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;
typedef long long ll;
typedef vector<int> vint;
typedef pair<int,int> pint;
typedef vector<pint> vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=n-1;i>=(0);i--)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define all(v) (v).begin(),(v).end()
#define eall(v) unique(all(v), v.end())
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll INFF = 1e18;

const double EPS = 1e-10;
const double PI = 3.141592653589793;
template<class T> bool eq(T a, T b){ return abs(a - b) < EPS; }

class Point { // ???
public:
	double x, y;
	Point(double x = 0, double y = 0):x(x), y(y){}
	Point operator + (Point p) const { return Point(x + p.x, y + p.y); }
	Point operator - (Point p) const { return Point(x - p.x, y - p.y); }
	Point operator * (double a) const { return Point(a * x, a * y); }
	Point operator / (double a) const { return Point(x / a, y / a); }
	double abs() const { return sqrt(norm()); }
	double norm() const { return x * x + y * y; }
	// bool operator < (const Point &p) const { return x != p.x ? x < p.x : y < p.y; }
	bool operator < (const Point &p) const { // ??????????¨±???????????????
		return x + EPS < p.x || (eq<double>(x, p.x) && y + EPS < p.y);
	}
	bool operator == (const Point &p) const { return (eq<double>(x, p.x) && eq<double>(y, p.y)); }
};
using Vector = Point;
using Polygon = vector<Point>; // ????§???¢

double dot(const Vector& a, const Vector& b) { return a.x * b.x + a.y * b.y; } // ????????????a??¨b?????????
double cross(const Vector& a, const Vector& b) { return a.x * b.y - a.y * b.x; } // ????????????a??¨b?????????
double length2(const Point& a) { return a.norm(); } // ??????????????????2???
double length(const Point& a) { return a.abs(); } // ???????????????
Point rotationalTransfer(Point c, double r, double deg) { // c???????????¨????????????r????????¨??????deg??????????????§?¨?
	double rad = PI * deg / 180.0; return c + Point(cos(rad), sin(rad)) * r;
}
// (x, y, z) ???????????????(xy??§?¨???§????§???????theta???, xy?????¢??????z?????????????§???????phi????????????)??????????????????????????±???xy??§?¨?
Point Shadow(double x, double y, double z, double theta, double phi) {
	theta = PI * theta / 180.0, phi = PI * phi / 180.0;
	return Point(x - z / tan(phi) * cos(theta), y - z / tan(phi) * sin(theta));
}

enum ccw_t {
	COUNTER_CLOCKWISE = 1, // p0->p1 ???????¨???????????????????p2
	CLOCKWISE = -1, // p0->p1 ????¨???????????????????p2
	ONLINE_BACK = 2, // p2->p0->p1 ????????§??´????????§p2
	ONLINE_FRONT = -2, // p0->p1->p2 ????????§??´??????p2
	ON_SEGMENT = 0, // p0->p2->p1 ????????§??????p0p1??????p2
};
ccw_t ccw(Point p0, Point p1, Point p2) {
	Vector a = p1 - p0, b = p2 - p0;
	if ( cross(a, b) > EPS  )  return COUNTER_CLOCKWISE;
	if ( cross(a, b) < -EPS )  return CLOCKWISE;
	if ( dot(a, b) < -EPS )    return ONLINE_BACK;
	if ( a.norm() < b.norm() ) return ONLINE_FRONT;
	return ON_SEGMENT;
}

class Segment { //??????
public:
	Point p1, p2;
	Segment(){}
	Segment(Point p1, Point p2):p1(p1), p2(p2){}
};
using Line = Segment;

