Your task is to write a program, which, given two circles, calculates the area of their intersection with the accuracy of three digits after decimal point.

Input

In the single line of input file there are space-separated real numbers x1 y1 r1 x2 y2 r2. They represent center coordinates and radii of two circles.

Process to the end of input.

Output

The output file must contain single real number - the area.

Sample Input

20.0 30.0 15.0 40.0 30.0 30.0

Sample Output

608.366

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>

using namespace std;

const double eps=1e-9;

int dcmp(double x){if(fabs(x)<eps) return 0; return (x<0)?-1:1;}

struct Point
{
    double x,y;
    Point(double _x=0,double _y=0):x(_x),y(_y){};
};

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

bool operator<(const Point&a,const Point&b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}

bool operator==(const Point&a,const Point&b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}

double Dot(Point A,Point B) {return A.x*B.x+A.y*B.y;}
double Length(Point A) {return sqrt(Dot(A,A));}
double Angle(Point A,Point B) {return acos(Dot(A,B)/Length(A)/Length(B));}
double Angle(Point v) {return atan2(v.y,v.x);}
double Cross(Point A,Point B) {return A.x*B.y-A.y*B.x;}

/**Cross
    P*Q > 0 P在Q的顺时针方向
    P*Q < 0 P在Q的逆时针方向
    P*Q = 0 PQ共线
*/

Point Horunit(Point x) {return x/Length(x);}///单位向量
Point Verunit(Point x) {return Point(-x.y,x.x)/Length(x);}///单位法向量

Point Rotate(Point A,double rad)///逆时针旋转(围绕原点)
{
    return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}

double Area2(const Point A,const Point B,const Point C)
{
    return Cross(B-A,C-A);
}

/// 过两点p1, p2的直线一般方程ax+by+c=0  (x2-x1)(y-y1) = (y2-y1)(x-x1)
void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double&b, double &c)
{
    a = p2.y-p1.y;
    b = p1.x-p2.x;
    c = -a*p1.x - b*p1.y;
}

///P+t*v Q+w*t的焦点
Point GetLineIntersection(Point P,Point v,Point Q,Point w)
{
    Point u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}


///点到直线距离
double DistanceToLine(Point P,Point A,Point B)
{
    Point v1=B-A,v2=P-A;
    return fabs(Cross(v1,v2))/Length(v1);
}

///点到线段距离
double DistanceToSegment(Point P,Point A,Point B)
{
    if(A==B) return Length(P-A);
    Point v1=B-A,v2=P-A,v3=P-B;
    if(dcmp(Dot(v1,v2))<0) return Length(v2);
    else if(dcmp(Dot(v1,v3))>0) return Length(v3);
    else return fabs(Cross(v1,v2))/Length(v1);
}

///点到直线投影
Point GetLineProjection(Point P,Point A,Point B)
{
    Point v=B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}

///判断规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
    double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);

    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}

///一个点是否在直线端点上
bool OnSegment(Point p,Point a1,Point a2)
{
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

///多边形有向面积
double PolygonArea(Point* p,int n)
{
    double area=0;
    for(int i=1;i<n-1;i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
    return area/2;
}

///有向直线
struct Line
{
    Point p;
    Point v;
    double ang;
    Line(Point _p,Point _v):p(_p),v(_v){ang=atan2(v.y,v.x);}
    Point point(double a) {return p+(v*a);}
    bool operator<(const Line& L)const
        {
            return ang<L.ang;
        }
};

///直线平移距离d
Line LineTransHor(Line l,int d)
{
    Point vl=Verunit(l.v);
    Point p1=l.p+vl*d,p2=l.p-vl*d;
    Line ll=Line(p1,l.v);
    return ll;
}

///直线交点（假设存在）
Point GetLineIntersection(Line a,Line b)
{
    return GetLineIntersection(a.p,a.v,b.p,b.v);
}

///点p在有向直线的左边
bool OnLeft(const Line& L,const Point& p)
{
    return Cross(L.v,p-L.p)>=0;
}

///圆
const double pi=3.1415926;

struct Circle
{
    Point c;
    double r;
    Circle(Point _c=0,double _r=0):c(_c),r(_r){}
    Point point(double a)///根据圆心角算圆上的点
    {
        return Point(c.x+cos(a)*r,c.y+sin(a)*r);
    }
};

///a点到b点（逆时针）在圆上的圆弧长度
double D(Point a,Point b,Circle C)
{
    double ang1,ang2;
    Point v1,v2;
    v1=a-C.c; v2=b-C.c;
    ang1=atan2(v1.y,v1.x);
    ang2=atan2(v2.y,v2.x);
    if(ang2<ang1) ang2+=2*pi;
    return C.r*(ang2-ang1);
}

