Problem: Suppose that there are some light sources and many spherical balloons. All light sources have sizes small enough to be modeled as point light sources, and they emit light in all directions. The surfaces of the balloons absorb light and do not reflect light. Surprisingly in this world, balloons may overlap.

You want the total illumination intensity at an objective point as high as possible. For this purpose, some of the balloons obstructing lights can be removed. Because of the removal costs, however, there is a certain limit on the number of balloons to be removed. Thus, you would like to remove an appropriate set of balloons so as to maximize the illumination intensity at the objective point.

The following figure illustrates the configuration specified in the first dataset of the sample input given below. The figure shows the xy-plane, which is enough because, in this dataset, the z-coordinates of all the light sources, balloon centers, and the objective point are zero. In the figure, light sources are shown as stars and balloons as circles. The objective point is at the origin, and you may remove up to 4 balloons. In this case, the dashed circles in the figure correspond to the balloons to be removed.

<image>

Figure G.1: First dataset of the sample input.



Input

The input is a sequence of datasets. Each dataset is formatted as follows.

N M R
S1x S1y S1z S1r
...
SNx SNy SNz SNr
T1x T1y T1z T1b
...
TMx TMy TMz TMb
Ex Ey Ez


The first line of a dataset contains three positive integers, N, M and R, separated by a single space. N means the number of balloons that does not exceed 2000. M means the number of light sources that does not exceed 15. R means the number of balloons that may be removed, which does not exceed N.

Each of the N lines following the first line contains four integers separated by a single space. (Six, Siy, Siz) means the center position of the i-th balloon and Sir means its radius.

Each of the following M lines contains four integers separated by a single space. (Tjx, Tjy, Tjz) means the position of the j-th light source and Tjb means its brightness.

The last line of a dataset contains three integers separated by a single space. (Ex, Ey, Ez) means the position of the objective point.

Six, Siy, Siz, Tjx, Tjy, Tjz, Ex, Ey and Ez are greater than -500, and less than 500. Sir is greater than 0, and less than 500. Tjb is greater than 0, and less than 80000.

At the objective point, the intensity of the light from the j-th light source is in inverse proportion to the square of the distance, namely

Tjb / { (Tjx − Ex)2 + (Tjy − Ey)2 + (Tjz − Ez)2 },

if there is no balloon interrupting the light. The total illumination intensity is the sum of the above.

You may assume the following.

1. The distance between the objective point and any light source is not less than 1.
2. For every i and j, even if Sir changes by ε (|ε| < 0.01), whether the i-th balloon hides the j-th light or not does not change.


The end of the input is indicated by a line of three zeros.

Output

For each dataset, output a line containing a decimal fraction which means the highest possible illumination intensity at the objective point after removing R balloons. The output should not contain an error greater than 0.0001.

Example

Input

12 5 4
0 10 0 1
1 5 0 2
1 4 0 2
0 0 0 2
10 0 0 1
3 -1 0 2
5 -1 0 2
10 10 0 15
0 -10 0 1
10 -10 0 1
-10 -10 0 1
10 10 0 1
0 10 0 240
10 0 0 200
10 -2 0 52
-10 0 0 100
1 1 0 2
0 0 0
12 5 4
0 10 0 1
1 5 0 2
1 4 0 2
0 0 0 2
10 0 0 1
3 -1 0 2
5 -1 0 2
10 10 0 15
0 -10 0 1
10 -10 0 1
-10 -10 0 1
10 10 0 1
0 10 0 260
10 0 0 200
10 -2 0 52
-10 0 0 100
1 1 0 2
0 0 0
5 1 3
1 2 0 2
-1 8 -1 8
-2 -3 5 6
-2 1 3 3
-4 2 3 5
1 1 2 7
0 0 0
5 1 2
1 2 0 2
-1 8 -1 8
-2 -3 5 6
-2 1 3 3
-4 2 3 5
1 1 2 7
0 0 0
0 0 0


Output

3.5
3.6
1.1666666666666667
0.0

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
#include<unordered_map>
#pragma warning(disable:4996)
using namespace std;

struct Point3 {
	long double x;
	long double y;
	long double z;
	Point3() :x(0), y(0), z(0) {}
	Point3(const long double _x, const long double _y, const long double _z):x(_x),y(_y),z(_z) {}
};

Point3 operator+(const Point3&l,const Point3& r) {
	return Point3(l.x + r.x, l.y + r.y, l.z + r.z);
}
Point3 operator+=(Point3&l, const Point3& r) {
	return l = Point3(l.x + r.x, l.y + r.y, l.z + r.z);
}
Point3 operator-(const Point3&l, const Point3& r) {
	return Point3(l.x - r.x, l.y - r.y, l.z - r.z);
}
Point3 operator-=(Point3&l, const Point3& r) {
	return l=Point3(l.x - r.x, l.y - r.y, l.z - r.z);
}
Point3 operator*(const Point3&l, const long double r) {
	return Point3(l.x * r, l.y * r, l.z * r);
}
Point3 operator*(const long double r, const Point3&l) {
	return l*r;
}
Point3 operator*=(Point3&l, const long double r) {
	return l=Point3(l.x * r, l.y * r, l.z * r);
}
Point3 operator/(const Point3&l, const long double r) {
	return Point3(l.x / r, l.y / r, l.z / r);
}
Point3 operator/=(Point3&l, const long double r) {
	return l=Point3(l.x / r, l.y / r, l.z / r);
}
const long double eps = 1e-9, pi = acos(-1.0);
const long double dtop = pi / 180.;
const long double ptod = 1. / dtop;

