Problem: Dima is living in a dormitory, as well as some cockroaches.

At the moment 0 Dima saw a cockroach running on a table and decided to kill it. Dima needs exactly T seconds for aiming, and after that he will precisely strike the cockroach and finish it.

To survive the cockroach has to run into a shadow, cast by round plates standing on the table, in T seconds. Shadow casted by any of the plates has the shape of a circle. Shadow circles may intersect, nest or overlap arbitrarily.

The cockroach uses the following strategy: first he equiprobably picks a direction to run towards and then runs towards it with the constant speed v. If at some moment t ≤ T it reaches any shadow circle, it immediately stops in the shadow and thus will stay alive. Otherwise the cockroach is killed by the Dima's precise strike. Consider that the Dima's precise strike is instant.

Determine the probability of that the cockroach will stay alive.

Input

In the first line of the input the four integers x0, y0, v, T (|x0|, |y0| ≤ 109, 0 ≤ v, T ≤ 109) are given — the cockroach initial position on the table in the Cartesian system at the moment 0, the cockroach's constant speed and the time in seconds Dima needs for aiming respectively.

In the next line the only number n (1 ≤ n ≤ 100 000) is given — the number of shadow circles casted by plates.

In the next n lines shadow circle description is given: the ith of them consists of three integers xi, yi, ri (|xi|, |yi| ≤ 109, 0 ≤ r ≤ 109) — the ith shadow circle on-table position in the Cartesian system and its radius respectively.

Consider that the table is big enough for the cockroach not to run to the table edges and avoid Dima's precise strike.

Output

Print the only real number p — the probability of that the cockroach will stay alive.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Examples

Input

0 0 1 1
3
1 1 1
-1 -1 1
-2 2 1


Output

0.50000000000

Input

0 0 1 0
1
1 0 1


Output

1.00000000000

Note

The picture for the first sample is given below. 

