Problem: For given two circles $c1$ and $c2$, print the coordinates of the cross points of them.

Constraints

* The given circle have at least one cross point and have different center coordinates.
* $-10,000 \leq c1x, c1y, c2x, c2y \leq 10,000$
* $1 \leq c1r, c2r \leq 10,000$

Input

The input is given in the following format.

$c1x\; c1y\; c1r$
$c2x\; c2y\; c2r$


$c1x$, $c1y$ and $c1r$ represent the coordinate and radius of the first circle. $c2x$, $c2y$ and $c2r$ represent the coordinate and radius of the second circle. All input values are given in integers.

Output

Print the coordinates ($x1$, $y1$) and ($x2$, $y2$) of the cross points $p1$ and $p2$ respectively in the following rules.

* If there is one cross point, print two coordinates with the same values.
* Print the coordinate with smaller $x$ first. In case of a tie, print the coordinate with smaller $y$ first.



The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

0 0 2
2 0 2


Output

1.00000000 -1.73205080 1.00000000 1.73205080


Input

0 0 2
0 3 1


Output

0.00000000 2.00000000 0.00000000 2.00000000

Solution (in CPP):

Solution:
//计算几何 板子
#include<bits/stdc++.h>
using namespace std ;
#define eps 1e-10
#define equals(a , b) (fabs((a) - (b)) < eps)

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

class Point
{
public:
    double x , y;

    Point(double x = 0 ,double y = 0):x(x) , y(y) {}

    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 ;
    }

};

typedef Point Vector ;

struct Segment  //表示线段结构体
{
	Point p1,p2;
	Segment(){}
	Segment(Point _p1,Point _p2):p1(_p1),p2(_p2){}
};

typedef Segment Line ;

class Circle{  //表示圆的结构体
public:
    Point c ;
    double r ;
    Circle(Point c = Point() , double r = 0.0):c(c) ,r(r){}
};

double dot(Vector a , Vector b)  // 向量a,b的内积,就是向量a*b
{
    return a.x * b.x + a.y * b.y ;
}

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

bool isOrthogonal(Vector a , Vector b)  //判断是否正交
{
    return equals(dot(a , b) , 0.0) ;
}

bool isOrthogonal(Point a1 , Point a2 , Point b1 , Point b2)
{
    return isOrthogonal(a1 - a2 , b1 - b2) ;
}

bool  isOrthogonal(Segment s1 , Segment s2)
{
    return equals(dot(s1.p2 - s1.p1 , s2.p2 - s2.p1) , 0.0) ;
}

bool isParallel(Vector a, Vector b)  //判断是否平行
{
	return equals(cross(a, b), 0.0) ;
}

bool isParallel(Point a1, Point a2, Point b1, Point b2)
{
	return isParallel(a1 - a2, b1 - b2) ;
}

bool isParallel(Segment s1, Segment s2)
{
	return equals(cross(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0) ;
}

Point project(Segment s, Point p)  //p在s上的投影
{
    Point base = s.p2 - s.p1 ;
    double r = dot(p - s.p1 , base) / base.norm() ;
    return s.p1 +base * r ;
}
/*
求垂足x：对于给定的三点p1,p2,p从点p向通过p1,p2的直线引一条垂线
base=p2-p1;
hypo=p-p1;
 x=s.p1+base*(hypo*base/|base|^2); hypo*base 可以用向量内积求
*/

Point reflect(Segment s , Point p)
{
    return p + (project(s , p) - p) * 2.0 ;
}

int ccw(Point p0 , Point p1 , Point p2)//判断三个点相对位置  // 判断逆时针方向的时候也可以使用
{
	Point a = p1 - p0 ;
	Point b = p2 - p0 ;
	if(cross(a , b) > eps) return 1 ;  //p0,p1,p2成逆时针方向
	if(cross(a , b) < -eps) return -1 ;  //p0,p1,p2成顺时针方向
	if(dot(a , b) < -eps) return 2 ;  //p2 p0 p1一次排列在同一直线上
	if(a.norm() < b.norm()) return -2 ;  // p0 p1 p2一次排列在同一直线上
	return 0 ;  //p2在线段p0p1上
}

bool intersect(Point p1,Point p2,Point p3,Point p4)  //判断是否相交 通过ccw来判断
{
	return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0&&ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0);
}

bool intersect(Segment s1 , Segment s2)
{
    return intersect(s1.p1 , s1.p2 , s2.p1 , s2.p2) ;
}

double getdis(Point a , Point b)  //求俩点间的距离
{
	return (a - b) . abs() ; //点a与点b之间的距离等于向量a-b或b-a的绝对值。
}

