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

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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 pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }

/*
 <url:>
 問題文============================================================
 =================================================================
 解説=============================================================
 ================================================================
 */

typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-9, pi = acos(-1.0);
namespace std {
    bool operator<(const Point &lhs, const Point &rhs) {
        if (lhs.real() < rhs.real() - eps) return true;
        if (lhs.real() > rhs.real() + eps) return false;
        return lhs.imag() < rhs.imag();
    }
}
Point input_point() { ld x, y; cin >> x >> y; return Point(x, y); } // 点の入力
bool eq(ld a, ld b) { return (abs(a - b) < eps); } // 誤差つき等号判定
ld dot(Point a, Point b) { return real(conj(a) * b); } // 内積
ld cross(Point a, Point b) { return imag(conj(a) * b); } // 外積

// 直線の定義
class Line {
public:
    Point a, b;
    Line() : a(Point(0, 0)), b(Point(0, 0)) {}
    Line(Point a, Point b) : a(a), b(b) {}
    Point operator[](const int _num) {
        if (_num == 0)return a;
        else if (_num == 1)return b;
        else assert(false);
    }
};

// 円の定義
class Circle {
public:
    Point p;
    ld r;
    Circle() : p(Point(0, 0)), r(0) {}
    Circle(Point p, ld r) : p(p), r(r) {}
};

// 垂線の足
Point proj(Line l, Point p) {
    ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
    return l.a + t * (l.a - l.b);
}

// CCW
int ccw(Point a, Point b, Point c) {
    b -= a; c -= a;
    if (cross(b, c) > eps) return 1;   // a,b,cが反時計周りの順に並ぶ
    if (cross(b, c) < -eps) return -1; // a,b,cが時計周りの順に並ぶ
    if (dot(b, c) < 0) return 2;       // c,a,bの順に直線に並ぶ
    if (norm(b) < norm(c)) return -2;  // a,b,cの順に直線に並ぶ
    return 0;                          // a,c,bの順に直線に並ぶ
}

/* 交差判定 */
// 直線と直線の交差判定
bool isis_ll(Line l, Line m) { return !eq(cross(l.b - l.a, m.b - m.a), 0); }
// 直線と線分の交差判定
bool isis_ls(Line l, Line s) {
    return isis_ll(l, s) &&
    (cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < eps);
}
// 線分と線分の交差判定
bool isis_ss(Line s, Line 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;
}
// 点の直線上判定
bool isis_lp(Line l, Point p) { return (abs(cross(l.b - p, l.a - p)) < eps); }
// 点の線分上判定
bool isis_sp(Line s, Point p) { return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps); }

/* 距離 */
// 直線と直線の交点
Point is_ll(Line s, Line t) {
    Point sv = s.b - s.a, tv = t.b - t.a;
    assert(cross(sv, tv) != 0);
    return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}

// 直線と点の距離
ld dist_lp(Line l, Point p) {
    return abs(p - proj(l, p));
}

// 直線と直線の距離
ld dist_ll(Line l, Line m) {
    return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

// 直線と線分の距離
ld dist_ls(Line l, Line s) {
    return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}

// 線分と点の距離
ld dist_sp(Line s, Point p) {
    Point r = proj(s, p);
    return isis_sp(s, r) ? abs(r - p) : min(abs(s.a - p), abs(s.b - p));
}

// 線分と線分の距離
ld dist_ss(Line s, Line t) {
    if (isis_ss(s, t)) return 0;
    return min({ dist_sp(s, t.a), dist_sp(s, t.b), dist_sp(t, s.a), dist_sp(t, s.b) });
}

/* 円 */
// 円と円の交点
vector<Point> is_cc(Circle c1, Circle c2) {
    vector<Point> res;
    ld d = abs(c1.p - c2.p);
    ld rc = (d * d + c1.r * c1.r - c2.r * c2.r) / (2 * d);
    ld dfr = c1.r * c1.r - rc * rc;
    if (abs(dfr) < eps) dfr = 0.0;
    else if (dfr < 0.0) return res; // no intersection
    ld rs = sqrt(dfr);
    Point diff = (c2.p - c1.p) / d;
    res.push_back(c1.p + diff * Point(rc, rs));
    if (dfr != 0.0) res.push_back(c1.p + diff * Point(rc, -rs));
    return res;
}
// 円と直線の交点
vector<Point> is_lc(Circle c, Line l) {
    vector<Point> res;
    ld d = dist_lp(l, c.p);
    if (d < c.r + eps) {
        ld len = (d > c.r) ? 0.0 : sqrt(c.r * c.r - d * d); //safety;
        Point nor = (l.a - l.b) / abs(l.a - l.b);
        res.push_back(proj(l, c.p) + len * nor);
        res.push_back(proj(l, c.p) - len * nor);
    }
    return res;
}
// 円と線分の距離
vector<Point> is_sc(Circle c, Line l) {
    vector<Point> v = is_lc(c, l), res;
    for (Point p : v)
        if (isis_sp(l, p)) res.push_back(p);
    return res;
}
// 円と点の接線
vector<Line> tangent_cp(Circle c, Point p) {
    vector<Line> ret;
    Point v = c.p - p;
    ld d = abs(v);
    ld l = sqrt(norm(v) - c.r * c.r);
    if (isnan(l)) { return ret; }
    Point v1 = v * Point(l / d, c.r / d);
    Point v2 = v * Point(l / d, -c.r / d);
    ret.push_back(Line(p, p + v1));
    if (l < eps) return ret;
    ret.push_back(Line(p, p + v2));
    return ret;
}
// 円と円の接線
vector<Line> tangent_cc(Circle c1, Circle c2) {
    vector<Line> ret;
    if (abs(c1.p - c2.p) - (c1.r + c2.r) > -eps) {
        Point center = (c1.p * c2.r + c2.p * c1.r) / (c1.r + c2.r);
        ret = tangent_cp(c1, center);
    }
    if (abs(c1.r - c2.r) > eps) {
        Point out = (-c1.p * c2.r + c2.p * c1.r) / (c1.r - c2.r);
        vector<Line> nret = tangent_cp(c1, out);
        ret.insert(ret.end(), nret.begin(), nret.end());
    }
    else {
        Point v = c2.p - c1.p;
        v /= abs(v);
        Point q1 = c1.p + v * Point(0, 1) * c1.r;
        Point q2 = c1.p + v * Point(0, -1) * c1.r;
        ret.push_back(Line(q1, q1 + v));
        ret.push_back(Line(q2, q2 + v));
    }
    return ret;
}

