Problem: Petya has recently started working as a programmer in the IT city company that develops computer games.

Besides game mechanics implementation to create a game it is necessary to create tool programs that can be used by game designers to create game levels. Petya's first assignment is to create a tool that allows to paint different arrows on the screen.

A user of this tool will choose a point on the screen, specify a vector (the arrow direction) and vary several parameters to get the required graphical effect. In the first version of the program Petya decided to limit parameters of the arrow by the following: a point with coordinates (px, py), a nonzero vector with coordinates (vx, vy), positive scalars a, b, c, d, a > c.

The produced arrow should have the following properties. The arrow consists of a triangle and a rectangle. The triangle is isosceles with base of length a and altitude of length b perpendicular to the base. The rectangle sides lengths are c and d. Point (px, py) is situated in the middle of the triangle base and in the middle of side of rectangle that has length c. Area of intersection of the triangle and the rectangle is zero. The direction from (px, py) point to the triangle vertex opposite to base containing the point coincides with direction of (vx, vy) vector.

Enumerate the arrow points coordinates in counter-clockwise order starting from the tip.

<image>

Input

The only line of the input contains eight integers px, py, vx, vy ( - 1000 ≤ px, py, vx, vy ≤ 1000, vx2 + vy2 > 0), a, b, c, d (1 ≤ a, b, c, d ≤ 1000, a > c).

Output

Output coordinates of the arrow points in counter-clockwise order. Each line should contain two coordinates, first x, then y. Relative or absolute error should not be greater than 10 - 9.

Examples

Input

8 8 0 2 8 3 4 5


Output

8.000000000000 11.000000000000
4.000000000000 8.000000000000
6.000000000000 8.000000000000
6.000000000000 3.000000000000
10.000000000000 3.000000000000
10.000000000000 8.000000000000
12.000000000000 8.000000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct outFloat {
  int precision;
  long double value;
  outFloat(long double v, int p) : precision(p), value(v) {}
};
const int inf = 123456789;
const long long llInf = 123456789012345678;
const long double pi = 3.14159265358979323846;
const int maxN = 2000000;
int init() {
  ios_base::sync_with_stdio(0);
  srand(time(NULL));
  return 0;
}
int initializer = init();
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &p) {
  for (int i = 0; i < p.size(); i++) is >> p[i];
  return is;
}
class inputer {
 private:
  istream *istr;

 public:
  inputer(istream &is) : istr(&is) {}
  template <typename T>
  void get(T &n) {
    (*istr) >> n;
  }
  template <typename T, typename... Args>
  void get(T &n, Args &...args) {
    get(n);
    get(args...);
  }
  template <typename... Args>
  void getLn(Args &...args) {
    get(args...);
    getLn();
  }
  template <typename T>
  operator T() {
    T n;
    get(n);
    return n;
  }
  void getLn() {
    string s;
    getline(cin, s);
  }
  template <typename T>
  void getv(vector<T> &v, int size) {
    v.resize(size);
    (*istr) >> v;
  }
};
string delim = " ";
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << p.first << delim << p.second;
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  for (int i = 0; i < p.size(); i++) {
    if (i != 0) os << delim;
    os << p[i];
  }
  return os;
}
ostream &operator<<(ostream &os, const outFloat &of) {
  os << setprecision(of.precision) << of.value;
  return os;
}
class outputer {
 private:
  ostream *ostr;

 public:
  outputer(ostream &os) : ostr(&os) {}
  template <typename T>
  void put(const T &n) {
    (*ostr) << n;
  }
  template <typename T, typename... Args>
  void put(const T &n, const Args &...args) {
    put(n);
    put(args...);
  }
  template <typename... Args>
  void putLn(const Args &...args) {
    put(args...);
    putLn();
  }
  template <typename T>
  outputer &operator=(const T &n) {
    put(n);
    return *this;
  }
  void putLn() { (*ostr) << "\n"; }
  template <typename T>
  void putv(vector<T> v, int size, string delim) {
    for (int i = 0; i < size; i++) {
      if (i != 0) (*ostr) << delim;
      (*ostr) << v[i];
    }
  }
};
inputer in(cin);
outputer out(cout);
char getChar() {
  char c = 0;
  while (c <= ' ') cin >> c;
  return c;
}
bool willItBeAC() {
  return true;
  ;
}
int getMyRating() { return 2121; }
bool prime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
vector<long long> divs(long long n) {
  vector<long long> v;
  long long i = 2;
  while (i * i <= n)
    if (n % i == 0) {
      n /= i;
      v.push_back(i);
    } else
      i++;
  v.push_back(n);
  return v;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long extGcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1, d = extGcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
template <typename T>
class fenwick {
 private:
  vector<T> t;
  int n;

 public:
  fenwick(int nn) { init(nn); }
  void init(int nn) {
    n = nn;
    t.assign(n, 0);
  }
  T sum(int r) {
    T result = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
    return result;
  }
  void inc(int i, T delta) {
    for (; i < n; i = (i | (i + 1))) t[i] += delta;
  }
  T sum(int l, int r) { return sum(r) - sum(l - 1); }
  T get(int i) { return sum(i, i); }
  void put(int i, T v) { inc(i, v - sum(i, i)); }
  int size() { return n; }
};
int main() {
  long double px = in, py = in, vx = in, vy = in, a = in, b = in, c = in,
              d = in;
  long double k = sqrt(vy * vy + vx * vx);
  out.putLn(outFloat(px + vx * b / k, 10), ' ', outFloat(py + vy * b / k, 10));
  out.putLn(outFloat(px - vy * a / 2 / k, 10), ' ',
            outFloat(py + vx * a / 2 / k, 10));
  out.putLn(outFloat(px - vy * c / 2 / k, 10), ' ',
            outFloat(py + vx * c / 2 / k, 10));
  out.putLn(outFloat(px - vy * c / 2 / k - vx * d / k, 10), ' ',
            outFloat(py + vx * c / 2 / k - vy * d / k, 10));
  out.putLn(outFloat(px + vy * c / 2 / k - vx * d / k, 10), ' ',
            outFloat(py - vx * c / 2 / k - vy * d / k, 10));
  out.putLn(outFloat(px + vy * c / 2 / k, 10), ' ',
            outFloat(py - vx * c / 2 / k, 10));
  out.putLn(outFloat(px + vy * a / 2 / k, 10), ' ',
            outFloat(py - vx * a / 2 / k, 10));
  return 0;
}