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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <complex>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cmath>
#include <cassert>
#include <queue>
#include <set>
#include <map>
#include <valarray>
#include <bitset>
#include <stack>
#include <iomanip>
#include <fstream>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
#define chmax(a,b) (a<b?(a=b,1):0)
#define chmin(a,b) (a>b?(a=b,1):0)
#define valid(y,x,h,w) (0<=y&&y<h&&0<=x&&x<w)
const int INF = 1<<29;
const double EPS = 1e-8;
const double PI = acos(-1);
typedef long long ll;
typedef pair<int,int> pii;

typedef complex<double> P;
P pIN() { double x,y; cin >> x >> y; return P(x,y); }
namespace std {
  bool operator < (const P& a, const P& b) {
    // if (abs(a-b)<EPS) return 0;
    return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
  }
bool operator==(const P &a, const P &b) {
  return abs(a-b)<EPS;
}
}
double cross(const P& a, const P& b) {
  return imag(conj(a)*b);
}
double dot(const P& a, const P& b) {
  return real(conj(a)*b);
}
int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > EPS)   return +1;       // counter clockwise
  if (cross(b, c) < -EPS)   return -1;       // clockwise
  if (dot(b, c) < -EPS)     return +2;       // c--a--b on line
  if (EPS < norm(b) && norm(b) < norm(c)-EPS) return -2; // a--b--c on line
  return 0;
}
struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
  L() { resize(2); }
};
ostream &operator<<(ostream &os, const L &a) {
  os << a[0] << " -> " << a[1];
  return os;
}
typedef vector<P> G;
#define curr(g, i) g[i]
#define next(g, i) g[(i+1)%g.size()]
#define prev(g, i) g[(i+g.size()-1)%g.size()]
L line(const G &g, int i) {
  return L(g[i],g[(i+1)%g.size()]);
}
struct C {
  P p; double r;
  C(const P &p, double r) : p(p), r(r) { }
};
bool contain(const C &c1, const C &c2) {
  double d = abs(c1.p - c2.p);
  return d < c1.r - c2.r + EPS;
}
vector<L> tangentCP(const C &c, const P &p) {
  vector<L> ret;
  P vect = c.p - p;
  double d = abs(vect);
  double l = sqrt(d*d-c.r*c.r);
  if (::isnan(l)) { return ret; }
  P v1 = vect * P(l / d,  c.r / d);
  P v2 = vect * P(l / d, -c.r / d);
  ret.push_back(L(p, p+v1));
  if (l > EPS)
    ret.push_back(L(p, p+v2));
  return ret;
}
vector<L> tangentCC(const C &c1, const C &c2) {
  vector<L> ret;
  if (abs(c1.p - c2.p) - (c1.r + c2.r) > -EPS) {
    P center = (c1.p * c2.r + c2.p * c1.r) / (c1.r + c2.r);
    ret = tangentCP(c1, center);
  }
  if (fabs(c1.r - c2.r) > EPS) {
    P out = (-c1.p * c2.r + c2.p * c1.r) / (c1.r - c2.r);
    vector<L> nret = tangentCP(c1, out);
    ret.insert(ret.end(), nret.begin(), nret.end());
  } else {
    P vect = c2.p - c1.p;
    vect /= abs(vect);
    P q1 = c1.p + vect * P(0, 1) * c1.r;
    P q2 = c1.p + vect * P(0, -1) * c1.r;
    ret.push_back(L(q1, q1 + vect));
    ret.push_back(L(q2, q2 + vect));
  }
  return ret;
}
P projection(const L &l, const P &p) {
  double t = dot(p-l[0], l[0]-l[1]) / norm(l[0]-l[1]);
  return l[0] + t*(l[0]-l[1]);
}
P reflection(const L &l, const P &p) {
  return p + P(2,0) * (projection(l, p) - p);
}

vector<P> crosspointLC(const L &l, const C &c) {
  vector<P> ret;
  P center = projection(l, c.p);
  double d = abs(center - c.p);
  double t = sqrt(c.r * c.r - d * d);
  if (::isnan(t)) { return ret; }
  P vect = (l[1] - l[0]);
  vect /= abs(vect);
  ret.push_back(center - vect * t);
  if (t > EPS) {
    ret.push_back(center + vect * t);
  }
  return ret;
}
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
}

vector<P> crosspointSC(const L &s, const C &c) {
  vector<P> ret;
  vector<P> nret = crosspointLC(s, c);
  FOR(it, nret)
    if (intersectSP(s, *it))
      ret.push_back(*it);
   return ret;
}


