Problem: Problem Statement

One day, you found an old scroll with strange texts on it.

You revealed that the text was actually an expression denoting the position of treasure. The expression consists of following three operations:

* From two points, yield a line on which the points lie.
* From a point and a line, yield a point that is symmetric to the given point with respect to the line.
* From two lines, yield a point that is the intersection of the lines.



The syntax of the expression is denoted by following BNF:


<expression>      ::= <point>
<point>       	  ::= <point-factor> | <line> "@" <line-factor> | <line> "@" <point-factor> | <point> "@" <line-factor>
<point-factor>    ::= "(" <number> "," <number> ")" | "(" <point> ")"
<line>            ::= <line-factor> | <point> "@" <point-factor>
<line-factor>     ::= "(" <line> ")"
<number>          ::= <zero-digit> | <positive-number> | <negative-number>
<positive-number> ::= <nonzero-digit> | <positive-number> <digit>
<negative-number> ::= "-" <positive-number>
<digit>           ::= <zero-digit> | <nonzero-digit>
<zero-digit>      ::= "0"
<nonzero-digit>   ::= "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9"

Each <point> or <point-factor> denotes a point, whereas each <line> or <line-factor> denotes a line. The former notion of <point-factor> $(X,Y)$ represents a point which has $X$ for $x$-coordinate and $Y$ for $y$-coordinate on the $2$-dimensional plane. "@" indicates the operations on two operands. Since each operation is distinguishable from others by its operands' types (i.e. a point or a line), all of these operations are denoted by the same character "@". Note that "@" is left-associative, as can be seen from the BNF.

Your task is to determine where the treasure is placed.

Input

The input consists of multiple datasets. Each dataset is a single line which contains an expression denoting the position of treasure.

It is guaranteed that each dataset satisfies the following conditions:

* The length of the string never exceeds $10^2$.
* If both operands of "@" are points, their distance is greater than $1$.
* If both operands of "@" are lines, they are never parallel.
* The absolute values of points' coordinates never exceed $10^2$ at any point of evaluation.



You can also assume that there are at most $100$ datasets.

The input ends with a line that contains only a single "#".

Output

For each dataset, print the $X$ and $Y$ coordinates of the point, denoted by the expression, in this order.

The output will be considered correct if its absolute or relative error is at most $10^{-2}$.

Sample Input


((0,0)@(1,1))@((4,1)@(2,5))
((0,0)@(3,1))@((1,-3)@(2,-1))
(0,0)@(1,1)@(4,1)
(0,0)@((1,1)@(4,1))
(((0,0)@((10,20)@(((30,40))))))
((0,0)@(3,1))@((1,-3)@(2,-1))@(100,-100)@(100,100)


Output for the Sample Input


3.00000000 3.00000000
3.00000000 1.00000000
1.00000000 4.00000000
0.00000000 2.00000000
-10.00000000 10.00000000
-99.83681795 -91.92248853





Example

Input

((0,0)@(1,1))@((4,1)@(2,5))
((0,0)@(3,1))@((1,-3)@(2,-1))
(0,0)@(1,1)@(4,1)
(0,0)@((1,1)@(4,1))
(((0,0)@((10,20)@(((30,40))))))
((0,0)@(3,1))@((1,-3)@(2,-1))@(100,-100)@(100,100)
#


Output

3.00000000 3.00000000
3.00000000 1.00000000
1.00000000 4.00000000
0.00000000 2.00000000
-10.00000000 10.00000000
-99.83681795 -91.92248853

Solution (in CPP):

