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<bits/stdc++.h>
using namespace std;
#define ALL(obj) (obj).begin(),(obj).end()
#define RALL(obj) (obj).rbegin(),(obj).rend()
#define REP(i, n) for(int i=0;i<(n);++i)
#define RANGE(i, a, b) for(int i=(a);i<(b);++i)
#define RREP(i, n) for(int i=(n)-1;i>= 0;--i)

#define endl '\n'
#define pb push_back
#define eb emplace_back
#define pque priority_queue
#define umap unordered_map
#define uset unordered_set
#define BIG 2000000000
#define VERYBIG 1000000000000000ll
#define PI 3.141592653589793238
#define dcout cout<<fixed<<setprecision(50)
#define _popcnt __builtin_popcount

constexpr long long dx[]={1,0,-1,0,1,1,-1,-1}, dy[]={0,-1,0,1,1,-1,1,-1};
constexpr long long MOD = 1e9+7;

// typedef long long ll;
#define int long long
// #define double long double

template<typename T> inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template<typename T> inline T LCM(T a,T b){T c=GCD(a,b);a/=c;return a*b;}
template<typename T> inline T nCr(T a,T b){T i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
template<typename T> inline T nHr(T a,T b){return nCr(a+b-1,b);}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}

typedef pair<int, int> INTP;
typedef vector<int> INTV;
struct edge2 {int to, cost;};
struct edge3 {int from, to, cost;};

const double eps = 1e-8, pi = acos(-1);
using DD = long double;

struct Point{
  DD x, y;
  Point(DD x = 0.0, DD y = 0.0) : x(x), y(y){}
  bool operator<(const Point &another) const{
  	if(x != another.x) return x < another.x;
  	return y < another.y;
  }
};

inline Point operator+ (const Point &p, const Point &q) {return Point(p.x+q.x, p.y+q.y);}
inline Point operator- (const Point &p, const Point &q) {return Point(p.x-q.x, p.y-q.y);}
inline Point operator* (const Point &p, DD a) {return Point(p.x*a, p.y*a);}
inline Point operator* (DD a, const Point &p) {return Point(p.x*a, p.y*a);}
inline Point operator/ (const Point &p, DD a) {return Point(p.x/a, p.y/a);}
inline DD cross(const Point &p, const Point &q) {return p.x*q.y - p.y*q.x;}
inline DD dot(const Point &p, const Point &q) {return p.x*q.x + p.y*q.y;}
inline DD abs(const Point &p) {return sqrt(dot(p, p));}
inline bool eq(const Point &p, const Point &q) {return abs(p-q) < eps;}

/*
istream &operator>> (istream &is, Point &p){
  DD a, b;
  is >> a >> b;
  p = Point(a, b);
  return is;
}
*/

struct Circle : Point {
  DD r;
  Circle(Point p = Point(0.0, 0.0), DD r = 0.0) : Point(p), r(r){}
};

using Points = vector<Point>;
using Circles = vector<Circle>;

Points crosspoint(const Circle &c1, const Circle &c2){
  Points res;
  DD d = abs(c1-c2);
  if(d < eps) return Points();
  if(d > c1.r + c2.r + eps) return Points();
  DD a = acos((c1.r*c1.r + d*d - c2.r*c2.r) / (2 * c1.r * d)); // 余弦定理
  DD t = atan2(c2.y - c1.y, c2.x - c1.x);
  Point p1 = c1 + Point(cos(t+a) * c1.r, sin(t+a) * c1.r);
  Point p2 = c1 + Point(cos(t-a) * c1.r, sin(t-a) * c1.r);
  res.pb(p1); res.pb(p2);
  // if(!eq(p1, p2)) res.pb(p2);
  return res;
}

signed main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);

	Circles c(2);
	REP(i, 2){
		int x, y, r; cin >> x >> y >> r;
		c[i] = Circle(Point(x, y), r);
	}
	Points ps = crosspoint(c[0], c[1]);
	sort(ALL(ps));
	for(auto p : ps){
		dcout << p.x << ' ' << p.y << ' ';
	}
	cout << endl;
}