/* 多角形 */
typedef vector<Point> Polygon;
// 面積
ld area(const Polygon &p) {
    ld res = 0;
    int n = (int)p.size();
    for (int j = 0;j < n;j++) res += cross(p[j], p[(j + 1) % n]);
    return res / 2;
}
// 多角形の回転方向
bool is_counter_clockwise(const Polygon &poly) {
    ld angle = 0;
    int n = (int)poly.size();
    for (int i = 0;i < n;i++) {
        Point a = poly[i], b = poly[(i + 1) % n], c = poly[(i + 2) % n];
        angle += arg((c - b) / (b - a));
    }
    return angle > eps;
}
// 凸性判定
bool isConvex(const Polygon &poly){
    int n = (int)poly.size();
    if(n < 3) return false;
    int s = -3;
    for(int i = 0; i < n;i++){
        int r = ccw(poly[(i+n-1)%n],poly[i%n],poly[(i+1)%n]);
        if(r == 1 && s == -3) s = r;
        if(s*r == -1) return false;
    }
    return true;
}
// 点の内外判定
//  0 => out : 1 => on : 2 => in
int is_in_polygon(const Polygon &poly, Point p) {
    ld angle = 0;
    int n = (int)poly.size();
    for (int i = 0;i < n;i++) {
        Point a = poly[i], b = poly[(i + 1) % n];
        if (isis_sp(Line(a, b), p)) return 1;
        angle += arg((b - p) / (a - p));
    }
    return eq(angle, 0) ? 0 : 2;
}

// 凸包 : 凸多角形のある一辺上にある点を含まない
Polygon convex_hull(vector<Point> ps) {
    int n = (int)ps.size();
    int k = 0;
    sort(ps.begin(), ps.end());
    Polygon ch(2 * n);
    for (int i = 0; i < n; ch[k++] = ps[i++])
        while (k >= 2 && ccw(ch[k - 2], ch[k - 1], ps[i]) <= 0) --k;
    for (int i = n - 2, t = k + 1; i >= 0; ch[k++] = ps[i--])
        while (k >= t && ccw(ch[k - 2], ch[k - 1], ps[i]) <= 0) --k;
    ch.resize(k - 1);
    return ch;
}
// 凸包 : 凸多角形のある一辺上にある点も含む
Polygon convex_hull2(vector<Point> ps) {
    int n = (int)ps.size();
    if (n < 3) return ps;
    sort(ps.begin(), ps.end());
    Polygon u = { ps[0], ps[1] }, l = { ps[n - 1],ps[n - 2] };
    for (int i = 2; i < n; i++) {
        for (int j = (int)u.size(); j >= 2 && ccw(u[j - 2], u[j - 1], ps[i]) >= 0;j--)u.pop_back();
        u.push_back(ps[i]);
    }
    for (int i = n - 3;i >= 0;i--) {
        for (int j = (int)l.size(); j >= 2 && ccw(l[j - 2], l[j - 1], ps[i]) >= 0;j--)l.pop_back();
        l.push_back(ps[i]);
    }
    reverse(l.begin(), l.end());
    for (int i = (int)u.size() - 2; i >= 1; i--)l.push_back(u[i]);
    return l;
}

