The configuration of three circles packed inside a triangle such that each circle is tangent to the other two circles and to two of the edges of the triangle has been studied by many mathematicians for more than two centuries. Existence and uniqueness of such circles for an arbitrary triangle are easy to prove. Many methods of numerical calculation or geometric construction of such circles from an arbitrarily given triangle have been discovered. Today, such circles are called the Malfatti circles.

Figure 7 illustrates an example. The Malfatti circles of the triangle with the vertices (20, 80), (-40, -20) and (120, -20) are approximately

* the circle with the center (24.281677, 45.219486) and the radius 21.565935,
* the circle with the center (3.110950, 4.409005) and the radius 24.409005, and
* the circle with the center (54.556724, 7.107493) and the radius 27.107493.



Figure 8 illustrates another example. The Malfatti circles of the triangle with the vertices (20, -20), (120, -20) and (-40, 80) are approximately

* the circle with the center (25.629089, −10.057956) and the radius 9.942044,
* the circle with the center (53.225883, −0.849435) and the radius 19.150565, and
* the circle with the center (19.701191, 19.203466) and the radius 19.913790.



Your mission is to write a program to calculate the radii of the Malfatti circles of the given triangles.

<image>



Input

The input is a sequence of datasets. A dataset is a line containing six integers x1, y1 , x2 , y2, x3 and y3 in this order, separated by a space. The coordinates of the vertices of the given triangle are (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ), respectively. You can assume that the vertices form a triangle counterclockwise. You can also assume that the following two conditions hold.

* All of the coordinate values are greater than −1000 and less than 1000.
* None of the Malfatti circles of the triangle has a radius less than 0.1.



The end of the input is indicated by a line containing six zeros separated by a space.

Output

For each input dataset, three decimal fractions r1 , r2 and r3 should be printed in a line in this order separated by a space. The radii of the Malfatti circles nearest to the vertices with the coordinates (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) should be r1 , r2 and r3 , respectively.

None of the output values may have an error greater than 0.0001. No extra character should appear in the output.

Example

Input

20 80 -40 -20 120 -20
20 -20 120 -20 -40 80
0 0 1 0 0 1
0 0 999 1 -999 1
897 -916 847 -972 890 -925
999 999 -999 -998 -998 -999
-999 -999 999 -999 0 731
-999 -999 999 -464 -464 999
979 -436 -955 -337 157 -439
0 0 0 0 0 0


Output

21.565935 24.409005 27.107493
9.942044 19.150565 19.913790
0.148847 0.207107 0.207107
0.125125 0.499750 0.499750
0.373458 0.383897 0.100456
0.706768 0.353509 0.353509
365.638023 365.638023 365.601038
378.524085 378.605339 378.605339
21.895803 22.052921 5.895714
#include<complex>
#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<cassert>
#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define EPS (1e-10)
#define COUNTER_CLOCKWISE 1
#define CLOCKWISE -1 
#define ONLINE_BACK 2
#define ONLINE_FRONT -2
#define ON_SEGMENT 0
#define equals(a,b) (fabs((a)-(b)) < EPS)
#define inf (1<<29)

using namespace std;

class Point
{
  public:
  double x,y;

  Point(double x = -1,double y = -1): 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);}//※イケメンに限る

  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;
  }

//必要に応じて
double norm()
{
return x*x+y*y;
}

};

struct Segment
{
  Point p1,p2;
  Segment(Point p1 = Point(-1,-1),Point p2 = Point(-1,-1)):p1(p1),p2(p2){}
};
typedef Point Vector;
typedef Segment Line;
typedef vector<Point> Polygon;


double dot(Point a,Point b)
{
  return a.x*b.x + a.y*b.y;
}
double cross(Point a,Point b)
{
  return a.x*b.y - a.y*b.x;
}

double norm(Point a)
{
  return a.x*a.x+a.y*a.y;
}

double abs(Point a)
{
  return sqrt(norm(a));
}

bool pequals(Point a,Point b)
{
  return equals(a.x,b.x) && equals(a.y,b.y);
}

//rad は角度をラジアンで持たせること
Point rotate(Point a,double rad)
{
  return Point(cos(rad)*a.x - sin(rad)*a.y,sin(rad)*a.x + cos(rad)*a.y);
}

// 度をラジアンに変換
double toRad(double agl)
{
  return agl*M_PI/180.0;
}

double getArea(vector<Point>& vec)
{
  double sum = 0;
  for(int i=0;i<vec.size();i++)
    sum += cross(vec[i],vec[(i+1)%vec.size()]);
  return fabs(sum)/2.0;
}

double getDist(Point a,Point b)
{
  return sqrt(norm(a-b));
}

int ccw(Point p0,Point p1,Point p2)
{
  Point a = p1-p0;
  Point b = p2-p0;
  if(cross(a,b) > EPS)return COUNTER_CLOCKWISE;
  if(cross(a,b) < -EPS)return CLOCKWISE;
  if(dot(a,b) < -EPS)return ONLINE_BACK;
  if(norm(a) < norm(b))return ONLINE_FRONT;
  return ON_SEGMENT;
}

struct P
{

};

