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>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define SIZE 105

struct Point{
	Point(double arg_x,double arg_y){
		x = arg_x;
		y = arg_y;
	}

	Point(){
		x = y = 0.0;
	}

	Point operator + (Point p){ return Point(x+p.x,y+p.y); }
	Point operator - (Point p){ return Point(x-p.x,y-p.y);}
	Point operator * (double a){ return Point(a*x,a*y); }
	Point operator / (double a){ return Point(x/a,y/a); }

	double abs(){ return sqrt(norm()); }
	double norm(){ 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 fabs(x-p.x) < EPS && fabs(y-p.y) < EPS;
	}

	double x,y;
};

typedef Point Vector;
typedef vector<Point> Polygon;

struct Line{
	Point p[2];
	Line(Point p1,Point p2){
		p[0] = p1;
		p[1] = p2;
	}
	Line(){

	}
};

struct Info{

	int num_points;
	double height;
	Polygon polygon;
};

int N;
Info info[SIZE];
double THETA,FAI;
double min_dist[SIZE][SIZE];
Point start,goal;


double norm(Vector a){
	return a.x*a.x+a.y*a.y;
}

double abs(Vector a){
	return sqrt(norm(a));
}

double cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}

double dot(Vector a,Vector b){
    return a.x*b.x + a.y*b.y;
}

static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -1;
static const int ON_SEGMENT = 0;

int ccw(Point p0,Point p1,Point p2){

	Vector a = p1 - p0;
	Vector 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;
}


Polygon ConvexHull(Polygon V){

	sort(V.begin(),V.end());

	vector<Point> UP,DOWN;

	UP.push_back(V[0]);
	DOWN.push_back(V[0]);

	for(int i = 1; i < V.size(); i++){

		while(UP.size() > 1 && ccw(UP[UP.size()-2],UP[UP.size()-1],V[i]) == COUNTER_CLOCKWISE){

			UP.pop_back();
		}

		while(DOWN.size() > 1 && ccw(DOWN[DOWN.size()-2],DOWN[DOWN.size()-1],V[i]) ==  CLOCKWISE){

			DOWN.pop_back();
		}

		UP.push_back(V[i]);
		DOWN.push_back(V[i]);
	}

	Polygon ret;

	for(int i = 0; i < UP.size(); i++){

		ret.push_back(UP[i]);
	}

	for(int i = DOWN.size()-1; i >= 0; i--){

		ret.push_back(DOWN[i]);
	}

	return ret;
}



Point calc_minus(Point a,Point b){
    Point ret;

    ret.x = a.x-b.x;
    ret.y = a.y-b.y;

    return ret;
}

double calc_len(Vector a){
    return sqrt(a.x*a.x+a.y*a.y);
}

double calc_slope(Line A){

	if(fabs(A.p[0].x-A.p[1].x) < EPS){

		return DBL_MAX;

	}else if(fabs(A.p[0].y-A.p[1].y) < EPS){

		return 0;

	}else{

		return (A.p[0].y-A.p[1].y)/(A.p[0].x-A.p[1].x);
	}
}

int func(double x1,double y1,double x2, double y2, double xp, double yp){
	double naiseki,norm1,norm2,gaiseki;
	norm1 = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	norm2 = sqrt((xp-x1)*(xp-x1)+(yp-y1)*(yp-y1));
	naiseki = (xp-x1)*(x2-x1)+(yp-y1)*(y2-y1);
	gaiseki = (x2-x1)*(yp-y1)-(xp-x1)*(y2-y1);
	if(gaiseki > EPS){
		return 1;
	}else if(gaiseki < -EPS){
		return -1;
	}
	if(naiseki < -EPS){
		return 2;
	}

	if(norm1 < norm2){
		return -2;
	}
	return 0;
}


//★★直線ではなく、線分の交差判定★★
bool is_Cross(Line a,Line b){

	if(func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[0].x,b.p[0].y)*
			func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[1].x,b.p[1].y) <= 0 &&
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[0].x,a.p[0].y)*
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[1].x,a.p[1].y) <= 0){
		return true;
	}
	return false;
}

//★★線分ではなく直線と点の距離★★
double getDistanceLP(Line l,Point p){
    return fabs(cross(calc_minus(l.p[1],l.p[0]),calc_minus(p,l.p[0]))/calc_len(calc_minus(l.p[1],l.p[0])));
}