namespace std {
	bool operator<(const Point3 &lhs, const Point3 &rhs) {
		if (lhs.x< rhs.x - eps) return true;
		if (lhs.x > rhs.x + eps) return false;
		if (lhs.y< rhs.y- eps) return true;
		if (lhs.y > rhs.y + eps) return false;
		return lhs.z < rhs.z;
	}
}

// 点の入力
Point3 input_point() {
	long double x, y,z;
	cin >> x >> y>>z;
	return Point3(x, y,z);
}

// 誤差つき等号判定
bool eq(const long double a, const long double b) {
	return (abs(a - b) < eps);
}

// 内積
long double dot(const Point3& a, const Point3& b) {
	return a.x*b.x+a.y*b.y+a.z*b.z;
}

// 外積
Point3 cross(const Point3& a, const Point3& b) {
	return Point3(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x);
}

//ノルム
long double norm(const Point3&p) {
	return (p.x*p.x + p.y*p.y + p.z*p.z);
}

long double abs(const Point3&p) {

	return sqrt(p.x*p.x + p.y*p.y + p.z*p.z);
}



//球
struct Sphere {
	Point3 p;
	long double r;
	Sphere() :p(), r(0) {}
	Sphere(const Point3 _point, const long double _radius) :p(_point), r(_radius) {}
};

// 直線の定義
class Line3 {
public:
	Point3 a, b;
	Line3() : a(Point3(0, 0,0)), b(Point3(0,0,0)) {}
	Line3(Point3 a, Point3 b) : a(a), b(b) {}
	Point3 operator[](const int _num)const {
		if (_num == 0)return a;
		else if (_num == 1)return b;
		else assert(false);
	}
};

// 点の線分上判定
bool isis_sp(const Line3& s, const Point3& p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}

// 垂線の足
Point3 proj(const Line3 &l, const Point3& p) {
	long double t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}
// 直線と点の距離
long double dist_lp3(const Line3& l, const Point3& p) {
	return abs(p - proj(l, p));
}
// 球と直線の交点
vector<Point3> is_lsp(const Sphere& c, const Line3& l) {
	vector<Point3> res;
	long double d = dist_lp3(l, c.p);
	if (d < c.r + eps) {
		long double len = (d > c.r) ? 0.0 : sqrt(c.r * c.r - d * d); //safety;
		Point3 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;
}
// 球と線分の交点
vector<Point3> is_ssp(const Sphere& c, const Line3& l) {
	vector<Point3> res(is_lsp(c, l));
	vector<Point3> nres;
	for (auto p : res) {
		if (isis_sp(l, p))nres.emplace_back(p);
	}
	return nres;
}
int main() {
	while (1) {

		int N, M, R; cin >> N >> M >> R;
		if (!N)break;
		vector<Sphere>obs;
		for (int i = 0; i < N; ++i) {
			long double x, y, z, r; cin >> x >> y >> z >> r;
			obs.emplace_back(Point3(x, y, z), r);
		}
		vector<Point3>lis;
		vector<long double>lumis;
		for (int i = 0; i < M; ++i) {
			long double x, y, z, b; cin >> x >> y >> z >> b;
			lis.emplace_back(x, y, z);
			lumis.emplace_back(b);
		}
		long double x, y, z;
		cin >> x >> y >> z;
		Point3 goal(x, y, z);
		for (int i = 0; i < M; ++i) {
			lumis[i] = lumis[i] / norm(goal - lis[i]);
		}
		vector<vector<int>>blocks(lis.size());
		for (int i = 0; i < obs.size(); ++i) {
			for (int j = 0; j < lis.size(); ++j) {
				Point3 li(lis[j]);
				const Line3 ltoob(li, goal);
				if (!is_ssp(obs[i], ltoob).empty()) {
					blocks[j].push_back(i);
				}
			}
		}
		long double ans = 0;
		for (int i = 0; i < (1 << M); ++i) {
			bitset<15>bs(i);
			vector<bool>need(N,false);
			set<int>needremove;
			for (int j = 0; j < M; ++j) {
				if (bs[j]) {
					for (int k = 0; k < blocks[j].size(); ++k) {
						need[blocks[j][k]] = true;
					}
				}
			}
			int num = 0;
			if (count(need.begin(), need.end(), true) <= R) {
				long double nowans = 0;
				for (int j = 0; j < M; ++j) {
					if (bs[j]) {
						nowans += lumis[j];
					}
				}
				ans = max(ans, nowans);

			}
		}
		cout <<setprecision(22)<<fixed<< ans << endl;
	}
	return 0;
}