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.

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Figure E1: For the first input

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Figure E2: For the second input

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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;

// conj(x) : complex conjugate,(0,1)->(0,-1)
// abs(x) : dist between(0,0) and x
// norm(x) : abs(x) * abs(x)
// arg(x) : argment,[-PI,PI]
using CP = complex<long double>;
#define X real()
#define Y imag()
const long double PI = acos(-1.0L);
const long double EPS = 1e-10;
// cos a
long double costh(long double a, long double b,
                  long double c) {
  return (b * b + c * c - a * a) / (2.0L * b * c);
}

// dot(a,b) = |a||b|cos x
long double dot(CP a, CP b) { return (conj(a) * b).X; }
// cross(a,b) : area of parallelogram
// sign : a-> b ,counter clockwise? + : -
long double cross(CP a, CP b) { return (conj(a) * b).Y; }
long double corner(CP a, CP b) {
  //[0,PI]
  return acos(dot(a, b) / (abs(a) * abs(b)));
}

CP projection(CP s, CP t, CP p) {
  CP base = t - s;
  long double r = dot(p - s, base) / norm(base);
  return s + base * r;
}

CP reflection(CP s, CP t, CP p) {
  CP tmp = (projection(s, t, p) - p);
  tmp *= 2;
  return p + tmp;
}

CP intersectionLL(CP a, CP b, CP c, CP d) {
  return a + (b - a) * (cross(d - c, c - a) /
                        cross(d - c, b - a));
}

bool on_seg(CP s, CP t, CP p) {
  // if not use end point, dot(s - p, t - p) < 0
  return abs(cross(s - p, t - p)) <= 1e-10 &&
         dot(s - p, t - p) <= 0;
}

// crossing lines? (a,b) and (c,d)
bool iscross(CP a, CP b, CP c, CP d) {
  // parallel
  if(abs(cross(a - b, c - d)) <= 1e-10) {
    return on_seg(a, b, c) || on_seg(a, b, d) ||
           on_seg(c, d, a) || on_seg(c, d, b);
  }
  CP isp = intersectionLL(a, b, c, d);
  return on_seg(a, b, isp) && on_seg(c, d, isp);
}

long double distLP(CP s, CP t, CP p) {
  return abs(cross(t - s, p - s) / abs(t - s));
}

// segmentver.
long double distSP(CP s, CP t, CP p) {
  if(dot(t - s, p - s) < 0) return abs(p - s);
  if(dot(s - t, p - t) < 0) return abs(p - t);
  return distLP(s, t, p);
}

// segment and segment
long double distSS(CP a, CP b, CP c, CP d) {
  long double res = 1e18;
  if(iscross(a, b, c, d)) return 0.0L;
  res = min(res, distSP(a, b, c));
  res = min(res, distSP(a, b, d));
  res = min(res, distSP(c, d, a));
  res = min(res, distSP(c, d, b));
  return res;
}

// counter clockwise
bool is_convex(const vector<CP> &v) {
  int n = v.size();
  for(int i = 0; i < n; ++i)
    if(cross(v[(i + 1) % n] - v[i],
             v[(i + 2) % n] - v[(i + 1) % n]) < -EPS)
      return 0;
  return 1;
}

vector<CP> convex_hull(vector<CP> &ps) {
  auto lmd = [&](const CP &l, const CP &r) {
    if(l.X != r.X) return l.X < r.X;
    return l.Y < r.Y;
  };
  vector<CP> qs;
  int psize = ps.size();
  sort(ps.begin(), ps.end(), lmd);
  int k = 0;
  qs.resize(psize * 2);
  for(int i = 0; i < psize; ++i) {
    while(k > 1 && cross(qs[k - 1] - qs[k - 2],
                         ps[i] - qs[k - 1]) <= 0)
      --k;
    qs[k++] = ps[i];
  }
  for(int i = psize - 2, t = k; i >= 0; --i) {
    while(k > t && cross(qs[k - 1] - qs[k - 2],
                         ps[i] - qs[k - 1]) <= 0)
      --k;
    qs[k++] = ps[i];
  }
  qs.resize(k - 1);
  return qs;
}

long double convex_diameter(vector<CP> &newv) {
  vector<CP> v = convex_hull(newv);
  int n = v.size(), i = 0, j = 0;
  if(n == 2) return abs(v[0] - v[1]);
  for(int k = 0; k < n; ++k) {
    if(v[k].X < v[i].X) i = k;
    if(v[k].X > v[j].X) j = k;
  }
  long double res = 0;
  int si = i, sj = j;
  while(i != sj || j != si) {
    res = max(res, abs(v[i] - v[j]));
    if(cross(v[(i + 1) % n] - v[i], v[(j + 1) % n] - v[j]) <
       0)
      (++i) %= n;
    else
      (++j) %= n;
  }
  return res;
}

vector<CP> convex_cut(const vector<CP> &v, const CP &s,
                      const CP &t) {
  vector<CP> res;
  int n = v.size();
  for(int i = 0; i < n; ++i) {
    CP nows = v[i], nowt = v[(i + 1) % n];
    if(cross(t - s, nows - s) >= -EPS) res.push_back(nows);
    if(cross(t - s, nows - s) * cross(t - s, nowt - s) < 0)
      res.push_back(intersectionLL(s, t, nows, nowt));
  }
  return res;
}

struct Circle {
  CP o;
  long double r;
  Circle(long double _x = 0.0L, long double _y = 0.0L,
         long double _r = 0.0L)
      : o(CP(_x, _y)), r(_r) {}
  Circle(CP _o, long double _r = 0.0) : o(_o), r(_r) {}
};

