/*

考验观察法？？ 

可以发现最终答案等于所有作为圆心的点求出凸包的周长加上一个圆的周长

向量旋转

(x1, y1) 相较于 (x2, y2) 旋转角c 答案是

(dtx * cosc - dty * sinc + x2, dty * cosc + dtx * sinc + y2) 

貌似我的凸包也不鲁棒耶 

*/

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#include<cmath>

#include<queue>

#define ll long long

#define M 400010

#define mmp make_pair

using namespace std;

int read()

{

	int nm = 0, f = 1;

	char c = getchar();

	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';

	return nm * f;

}

const double pi = acos(-1), eps = 1e-12;

struct Vec

{

	double x, y;

}pt[M];

double operator * (const Vec &a, const Vec &b)

{

	return a.x * b.y - a.y * b.x;

}

Vec operator + (Vec a, Vec b)

{

	return Vec{a.x + b.x, a.y + b.y};

}

Vec operator - (Vec a, Vec b)

{

	return Vec{a.x - b.x, a.y - b.y};

}

double cross(Vec a, Vec b, Vec c)

{

	return (a - c) * (b - c);

}

bool cmp1(Vec a, Vec b)

{

	return a.y == b.y ? a.x < b.x : a.y < b.y;

}

bool cmp(Vec a, Vec b)

{

	return cross(a, b, pt[1]) >= 0;

}

Vec rotate(Vec a, double p)

{

	return Vec{a.x * cos(p) - a.y * sin(p), a.y * cos(p) + a.x * sin(p)};

}

double diss(Vec a, Vec b)

{

	double x = a.x - b.x, y = a.y - b.y;

	return sqrt(x * x + y * y);

}

double r, h, l, a, b, ap;

int n, tp, sta[M]; 

int main()

{

	n = read();

	scanf("%lf%lf%lf", &l, &h, &r);

	h = (h - r * 2) / 2.0, l = (l - r * 2) / 2.0;

	for(int i = 1; i <= n; i++)

	{

		scanf("%lf%lf%lf", &a, &b, &ap);

		pt[i] = rotate(Vec{h, l}, ap) + Vec{a, b};

		pt[i + 2 * n] = rotate(Vec{-h, l}, ap) + Vec{a, b};

		pt[i + 3 * n] = rotate(Vec{h, -l}, ap) + Vec{a, b};

		pt[i + n] = rotate(Vec{-h, -l}, ap) + Vec{a, b};

	}

	n *= 4;

	sort(pt + 1, pt + n + 1, cmp1);

	sta[++tp] = 1;

	sort(pt + 2, pt + n + 1, cmp);

	sta[++tp] = 2;

	sta[++tp] = 3;

	for(int i = 4; i <= n; i++)

	{

		while(tp > 1 && cross(pt[i], pt[sta[tp]], pt[sta[tp - 1]]) >= 0) tp--;

		sta[++tp] = i;

	}

	double ans = 0;

	for(int i = 1; i < tp; i++) ans += diss(pt[sta[i]], pt[sta[i + 1]]);

	ans += diss(pt[sta[1]], pt[sta[tp]]);

	ans += r * pi * 2;

	printf("%.2lf\n", ans);

	return 0;

}