Solution:
#include<iostream>
#include<complex>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
#define show1d(v) rep(_,v.size())cout<<" "<<v[_];cout<<endl;
#define show2d(v) rep(_,v.size()){rep(__,v[_].size())cout<<" "<<v[_][__];cout<<endl;}cout<<endl;
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
//#define mt make_tuple
//typedef tuple<double,int,double> tp;
//typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-8;
Def inf = sizeof(Def) == sizeof(long long) ? 2e18 : 1e9;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
typedef complex<double> P;
typedef vector<P> G;
typedef vector<double> vd;
typedef vector<vd> vvd;

struct L : public vector<P> {
	L(){
		pb(0);
		pb(0);
	}
	L(const P &a, const P &b) {
	push_back(a); push_back(b);
  }
};
struct C{
	P c;double r;
	C(const P &c,double r):c(c),r(r){}
};
#define curr(P, i) P[i]
#define next(P, i) P[(i+1)%P.size()]
#define diff(P, i) (next(P, i) - curr(P, i))
namespace std {
	bool operator < (const P& a, const P& b) {
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
		//return imag(a) != imag(b) ? imag(a) < imag(b) : real(a) < real(b);  
	}
	bool operator == (const P& a, const P& b) {
		return a.real()==b.real()&&a.imag()==b.imag();
	}
}
P pin(){
	double x,y;
	char d;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int n){
	rep(i,n)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > 0)   return +1;       // counter clockwise
  if (cross(b, c) < 0)   return -1;       // clockwise
  if (dot(b, c) < 0)     return +2;       // c--a--b on line
  if (norm(b) < norm(c)) return -2;       // a--b--c on line
  return 0;                               // a--c--b on line
}
P projection(L a,P p){
	double t=dot(p-a[0],a[0]-a[1])/norm(a[0]-a[1]);
	return a[0]+t*(a[0]-a[1]);
}
P reflection(L a,P p){
	return p+2.0*(projection(a,p)-p);
}
bool intersectLL(const L &l, const L &m) {
  return abs(cross(l[1]-l[0], m[1]-m[0])) > EPS || // non-parallel
		 abs(cross(l[1]-l[0], m[0]-l[0])) < EPS;   // same line
}
bool intersectLS(const L &l, const L &s) {
  return cross(l[1]-l[0], s[0]-l[0])*       // s[0] is left of l
		 cross(l[1]-l[0], s[1]-l[0]) < EPS; // s[1] is right of l
}
bool intersectLP(const L &l, const P &p) {
  return abs(cross(l[1]-p, l[0]-p)) < EPS;
}
bool intersectSS(const L &s, const L &t) {
  return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
		 ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
bool intersectSP(const L &s, const P &p) {
  return abs(s[0]-p)+abs(s[1]-p)-abs(s[1]-s[0]) < EPS; // triangle inequality
}
double distanceLP(const L &l, const P &p) {
	return abs(p - projection(l, p));
}
double distanceLL(const L &l, const L &m) {
  return intersectLL(l, m) ? 0 : distanceLP(l, m[0]);
}
double distanceLS(const L &l, const L &s) {
  if (intersectLS(l, s)) return 0;
  return min(distanceLP(l, s[0]), distanceLP(l, s[1]));
}
double distanceSP(const L &s, const P &p) {
  const P r = projection(s, p);
  if (intersectSP(s, r)) return abs(r - p);
  return min(abs(s[0] - p), abs(s[1] - p));
}
double distanceSS(const L &s, const L &t) {
  if (intersectSS(s, t)) return 0;
  return min(min(distanceSP(s, t[0]), distanceSP(s, t[1])),
			 min(distanceSP(t, s[0]), distanceSP(t, s[1])));
}
/*bool intersectCS(C c,const L &l){
	return (distanceLP(l,c.c) < c.r+EPS &&
			(c.r < abs(c.c-l[0]) + EPS || c.r < abs(c.c-l[1]) + EPS));
}*/
int intersectCS(C c,L l){
	if(norm(projection(l,c.c)-c.c)-c.r*c.r>EPS)return 0;
	const double d1=abs(c.c-l[0]),d2=abs(c.c-l[1]);
	if(d1<c.r+EPS&&d2<c.r+EPS)return 0;
	if(d1<c.r-EPS&&d2>c.r+EPS||d1>c.r+EPS&&d2<c.r-EPS)return 1;
	const P h=projection(l,c.c);
	if(dot(l[0]-h,l[1]-h)<0)return 2;
	return 0;
}
P crosspointSS(L a,L b){
	double t1=abs(cross(a[1]-a[0],b[0]-a[0]));
	double t2=abs(cross(a[1]-a[0],b[1]-a[0]));
	return b[0]+(b[1]-b[0])*t1/(t1+t2);
}
L crosspointCL(C c,L l){
	P pr=projection(l,c.c);
	P e=(l[1]-l[0])/abs(l[1]-l[0]);
	double tmp=c.r*c.r-norm(pr-c.c);
	if(abs(tmp)<EPS)tmp=0;
	double t=sqrt(tmp);
	P a=pr+t*e;
	P b=pr-t*e;
	if(b<a)swap(a,b);
	return L(a,b);
}
L crosspointCS(C c,L l){
	if(intersectCS(c,l)==2)return crosspointCL(c,l);
	L ret=crosspointCL(c,l);
	if(dot(l[0]-ret[0],l[1]-ret[0])<0)ret[1]=ret[0];
	else ret[0]=ret[1];
	return ret;
}
L crosspointCC(C a,C b){
	P tmp=b.c-a.c;
	double d=abs(tmp);
	double q=acos((a.r*a.r+d*d-b.r*b.r)/(2*a.r*d));
	double t=arg(tmp);//atan(tmp.imag()/tmp.real());
	P p1=a.c+polar(a.r,t+q);
	P p2=a.c+polar(a.r,t-q);
	if(p2<p1)swap(p1,p2);
	return L(p1,p2);
}
P crosspointLL(const L &l, const L &m) {
  double A = cross(l[1] - l[0], m[1] - m[0]);
  double B = cross(l[1] - l[0], l[1] - m[0]);
  if (abs(A) < EPS && abs(B) < EPS) return m[0]; // same line
  return m[0] + B / A * (m[1] - m[0]);
}
double area(const G &g){
	double S =0;
	for(int i =0;i <g.size();i++){
		S +=(cross(g[i],g[(i+1)%g.size()]));
	}
	return abs(S/2.0);
}
bool isconvex(const G &g){
	int n=g.size();
	rep(i,n)if(ccw(g[(i+n-1)%n],g[i%n],g[(i+1)%n])==-1)return false;
	return true;
}
int inconvex(const G& g, const P& p) {
	bool in = false;
	int n=g.size();
	rep(i,n){
		P a=g[i%n]-p;
		P b=g[(i+1)%n]-p;
		if(imag(a)>imag(b))swap(a, b);
		if(imag(a)<EPS&&0<imag(b))if(cross(a,b)<0)in=!in;
		if(abs(cross(a,b))<EPS&&dot(a,b)<EPS)return 1;//ON
	}
	return in?2:0;//IN : OUT;
}
G convex_hull(G &ps) {
	int n=ps.size(),k=0;
	sort(ps.begin(), ps.end());
	G ch(2*n);
	for(int i=0;i<n;ch[k++]=ps[i++])//lower-hull
		while(k>=2&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//<=0  ->  ==-1
	for(int i=n-2,t=k+1;i>=0;ch[k++]=ps[i--])//upper-hull
		while(k>=t&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//
	ch.resize(k-1);
	return ch;
}
double convex_diameter(const G &pt) {
  const int n = pt.size();
  int is = 0, js = 0;
  for (int i = 1; i < n; ++i) {
	if (imag(pt[i]) > imag(pt[is])) is = i;
	if (imag(pt[i]) < imag(pt[js])) js = i;
  }
  double maxd = norm(pt[is]-pt[js]);

