.b, p)) != -1;
  }
 
  bool Intersect(const Segment &s, const Segment &t)
  {
    return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0 && ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
  }
 
  bool Intersect(const Segment &s, const Point &p)
  {
    return ccw(s.a, s.b, p) == 0;
  }
 
  bool Intersect(const Circle &c, const Line &l)
  {
    return Distance(l, c.p) <= c.r + EPS;
  }
 
  bool Intersect(const Circle &c, const Point &p)
  {
    return abs((p - c.p).abs() - c.r) < EPS;
  }
 
  int Intersect(const Circle &c, const Segment &l)
  {
    if((Projection(l, c.p) - c.p).norm() - c.r * c.r > EPS) return 0;
    const double d1 = (c.p - l.a).abs(), d2 = (c.p - l.b).abs();
    if(d1 < c.r + EPS && d2 < c.r + EPS) return 0;
    if(d1 < c.r - EPS && d2 > c.r + EPS || d1 > c.r + EPS && d2 < c.r - EPS) return 1;
    const Point h = Projection(l, c.p);
    if(dot(l.a - h, l.b - h) < 0) return 2;
    return 0;
  }
 
  bool Intersect(const Circle &a, const Circle &b)
  {
    return ((a.p - b.p).norm() - (a.r + b.r) * (a.r + b.r) < EPS) &&
           ((a.p - b.p).norm() - (a.r - b.r) * (a.r - b.r) > -EPS);
  }
 
  double Distance(const Segment &s, const Point &p)
  {
    Point r = Projection(s, p);
    if(Intersect(s, r)) return (r - p).abs();
    return min((s.a - p).abs(), (s.b - p).abs());
  }
 
  double Distance(const Segment &a, const Segment &b)
  {
    if(Intersect(a, b)) return 0;
    return min(min(Distance(a, b.a), Distance(a, b.b)), min(Distance(b, a.a), Distance(b, a.b)));
  }
 
  double Distance(const Line &l, const Line &m)
  {
    return Intersect(l, m) ? 0 : Distance(l, m.a);
  }
 
  double Distance(const Line &l, const Segment &s)
  { //OK
    if(Intersect(l, s)) return 0;
    return min(Distance(l, s.a), Distance(l, s.b));
  }
 
  double Distance(const Point &a, const Point &b)
  { //OK
    return (a - b).abs();
  }
 
  Point Crosspoint(const Segment &l, const Segment &m)
  { //OK
    double A = cross(l.b - l.a, m.b - m.a);
    double B = cross(l.b - l.a, l.b - m.a);
    if(abs(A) < EPS && abs(B) < EPS) return m.a; // same line
    return m.a + (m.b - m.a) * B / A;
  }
 
  PointPoint Crosspoint(const Circle &c, const Line l)
  {
    Point hp = Projection(l, c.p), h = hp - c.p;
    const double d2 = h.norm();
    Point v = (l.b - l.a) * sqrt(c.r * c.r - d2) / (l.b - l.a).abs();
    return PointPoint(hp - v, hp + v);
  }
 
  PointPoint Crosspoint(const Circle &c, const Segment &l)
  {
    Line aa = Line(l.a, l.b);
    if(Intersect(c, l) == 2) return Crosspoint(c, aa);
    PointPoint ret = Crosspoint(c, aa);
    if(dot(l.a - ret.first, l.b - ret.first) < 0) ret.second = ret.first;
    else ret.first = ret.second;
    return ret;
  }
 
  PointPoint Crosspoint(const Circle &c1, const Circle &c2)
  {
    double d = (c1.p - c2.p).abs();
    double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));
    double t = atan2(c2.p.y - c1.p.y, c2.p.x - c1.p.x);
    Point p1 = c1.p + Point(cos(t + a) * c1.r, sin(t + a) * c1.r);
    Point p2 = c1.p + Point(cos(t - a) * c1.r, sin(t - a) * c1.r);
    return (PointPoint(p1, p2));
  }
 
  bool parallel(const Line &a, const Line &b)
  {
    return abs(cross(a.b - a.a, b.b - b.a)) < EPS;
  }
 
  bool orthogonal(const Line &a, const Line &b)
  {
    return abs(dot(a.a - a.b, b.a - b.b)) < EPS;
  }
 
  int Contains(const Polygon &Q, const Point &p)
  {
    bool in = false;
    for(int i = 0; i < Q.size(); i++) {
      Point a = curr(Q, i) - p, b = next(Q, i) - p;
      if(a.y > b.y) swap(a, b);
      if(a.y <= 0 && 0 < b.y && cross(a, b) < 0) in = !in;
      if(cross(a, b) == 0 && dot(a, b) <= 0) return ON;
    }
    return in ? IN : OUT;
  }
 
  bool Contains(const Circle &c, const Point &p)
  {
    return (c.p - p).abs() < c.r + EPS;
  }
 
  double Area2(const Polygon &p)
  { //OK
    double A = 0;
    for(int i = 0; i < p.size(); ++i) {
      A += cross(curr(p, i), next(p, i));
    }
    return A;
  }
 
  bool IsConvex(const Polygon &p)
  {
    for(int i = 0; i < p.size(); i++) {
      if(ccw(prev(p, i), curr(p, i), next(p, i)) == -1) return false;
    }
    return true;
  }
 