<image>

Red color stands for points which being chosen as the cockroach's running direction will cause him being killed, green color for those standing for survival directions. Please note that despite containing a circle centered in ( - 2, 2) a part of zone is colored red because the cockroach is not able to reach it in one second.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double safe_sqrt(long double x) { return sqrt(max((long double)0.0, x)); }
long long GI(long long& x) { return scanf("%lld", &x); }
void GN(long double& x) {
  long long t;
  GI(t);
  x = t;
}
const long double PI = acos(-1.0);
long double DEG_to_RAD(long double d) { return d * PI / 180.0; }
long double RAD_to_DEG(long double r) { return r * 180.0 / PI; }
inline long long cmp(long double a, long double b) {
  return (a < b - 1e-6) ? -1 : ((a > b + 1e-6) ? 1 : 0);
}
struct Point {
  long double x, y;
  Point(long double x = 0.0, long double y = 0.0) : x(x), y(y) {}
  Point operator+(const Point& a) const { return Point(x + a.x, y + a.y); }
  Point operator-(const Point& a) const { return Point(x - a.x, y - a.y); }
  Point operator*(long double k) const { return Point(x * k, y * k); }
  Point operator/(long double k) const { return Point(x / k, y / k); }
  long double operator*(const Point& a) const { return x * a.x + y * a.y; }
  long double operator%(const Point& a) const { return x * a.y - y * a.x; }
  long long cmp(Point q) const {
    if (long long t = ::cmp(x, q.x)) return t;
    return ::cmp(y, q.y);
  }
  bool operator>(Point q) const { return cmp(q) > 0; }
  bool operator<(Point q) const { return cmp(q) < 0; }
  bool operator==(Point q) const { return cmp(q) == 0; }
  bool operator>=(Point q) const { return cmp(q) >= 0; }
  bool operator<=(Point q) const { return cmp(q) <= 0; }
  bool operator!=(Point q) const { return cmp(q) != 0; }
  Point conj() { return Point(x, -y); }
  long double norm() { return x * x + y * y; }
  Point normalizeTo(long double len) { return Point(x, y) / norm() * len; }
  long double len() { return sqrt(norm()); }
  Point rotate(long double alpha) {
    long double cosa = cos(alpha), sina = sin(alpha);
    return Point(x * cosa - y * sina, x * sina + y * cosa);
  }
};
long long ccw(Point a, Point b, Point c) { return cmp((b - a) % (c - a), 0); }
long long RE_TRAI = ccw(Point(0, 0), Point(0, 1), Point(-1, 1));
long long RE_PHAI = ccw(Point(0, 0), Point(0, 1), Point(1, 1));
istream& operator>>(istream& cin, Point& p) {
  cin >> p.x >> p.y;
  return cin;
}
ostream& operator<<(ostream& cout, Point& p) {
  cout << p.x << ' ' << p.y;
  return cout;
}
struct Circle : Point {
  long double r;
  Circle(long double x = 0, long double y = 0, long double r = 0)
      : Point(x, y), r(r) {}
  Circle(Point p, long double r) : Point(p), r(r) {}
  bool contains(Point p) { return (*this - p).len() <= r + 1e-6; }
};
bool areIntersect(Circle u, Circle v) {
  if (cmp((u - v).len(), u.r + v.r) > 0) return false;
  if (cmp((u - v).len() + v.r, u.r) < 0) return false;
  if (cmp((u - v).len() + u.r, v.r) < 0) return false;
  return true;
}
vector<Point> circleIntersect(Circle u, Circle v) {
  vector<Point> res;
  if (!areIntersect(u, v)) return res;
  long double d = (u - v).len();
  long double alpha = acos((u.r * u.r + d * d - v.r * v.r) / 2.0 / u.r / d);
  Point p1 = (v - u).rotate(alpha);
  Point p2 = (v - u).rotate(-alpha);
  res.push_back(p1 / p1.len() * u.r + u);
  res.push_back(p2 / p2.len() * u.r + u);
  return res;
}
const long long MN = 200111;
Circle a[MN];
long long n;
Point start;
long double R;
long double segment_union(vector<pair<long double, long double> > segs) {
  long long n = ((long long)(segs).size());
  vector<pair<long double, bool> > x(n * 2);
  for (long long i = 0, _a = (n); i < _a; ++i) {
    x[i * 2] = make_pair(segs[i].first, false);
    x[i * 2 + 1] = make_pair(segs[i].second, true);
  }
  sort(x.begin(), x.end());
  long double res = 0.0;
  long long c = 0;
  for (long long i = 0, _a = (n * 2); i < _a; ++i) {
    if (c && i) res += x[i].first - x[i - 1].first;
    if (x[i].second)
      ++c;
    else
      --c;
  }
  return res;
}
long double directed_angle(Point a, Point o, Point b) {
  long double t = -atan2(a.y - o.y, a.x - o.x) + atan2(b.y - o.y, b.x - o.x);
  while (t < 0) t += 2 * PI;
  return t;
}
long double range(long double l, long double r) {
  if (l < r)
    return r - l;
  else
    return r - l + 2 * PI;
}
void solve() {
  for (long long i = (1), _b = (n); i <= _b; ++i)
    if (((a[i].x) * (a[i].x)) + ((a[i].y) * (a[i].y)) <
        ((a[i].r) * (a[i].r)) + 1e-10) {
      cout << 1.0 << endl;
      return;
    }
  Circle C;
  C.x = start.x;
  C.y = start.y;
  C.r = R;
  Point hor(1, 0);
  vector<pair<long double, long double> > segs;
  for (long long i = (1), _b = (n); i <= _b; ++i) {
    auto t = circleIntersect(C, a[i]);
    long double a1, a2;
    if (C.contains(a[i])) {
      long double mid = atan2(a[i].x, a[i].y);
      long double half = asin(a[i].r / a[i].len());
      a1 = mid - half;
      a2 = mid + half;
    } else if (((long long)(t).size()) == 2) {
      auto u = t[0];
      auto v = t[1];
      a1 = atan2(u.x, u.y);
      a2 = atan2(v.x, v.y);
    } else
      continue;
    while (a1 < 0) a1 += 2 * PI;
    while (a2 < 0) a2 += 2 * PI;
    while (a1 > 2 * PI) a1 -= 2 * PI;
    while (a2 > 2 * PI) a2 -= 2 * PI;
    if (fabs(a1 - a2) < 1e-10) continue;
    if (range(a1, a2) > PI) swap(a1, a2);
    if (a1 <= a2) {
      segs.emplace_back(a1, a2);
    } else {
      segs.emplace_back(a1, 2 * PI);
      segs.emplace_back(0, a2);
    }
  }
  long double sum = segment_union(segs);
  cout << sum / (long double)(2 * PI) << endl;
}
int main() {
  cout << (fixed) << setprecision(14);
  long double v, T;
  GN(start.x);
  GN(start.y);
  GN(v);
  GN(T);
  R = v * T;
  GI(n);
  for (long long i = (1), _b = (n); i <= _b; ++i) {
    GN(a[i].x);
    GN(a[i].y);
    GN(a[i].r);
    a[i].x -= start.x;
    a[i].y -= start.y;
  }
  start = Point(0, 0);
  solve();
}