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
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <ctime>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <iostream>
#include <deque>
#include <complex>
#include <string>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <valarray>
#include <unordered_map>
#include <iterator>
#include <assert.h>
using namespace std;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const set<T> &t) {
os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

namespace geom{
#define X real()
#define Y imag()
#define at(i) ((*this)[i])
#define SELF (*this)
	enum {TRUE = 1, FALSE = 0, BORDER = -1};
	typedef int BOOL;
	typedef double R;
	const R INF = 1e8;
	const R EPS = 1e-8;
	const R PI = 3.1415926535897932384626;
	inline int sig(const R &x) { return (abs(x) < EPS ? 0 : x > 0 ? 1 : -1); }
	inline BOOL less(const R &x, const R &y) {return sig(x-y) ? x < y : BORDER;}
	typedef complex<R> P;
	inline R norm(const P &p){return p.X*p.X+p.Y*p.Y;}
	inline R inp(const P& a, const P& b){return (conj(a)*b).X;}
	inline R outp(const P& a, const P& b){return (conj(a)*b).Y;}
	inline P unit(const P& p){return p/abs(p);}
	inline P proj(const P &s, const P &t){return t*inp(s, t)/norm(t);}
	inline int ccw(const P &s, const P &t, const P &p, int adv=0){
		int res = sig(outp(t-s, p-s));
		if(res || !adv) return res;
		if(sig(inp(t-s, p-s)) < 0) return -2;	// p-s-t
		if(sig(inp(s-t, p-t)) < 0) return 2;	// s-t-p
		return 0;								// s-p-t
	}
	
	
	struct L : public vector<P>{	// line
		L(const P &p1, const P &p2){this->push_back(p1);this->push_back(p2);}
		L(){}
		P dir()const {return at(1) - at(0);}
		BOOL online(const P &p)const {return !sig(outp(p-at(0), dir()));}
	};
	struct S : public L{	// segment
		S(const P &p1, const P &p2):L(p1, p2){}
		S(){}
		BOOL online(const P &p)const {
			if(!sig(norm(p - at(0))) || !sig(norm(p - at(1)))) return BORDER;
			return !sig(abs(at(0)-p) + abs(at(1) - p) - abs(at(0) - at(1)));
		}
	};
	struct C : public P{
		C(){}
		C(const P& p, const R r):P(p), r(r){}
		R r;
		BOOL inside(const P& p)const { return less(norm(p-SELF), r*r);}
	};
	struct F : public C{
		R s, t;
		F(const C &c, R ss, R tt):C(c), s(ss), t(tt){
			if(PI < s) s -= 2*PI;
			if(PI < t) t -= 2*PI;
		}
		BOOL inside(const P& p)const {
			P v = p - SELF;
			if(!sig(norm(v))) return BORDER;
			R a = arg(v);
			if(t < s){
				if((!less(s, a) && !less(a, t)) || !less(norm(v), r*r)) return FALSE;
				return less(s, a) | less(a, t) | less(norm(v), r*r);
			}else{
				if(!less(s, a) || !less(a, t) || !less(norm(v), r*r)) return FALSE;
				return less(s, a) | less(a, t) | less(norm(v), r*r);
			}
		}
	};
	P crosspoint(const L &l, const L &m);
	struct G : public vector<P>{
		G(size_type size=0):vector(size){}
		S edge(int i)const {return S(at(i), at(i+1 == size() ? 0 : i+1));}
		BOOL contains(const P &p)const {
			R sum = .0;
			REP(i, size()){
				if(S(at(i), at((i+1)%size())).online(p)) return BORDER;	// online
				sum += arg((at(i) - p) / (at((i+1)%size()) - p));
			}
			return !!sig(sum);
		}
		R area()const {
			R sum = 0;
			REP(i, size()) sum += outp(at(i), at((i+1)%size()));
			return abs(sum / 2.);
		}
		