G g;
double l,r;
int N;

struct Res {
  P ct;
  L s;
  double ang;
  Res(const P &ct, const L &s, double ang) :
    ct(ct),s(s),ang(ang) {}
  Res() {}
};

bool win(const Res &a, const Res &b, const P &ct) {
  if (abs(a.ang-b.ang)>EPS) return a.ang > b.ang;
  return abs(a.ct-ct) < abs(b.ct-ct);
}
bool sameang(const P &a, const P &b) {
  return abs(cross(a,b))<EPS && dot(a,b)>0;
}
double angle(const P &a, const P &b) {
  double ret = arg(b)-arg(a);
  return (ret>=0) ? ret : ret + 2*PI;
}
// テ」ツδ凖」ツつッテ」ツδ暗」ツδォaテ」ツ?ィテ」ツδ凖」ツつッテ」ツδ暗」ツδォbテ」ツ?ョテゥツ鳴禿」ツ?ョティツァツ津・ツコツヲ
double angle2(const P &a, const P &b) {
  return min(angle(a,b), angle(b,a));
}
double angle2_2(const P &a, const P &b) {
  double ret = abs(arg(b)-arg(a));
  return ret>PI ? 2*PI-ret : ret;
}
// テヲツ卍づィツィツ暗・ツ崢榲」ツつ甘・ツ?エテ」ツ?ッテ・ツ青ォテ」ツ?ソテ」ツ??・ツ債甘ヲツ卍づィツィツ暗・ツ崢榲」ツつ甘・ツ?エテ」ツ?ッテ・ツ青ォテ」ツ?セテ」ツ?ェテ」ツ??
int quadrant(const P &a) {
  if (a.real() > 0 && a.imag() >= 0) return 0;
  if (a.real() <= 0 && a.imag() > 0) return 1;
  if (a.real() < 0 && a.imag() <= 0) return 2;
  return 3;
}
// テ」ツδュテ」ツδ静」ツつケテ」ツδ暗」ツ?ェティツァツ津・ツコツヲテヲツッツ氾ィツシツ?
bool cmpAngle(const P &a, const P&b) {
  int q1 = quadrant(a);
  int q2 = quadrant(b);
  if (q1 != q2) return q1 < q2;
  return cross(a,b)>0;
}
// テ」ツδ凖」ツつッテ」ツδ暗」ツδォテ」ツ?ョテ・ツ崢榲ィツサツ「
P rotate(P p, double ang) {
  return p * P(cos(ang), sin(ang));
}
// テ・ツ篠淌ァツつケテ・ツ堕ィテ」ツつ甘」ツ?ョテァツ崢エテァツキツ堙」ツ?ョテ・ツ崢榲ィツサツ「
L rotate(L l, double ang) {
  return L(rotate(l[0], ang),rotate(l[1], ang));
}

double signedLength(const P &p1, const P &p2, const P &p) {
  P pc = projection(L(p1,p2),p);
  double res = abs(pc-p1);
  if (dot(pc-p1,p2-p1)>0) return res;
  else return -res;
}

