Problem: Read a set of coordinates of three different points P1 (x1, y1), P2 (x2, y2), Q (xq, yq) on the plane, and line symmetry with point Q with the straight line passing through point P1 point P2 as the axis of symmetry. Create a program that outputs the point R (x, y) at the position of. Note that the point Q is not on its axis of symmetry.

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input

The input consists of multiple datasets. Each dataset is given in the following format:


x1, y1, x2, y2, xq, yq


x1, y1, x2, y2, xq, yq (real numbers between -100 and 100) are given on one line separated by commas.

The number of datasets does not exceed 50.

output

Print x, y on a single line, separated by blanks, for each dataset. The output may contain an error of 0.0001 or less in real numbers.

Example

Input

1.0,0.0,-1.0,0.0,1.0,1.0
1.0,0.0,0.0,-1.0,3.0,0.0
0.0,1.0,0.0,-1.0,1.0,1.0


Output

1.000000 -1.000000
1.000000 2.000000
-1.000000 1.000000

Solution (in CPP):

Solution:
//幾何ライブラリ 2013/6/23
//蟻本ベース

#include<iostream>
#include<cstdio>
#define _USE_MATH_DEFINES
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;


#define INF 1e+10
#define EPS 1e-10
#define EQ(a,b) (abs(a-b)<EPS)

//誤差を考慮して足し算
double add(double a,double b){
	if(abs(a+b) < EPS*(abs(a)+abs(b)))return 0;
	return a+b;
}
struct P{//2次元ベクトル
	double x,y;
	P(){}
	P(double x,double y):x(x),y(y){}
	P operator + (P p){
		return P(add(x,p.x),add(y,p.y));
	}
	P operator - (P p){
		return P(add(x,-p.x),add(y,-p.y));
	}
	P operator * (double d){
		return P(x*d,y*d);
	}
	P operator / (double d){
		return P(x/d,y/d);
	}
	double dot(P p){//内積
		return add(x*p.x,y*p.y);
	}
	double det(P p){//外積
		return add(x*p.y,-y*p.x);
	}
	bool equal(P p){
		return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y) < EPS*EPS;
	}
};
//typedef pair<P,P> Line;

//線分p1-p2上に点qがあるか判定
bool on_seg(P p1,P p2,P q){
	return (p1-q).det(p2-q)==0 && (p1-q).dot(p2-q)<=0;
}
//直線p1-p2と直線q1-q2の交点
P intersection(P p1,P p2,P q1,P q2){
	return p1+(p2-p1)*( (q2-q1).det(q1-p1)/(q2-q1).det(p2-p1) );
}
//直線p1-p2でqに線対称な点
P line_symmetry(P p1,P p2,P q){
	double dx,dy;
	dx=p2.x-p1.x;
	dy=p2.y-p1.y;
	P inter=intersection(p1,p2,q,q+P(dy,-dx));
	return q+(inter-q)*2;
}


int main(){
	double x1,y1,x2,y2,xq,yq;
	char _;
	while(cin>>x1>>_>>y1>>_>>x2>>_>>y2>>_>>xq>>_>>yq){
		P ans=line_symmetry(P(x1,y1),P(x2,y2),P(xq,yq));
		printf("%.5f %.5f\n",ans.x,ans.y);
	}
	return 0;
}