		G convex_hull(bool online = false) {
			if(size() < 2) return *this;
			sort(ALL(*this));
			G r;
			r.resize((int)size()*2);
			int k=0;
			for(int i=0;i<size();r[k++]=at(i++))
				while(k>1 && ccw(r[k-2], r[k-1], at(i)) < 1-online) k--;
			int t = k;
			for(int i=(int)size()-1;i>=0;r[k++]=at(i--))
				while(k>t && ccw(r[k-2], r[k-1], at(i)) < 1-online) k--;
			r.resize(k-1);
			return r;
		}
		G cut(const L &l)const {
			G g;
			REP(i, size()){
				const S &s = edge(i);
				if(ccw(l[0], l[1], s[0], 0) >= 0) g.push_back(s[0]);
				if(ccw(l[0], l[1], s[0], 0) * ccw(l[0], l[1], s[1], 0) < 0)
					g.push_back(crosspoint(s, l));
			}
			return g;
		}
		G Voronoi(const vector<P> &p, const int t)const {
			G g = *this;
			REP(i, p.size())if(i!=t){
				const P m = (p[t]+p[i])*0.5;
				g = g.cut(L(m, m+(p[i]-p[t])*P(0, 1)));
			}
			return g;
		}
	};

	inline P proj(const P &s, const L &t){return t[0] + proj(s-t[0], t[1]-t[0]);}
	inline P reflect(const P &s, const L &t){return 2.*proj(s, t) - s;}
	inline S reflect(const S &s, const L &t){return S(reflect(s[0], t), reflect(s[1], t));}
	inline P crosspoint(const L &l, const L &m){
		R A = outp(l.dir(), m.dir()), B = outp(l.dir(), l[1] - m[0]);
		if(!sig(abs(A)) && !sig(abs(B))) return m[0]; // same line
		if(abs(A) < EPS) assert(false); // !!!PRECONDITION NOT SATISFIED!!!
		return m[0] + B / A * (m[1] - m[0]);
	}
#undef SELF
#undef at
}

using namespace geom;



namespace std{
	bool operator<(const P &a, const P &b){return sig(a.X-b.X) ? a.X < b.X : a.Y+EPS < b.Y;}
	bool operator==(const P &a, const P &b){return !sig(norm(a-b));}
/*	istream& operator>>(istream &is, P &p){R x,y;is>>x>>y;p=P(x, y);return is;}
	istream& operator>>(istream &is, L &l){l.resize(2);return is >> l[0] >> l[1];}
	istream& operator>>(istream &is, C &c){return is >> (P &)c >> c.r;}
	const double B = 500;
	const double Z = 1;
	ostream& operator<<(ostream &os, const C &c){return os << "circle("<<B+Z*(c.X)<<", "<<1000-B-Z*(c.Y)<<", "<<Z*(c.r)<<")";}
	ostream& operator<<(ostream &os, const P &p){return os << C(p, 2.);}
	ostream& operator<<(ostream &os, const S &s){return os << "line("<<B+Z*(s[0].X)<<", "<<1000-B-Z*(s[0].Y)<<", "<<B+Z*(s[1].X)<<", "<<1000-B-Z*(s[1].Y)<<")";}
	ostream& operator<<(ostream &os, const G &g){REP(i, g.size()) cout << g.edge(i) << endl;return os;}
*/
}

int n;
string s;

struct PL{
	int t;
	P p;
	L l;
	PL(){}
	PL(P p):p(p),t(0){}
	PL(L l):l(l),t(1){}
	PL operator*(const PL &opp){
		if(!t && !opp.t) return PL(L(p, opp.p));
		if( t &&  opp.t) return PL(crosspoint(l, opp.l));
		if(!t &&  opp.t) return PL(reflect(p, opp.l));
		if( t && !opp.t) return PL(reflect(opp.p, l));
	}
};

R number(int &p){
	int sig = 1;
	if(s[p] == '-'){
		sig = -1;
		p++;
	}
	int res = 0;
	while(isdigit(s[p])) res = res*10 + s[p++]-'0';
	return sig * res;
}

PL point(int &p);
PL fact(int &p){
	assert(s[p] == '(');
	p++;
	PL res;
	if(s[p] == '('){
		res =  point(p);
		p++;
		return res;
	}else if(isdigit(s[p]) || s[p] == '-'){
		R x = number(p);
		assert(s[p] == ',');
		p++;
		R y = number(p);
		res = PL(P(x, y));
	}else assert(false);
	assert(s[p] == ')');
	p++;
	return res;
}

PL point(int &p){
	PL res = fact(p);
	while(p < s.size() && s[p] == '@'){
		res = res * fact(++p);
	}
	return res;
}

int main(){
	ios::sync_with_stdio(false);

	while(cin >> s, s!="#"){
		int p = 0;
		P res = point(p).p;
		printf("%.10f %.10f\n", res.X, res.Y);
	}
	return 0;
}