#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+100;
const double PI = acos(-1);
const double eps=1e-8;
int sgn(double x){///判断x是否等于0
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}
struct point{
    double x,y;
    point(){}
    point(double x,double y):x(x),y(y){}
    point operator+(point b){
        return point(x+b.x,y+b.y);
    }
    point operator-(point b){
        return point(x-b.x,y-b.y);
    }
    bool operator == (point b){
        return sgn(x-b.x)==0&&sgn(y-b.y)==0;
    }
    bool operator<(point b)const{
        if(sgn(x-b.x)==0) return sgn(y-b.y)<0;
        return sgn(x-b.x)<0;
    }
};
point s[maxn],g[maxn],h[maxn];
int n;
double dot(point a,point b){return a.x*b.x+a.y*b.y;}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
double cul(point a,point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
struct line{
    point p1,p2;
    line(){}
    line(point p1,point p2):p1(p1),p2(p2){}
};
double mult(point a,point b,point o){
    ///计算叉乘ao和bo
    return (a.x-o.x)*(b.y-o.y)>=(b.x-o.x)*(a.y-o.y);
}
int Graham(int n){
    int idx=1;
    sort(s,s+n);
    if(!n) return 0;
    h[0]=s[0];
    if(n==1) return 0;
    h[1]=s[1];
    if(n==2) return 0;
    h[2]=s[2];
    ///求凸包的上半部分
    for(int i=2;i<n;i++){
        while(idx&&(mult(s[i],h[idx],h[idx-1]))) idx--;
        h[++idx]=s[i];
    }
    int tmp=idx;
    h[++idx]=s[n-2];
    for(int i=n-3;i>=0;i--){
        while(idx!=tmp&&(mult(s[i],h[idx],h[idx-1]))) idx--;
        h[++idx]=s[i];
    }
    return idx;
}
//说明 返回值为凸包点的个数,h数组存储的是凸包中的点。
point rotate(point a,double t){
    ///旋转
    double c=cos(t),s=sin(t);
    return point{a.x*c-a.y*s,a.x*s+a.y*c};
}
int main(){
    int m;cin>>m;
    double a,b,r;
    cin>>b>>a>>r;
    a/=2.0;b/=2.0;
    for(int i=1;i<=m;i++){
        double x,y,z;cin>>x>>y>>z;
        point t={x,y};
        s[n++]=rotate({a-r,b-r},z)+t;
        s[n++]=rotate({r-a,b-r},z)+t;
        s[n++]=rotate({r-a,r-b},z)+t;
        s[n++]=rotate({a-r,r-b},z)+t;
    }
    int res=Graham(n);
    double ans=2*PI*r;
    for(int i=0;i<res-1;i++)
        ans=ans+cul(h[i],h[i+1]);
    ans+=(cul(h[res-1],h[0]));
    printf("%.2f\n",ans);
    return 0;
}