  Polygon Convex_Hull(Polygon &p)
  {
    int n = p.size(), k = 0;
    if(n >= 3) {
      sort(p.begin(), p.end());
      vector< Point > ch(2 * n);
      for(int i = 0; i < n; ch[k++] = p[i++]) {
        while(k >= 2 && cross(ch[k - 1] - ch[k - 2], p[i] - ch[k - 1]) < 0) --k;
      }
      for(int i = n - 2, t = k + 1; i >= 0; ch[k++] = p[i--]) {
        while(k >= t && cross(ch[k - 1] - ch[k - 2], p[i] - ch[k - 1]) < 0) --k;
      }
      ch.resize(k - 1);
      return ch;
    } else {
      return p;
    }
  }
 
  double Convex_Diameter(Polygon &p)
  {
    int n = p.size();
    int is = 0, js = 0;
    for(int i = 1; i < n; i++) {
      if(p[i].y > p[is].y) is = i;
      if(p[i].y < p[js].y) js = i;
    }
    double maxdis = (p[is] - p[js]).norm();
 
    int maxi, maxj, i, j;
    i = maxi = is;
    j = maxj = js;
    do {
      if(cross(next(p, i) - curr(p, i), next(p, j) - curr(p, j)) >= 0) {
        j = (j + 1) % n;
      } else {
        i = (i + 1) % n;
      }
      if((p[i] - p[j]).norm() > maxdis) {
        maxdis = (p[i] - p[j]).norm();
        maxi = i;
        maxj = j;
      }
    } while(i != is || j != js);
    return maxdis;
  }
};
 
using namespace Geometry;
 
bool solve()
{
  int N, H[100];
  vector< Point > P[100];
  double theta, phi;
  Point S, T;
 
  cin >> N;
  if(N == 0) return (false);
  for(int i = 0; i < N; i++) {
    int sz;
    cin >> sz >> H[i];
    while(sz--) {
      Point s;
      cin >> s;
      P[i].emplace_back(s);
    }
  }
  cin >> theta >> phi;
  theta *= PI / 180;
  phi *= PI / 180;
  cin >> S >> T;
 
  Polygon all[102];
  for(int i = 0; i < N; i++) {
    Point ident(-H[i] / tan(phi), 0);
    ident = ident.rotate(theta);
    for(auto &p : P[i])all[i].emplace_back(p + ident);
    for(auto &p : P[i]) all[i].emplace_back(p);
    all[i] = Convex_Hull(all[i]);
  }
 
  double mat[100][100];
  fill_n(*mat, 100 * 100, 1e18);
 
  for(int i = 0; i < N; i++) {
    mat[i][i] = 0.0;
  }
  for(int i = 0; i < N; i++) {
    for(int j = 0; j < N; j++) {
      if(i == j) continue;
      for(int k = 0; k < all[i].size(); k++) {
        for(int l = 0; l < all[j].size(); l++) {
          Segment latte = Segment(all[i][k], all[i][(k + 1) % all[i].size()]);
          Segment malta = Segment(all[j][l], all[j][(l + 1) % all[j].size()]);
          if(Intersect(latte, malta)) mat[i][j] = 0.0;
          else mat[i][j] = min(mat[i][j], Distance(latte, malta));
        }
      }
    }
  }
 
  for(int k = 0; k < N; k++) {
    for(int i = 0; i < N; i++) {
      for(int j = 0; j < N; j++) {
        mat[i][j] = min(mat[i][j], mat[i][k] + mat[k][j]);
      }
    }
  }
 
  double latte[100], malta[100];
  fill_n(latte, 100, 1e18);
  fill_n(malta, 100, 1e18);
  for(int i = 0; i < N; i++) {
    if(Contains(all[i], S)) {
      latte[i] = 0.0;
    } else {
      for(int j = 0; j < all[i].size(); j++) {
        Segment beet = Segment(all[i][j], all[i][(j + 1) % all[i].size()]);
        latte[i] = min(latte[i], Distance(beet, S));
      }
    }
    if(Contains(all[i], T)) {
      malta[i] = 0.0;
    } else {
      for(int j = 0; j < all[i].size(); j++) {
        Segment beet = Segment(all[i][j], all[i][(j + 1) % all[i].size()]);
        malta[i] = min(malta[i], Distance(beet, T));
      }
    }
  }
 
  double ret = (S - T).abs();
  for(int i = 0; i < N; i++) {
    for(int j = 0; j < N; j++) {
      ret = min(ret, mat[i][j] + latte[i] + malta[j]);
    }
  }
  cout << fixed << setprecision(10) << ret << endl;
 
  return (true);
}
 
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  while(solve());
}