// *** ????§???¢ ***
// IN := 2, ON := 1, OUT := 0
vector<Segment> getPolygonSegument(const Polygon& p) { //????§???¢????????????????§???¢??????????±???????
	vector<Segment> ret;
	rep(i, p.size() - 1) ret.push_back(Segment(p[i], p[i + 1]));
	ret.push_back(Segment(p[p.size() - 1], p[0]));
	return ret;
}
int contains(const Polygon& g, const Point& p){ // ????§???¢g????????????p????????????????????????
	int n = g.size(); bool x = false;
	for (int i = 0; i < n; ++i) {
		Point a = g[i] - p, b = g[(i + 1) % n] - p;
		if ( abs(cross(a, b)) < EPS && dot(a, b) < EPS ) return 1;
		if (a.y > b.y) swap(a, b);
		if (a.y < EPS && EPS < b.y && cross(a, b) > EPS ) x = !x;
	}
	return (x ? 2 : 0);
}
Polygon andrewScan(Polygon s) { // ??????(??????????????????????????????)
	Polygon u, l;
	if (s.size() < 3) return s;
	sort(s.begin(), s.end()); // x, y???????????????????????????
	// x????°???????????????????2??? u ?????????
	u.push_back(s[0]), u.push_back(s[1]);
	// x ?????§??????????????????2???1?????????
	l.push_back(s[s.size() - 1]), l.push_back(s[s.size() - 2]);
	// ??????????????¨?????????
	for (int i = 2; i < s.size(); i++) {
		for (int n = u.size(); n >= 2 && ccw(u[n - 2], u[n - 1], s[i]) != CLOCKWISE; n--){
			u.pop_back();
		}
		u.push_back(s[i]);
	}
	// ??????????????¨?????????
	for (int i = s.size() - 3; i >= 0; i--) {
		for (int n = l.size(); n >= 2 && ccw(l[n - 2], l[n - 1], s[i]) != CLOCKWISE; n--){
			l.pop_back();
		}
		l.push_back(s[i]);
	}
	// ????¨????????????????????????????????????????????????????
	reverse(l.begin(), l.end());
	for (int i = u.size() - 2; i >= 1; i--) l.push_back(u[i]);
	return l;
}


// *** ????????????????????? ***
bool intersect(const Point& p1, const Point& p2, const Point& p3, const Point& p4) {
	return ( ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 && 
			 ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0 );
}
bool intersect(const Segment& s1, const Segment& s2) { // ?????????????????????true
	return intersect(s1.p1, s1.p2, s2.p1, s2.p2);
}
//*** ??????????????? ***
Point getCrossPoint(Segment s1, Segment s2) { // ???????????????????????¨??????????????????????????????????????¨
	Vector base = s2.p2 - s2.p1;
	double d1 = abs(cross(base, s1.p1 - s2.p1)), d2 = abs(cross(base, s1.p2 - s2.p1));
	double t  = d1 / (d1 + d2);
	return s1.p1 + (s1.p2 - s1.p1) * t;
}
// *** ?????¢ ***
double getDistance(Point& a, Point& b) { // ???a??¨???b????????¢
	return length(a - b);
}
double getDistanceLP(Line& l, Point& p) { // ??´???s??¨???p????????¢
	return length(cross(l.p2 - l.p1, p - l.p1) / length(l.p2 - l.p1));
}
double getDistanceSP(Segment s, Point p) { // ??????s??¨???p????????¢
	if( dot(s.p2 - s.p1, p - s.p1) < EPS ) return length(p - s.p1);
	if( dot(s.p1 - s.p2, p - s.p2) < EPS ) return length(p - s.p2);
	return getDistanceLP(s, p);
}
double getDistanceSS(const Segment& s1, const Segment& s2) { // ??????s1??¨??????s2?????????
	if( intersect(s1, s2) ) return 0.0; //????????£???????????¨???
	return min(min(getDistanceSP(s1, s2.p1), getDistanceSP(s1, s2.p2)),
			   min(getDistanceSP(s2, s1.p1), getDistanceSP(s2, s1.p2)));
}
double getDistancePolP(const Polygon& pol, const Point& p) { // ????§???¢pol??¨???p????????¢
	if(contains(pol, p) != 0) return 0.0; // ??????????§???¢????????¨?????????????????????
	double ret = 1e9;
	for(Segment& u : getPolygonSegument(pol)) ret = min(ret, getDistanceSP(u, p));
	return ret;
}
double getDistancePolPol(const Polygon& p1, const Polygon& p2) { // ????§???¢p1??¨p2????????¢
	//for(const Point& p : p1) if(contains(p2, p) != 0) return 0.0; // p1?????????????§???¢p2???????????????????????????
	//for(const Point& p : p2) if(contains(p1, p) != 0) return 0.0; // p2?????????????§???¢p1???????????????????????????
	double ret = 1e9;
	for(const Segment& u : getPolygonSegument(p1))for(const Segment& v : getPolygonSegument(p2)) {
		ret = min(ret, getDistanceSS(u, v));
	}
	return ret;
}


