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<bits/stdc++.h>

#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 equals(a,b) (fabs((a)-(b)) < EPS)
#define COUNTER_CLOCKWISE 1
#define CLOCKWISE -1 
#define ONLINE_BACK 2
#define ONLINE_FRONT -2
#define ON_SEGMENT 0

using namespace std;

class Point{
public:
  double x,y;

  Point(double x = 0,double y = 0): 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);}
  Point operator * (Point a){ return Point(x * a.x - y * a.y, x * a.y + y * a.x); }

  bool operator < (const Point& p) const{ return !equals(x,p.x)?x<p.x:(!equals(y,p.y)&&y<p.y); }

  bool operator == (const Point& p)const{ return fabs(x-p.x) < EPS && fabs(y-p.y) < EPS; }

};

struct Segment{
  Point p1,p2;
  Segment(Point p1 = Point(),Point p2 = Point()):p1(p1),p2(p2){}
  bool operator <  (const Segment& s) const { return ( p2 == s.p2 ) ? p1 < s.p1 : p2 < s.p2; }
  bool operator == (const Segment& s) const { return ( s.p1 == p1 && s.p2 == p2 ) || ( s.p1 == p2 && s.p2 == p1 ); }

};

typedef Point Vector;
typedef Segment Line;
typedef vector<Point> Polygon;

ostream& operator << (ostream& os,const Point& a){ os << "(" << a.x << "," << a.y << ")"; }

ostream& operator << (ostream& os,const Segment& a){ os << "( " << a.p1 << " , " << a.p2 << " )"; }

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)); }

//rad ????§???????????????¢?????§?????????????????¨
Point rotate(Point a,double rad){ return Point(cos(rad)*a.x - sin(rad)*a.y,sin(rad)*a.x + cos(rad)*a.y); }

// ??????????????¢????????????
double toRad(double agl){ return agl*M_PI/180.0; }

// a => prev, b => cur, c=> next
// prev ?????? cur ????????£??? next ????????????????§????????±???????
double getArg(Point a,Point b,Point c){
  double arg1 = atan2(b.y-a.y,b.x-a.x);
  double arg2 = atan2(c.y-b.y,c.x-b.x);
  double arg = fabs( arg1 - arg2 );
  while( arg > M_PI ) arg -= 2.0 * M_PI;
  return fabs(arg);
}

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;
}

bool intersectLL(Line l, Line m) {
  return abs(cross(l.p2-l.p1, m.p2-m.p1)) > EPS || // non-parallel
         abs(cross(l.p2-l.p1, m.p1-l.p1)) < EPS;   // same line
}
bool intersectLS(Line l, Line s) {
  return cross(l.p2-l.p1, s.p1-l.p1)*       // s[0] is left of l
         cross(l.p2-l.p1, s.p2-l.p1) < EPS; // s[1] is right of l
}
bool intersectLP(Line l,Point p) {
  return abs(cross(l.p2-p, l.p1-p)) < EPS;
}
bool intersectSS(Line s, Line t) {
  return ccw(s.p1,s.p2,t.p1)*ccw(s.p1,s.p2,t.p2) <= 0 &&
         ccw(t.p1,t.p2,s.p1)*ccw(t.p1,t.p2,s.p2) <= 0;
}
bool intersectSP(Line s, Point p) {
  return abs(s.p1-p)+abs(s.p2-p)-abs(s.p2-s.p1) < EPS; // triangle inequality
}

Point projection(Line l,Point p) {
  double t = dot(p-l.p1, l.p1-l.p2) / norm(l.p1-l.p2);
  return l.p1 + (l.p1-l.p2)*t;
}
Point reflection(Line l,Point p) {
  return p + (projection(l, p) - p) * 2;
}
double distanceLP(Line l, Point p) {
  return abs(p - projection(l, p));
}
double distanceLL(Line l, Line m) {
  return intersectLL(l, m) ? 0 : distanceLP(l, m.p1);
}

double distanceLS(Line l, Line s) {
  if (intersectLS(l, s)) return 0;
  return min(distanceLP(l, s.p1), distanceLP(l, s.p2));
}
double distanceSP(Line s, Point p) {
  Point r = projection(s, p);
  if (intersectSP(s, r)) return abs(r - p);
  return min(abs(s.p1 - p), abs(s.p2 - p));
}

double distanceSS(Line s, Line t) {
  if (intersectSS(s, t)) return 0;
  return min(min(distanceSP(s, t.p1), distanceSP(s, t.p2)),
             min(distanceSP(t, s.p1), distanceSP(t, s.p2)));
}

Point crosspoint(Line l,Line m){
  double A = cross(l.p2-l.p1,m.p2-m.p1);
  double B = cross(l.p2-l.p1,l.p2-m.p1);
  if(abs(A) < EPS && abs(B) < EPS){
    vector<Point> vec;
    vec.push_back(l.p1),vec.push_back(l.p2),vec.push_back(m.p1),vec.push_back(m.p2);
    sort(vec.begin(),vec.end());
    assert(vec[1] == vec[2]); //???????????°??????????????????
    return vec[1];
    //return m.p1;
  }
  if(abs(A) < EPS)assert(false);
  return m.p1 + (m.p2-m.p1)*(B/A);
}

// ---------------------------------------------------------------

struct Data {
  Line value;
  bool state; // false -> point, true -> line
};

string context;
int ptr,n;

Data expression();

Data term(){
  Data data;
  if( context[ptr] == '(' && context[ptr+1] == '(' ) {
    ++ptr;
    data = expression();
    ++ptr;
  } else {
    assert( context[ptr+1] == '-' || isdigit(context[ptr+1]) );
    data.state = false;
    ++ptr;
    string v = "";
    bool minus = false;
    if( context[ptr] == '-' ) minus = true, ++ptr;
    while( isdigit(context[ptr]) ) v += context[ptr++];
    assert( context[ptr] == ',' );
    data.value.p1.x = atoi(v.c_str());
    if( minus ) data.value.p1.x *= -1;
    ++ptr;
    v = "";
    minus = false;
    if( context[ptr] == '-' ) minus = true, ++ptr;
    while( isdigit(context[ptr]) ) v += context[ptr++];
    data.value.p1.y = atoi(v.c_str());
    if( minus ) data.value.p1.y *= -1;
    assert( context[ptr] == ')' );
    ++ptr;
  }
  return data;
}

Data expression(){
  Data data = term();
  while( ptr < n && context[ptr] == '@' ){
    ++ptr;
    Data tmp = term();
    if( !data.state && !tmp.state ) {
      data.state = true;
      data.value.p2 = tmp.value.p1;
    } else if( !data.state && tmp.state ) {
      data.value.p1 = reflection(tmp.value,data.value.p1);
    } else if( data.state && !tmp.state ) {
      data.state = false;
      data.value.p1 = reflection(data.value,tmp.value.p1);
    } else {
      data.state = false;
      data.value.p1 = crosspoint(data.value,tmp.value);
    }
  }
  return data;
}

int main(){
  while( getline(cin,context), context[0] != '#' ){
    ptr = 0, n = context.size();
    Point p = expression().value.p1;
    printf("%.10f %.10f\n",p.x,p.y);

  }
  return 0;
}