Problem: I'm traveling to a country with a rabbit. There are n cities in this country numbered from 1 to n, and the rabbit is now in city 1. City i is a point on the coordinate plane (xi, yi) ).

Rabbits travel to meet the following conditions.

* The travel path is a polygonal line, each part of which must be a line segment connecting two different cities.
* The total length of the travel route must be r or less. The overlapping part of the route is also counted for the number of times it has passed.
* When the direction of movement changes, the bending angle must be less than or equal to θ. There is no limit to the initial direction of movement.



If you move from one city to another, you will get one carrot in the destination city. You can visit the same city multiple times, and you will get a carrot each time you visit. Find the maximum number of carrots you can put in.



Input

One integer n is given on the first line of the input, and two real numbers r and θ are given on the second line, separated by a space.

1 ≤ n ≤ 20
0 <r <104
0 ° <θ <180 °


The following n lines are given the integers xi and yi separated by spaces.

-10 000 ≤ xi, yi ≤ 10 000

The answer does not change even if r and θ are changed within ± 10-3.
The locations of both cities are different.

Output

Output the maximum number of carrots that the rabbit can get on this trip in one line.

Example

Input

5
100.1 90.1
0 0
0 10
5 5
10 0
10 10


Output

10

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
#include<complex>

using namespace std;

#define ll long long
#define ld long double
#define INF 1e9
#define LINF (ll)INF*INF
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(int i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

#define int ll //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef vector<pii> vp;
typedef vector<vi> vvi;
typedef pair<pii,pii> data;

#define in(a,b,c,d) make_pair(pii(a,b),pii(c,d))

int gcd(int a, int b){
    if(b==0) return a;
    return gcd(b,a%b);
}
int lcm(int a, int b){
    return a/gcd(a,b)*b;
}


const double PI=acos(-1);
const double EPS=1e-5; //うまくいかなかったらゆるめる
typedef complex<double> P;
typedef vector<P> G;
#define shosu(x) fixed<<setprecision(x)
#define diff(P,i) (P[(i+1)%P.size()] - P[i])

struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
  L(){;}
};
struct C {
  P c;double r;
  C(const P &c,double r):c(c),r(r){}
};

namespace std{//演算子の定義
  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();
  }
}


void printP(const P &p){
  cout << shosu(10) << p.real() << " " << p.imag() << endl;
}

void printL(const L &l) {
  	cout << shosu(10) << l[0].real() << " " << l[0].imag() << " " << l[1].real() << " " << l[1].imag() << endl;
}

void printG(const G &g) {
  for(int i = 0; i < g.size(); i++){
    cout << shosu(10) << g[i].real() << " " << g[i].imag() << endl;
  }
}

double distancePP(const P &p, const P &q){ //2点間の距離
  return hypot(p.real()-q.real(), p.imag()-q.imag());
}

int n;
double r,si;
G g;

bool f(int i, int j, int k){
    double rad = abs(arg((g[i]-g[j])/(g[j]-g[k])));
    return rad < si + EPS;
}


P inP(){
  double x,y;
//  scanf("%lf,%lf", &x, &y);
  cin >> x >> y;
  P p(x,y);
  return p;
}

G inG(int n){
  G g(n);
  for(int i = 0; i < n; i++){
    g[i] = inP();
  }
  return g;
}

#define MAX 10005

double dist[20][20];
double memo[20][20][MAX] = {};

int ans = 0;
vector<vvi> v;


signed main(void) {
    cin >> n;
    cin >> r >> si;
    si *= acos(-1)/180;
    g = inG(n);
    v = vector<vvi> (n,vvi(n));
    rep(i,n)rep(j,n)if(i != j){
        rep(k,n)if(k != i && k != j){
            if(f(i,j,k))v[i][j].push_back(k);
        }
    }
    rep(i,n)rep(j,n)dist[i][j] = distancePP(g[i],g[j]);
    rep(i,n)rep(j,n)rep(k,MAX) memo[i][j][k] = INF;
    priority_queue<data> q;
    loop(i,1,n)if(dist[0][i] < r + EPS){
        memo[0][i][1] = dist[0][i];
        q.push(in(-dist[0][i],0,i,1));
    }
    int ans = 0;
    while(q.size()){
        data d = q.top();
        q.pop();
        int prev = d.first.second;
        int now = d.second.first;
        int num = d.second.second;
        if(memo[prev][now][num] < -d.first.first)continue;
        //cout << prev << " " << now << " " << num << endl;
        ans = max(ans, num);
        rep(i,v[prev][now].size()){
            int to = v[prev][now][i];
            double cost = memo[prev][now][num] + dist[now][to];
            if(memo[now][to][num+1] < cost + EPS)continue;
            if(cost > r + EPS)continue;
            memo[now][to][num+1] = min(memo[now][to][num+1], cost);
            q.push(in(-cost,now,to,num+1));
        }
    }
    cout << ans << endl;
}