Problem: Let's think about a bar rotating clockwise as if it were a twirling baton moving on a planar surface surrounded by a polygonal wall (see Figure 1).

<image>


Figure 1. A bar rotating in a polygon

Initially, an end of the bar (called "end A") is at (0,0), and the other end (called "end B") is at (0,L) where L is the length of the bar. Initially, the bar is touching the wall only at the end A.

The bar turns fixing a touching point as the center. The center changes as a new point touches the wall.

Your task is to calculate the coordinates of the end A when the bar has fully turned by the given count R.

<image>


Figure 2. Examples of turning bars

In Figure 2, some examples are shown. In cases (D) and (E), the bar is stuck prematurely (cannot rotate clockwise anymore with any point touching the wall as the center) before R rotations. In such cases, you should answer the coordinates of the end A in that (stuck) position.

You can assume the following:

> When the bar's length L changes by ε (|ε| < 0.00001), the final (x,y) coordinates will not change more than 0.0005.

Input

The input consists of multiple datasets. The number of datasets is no more than 100. The end of the input is represented by "0 0 0".

The format of each dataset is as follows:


L R N
X1 Y1
X2 Y2
...
XN YN


L is the length of the bar. The bar rotates 2π× R radians (if it is not stuck prematurely). N is the number of vertices which make the polygon.

The vertices of the polygon are arranged in a counter-clockwise order. You may assume that the polygon is simple, that is, its border never crosses or touches itself.

N, Xi and Yi are integer numbers; R and L are decimal fractions. Ranges of those values are as follows:


1.0 ≤ L ≤ 500.0,
1.0 ≤ R ≤ 10.0,
3 ≤ N ≤ 100,
-1000 ≤ Xi ≤ 1000,
-1000 ≤ Yi ≤ 1000,



X1 ≤ -1,  Y1 = 0,
X2 ≥ 1,  Y2 = 0.


Output

For each dataset, print one line containing x- and y-coordinates of the final position of the end A, separated by a space. The value may contain an error less than or equal to 0.001. You may print any number of digits after the decimal point.

Sample Input


4.0 2.0 8
-1 0
5 0
5 -2
7 -2
7 0
18 0
18 6
-1 6
4.0 2.0 4
-1 0
10 0
10 12
-1 12
4.0 1.0 7
-1 0
2 0
-1 -3
-1 -8
6 -8
6 6
-1 6
4.0 2.0 6
-1 0
10 0
10 3
7 3
7 5
-1 5
5.0 2.0 6
-1 0
2 0
2 -4
6 -4
6 6
-1 6
6.0 1.0 8
-1 0
8 0
7 2
9 2
8 4
11 4
11 12
-1 12
0 0 0


Output for the Sample Input


16.0 0.0
9.999999999999998 7.4641016151377535
0.585786437626906 -5.414213562373095
8.0 0.0
6.0 0.0
9.52786404500042 4.0


Note that the above sample input corresponds to the cases in Figure 2. For convenience, in Figure 3, we will show an animation and corresponding photographic playback for the case (C).

<image>
<image> | <image> | <image> | <image> | <image> | <image> | <image> | <image> | <image>
---|---|---|---|---|---|---|---|---

Figure 3. Animation and photographic playback for case (C)





Example

Input

4.0 2.0 8
-1 0
5 0
5 -2
7 -2
7 0
18 0
18 6
-1 6
4.0 2.0 4
-1 0
10 0
10 12
-1 12
4.0 1.0 7
-1 0
2 0
-1 -3
-1 -8
6 -8
6 6
-1 6
4.0 2.0 6
-1 0
10 0
10 3
7 3
7 5
-1 5
5.0 2.0 6
-1 0
2 0
2 -4
6 -4
6 6
-1 6
6.0 1.0 8
-1 0
8 0
7 2
9 2
8 4
11 4
11 12
-1 12
0 0 0


Output

16.0 0.0
9.999999999999998 7.4641016151377535
0.585786437626906 -5.414213562373095
8.0 0.0
6.0 0.0
9.52786404500042 4.0

Solution (in CPP):

Solution:
#include <cassert>// c
#include <ctime>
#include <iostream>// io
#include <iomanip>
#include <fstream>
#include <sstream>
#include <vector>// container
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <algorithm>// other
#include <complex>
#include <numeric>
#include <functional>
#include <random>
#include <regex>

using namespace std;
typedef long long ll;

