Problem: Problem

In this problem, a straight line passing through a certain point X and a certain point Y in the three-dimensional space is written as a straight line XY.

A cone and a point P inside the cone are given in three-dimensional space. Let point Q be the intersection of a straight line passing through point P and perpendicular to the bottom of the cone and the bottom of the cone. At this time, the center of the bottom surface of the cone is O, and the intersection of the straight line OQ and the circumference of the bottom surface of the cone is point A and point B from the side closer to point Q (however, point O and point Q are the same point). In the case of, let points A and B be the intersections of any straight line passing through point O and the circumference of the bottom of the cone.) Also, in a plane that includes the bottom surface of the cone, let points C and D be the intersections of the straight line that passes through point O and intersects the straight line AB perpendicularly with the circumference of the bottom surface of the cone (however, points C and D). Can be replaced).


The figure below is a view of the bottom surface of the cone as seen from the direction of the apex of the cone.

View of the bottom of the cone as seen from the apex of the cone


Let the intersection of the straight line AP and the cone of the straight line BP (in the case of the straight line AP, the intersection that is not the point A. The same applies to the straight line BP) be the point A'and the point B', and pass through the points A'and B'. Cut the cone at the plane S parallel to the straight line CD. At this time, output the volume of the figure including the vertices after cutting and the volume of the other figure separated by blanks.

Input

X coordinate y coordinate z coordinate of the apex of the cone
X coordinate y coordinate z coordinate of the center of the base of the cone Radius of the base of the cone
X coordinate y coordinate z coordinate of point P

Each value on each row is given separated by spaces.

Constraint

All input values ​​are integers, the absolute value of which is less than or equal to 1000.
The height of the cone and the radius of its base are greater than or equal to 1, and the distance between point P and the cone is greater than or equal to 10-5.

Output

Output the volume of the figure including the vertices after cutting and the volume of the other figure separated by blanks.
However, it must not contain an error greater than 10-5.

Sample Input 1


0 0 10
0 0 0 4
0 0 1


Sample Output 1


91.769438 75.782170


Sample Input 2


0 0 10
0 0 0 4
1 1 1


Sample Output 2


84.050413 83.501195


Sample Input 3


0 0 0
8 9 10 5
1 1 1


Sample Output 3


0.071663 409.709196






Example

Input

0 0 10
0 0 0 4
0 0 1


Output

91.769438 75.782170

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

#define double long double

/* 基本要素 */
typedef complex<double> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> VP;
const double PI = acos(-1);
const double EPS = 1e-5; // 許容誤差^2
const double INF = 1e9;
#define X real()
#define Y imag()
// #define LE(n,m) ((n) < (m) + EPS)
#define LE(n,m) ((n) - (m) < EPS)
// #define GE(n,m) ((n) + EPS > (m))
#define GE(n,m) (EPS > (m) - (n))
#define EQ(n,m) (abs((n)-(m)) < EPS)

// 内積　dot(a,b) = |a||b|cosθ
double dot(Point a, Point b) {
  return a.X*b.X + a.Y*b.Y;
}

// 外積　cross(a,b) = |a||b|sinθ
double cross(Point a, Point b) {
  return a.X*b.Y - a.Y*b.X;
}

// 点の進行方向
// !!! 誤差に注意 !!! (掛け算したものとかなり小さいものを比べているので)
int ccw(Point a, Point b, Point c) {
  b -= a;  c -= a;
  if (cross(b,c) >  EPS) return +1;  // counter clockwise
  if (cross(b,c) < -EPS) return -1;  // clockwise
  if (dot(b,c)   < -EPS) return +2;  // c--a--b on line
  if (norm(b) < norm(c)) return -2;  // a--b--c on line or a==b
  return 0;                          // a--c--b on line or a==c or b==c
}

// 点pの直線aへの射影点を返す
Point proj(Point a1, Point a2, Point p) {
  return a1 + dot(a2-a1, p-a1)/norm(a2-a1) * (a2-a1);
}

double distLP(Point a1, Point a2, Point p) {
  return abs(proj(a1, a2, p) - p);
}

// 2直線の交点
Point crosspointLL(Point a1, Point a2, Point b1, Point b2) {
  double d1 = cross(b2-b1, b1-a1);
  double d2 = cross(b2-b1, a2-a1);
  if (EQ(d1, 0) && EQ(d2, 0)) return a1;  // same line
  assert(!EQ(d2, 0)); // 交点がない
  return a1 + d1/d2 * (a2-a1);
}


struct P{
    double x,y,z;
    P(){}
    P(double _x, double _y, double _z){
        x = _x;
        y = _y;
        z = _z;
    }
};

P READ(){
    int x,y,z;
    cin >>x >>y >>z;
    return P(x,y,z);
}

int main(){
    P top,c,p;
    int r;

    top = READ();
    c = READ();
    cin >>r;
    p = READ();

    // 中心を(0,0,0)に移動
    top.x -= c.x;
    top.y -= c.y;
    top.z -= c.z;
    p.x -= c.x;
    p.y -= c.y;
    p.z -= c.z;
    c.x -= c.x;
    c.y -= c.y;
    c.z -= c.z;

    // topをx軸正方向に来るように回転
    if( !EQ(abs(Point(top.x,top.y)-Point(0,0)),0) ){
        double theta = acos(top.x/hypot(top.x,top.y));
        if(top.y<0) theta = theta + 2*(PI-theta);

        Point tmp = Point(p.x,p.y) * polar((double)1.0, -theta);
        p.x = tmp.X;
        p.y = tmp.Y;

        tmp = Point(top.x,top.y) * polar((double)1.0, -theta);
        top.x = tmp.X;
        top.y = tmp.Y;
    }
    if(top.z<0){
        top.z = -top.z;
        p.z = -p.z;
    }

    // y軸方向に回転させて、頂点をz軸上へ
    if( !EQ(top.x,0) ){
        double tz = PI/2 - acos(top.x/hypot(top.x,top.z));

        Point tmp = Point(p.x,p.z) * polar((double)1.0,tz);
        p.x = tmp.X;
        p.z = tmp.Y;

        tmp = Point(top.x,top.z) * polar((double)1.0,tz);
        top.x = tmp.X;
        top.z = tmp.Y;
    }

    // pをx軸上へ
    if( !EQ(p.y,0) ){
        double theta = acos(p.x/hypot(p.x,p.y));
        if(p.y<0) theta = theta + 2*(PI-theta);

        Point tmp = Point(p.x,p.y) * polar((double)1.0, -theta);
        p.x = tmp.X;
        p.y = tmp.Y;
    }

    // printf(" top %Lf %Lf %Lf\n", top.x,top.y,top.z);
    // printf("   c %Lf %Lf %Lf\n", c.x,c.y,c.z);
    // printf("   p %Lf %Lf %Lf\n", p.x,p.y,p.z);

    double V = PI*r*r*top.z/3;

    Point O(0,0), A(r,0), B(-r,0), T(0,top.z), P(p.x,p.z);
    Point AA = crosspointLL(A,P,B,T);
    Point BB = crosspointLL(B,P,A,T);

    Point M = (AA+BB)*(double)0.5;
    // dbg(M);

    double h = distLP(AA,BB,T);

    double a = abs(AA-M);
    double xx = r*(T.Y-M.Y)/T.Y;
    double b = sqrt(xx*xx - M.X*M.X);

    double U = PI*a*b*h/3;
    printf("%.10Lf %.10Lf\n", U, V-U);
    return 0;
}