///直线与圆交点 返回交点个数
int getLineCircleIntersection(Line L,Circle C,double& t1,double& t2,vector<Point>& sol)
{
    double a=L.v.x,b=L.p.x-C.c.x,c=L.v.y,d=L.p.y-C.c.y;
    double e=a*a+c*c,f=2*(a*b+c*d),g=b*b+d*d-C.r*C.r;
    double delta=f*f-4.*e*g;
    if(dcmp(delta)<0) return 0;//相离
    if(dcmp(delta)==0)//相切
    {
        t1=t2=-f/(2.*e); sol.push_back(L.point(t1));
        return 1;
    }
    //相切
    t1=(-f-sqrt(delta))/(2.*e); sol.push_back(L.point(t1));
    t2=(-f+sqrt(delta))/(2.*e); sol.push_back(L.point(t2));
    return 2;
}


///圆与圆交点 返回交点个数
int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& Sol)
{
    double d=Length(C1.c-C2.c);
    if(dcmp(d)==0)
    {
        if(dcmp(C1.r-C2.r)==0) return -1;//重合
        return 0;
    }
    if(dcmp(C1.r+C2.r-d)<0) return 0;
    if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0;

    double a=Angle(C2.c-C1.c);
    double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));

    Point p1=C1.point(a-da),p2=C1.point(a+da);

    Sol.push_back(p1);
    if(p1==p2) return 1;

    Sol.push_back(p2);
    return 2;
}

///P到圆的切线 v[] 储存切线向量
int getTangents(Point p,Circle C,Point* v)
{
    Point u=C.c-p;
    double dist=Length(u);
    if(dist<C.r) return 0;
    else if(dcmp(dist-C.r)==0)
    {
        ///p在圆上只有一条切线
        v[0]=Rotate(u,pi/2);
        return 1;
    }
    else
    {
        double ang=asin(C.r/dist);
        v[0]=Rotate(u,-ang);
        v[1]=Rotate(u,ang);
        return 2;
    }
}

//两圆公切线 a,b  公切线再 圆 A B 上的切点
int getTengents(Circle A,Circle B,Point* a,Point* b)
{
    int cnt=0;
    if(A.r<B.r) { swap(A,B); swap(a,b); }
    int d2=(A.c.x-B.c.x)*(A.c.x-B.c.x)+(A.c.y-B.c.y)*(A.c.y-B.c.y);
    int rdiff=A.r-B.r;
    int rsum=A.r+B.r;
    if(d2<rdiff*rdiff) return 0;///内含

    double base=atan2(B.c.y-A.c.y,B.c.x-A.c.x);
    if(d2==0&&A.r==B.r) return -1; ///无穷多
    if(d2==rdiff*rdiff)//内切 1条
    {
        a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++;
        return 1;
    }
    ///外切
    double ang=acos((A.r-B.r)/sqrt(1.*d2));
    a[cnt]=A.point(base+ang); b[cnt]=B.point(base+ang); cnt++;
    a[cnt]=A.point(base-ang); b[cnt]=B.point(base-ang); cnt++;
    if(d2==rsum*rsum)// one
    {
        a[cnt]=A.point(base); b[cnt]=B.point(pi+base); cnt++;
    }
    else if(d2>rsum*rsum)// two
    {
        double ang=acos((A.r-B.r)/sqrt(1.*d2));
        a[cnt]=A.point(base+ang); b[cnt]=B.point(pi+base+ang); cnt++;
        a[cnt]=A.point(base-ang); b[cnt]=B.point(pi+base-ang); cnt++;
    }
    return cnt;
}

///三角形外接圆
Circle CircumscribedCircle(Point p1,Point p2,Point p3)
{
    double Bx=p2.x-p1.x,By=p2.y-p1.y;
    double Cx=p3.x-p1.x,Cy=p3.y-p1.y;
    double D=2*(Bx*Cy-By*Cx);
    double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x;
    double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y;
    Point p=Point(cx,cy);
    return Circle(p,Length(p1-p));
}

///三角形内切圆
Circle InscribedCircle(Point p1,Point p2,Point p3)
{
    double a=Length(p2-p3);
    double b=Length(p3-p1);
    double c=Length(p1-p2);
    Point p=(p1*a+p2*b+p3*c)/(a+b+c);
    return Circle(p,DistanceToLine(p,p1,p2));
}

double RtoDegree(double x) {return x/pi*180.;}

double R,r;
Point O1,O2;

/// 两圆的面积交
double Circle_Circle_Area_of_overlap(Circle c1, Circle c2)
{
    double d=Length(c1.c-c2.c);
    if(dcmp(c1.r+c2.r-d)<=0) return 0.;
    if(dcmp(fabs(c1.r-c2.r)-d)>=0)
    {
        double minr = min(c1.r,c2.r);
        return pi*minr*minr;
    }
    double x = (d*d + c1.r*c1.r - c2.r*c2.r)/(2*d);
    double t1 = acos(x/c1.r);
    double t2 = acos((d-x)/c2.r);
    return c1.r*c1.r*t1 + c2.r*c2.r*t2 - d*c1.r*sin(t1);
}

int main()
{
    double x1,y1,r1,x2,y2,r2;
    while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&r1,&x2,&y2,&r2)!=EOF)
    {
        printf("%.3lf\n",Circle_Circle_Area_of_overlap(Circle(Point(x1,y1),r1),Circle(Point(x2,y2),r2)));
    }
}

POJ 2546 & ZOJ 1597 Circular Area 两圆的面积交