//注意：这里Segmet s是不规范的应该是 Line s 才对
double getdis_Li(Segment s , Point p)  //点与直线的距离
{
	return fabs(cross(s.p2 - s.p1 , p - s.p1)/(s.p2-s.p1).abs());
}

double getdis_sp(Segment s , Point p)  //点与线段的距离
{
	if(dot(s.p2 - s.p1 , p - s.p1) < 0.0) return (p-s.p1).abs();
	if(dot(s.p1 - s.p2 , p - s.p2) < 0.0) return (p-s.p2).abs();
	return getdis_Li(s , p) ;
}

double getdis_ss(Segment s1 , Segment s2)  //线段与线段之间的距离
{
	if(intersect(s1.p1 , s1.p2 , s2.p1 , s2.p2)) return 0.0 ;
	return min(min(getdis_sp(s1 , s2.p1) , getdis_sp(s1 , s2.p2)) , min(getdis_sp(s2 , s1.p1) , getdis_sp(s2 , s1.p2))) ;
}

Point getCrossPoint(Segment s1 , Segment s2) //线段的交点，返回的是点
{
    Vector base = s2.p2 - s2.p1 ;
    double d1 = abs(cross(base , s1.p1 - s2.p1)) ;
    double d2 = abs(cross(base , s1.p2 - s2.p2)) ;
    double t = d1 / (d1 + d2) ;
    return s1.p1 + (s1.p2 - s1.p1) * t ;
}

bool intersect(Circle c , Line l) //判断线和圆是否相交
{
    if(getdis_Li(l , c.c) > c.r)
    {
        return false ;
    }
    else
    {
        return true ;
    }
}

bool intersect(Circle c1, Circle c2) {//判断圆c1和圆c2是否相交
	if((c1.c - c2.c).abs() <= c1.r + c2.r)
    {
		return true;
	}
	else
    {
		return false;
	}
}

double arg(Vector p)
{
    return atan2(p.y , p.x) ;
}
Vector polar(double a , double r)
{
    return Point(cos(r) * a, sin(r) * a) ;
}

pair<Point, Point> getCrossPoints(Circle c, Line l)  //直线与圆的交点
{
    assert(intersect(c , l)) ;
    Vector pr = project(l , c.c) ;
    Vector e = (l.p2 - l.p1) / (l.p2 - l.p1).abs() ;
    double base = sqrt(c.r * c.r - (pr - c.c).norm()) ;
    return make_pair(pr + e * base, pr - e * base);
}

pair<Point, Point> getCrossPoints(Circle c1, Circle c2)  //圆与圆相交
{
    assert(intersect(c1, c2));
    double d = (c1.c -c2.c).abs() ;
    double a = acos((c1.r * c1.r + d * d - c2.r *c2.r) / (2 * c1.r *d)) ;
    double t = arg(c2.c - c1.c) ;
    return make_pair(c1.c + polar(c1.r  , t + a) , c1.c + polar(c1.r , t - a)) ;
}
int main()
{
    Circle c1 , c2 ;
    cin >> c1.c.x >> c1.c.y >> c1.r >> c2.c.x >> c2.c.y >>c2.r ;
    pair<Point , Point >p;
    p = getCrossPoints(c1 , c2) ;
    if(p.first.x < p.second.x || (p.first.x == p.second.x && p.first.y <= p.second.y))
    {
        printf("%.8f %.8f %.8f %.8f\n", p.first.x, p.first.y, p.second.x, p.second.y);
    }
    else
    {
        printf("%.8f %.8f %.8f %.8f\n", p.second.x, p.second.y, p.first.x, p.first.y);
    }
    return 0 ;
}