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>
double pi = acos(-1);
double eps = 1e-9;
long long z = 1000000007;
long long inf = 100000000000000000;
long long p1 = 37;
long long p2 = 53;
long long mod1 = 202976689;
long long mod2 = 203034253;
long long fact[100];
long long gdp(long long a, long long b) { return (a - (a % b)); }
long long ld(long long a, long long b) {
  if (a < 0) return -1 * gdp(abs(a), b);
  if (a % b == 0) return a;
  return (a + (b - a % b));
}
long long gd(long long a, long long b) {
  if (a < 0) return (-1 * ld(abs(a), b));
  return (a - (a % b));
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long e_gcd(long long a, long long b, long long& x, long long& y) {
  if (b > a) return e_gcd(b, a, y, x);
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  e_gcd(b, a % b, x1, y1);
  x = y1;
  y = (x1 - ((a / b) * y1));
  return e_gcd(b, a % b, x1, y1);
}
long long power(long long a, long long b, long long p) {
  if (b == 0) return 1;
  long long c = power(a, b / 2, p);
  if (b % 2 == 0)
    return ((c * c) % p);
  else
    return ((((c * c) % p) * a) % p);
}
long long inverse(long long a, long long n) { return power(a, n - 2, n); }
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
double min(double a, double b) {
  if (a < b) return a;
  return b;
}
double max(double a, double b) {
  if (a > b) return a;
  return b;
}
long long left(long long i) { return ((2 * i) + 1); }
long long right(long long i) { return ((2 * i) + 2); }
long long ncr(long long n, long long r) {
  if (n < r || (n < 0) || (r < 0)) return 0;
  return ((((fact[n] * inverse(fact[r], z)) % z) * inverse(fact[n - r], z)) %
          z);
}
void swap(long long& a, long long& b) {
  long long c = a;
  a = b;
  b = c;
  return;
}
using namespace std;
double angle(double x, double y) {
  double ang = atan2(y, x);
  if (ang < -eps) ang += (2 * pi);
  return ang;
}
double cross(double x1, double y1, double x2, double y2) {
  return (x1 * y2 - x2 * y1);
}
double dist_(double x1, double y1, double x2, double y2) {
  double x = x1 - x2;
  double y = y1 - y2;
  return sqrtl(x * x + y * y);
}
void get_pts(double a, double b, double c, double& x1, double& y1, double& x2,
             double& y2) {
  if (fabs(b) < eps) {
    x1 = x2 = (-c) / a;
    y1 = 1;
    y2 = 2;
    return;
  }
  x1 = 0, x2 = 1;
  y1 = ((-c) / b);
  y2 = (-(c + a)) / b;
  return;
}
double std_dist(double x, double y, double a, double b, double c) {
  double x1, y1, x2, y2;
  get_pts(a, b, c, x1, y1, x2, y2);
  double area = cross(x - x1, y - y1, x2 - x1, y2 - y1);
  double d = dist_(x1, y1, x2, y2);
  return fabs(area / d);
}
void line_circle(double a, double b, double c, double r,
                 vector<pair<double, double> >& v) {
  double x = (-a * c) / (a * a + b * b);
  double y = (-b * c) / (a * a + b * b);
  double dist = std_dist(0, 0, a, b, c);
  if (dist > r && fabs(dist - r) > eps) {
    return;
  }
  if (fabs(dist - r) < eps) {
    v.push_back(make_pair(x, y));
    return;
  }
  double d = sqrtl(r * r - dist * dist);
  double x1 = x - ((-b * d) / (sqrtl(a * a + b * b)));
  double y1 = y - ((a * d) / (sqrtl(a * a + b * b)));
  v.push_back(make_pair(x1, y1));
  x1 = x + ((-b * d) / (sqrtl(a * a + b * b)));
  y1 = y + ((a * d) / (sqrtl(a * a + b * b)));
  v.push_back(make_pair(x1, y1));
  return;
}
void circle_circle(double r1, double x, double y, double r2, double& a1,
                   double& a2) {
  if (r1 > sqrtl(x * x + y * y - r2 * r2 + eps)) {
    r1 = sqrtl(x * x + y * y - r2 * r2);
  }
  double len = sqrt(x * x + y * y);
  double c = (x * x + y * y + r1 * r1 - r2 * r2) / (2 * len * r1);
  double deta = acos(c);
  double ang = atan2(y, x);
  a1 = ang - deta, a2 = ang + deta;
  if (a1 < -eps) a1 += 2 * pi;
  if (a1 > 2 * pi + eps) a1 -= 2 * pi;
  if (a2 < -eps) a2 += 2 * pi;
  if (a2 > 2 * pi + eps) a2 -= 2 * pi;
  return;
}
void solve() {
  double x, y, vi, t;
  cin >> x >> y >> vi >> t;
  double r1 = vi * t;
  int n;
  cin >> n;
  vector<tuple<double, double, double> > v;
  for (long long i = 0; i < n; i++) {
    double x2, y2, r2;
    cin >> x2 >> y2 >> r2;
    x2 -= x;
    y2 -= y;
    if ((x2 * x2 + y2 * y2) <= (r2 * r2)) {
      cout << fixed << setprecision(10) << 1.0 << endl;
      return;
    }
    if (sqrtl(x2 * x2 + y2 * y2) > (r1 + r2 + eps)) continue;
    v.push_back(make_tuple(x2, y2, r2));
  }
  vector<pair<double, double> > ans;
  for (long long i = 0; i < v.size(); i++) {
    double x = get<0>(v[i]);
    double y = get<1>(v[i]);
    double r2 = get<2>(v[i]);
    double ang_1, ang_2;
    circle_circle(r1, x, y, r2, ang_1, ang_2);
    double a1 = min(ang_1, ang_2);
    double a2 = max(ang_1, ang_2);
    if (((a2 - a1) < (pi + eps))) {
      ans.push_back(make_pair(a1, a2));
    } else {
      ans.push_back(make_pair(0, a1));
      ans.push_back(make_pair(a2, 2 * pi));
    }
  }
  sort(ans.begin(), ans.end());
  vector<pair<double, double> > fin_ans;
  int s = 0;
  if (ans.size() == 0) {
    cout << fixed << setprecision(10) << 0.0 << endl;
    return;
  }
  fin_ans.push_back(ans[0]);
  for (int i = 1; i < ans.size(); i++) {
    if (ans[i].first > (fin_ans[s].second + eps)) {
      fin_ans.push_back(ans[i]);
      s++;
    } else {
      fin_ans[s].second = max(ans[i].second, fin_ans[s].second);
    }
  }
  double sum = 0;
  for (int i = 0; i < fin_ans.size(); ++i) {
    sum += (fin_ans[i].second - fin_ans[i].first);
  }
  cout << fixed << setprecision(10) << sum / (2 * pi) << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}