  int i, maxi, j, maxj;
  i = maxi = is;
  j = maxj = js;
  do {
	if (cross(diff(pt,i), diff(pt,j)) >= 0) j = (j+1) % n;
	else i = (i+1) % n;
	if (norm(pt[i]-pt[j]) > maxd) {
	  maxd = norm(pt[i]-pt[j]);
	  maxi = i; maxj = j;
	}
  } while (i != is || j != js);
  return sqrt(maxd); /* farthest pair is (maxi, maxj). */
}//convex_diameter(g)
G convex_cut(const G& g, const L& l) {
  G Q;
  for (int i = 0; i < g.size(); ++i) {
	P a= curr(g, i), b= next(g, i);
	if (ccw(l[0], l[1], a) != -1) Q.push_back(a);
	if (ccw(l[0], l[1], a)*ccw(l[0], l[1], b) < 0)
	  Q.push_back(crosspointLL(L(a,b), l));
  }
  return Q;
}
P turn(P p,double t){
	return p*exp(P(.0,t*PI/180.0));
}
P turn2(P p,double t){
	return p*exp(P(.0,t));
}
vector<L> tangentCC(C a,C b){
	if(a.r<b.r)swap(a,b);
	double d=abs(a.c-b.c);
	vector<L>l;
	if(d<EPS)return l;
	if(a.r+b.r<d-EPS){//hanareteiru
		double t=acos((a.r+b.r)/d);
		t=t*180/PI;
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),t),b.c+turn(b.r/d*(a.c-b.c),t)));
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),-t),b.c+turn(b.r/d*(a.c-b.c),-t)));
	}else if(a.r+b.r<d+EPS){//kuttuiteiru soto
		P p=a.c+a.r/d*(b.c-a.c);
		l.pb(L(p,p+turn(b.c-a.c,90)));
	}
	if(abs(a.r-b.r)<d-EPS){//majiwatteiru
		double t1=acos((a.r-b.r)/d);
		t1=t1*180/PI;
		double t2=180-t1;
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),t1),b.c+turn(b.r/d*(a.c-b.c),-t2)));
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),-t1),b.c+turn(b.r/d*(a.c-b.c),t2)));
	}else if(abs(a.r-b.r)<d+EPS){//kuttuiteiru uti
		P p=a.c+a.r/d*(b.c-a.c);
		l.pb(L(p,p+turn(b.c-a.c,90)));
	}
	return l;
}
void printL(const L &out){
	printf("%0.9f %0.9f %0.9f %0.9f\n",out[0].real(),out[0].imag(),out[1].real(),out[1].imag());
}
C CIN(){
	P p=pin();
	double r;
	cin>>r;
	return C(p,r);
}
bool para(L a,L b){
	return (abs(cross(a[1]-a[0],b[1]-b[0]))<EPS);
}
double min(double a,double b){return a<b?a:b;}
double max(double a,double b){return a>b?a:b;}


