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
#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=a;i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> void chmax(t&a,u b){if(a<b)a=b;}
template<class t,class u> void chmin(t&a,u b){if(a>b)a=b;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

template<class t> void mkuni(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

ll read(){ll i;cin>>i;return i;}

using ld=long double;
using cm=complex<ld>;
#define x real()
#define y imag()
const ld eps=1e-7;
const ld PI=acos(ld(-1));
int sgn(ld a){return a<-eps?-1:(a>eps?1:0);}
auto cmcmp=[](const cm&a,const cm&b){
	if(sgn(a.x-b.x))return a.x<b.x;
	else return sgn(a.y-b.y)<0;
};
ld dot(cm a,cm b){return a.x*b.x+a.y*b.y;}
ld crs(cm a,cm b){return a.x*b.y-a.y*b.x;}
int ccw(cm a,cm b){return sgn(crs(a,b));}
int ccw(cm a,cm b,cm c){return ccw(b-a,c-a);}
//AOJ1183
int qeq(ld a,ld b,ld c,ld&d,ld&e){
	ld f=b*b-4*a*c;
	if(sgn(f)<0)return 0;
	ld g=sqrt(max(f,ld(0)));
	d=(-b+g)/(2*a);
	e=(-b-g)/(2*a);
	return sgn(f)+1;
}
//(-2)[a,-1](0)[b,1](2)
int bet(cm a,cm b,cm c){
	cm d=b-a;
	ld e=dot(d,c-a);
	if(sgn(e)<=0)return sgn(e)-1;
	return sgn(e-norm(d))+1;
}
ld tri2(cm a,cm b,cm c){
	return crs(b-a,c-a);
}
//AOJ0153
//0-no,1-edge,2-in
int cont(cm a,cm b,cm c,cm d){
	if(ccw(a,b,c)==-1)
		swap(b,c);
	return min({ccw(a,b,d),ccw(b,c,d),ccw(c,a,d)})+1;
}
//AOJ1183
//arg between ab
//assume given lengths are valid
ld arg(ld a,ld b,ld c){
	return acos(min(max((a*a+b*b-c*c)/(2*a*b),ld(-1)),ld(1)));
}

//a->b->c と進むときに曲がる角度
//a-b-cが一直線上にあれば0が帰る
ld turn(cm a,cm b,cm c){
	return arg((c-b)/(b-a));
}

const int N=22;
const int R=10010;
bool can[N][N][N];
ld ds[N][N],dp[R][N][N];

signed main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(20);
	
	rep(i,R)rep(j,N)rep(k,N)
		dp[i][j][k]=1e9;
	
	int n=read();
	ld r,t;
	cin>>r>>t;
	t*=PI/180;
	vc<cm> a;
	rep(i,n){
		int b=read(),c=read();
		a.eb(b,c);
	}
	rep(i,n)rep(j,n)if(i!=j)rep(k,n)if(i!=k&&j!=k)
		can[i][j][k]=sgn(abs(turn(a[i],a[j],a[k]))-t)<=0;
	
	rep(i,n)rep(j,n)if(i!=j)
		ds[i][j]=abs(a[i]-a[j]);
	rng(i,1,n)
		dp[1][0][i]=ds[0][i];
	
	int ans=0;
	rng(z,1,R-1){
		rep(i,n)rep(j,n)if(sgn(dp[z][i][j]-r)<=0){
			ans=z;
			rep(k,n)if(can[i][j][k])
				chmin(dp[z+1][j][k],dp[z][i][j]+ds[j][k]);
		}
	}
	
	cout<<ans<<endl;
	
}

