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;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x + _y;
  if (_ >= _mod) _ -= _mod;
  return _;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x - _y;
  if (_ < 0) _ += _mod;
  return _;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x * _y;
  if (_ >= _mod) _ %= _mod;
  return _;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _tmp = mypow(_a, _x / 2, _mod);
  _tmp = mul(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = mul(_tmp, _a, _mod);
  return _tmp;
}
long long mymul(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 0ll;
  long long _tmp = mymul(_a, _x / 2, _mod);
  _tmp = add(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = add(_tmp, _a, _mod);
  return _tmp;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-12 && _x < _y + 1e-12;
}
int __ = 1, _cs;
void build() {}
inline pair<double, double> operator-(const pair<double, double>& p1,
                                      const pair<double, double>& p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
inline double operator*(const pair<double, double>& p1,
                        const pair<double, double>& p2) {
  return p1.first * p2.first + p1.second * p2.second;
}
inline double norm(const pair<double, double>& tp) { return sqrt(tp * tp); }
inline double angle(const pair<double, double>& tp) {
  return atan2(tp.second, tp.first);
}
const double pi = acos(-1.0);
inline void scan(pair<double, double>& _p) {
  scanf("%lf%lf", &_p.first, &_p.second);
}
inline void scan(pair<pair<double, double>, double>& _c) {
  scan(_c.first);
  scanf("%lf", &_c.second);
}
inline bool in(const pair<double, double>& tp,
               const pair<pair<double, double>, double>& tc) {
  return norm(tp - tc.first) < tc.second + 1e-12;
}
inline double sqr(const double& tk) { return tk * tk; }
int n;
pair<double, double> o;
double v, t;
pair<pair<double, double>, double> c[101010];
void init() {
  scan(o);
  scanf("%lf%lf", &v, &t);
  v *= t;
  n = getint();
  for (int i = 0; i < n; i++) scan(c[i]);
}
vector<pair<double, double> > ans;
void Add(const double& sang, const double& eang) {
  if (sang < -pi) {
    ans.push_back({sang + 2.0 * pi, pi});
    ans.push_back({-pi, eang});
  } else if (eang > pi) {
    ans.push_back({sang, pi});
    ans.push_back({-pi, eang - 2.0 * pi});
  } else
    ans.push_back({sang, eang});
}
void Add(const pair<pair<double, double>, double>& tc) {
  double dst = norm(tc.first - o);
  if (dst - tc.second > v) return;
  double tangd = sqrt(sqr(dst) - sqr(tc.second));
  double ang = angle(tc.first - o);
  double theta = 0.0;
  if (tangd < v + 1e-12)
    theta = asin(tc.second / dst);
  else
    theta = acos((sqr(v) + sqr(dst) - sqr(tc.second)) / (2.0 * v * dst));
  Add(ang - theta, ang + theta);
}
void solve() {
  for (int i = 0; i < n; i++)
    if (in(o, c[i])) {
      printf("%.12f\n", 1.0);
      exit(0);
    }
  for (int i = 0; i < n; i++) Add(c[i]);
  sort((ans).begin(), (ans).end());
  double sum = 0.0;
  if (ans.size()) {
    pair<double, double> pre = ans[0];
    for (size_t i = 1; i < ans.size(); i++) {
      if (ans[i].first < pre.second)
        pre.second = max(pre.second, ans[i].second);
      else {
        sum += pre.second - pre.first;
        pre = ans[i];
      }
    }
    sum += pre.second - pre.first;
  }
  printf("%.12f\n", sum / (2.0 * pi));
}
int main() {
  build();
  while (__--) {
    init();
    solve();
  }
}