void intersectionCC(Circle c1, Circle c2, CP &res1,
                    CP &res2) {
  long double d = abs(c1.o - c2.o);
  long double arg = acos(costh(c2.r, c1.r, d));
  res1 = (c2.o - c1.o) * CP(cos(arg), sin(arg));
  res1 *= c1.r / abs(res1);
  res1 += c1.o;
  res2 = reflection(c1.o, c2.o, res1);
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

void intersectionCL(Circle ci, CP s, CP t, CP &res1,
                    CP &res2) {
  res1 = res2 = projection(s, t, ci.o);
  long double r = sqrtl(ci.r * ci.r - norm(res1 - ci.o));
  t -= s;
  t *= r / abs(t);
  res1 += t;
  res2 -= t;
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

void contactCP(Circle ci, CP p, CP &res1, CP &res2) {
  long double d = abs(ci.o - p);
  if(abs(d - ci.r) <= EPS) {
    res1 = res2 = p;
    return;
  }
  else if(d < ci.r) {
    res1 = res2 = CP(-1e18, -1e18);
    return;
  }
  long double arg = asin(ci.r / d);
  res1 = (ci.o - p) * CP(cos(arg), sin(arg));
  res1 *= (d * cos(arg)) / abs(res1);
  res1 += p;
  res2 = reflection(p, ci.o, res1);
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

long double polygonarea(const vector<CP> &v) {
  int n = v.size();
  long double res = 0;
  for(int i = 0; i < n; ++i)
    res +=
        (v[(i - 1 + n) % n].X - v[(i + 1) % n].X) * v[i].Y;
  return res / 2.0L;
}

long long calc_clockwise(CP s, CP t, CP p) {
  t -= s;
  p -= s;
  if(cross(t, p) > EPS) return 1;   // "COUNTER_CLOCKWISE"
  if(cross(t, p) < -EPS) return -1; //"CLOCK_WISE"
  if(dot(t, p) < 0) return 2;       // ONLINE_BACK
  if(norm(t) < norm(p)) return -2;  // ONLINE_FRONT
  return 0;                         // ON_SEGMENT
}

struct buil {
  int n, h;
  vector<CP> v;
};

long long n;
long double theta, phi;
vector<buil> buildings;
vector<vector<long double>> dist;

long double solve();
bool iscontain(CP p, vector<CP> &v);

int main() {
  cout << fixed << setprecision(10);
  while(1) {
    int x, y;
    cin >> n;
    if(n == 0) break;
    buildings.clear();
    buildings.resize(n);
    for(int i = 0; i < n; ++i) {
      cin >> buildings[i].n >> buildings[i].h;
      buildings[i].v.resize(buildings[i].n);
      for(int j = 0; j < buildings[i].n; ++j) {
        cin >> x >> y;
        buildings[i].v[j] = CP(x, y);
      }
    }
    cin >> theta >> phi;
    theta *= PI / 180.0L;
    phi *= PI / 180.0L;
    theta += PI;
    cin >> x >> y;
    buildings.push_back({1, 0, {CP(x, y)}});
    cin >> x >> y;
    buildings.push_back({1, 0, {CP(x, y)}});
    cout << solve() << endl;
  }
  return 0;
}

long double solve() {
  CP e = CP(cos(theta), sin(theta));
  e /= abs(e);
  for(int i = 0; i < n; ++i) {
    vector<CP> tmp = buildings[i].v;
    int vsize = buildings[i].n, h = buildings[i].h;
    for(int j = 0; j < vsize; ++j)
      tmp.push_back(tmp[j] +
                    e * (h / (long double)tan(phi)));
    buildings[i].v = convex_hull(tmp);
    buildings[i].n = buildings[i].v.size();
  }
  n += 2;
  dist.assign(n, vector<long double>(n, 1e18));
  for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j)
      for(int k = 0; k < buildings[i].n; ++k) {
        if(iscontain(buildings[i].v[k], buildings[j].v)) {
          dist[i][j] = dist[j][i] = 0;
          break;
        }
        for(int l = 0; l < buildings[j].n; ++l) {
          CP sk = buildings[i].v[k],
             tk = buildings[i].v[(k + 1) % buildings[i].n],
             sl = buildings[j].v[l],
             tl = buildings[j].v[(l + 1) % buildings[j].n];
          if(i < (n - 2))
            dist[i][j] =
                min(dist[i][j], distSP(sk, tk, sl));
          if(j < (n - 2))
            dist[i][j] =
                min(dist[i][j], distSP(sl, tl, sk));
          dist[i][j] = min(dist[i][j], abs(sk - sl));
        }
      }
  for(int k = 0; k < n; ++k)
    for(int i = 0; i < n; ++i)
      for(int j = 0; j < n; ++j)
        dist[i][j] =
            min(dist[i][j], dist[i][k] + dist[k][j]);
  return dist[n - 2][n - 1];
}

bool iscontain(CP p, vector<CP> &v) {
  bool ch = 0, onseg = 0;
  int vsize = v.size();
  for(int i = 0; i < vsize; ++i) {
    onseg |= on_seg(v[i], v[(i + 1) % vsize], p);
    CP s = v[i] - p, t = v[(i + 1) % vsize] - p;
    if(s.Y > t.Y) swap(s, t);
    if(s.Y * t.Y <= 0 && t.Y > 0 && cross(s, t) < 0)
      ch = !ch;
  }
  return ch || onseg;
}