string s;
int t;
int digit(){
	int a=0;
	bool h=false;
	if(s[t]=='-')t++,h=true;
	while(isdigit(s[t])){
		a*=10;
		a+=s[t]-'0';
		t++;
	}
	if(h)a*=-1;
	return a;
}
L num(){
	t++;
	double a,b;
	a=digit();
	t++;
	b=digit();
	t++;
	return L(P(a,b),P(a,b));
}
L f();
L F(){
	L out;
	if(s[t]=='{')out=num();
	else{
		t++;
		out=f();
		t++;
	}
	return out;
}
L f(){
	L l,r;
	l=F();
	while(s[t]=='@'){
		t++;
		r=F();
		double dis1=abs(l[0]-l[1]),dis2=abs(r[0]-r[1]);
		if(dis1<EPS&&dis2<EPS)l=L(l[0],r[1]);
		else if(dis1<EPS){
			P p=reflection(r,l[0]);
			l=L(p,p);
		}
		else if(dis2<EPS){
			P p=reflection(l,r[0]);
			l=L(p,p);
		}else{
			P p=crosspointLL(l,r);
			l=L(p,p);
		}
	}
	return l;
}
int main(){
	while(cin>>s,s!="#"){
		rep(i,s.size())if(isdigit(s[i])||s[i]=='-'){
			if(s[i-1]=='(')s[i-1]='{';
			if(s[i+1]==')')s[i+1]='}';
		}
		t=0;
		L l=f();
		cout<<shosu(9)<<l[0].real()<<" "<<l[0].imag()<<endl;
	}
}