//★★点と線分の距離★★
double getDistanceSP(Line l,Point p){
	if(dot(calc_minus(l.p[1],l.p[0]),calc_minus(p,l.p[0])) < 0.0)return calc_len(calc_minus(p,l.p[0]));
	if(dot(calc_minus(l.p[0],l.p[1]),calc_minus(p,l.p[1])) < 0.0)return calc_len(calc_minus(p,l.p[1]));
	return getDistanceLP(l,p);
}

//★★線分と線分の距離★★
double getDistance(Line A,Line B){
	if(is_Cross(A,B))return 0.0;
	return min(min(getDistanceSP(A,B.p[0]),getDistanceSP(A,B.p[1])),
			min(getDistanceSP(B,A.p[0]),getDistanceSP(B,A.p[1])));
}

/*
 * IN 2
 * ON 1
 * OUT 0
 *
 */
int contains(Polygon g,Point 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);
}

double calc_dist(Polygon A,Polygon B){

	double ret = HUGE_NUM;

	for(int i = 0; i < A.size(); i++){
		Line a = Line(A[i],A[(i+1)%A.size()]);
		for(int k = 0; k < B.size(); k++){
			Line b = Line(B[k],B[(k+1)%B.size()]);
			ret = min(ret,getDistance(a,b));
		}
	}
	return ret;
}

double calc_dist(Polygon polygon,Point point){

	if(contains(polygon,point) == 2)return 0;

	double ret = HUGE_NUM;
	for(int i = 0; i < polygon.size(); i++){

		Line tmp_line = Line(polygon[i],polygon[(i+1)%polygon.size()]);
		ret = min(ret,getDistanceSP(tmp_line,point));
	}

	return ret;
}

double calc_dist(Point A,Point B){

	return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

void func(){

	for(int i = 0; i < N; i++){

		info[i].polygon.clear();
	}

	for(int i = 0; i < N; i++){

		scanf("%d %lf",&info[i].num_points,&info[i].height);

		for(int loop = 0; loop < info[i].num_points; loop++){

			double x,y;
			scanf("%lf %lf",&x,&y);

			info[i].polygon.push_back(Point(x,y));
		}
	}

	scanf("%lf %lf",&THETA,&FAI);
	double theta = THETA*M_PI/180; //ラジアンにする
	double fai = FAI*M_PI/180;

	scanf("%lf %lf %lf %lf",&start.x,&start.y,&goal.x,&goal.y);

	for(int i = 0; i < N; i++){

		Polygon tmp;

		for(int k = 0; k < info[i].polygon.size(); k++){

			double L = info[i].height/tan(fai);
			tmp.push_back(Point(info[i].polygon[k].x-L*cos(theta),info[i].polygon[k].y-L*sin(theta)));

		}

		for(int k = 0; k < tmp.size(); k++){

			info[i].polygon.push_back(tmp[k]);
		}

		info[i].polygon = ConvexHull(info[i].polygon);
	}

	int START = N,GOAL = N+1;

	for(int i = 0; i <= N+1; i++){
		for(int k = 0; k <= N+1; k++){
			if(k == i){

				min_dist[i][k] = 0;

			}else{

				min_dist[i][k] = HUGE_NUM;
			}
		}
	}

	//スタートとゴールの距離
	min_dist[START][GOAL] = calc_dist(start,goal);
	min_dist[GOAL][START] = calc_dist(start,goal);

	//スタートと各建物の距離
	for(int i = 0; i < N; i++){

		min_dist[START][i] = calc_dist(info[i].polygon,start);
		min_dist[i][START] = min_dist[START][i];
	}

	//ゴールと各建物の距離
	for(int i = 0; i < N; i++){

		min_dist[GOAL][i] = calc_dist(info[i].polygon,goal);
		min_dist[i][GOAL] = min_dist[GOAL][i];
	}

	//建物同士の距離
	for(int i = 0; i < N-1; i++){
		for(int k = i+1; k < N; k++){
			min_dist[i][k] = calc_dist(info[i].polygon,info[k].polygon);
			min_dist[k][i] = min_dist[i][k];
		}
	}

	for(int mid = 0; mid <= N+1; mid++){
		for(int st = 0; st <= N+1; st++){
			if(fabs(min_dist[st][mid]-HUGE_NUM) < EPS)continue;
			for(int gl = 0; gl <= N+1; gl++){
				if(fabs(min_dist[mid][gl]-HUGE_NUM) < EPS)continue;

				min_dist[st][gl] = min(min_dist[st][gl],min_dist[st][mid]+min_dist[mid][gl]);
			}
		}
	}

	printf("%.10lf\n",min_dist[START][GOAL]);
}

int main(){

	while(true){

		scanf("%d",&N);
		if(N == 0)break;

		func();
	}

	return 0;
}