#define ALL(c) (c).begin(),(c).end()
#define FOR(i,l,r) for(int i=(int)l;i<(int)r;++i)
#define REP(i,n) FOR(i,0,n)
#define FORr(i,l,r) for(int i=(int)r-1;i>=(int)l;--i)
#define REPr(i,n) FORr(i,0,n)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define IN(l,v,r) ((l)<=(v) && (v)<(r))
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end())
//debug
#define DUMP(x)  cerr << #x << " = " << (x)
#define LINE()    cerr<< " (L" << __LINE__ << ")"

template<typename T,typename U> T pmod(T x,U M){return (x%M+M)%M;}


//output
template<typename T> ostream& operator << (ostream& os, const vector<T>& as){REP(i,as.size()){if(i!=0)os<<" "; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const vector<vector<T>>& as){REP(i,as.size()){if(i!=0)os<<endl; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const set<T>& ss){for(auto a:ss){if(a!=ss.begin())os<<" "; os<<a;}return os;}
template<typename T1,typename T2> ostream& operator << (ostream& os, const pair<T1,T2>& p){os<<p.first<<" "<<p.second;return os;}
template<typename K,typename V> ostream& operator << (ostream& os, const map<K,V>& m){bool isF=true;for(auto& p:m){if(!isF)os<<endl;os<<p;isF=false;}return os;}
template<typename T1> ostream& operator << (ostream& os, const tuple<T1>& t){os << get<0>(t);return os;}
template<typename T1,typename T2> ostream& operator << (ostream& os, const tuple<T1,T2>& t){os << get<0>(t)<<" "<<get<1>(t);return os;}
template<typename T1,typename T2,typename T3> ostream& operator << (ostream& os, const tuple<T1,T2,T3>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t);return os;}
template<typename T1,typename T2,typename T3,typename T4> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t);return os;}
template<typename T1,typename T2,typename T3,typename T4,typename T5> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t);return os;}
template<typename T1,typename T2,typename T3,typename T4,typename T5,typename T6> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5,T6>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t)<<" "<<get<5>(t);return os;}
template<typename T1,typename T2,typename T3,typename T4,typename T5,typename T6,typename T7> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5,T6,T7>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t)<<" "<<get<5>(t)<<" "<<get<6>(t);return os;}

// must template
typedef long double D;
const D INF = 1e12,EPS = 1e-10;

typedef complex<D> P;
#define X real()
#define Y imag()
istream& operator >> (istream& is,complex<D>& p){D x,y;is >> x >> y; p=P(x,y); return is;}

int sig(D a,D b=0){return a<b-EPS?-1:a>b+EPS?1:0;}
template<typename T> bool eq(const T& a,const T& b){return sig(abs(a-b))==0;}
bool compX (const P& a,const P& b){return !eq(a.X,b.X)?sig(a.X,b.X)<0:sig(a.Y,b.Y)<0;}
namespace std{
	bool operator <  (const P& a,const P& b){return compX(a,b);}
  	bool operator == (const P& a,const P& b){return eq(a,b);}
};
// a×b
D cross(const P& a,const P& b){return imag(conj(a)*b);}
// a・b
D dot(const P& a,const P& b) {return real(conj(a)*b);}
int ccw(const P& a,P b,P c){
    b -= a; c -= a;
    if (sig(cross(b,c)) > 0)   return +1;       // counter clockwise
    if (sig(cross(b,c)) < 0)   return -1;       // clockwise
    if (sig(dot(b,c)) < 0)     return +2;       // c--a--b on line
    if (sig(norm(b),norm(c))<0) return -2;       // a--b--c on line
    return 0; //a--c--b on line (c==b,c==a)
}
// //must template

//直線・線分
struct L : public vector<P> {
	P vec() const {return this->at(1)-this->at(0);}
	L(const P &a, const P &b){push_back(a); push_back(b);}
	L(){push_back(P());push_back(P());}
};
istream& operator >> (istream& is,L& l){P s,t;is >> s >> t;l=L(s,t);return is;}

bool isIntersectLS(const L &l, const L &s) {
	return sig(cross(l.vec(), s[0]-l[0])*       // s[0] is left of l
		cross(l.vec(), s[1]-l[0]))<=0; // s[1] is right of l
}
// verified by ACAC003 B
// http://judge.u-aizu.ac.jp/onlinejudge/creview.jsp?rid=899178&cid=ACAC003
bool isIntersectSS(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 isStrictIntersectSS(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 isIntersectSP(const L &s, const P &p) {
	return sig(abs(s[0]-p)+abs(s[1]-p),abs(s[1]-s[0])) <=0; // triangle inequality
}