class Rectangle { // ????????¢
public:
	// 3 2
	// 0 1 (???????¨??????????????????¢??????????????????????????¨)
	vector<Point> p; // ?????????????????????????????¨
	Rectangle(vector<Point>&p):p(p) {
		rep(i, 3) reps(j, i + 1, 4) { //????????????????????????????????§???????????????????
			int cnt = 0;
			rep(k, 4) if(k != i && k != j) {
				cnt += ccw(p[i], p[j], p[k]) == COUNTER_CLOCKWISE;
			}
			if(cnt == 2) {
				swap(p[i + 1], p[j]);
				break;
			}
		}
	}
	bool intersect(const Segment& s) { // ??????s??¨????????¢????°?????????¨???1??????????????????????????°true
		bool flag = false;
		rep(i, 4) flag |= ::intersect(s, Segment(p[i], p[(i + 1) % 4]));
		return flag;
	}
	bool contain(const Point& pp) { // ???pp???????????¢????????????????????°(??????????????????)true
		bool flag = true;
		rep(i, 4) flag &= ccw(p[i], p[(i + 1) % 4], pp) == COUNTER_CLOCKWISE;
		return flag;
	}
	bool contain(const Segment& s) { // ??????s???????????¢????????????????????°(??????????????????)true
		return contain(s.p1) && contain(s.p2);
	}
};




const int MAX_N = 210;
using TYPE = double; // ?????¢???????????\??????
vector<pair<int, TYPE> > G[MAX_N];
vector<TYPE> dijkstra(int start){
	vector<TYPE> dist(MAX_N, INFF);
	dist[start] = 0;//dist[i] :=???start->i?????§??????????????¢
	priority_queue<pair<TYPE, int>, vector<pair<TYPE, int> >, greater<pair<TYPE, int> > >  que;
	que.push(make_pair(0, start));
	while(!que.empty()){
		TYPE cost; int u;//????????§???????????£????????? ?????¨?????????
		cost = que.top().first, u = que.top().second;
		que.pop();
		if(dist[u] < cost) continue;
		for (auto tmp : G[u]){
			int v = tmp.first; TYPE time = tmp.second;//??£??\???????????? ?????????????????§????????????
			if(dist[v] > dist[u] + time){//u->v
				dist[v] = dist[u] + time;
				que.push(make_pair(dist[v], v));
			}
		}
	}
	return dist;
}


int main(void) {
	

	while(1) {
		int N; scanf("%d", &N);
		if(N == 0) break;
		int NV[110], H[110];
		vector<vector<pair<double, double>>> p;
		rep(i, N){
			scanf("%d %d", &NV[i], &H[i]);
			vector<pair<double, double>> tmp;
			rep(j, NV[i]){
				double x, y; scanf("%lf %lf", &x, &y);
				tmp.pb(mp(x, y));
			}
			p.pb(tmp);
		}
		double theta, phi; scanf("%lf %lf", &theta, &phi);
		double Sx, Sy, Tx, Ty; scanf("%lf %lf %lf %lf", &Sx, &Sy, &Tx, &Ty);
		Point start(Sx, Sy), goal(Tx, Ty);

		vector<Polygon> v;
		rep(i, N){
			Polygon t;
			for(auto u : p[i]) t.pb(Shadow(u.fi, u.se, H[i], theta, phi));
			for(auto u : p[i]) t.pb(Point(u.fi, u.se));
			Polygon tt = andrewScan(t);
			v.pb(tt);
		}

		/*
		for(auto u : v) {
			for(auto k : u) printf("%f %f, ", k.x, k.y);
			printf("\n");
		}
		*/	

		rep(i, MAX_N) G[i].clear();

		rep(i, v.size())rep(j, v.size()){
			if(i == j) continue;
			int s = i + 1, t = j + 1;
			auto dis = getDistancePolPol(v[i], v[j]);
			G[s].pb(mp(t, dis)), G[t].pb(mp(s, dis));
		}

		rep(i, v.size()){
			auto dis = getDistancePolP(v[i], start);
			G[0].pb(mp(i + 1, dis)), G[i + 1].pb(mp(0, dis));
		}
		rep(i, v.size()){
			auto dis = getDistancePolP(v[i], goal);
			G[N + 1].pb(mp(i + 1, dis)), G[i + 1].pb(mp(N + 1, dis));
		}

		G[0].pb(mp(N + 1, getDistance(start, goal))), G[N + 1].pb(mp(0, getDistance(start, goal)));

		auto ret = dijkstra(0);
		printf("%.9f\n", ret[N + 1]);
	}


	return 0;
}