// 凸多角形の直径
pair<pll,ld> convex_diameter(const Polygon& poly){
    int n = (int)poly.size();
    if(n == 2) return make_pair(pll(0,1),abs(poly[0]-poly[1]));
    int ii = 0, jj = 0;
    for(int i = 1;i < n;i++){
        if(poly[i].imag() > poly[ii].imag())ii = i;
        if(poly[i].imag() < poly[jj].imag())jj = i;
    }
    pll resp = make_pair(-1,-1);
    ld resd = 0;
    
    int i, maxi,j,maxj;
    i = maxi = ii; j = maxj = jj;
    while(i != maxj || j != maxi){
        ld cur = abs(poly[i] - poly[j]);
        if(resd + eps < cur){ resd = cur; resp = pll(i,j); }
        int di = (i+1)%n, dj = (j+1)%n;
        if(cross(poly[di]-poly[i],poly[dj]-poly[j]) < 0) i = di;
        else j = dj;
    }
    return make_pair(resp,resd);
}
// 凸カット
Polygon convex_cut(const Polygon &ps, Line l) {
    int n = (int)ps.size();
    Polygon Q;
    for (int i = 0;i < n;i++) {
        Point A = ps[i], B = ps[(i + 1) % n];
        Line m = Line(A, B);
        if (ccw(l.a, l.b, A) != -1) Q.push_back(A);
        if (ccw(l.a, l.b, A) * ccw(l.a, l.b, B) < 0 && isis_ll(l, m))
            Q.push_back(is_ll(l, m));
    }
    return Q;
}

// 円と多角形の共通部分
double area_of_polygon_and_circle(const Polygon& poly, const Circle c) {
    int n = (int)poly.size();
    ld r = 0;
    for (int i = n - 1, j = 0; j < n; i=j++) {
        Point v = abs(poly[j] - poly[i]) / (poly[j] - poly[i]);
        if (poly[j] == poly[i])continue;
        assert(poly[j] != poly[i]);
        Point a = (poly[i] - c.p)*v, b = (poly[j] - c.p)*v;
        ld d = norm(c.r) - norm(a.imag());
        if (abs(a.imag()) < eps) continue;
        if (d < 0)d = 0;
        d = sqrt(d);
        double l, m;
        r += norm(c.r)*((l = atan2(b.imag(), min(b.real(), -d)) - atan2(a.imag(), min(a.real(), -d)))
                        + (m = atan2(b.imag(), max(b.real(), d)) - atan2(a.imag(), max(a.real(), d))))
        + a.imag()*(min(d, max(a.real(), -d)) - max(-d, min(b.real(), d)));
        assert(-pi < l && -pi < m && l < pi && m < pi);
    }
    return r / 2;
}

int N;
ld solve(){
    ld res = 0;
    
    vector<double> Hs(N);
    vector<Polygon> polys(N);
    for(int i = 0; i < N;i++){
        int n; cin >> n >> Hs[i];
        for(int j = 0; j < n;j++){
            polys[i].push_back(input_point());
        }
    }
    double theta,phi; cin >> theta >> phi;
    theta = theta*pi/180; phi = phi*pi/180;
    Point S = input_point(), T = input_point();
    for(int i = 0; i < N;i++){
        int sz = (int)polys[i].size();
        ld L = Hs[i]/tan(phi);
        for(int j = 0; j < sz;j++){
            ld x = polys[i][j].real(), y = polys[i][j].imag();
            polys[i].push_back(Point(x - L*cos(theta),y - L*sin(theta)));
        }
        polys[i] = convex_hull2(polys[i]);
    }
    
    
    vector<vector<ld>> G(N+2,vector<ld>(N+2,LINF));
    for(int i = 0; i < N+2;i++) G[i][i] = 0;
    
    for(int i = 0; i < N;i++){
        for(int j = i+1; j < N;j++){
            if(i == j) continue;
            ld D = LINF;
            for(int k = 0; k < polys[i].size(); k++){
                for(int l = 0; l < polys[j].size();l++){
                    Line l1(polys[i][k],polys[i][(k+1)%polys[i].size()]);
                    Line l2(polys[j][l],polys[j][(l+1)%polys[j].size()]);
                    D = min(D,dist_ss(l1,l2));
                }
            }
            
            G[i][j] = G[j][i] = D;
        }
    }
    for(int i = 0; i < N;i++){
        if(is_in_polygon(polys[i],S) > 0){
            G[N][i] = G[i][N] = 0;
        }else{
            ld D = LINF;
            for(int k = 0; k < polys[i].size(); k++){
                Line l1(polys[i][k],polys[i][(k+1)%polys[i].size()]);
                D = min(D,dist_sp(l1,S));
            }
            G[N][i] = G[i][N] = D;
        }
        
        if(is_in_polygon(polys[i], T) > 0){
            G[N+1][i] = G[i][N+1] = 0;
        }else{
            ld D = LINF;
            for(int k = 0; k < polys[i].size(); k++){
                Line l1(polys[i][k],polys[i][(k+1)%polys[i].size()]);
                D = min(D,dist_sp(l1,T));
            }
            G[N+1][i] = G[i][N+1] = D;
        }
    }
    G[N][N+1] = G[N+1][N] = abs(S-T);
    for(int i = 0; i < N+2;i++){
        for(int j = 0; j < N+2;j++){
            for(int k = 0; k < N+2;k++){
                G[j][k] = min(G[j][k],G[j][i] + G[i][k]);
            }
        }
    }
    res = G[N][N+1];
    return res;
}
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    while(cin >> N,N){
        cout << fixed << setprecision(12) << solve() << endl;
    }
    return 0;
}