// 直線へ射影した時の点
// verified by AOJLIB
// http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=1092212
P projection(const L &l, const P &p) {
	D t = dot(p-l[0],l.vec()) / norm(l.vec());
	return l[0] + t * l.vec();
}

D distanceLP(const L &l, const P &p) {return abs(p - projection(l, p));}

struct C {
	P o; D r;
	C(const P &o,D r) : o(o), r(r) { }
};

// verified by http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=1088814
vector<P> intersectionLC(L l,C c){
	D d = distanceLP(l,c.o);
	vector <P> res;
 	if(sig(d-c.r)==0){
		res.push_back(projection(l,c.o));
	}else if(sig(d,c.r)<0){
		P m = projection(l,c.o),u = (l[1]-l[0])/abs(l[1]-l[0]);
	 	D t = sqrt(c.r*c.r - d*d);
		res.push_back( m + t * u);res.push_back( m - t * u);
	}
	return res;
}

// verified by http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=1088814
vector<P> intersectionSC(L l,C c){
	vector<P> ps=intersectionLC(l,c),res;
	REP(i,ps.size()){
		if(ccw(ps[i],l[0],l[1])==+2 || ccw(ps[i],l[1],l[0])==+2){
			res.push_back(ps[i]);
		}
	}
	return res;
}

//浮動小数点 pmod
D pfmod(D v,D M=2*M_PI){return fmod(fmod(v,M)+M,M);}



class Main{
	public:

	// s(= o->a)を右回転(>0)させた時、t上の点pと衝突するか．
	bool canStrictRotR(const L& s,const L& t,const P& p){
		if(sig(abs(s.vec()),abs(p-s[0]))<0)return false;
		if(eq(s[1],p) && !isStrictIntersectSS(L(s[0],s[0]+(s[1]-s[0])*polar(1.0L,-1e-5L)),t))return false;//rot 0
		return true;
	}

	void run(){

		while(true){
			D l,r;cin >> l >> r;int N;cin >> N;if(N==0)break;
			r*=2*M_PI;
			
			vector<P> ps(N);REP(i,N) cin >> ps[i];
			vector<L> ls(N);REP(i,N) ls[i]=L(ps[i],ps[pmod(i+1,N)]);

			L seg(P(0,0),P(0,l));P c(0,0);

			int apos=0;
			auto r_arg=[&](const P& x,const P&y)->D{
				D res=-(arg(x-c)-arg(y-c));
				return sig(res)>=0?res:res+2*M_PI;
			};
			while(sig(r)>0){
				vector<tuple<D,D,P>> cps;

				REP(k,2){
					C cir(c,abs(seg[k]-c));
					L ac(c,seg[k]),car(c,c+(seg[k]-c)*polar(1.0L,-1e-5L));
					REP(i,N)if(sig(abs(ps[i]-cir.o),cir.r)<=0)if(!eq(c,ps[i])){
						if(isIntersectSS(car,ls[pmod(i-1,N)]) || !isIntersectSP(ac,ps[i])){// つっかかれる
							cps.emplace_back(r_arg(ps[i],seg[k]),-abs(ps[i]-c),ps[i]);
						}
					}
					REP(i,N){
						vector<P> _cps=intersectionSC(ls[i],cir);
						REP(j,_cps.size())if(canStrictRotR(ac,ls[i],_cps[j]))
							cps.emplace_back(r_arg(_cps[j],seg[k]),-abs(_cps[j]-c),_cps[j]);
					}
				}
				sort(ALL(cps));
				assert(cps.size()>0);
				D ar=get<0>(cps[0]);P cp=get<2>(cps[0]);
				
				// can't move
				if(sig(ar)<=0)break;
				//end
				if(sig(r,ar)<0)ar=r;

				//rotate
				REP(k,2)seg[k]=c + (seg[k]-c)*polar(1.0L,-ar);
				c=cp;
				r-=ar;
				bool rev=false;REP(i,ls.size())if(isIntersectSP(ls[i],seg[apos]))rev=true;
				if(rev){
					swap(seg[0],seg[1]);apos=1-apos;
				}
			}
			cout << seg[apos].X <<" " << seg[apos].Y <<endl;
		}
	}
};
int main(){
	cerr <<fixed<<setprecision(20);	
	cout <<fixed<<setprecision(20);	
	ios::sync_with_stdio(false);
	Main().run();
	return 0;
}