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<bits/stdc++.h>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//clock_t start=clock();clock_t end=clock();cout<<(double)(end-start)/CLOCKS_PER_SEC<<endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
typedef pair<double,double> pdd;
typedef vector<vector<ll>> mat;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll mod5=1000000009;
ll inf=1LL<<60;
double pi=3.141592653589793238462643383279L;
double eps=1e-8;
#define rep(i,m,n) for(int i=m;i<n;i++)
#define rrep(i,n,m) for(int i=n;i>=m;i--)
#define srep(itr,st) for(auto itr=st.begin();itr!=st.end();itr++)
#define mrep(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define Max(a,b) a=max(a,b)
#define Min(a,b) a=min(a,b)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
ll popcount(ll n){
    ll ret=0;
    ll u=n;
    while(u>0){
        ret+=u%2;
        u/=2;
        
    }
    return ret;
}
double dist(double a,double b,double c,double d){
    return pow(((a-c)*(a-c)+(b-d)*(b-d)),0.5);
}
typedef pair<double,pipi> P;
int main(){
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n;cin>>n;
    double r,si;cin>>r>>si;
    double x[n],y[n];
    rep(i,0,n)cin>>x[i]>>y[i];
    vector<int> v[n][n];
    rep(i,0,n){
        rep(j,0,n){
            if(i==j)continue;
            double nx=x[i]-x[j],ny=y[i]-y[j];
            rep(k,0,n){
                if(i==k||j==k)continue;
                double nex=x[k]-x[i],ney=y[k]-y[i];
                double da=dist(nx,ny,0,0),db=dist(nex,ney,0,0);
                double c=nx*nex+ny*ney;
                c/=da*db;
                if(c>=1)c=1;
                if(c<=-1)c=-1;
                double sii=acos(c)/pi*180;
                if(sii<=si+eps)v[i][j].push_back(k);
            }
        }
    }
    double dp[n][n][10001];
    rep(i,0,n)rep(j,0,n)rep(k,0,10001)dp[i][j][k]=1000000000;
    priority_queue<P,vector<P>,greater<P>> q;
    rep(i,1,n){
        dp[i][0][1]=dist(x[0],y[0],x[i],y[i]);
        q.push({dp[i][0][1],{1,{i,0}}});
    }
    int ans=0;
    while(q.size()>0){
        P p=q.top();q.pop();
        double nd=p.first;
        int ko=p.second.first;
        int pa=p.second.second.second;
        int now=p.second.second.first;
        if(nd>r+eps)break;
        if(dp[now][pa][ko]+eps<nd)continue;
        Max(ans,ko);
        rep(i,0,v[now][pa].size()){
            int ne=v[now][pa][i];
            double co=dist(x[now],y[now],x[ne],y[ne]);
            if(dp[ne][now][ko+1]>dp[now][pa][ko]+co){
                dp[ne][now][ko+1]=dp[now][pa][ko]+co;
                q.push({dp[ne][now][ko+1],{ko+1,{ne,now}}});
            }
        }
    }
    cout<<ans<<endl;
}