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:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cassert>
#pragma warning(disable: 4996)
#define EPS (1e-10)
#define equals(a,b) (fabs(a)-fabs(b)<EPS)

using namespace std;

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(x * a, y * a); }
	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;

///Program 16.3 表示线段的结构体
struct Segment {
	Point p1, p2;
};

///Program 16.4 表示直线的结构体
typedef Segment Line;

class Circle {
public:
	Point c;
	double r;
	Circle(Point c = Point(), double r = 0.0) :c(c), r(r) {}
};


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

int ccw(Point p0, Point p1, Point p2) {
	Vector a = p1 - p0;
	Vector b = p2 - p0;
	if (cross(a, b) > EPS) {//printf("COUNTER_CLOCKWISE\n");
		return 1;
	}///p2位于p0-->p1的逆时针方向
	if (cross(a, b) < -EPS) {//printf("CLOCKWISE\n");
		return -1;
	}///p2位于p0-->p1的顺时针方向
	if (dot(a, b) < -EPS) {//printf("ONLINE_BACK\n");
		return 2;
	}///p2位于线段p0p1后方
	if (a.norm() < b.norm()) { //printf("ONLINE_FRONT\n");
		return -2;
	}///p2位于线段p0p1前方

//  printf("ON_SEGMENT\n");///p2位于线段p0p1的中间
	return 0;
}

bool intersect(Point p1, Point p2, Point p3, Point p4) {
	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 norm(Point p) { return p.x * p.x + p.y * p.y; }
double abs(Point p) { return sqrt(norm(p)); }

Point project(Segment s, Point p) {
	Vector base = s.p2 - s.p1;
	double r = dot(p - s.p1, base) / base.norm();
	//   return s.p1 + (base*r);
	Point ans;
	ans.x = s.p1.x + (base * r).x;
	ans.y = s.p1.y + (base * r).y;
	return ans;
}

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 c1, Circle c2) {
	double d = abs(c1.c - c2.c);
	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;
	scanf("%lf%lf%lf", &c1.c.x, &c1.c.y, &c1.r);
	scanf("%lf%lf%lf", &c2.c.x, &c2.c.y, &c2.r);
	pair<Point, Point> 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;
}