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



Point getCrossPoint(Point a1,Point a2,Point b1,Point b2)
{
  assert(isIntersect(a1,a2,b1,b2));//交差していない　これを外してもある程度（またはしっかり？）はうごいてくれる
  Point base = b2 - b1;
  double d1 = abs(cross(base,a1-b1));
  double d2 = abs(cross(base,a2-b1));
  double t = d1/(d1+d2);
  return a1+(a2-a1)*t;
}

// Line Line 上と同じ assertなし
Point getCrossPointLines( Line s1, Line s2){
    Point a = s1.p2 - s1.p1;
    Point base = s2.p2 -s2.p1;
    return s1.p1 + a * (cross(base, s2.p1 - s1.p1)/cross(base, a));
}

//三角形ABCの内接円半径を求める
double getInradius(Point A,Point B,Point C)
{
  vector<Point> vec(3);
  vec[0] = A, vec[1] = B, vec[2] = C;
  double S = getArea(vec);
  double a = getDist(B,C);
  double b = getDist(A,C);
  double c = getDist(A,B);
  return 2.0*S/(a+b+c);
}

//segをrだけ平行にスライド
Segment slideSeg(Segment seg,double r)
{
  Point p = (seg.p2-seg.p1)*r/abs(seg.p2-seg.p1);
  p = rotate(p,toRad(90.0));
  seg = Segment(seg.p1+p,seg.p2+p);
  if(fabs(seg.p1.x) < EPS)seg.p1.x = 0;
  if(fabs(seg.p1.y) < EPS)seg.p1.y = 0;
  if(fabs(seg.p2.x) < EPS)seg.p2.x = 0;
  if(fabs(seg.p2.y) < EPS)seg.p2.y = 0;
  return seg;
}

//三角形ABCの内心を求める
Point getIncenter(Point A,Point B,Point C)
{
  double inradius = getInradius(A,B,C);

  Segment seg1 = Segment(A,B);
  Segment seg2 = Segment(B,C);
  seg1 = slideSeg(seg1,inradius);
  seg2 = slideSeg(seg2,inradius);
  return getCrossPointLines(seg1,seg2);
}


//vector<Point> getMalfattieCircles(Point A,Point B,Point C)
void getMalfattieCircles(Point A,Point B,Point C)
{
  Point I = getIncenter(A,B,C);
  double r = getInradius(A,B,C);
  double a = getDist(B,C);
  double b = getDist(C,A);
  double c = getDist(A,B);
  double s = (a+b+c)/2.0;
  double IA = getDist(I,A);
  double IB = getDist(I,B);
  double IC = getDist(I,C);
  /*
    r1,r2,r3はそれぞれ三角形内の円の半径
    チェック済み
   */
  double r1 = ( r/(2.0*(s-a)) )*( s - r - (IB + IC - IA));
  double r2 = ( r/(2.0*(s-b)) )*( s - r - (IC + IA - IB));
  double r3 = ( r/(2.0*(s-c)) )*( s - r - (IA + IB - IC));

  cout << setiosflags(ios::fixed) << setprecision(6) << r1 << " " << r2 << " " << r3 << endl;

  /*
    以下は未チェック（ちょっとだけチェック済み）
    三角形内の３つの円の中心点
  */
  /*
  vector<Point> vec(3);
  
  Segment seg1 = Segment(C,A);
  Segment seg2 = Segment(A,B);
  seg1 = slideSeg(seg1,r1);
  seg2 = slideSeg(seg2,r1);
  vec[0] = getCrossPointLines(seg1,seg2);
  //cout << vec[0].x << "," << vec[0].y << endl;

  seg1 = Segment(A,B);
  seg2 = Segment(B,C);
  seg1 = slideSeg(seg1,r2);
  seg2 = slideSeg(seg2,r2);
  vec[1] = getCrossPointLines(seg1,seg2);
  //cout << vec[1].x << "," << vec[1].y << endl;

  seg1 = Segment(B,C);
  seg2 = Segment(C,A);
  seg1 = slideSeg(seg1,r3);
  seg2 = slideSeg(seg2,r3);
  vec[2] = getCrossPointLines(seg1,seg2);
  //cout << vec[2].x << "," << vec[2].y << endl;

  return vec;
  */
}

int main()
{
  while(true)
    {
      Point A,B,C;
      cin >> A.x >> A.y >> B.x >> B.y >> C.x >> C.y;
      if(A.x+A.y+B.x+B.y+C.x+C.y == 0)break;
      getMalfattieCircles(A,B,C);
    }
  return 0;
}