Res find_next(const P &ct, const L& s) {
  Res res;
  res.ang = INF;
  // vertex
  REP(i,N) {
    if (ct==g[i]) continue;
    REP(j,2) {
      if (s[j]==ct) continue;
      if (abs(g[i]-ct)<abs(s[j]-ct)) {
        if (sameang(g[i]-ct,s[j]-ct)) {
          if (ccw(ct,s[j],next(g,i)) == -1 ||
              ccw(ct,s[j],prev(g,i)) == -1) {
            // stuck
            Res t(g[i], s, 0);
            if (win(res,t,ct)) res = t;
          }
        } else {
          double ang = angle(g[i]-ct,s[j]-ct);
          Res t(g[i],L(ct+rotate(s[0]-ct,-ang),ct+rotate(s[1]-ct,-ang)),ang);
          if (win(res,t,ct)) res = t;
        }
      }
    }
  }
  // edge
  REP(i,N) {
    REP(j,2) {
      if (s[j]==ct) continue;
      vector<P> ps = crosspointSC(line(g,i),C(ct,abs(s[j]-ct)));
      FOR(it, ps) {
        if (sameang(*it-ct,s[j]-ct)) {
          P a = g[i];
          P b = next(g,i);
          P c = prev(g,i);
          if (abs(signedLength(ct,s[j],a) - signedLength(ct,s[j],b)) < EPS) continue;
          if (signedLength(ct,s[j],a) > signedLength(ct,s[j],b)) {
            swap(a,b);
            c = g[(i+2)%g.size()];
          }
          if (ccw(ct,*it,a) == 1) continue;
          if (ccw(ct,*it,a) != -1) {
            continue;
            if (ccw(ct,*it,c) == 1) continue;
          }
          // cout << ct << " " << s[j] << " " << a << " " << b << " :" << signedLength(ct,s[j],a) << " " << signedLength(ct,s[j],b) << endl;
          // cout << *it << endl;
          // cout << ccw(ct,*it,a) << endl;
          
          // cout << *it << endl;
          // stuck
          Res t(*it, s, 0);
          if (win(res,t,ct)) res = t;
        } else {
          // cout << *it << endl;
          double ang = angle(*it-ct,s[j]-ct);
          Res t(*it,L(ct+rotate(s[0]-ct,-ang),ct+rotate(s[1]-ct,-ang)),ang);
          // cout << res.ang << " " << ang << endl;
          if (win(res,t,ct)) res = t;
        }      
      }
    }
  }
  return res;
}
FILE *fp;
double scale = 10;
P diff(300, 300);
double WH = 500;
void outputLine(const L &l) {
  if (!fp) return;
  P a = l[0];
  P b = l[1];
  a = a*P(scale,0)+diff;
  b = b*P(scale,0)+diff;
  fprintf(fp,"line(%f,%f,%f,%f);\n",a.real(),WH-a.imag(),b.real(),WH-b.imag());  
}
void outputCircle(C c) {
  if (!fp) return;
  c.p = c.p*P(scale,0) + diff;
  c.r *= scale;
  fprintf(fp, "circle(%f,%f,%f);\n", c.p.real(), WH-c.p.imag(), c.r);
}


int main() {
  fp = fopen("data.js","w");
  while(cin>>l>>r>>N, l||r||N) {
    g.clear();
    double tang = r*2*PI;
    REP(i,N) g.push_back(pIN());
    REP(i,N) outputLine(line(g,i));
    L s(P(0,0),P(0,l));
    P ct(0,0);
    double ang = 0;
    outputLine(s);
    REP(i,1000) {
      Res ret = find_next(ct,s);
      // cout << ang << " " << ret.ang << " " << tang << endl;
      if (ang+ret.ang>=tang) {
        double ag = tang-ang;
        s[0] = ct+rotate(s[0]-ct,-ag);
        s[1] = ct+rotate(s[1]-ct,-ag);
        outputLine(s);
        break;
      }
      // cout << ang << " " << s  << ":" << ret.ang << " " << ret.s<< " " << ret.ct << endl;
      outputLine(ret.s);
      outputCircle(C(ct,l));
      ang += ret.ang;
      s = ret.s;
      ct = ret.ct;
    }
    printf("%.10f %.10f\n", s[0].real(), s[0].imag());
  }
}