Problem: Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples

Input

0.000000 0.000000
1.000000 1.000000
0.000000 1.000000


Output

1.00000000

Solution (in PYTHON):

Solution:
import sys
import math

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition) :
  if condition :
    print ("YES")
  else :
    print ("NO")

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rvalues(t) :
  return map(t, sys.stdin.readline().split())

def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def vector_and_scalar_product(const, v) :
  r = [x * const for x in v]
  return tuple(r)


def triangle_R(A, B, C) :
  a = vector_distance(B, C) 
  b = vector_distance(A, C)
  c = vector_distance(A, B)
  
  p = (a + b + c) * 0.5 

  R = a * b * c 
  R /= 4 * (p * (p - a) * (p - b) * (p - c)) ** 0.5

  return R

def triangle_center(A, B, C) :
  R = triangle_R(A, B, C)

  a = vector_distance(B, C)
  b = vector_distance(A, C)
  c = vector_distance(A, B)
  

  S = a * b * c / 4 / R

  alpha_A = a * a * scalar_product(vector_diff(A, B), vector_diff(A, C)) / (8 * S ** 2)
  alpha_B = b * b * scalar_product(vector_diff(B, A), vector_diff(B, C)) / (8 * S ** 2)
  alpha_C = c * c * scalar_product(vector_diff(C, A), vector_diff(C, B)) / (8 * S ** 2)

  O_a = vector_and_scalar_product(alpha_A, A)
  O_b = vector_and_scalar_product(alpha_B, B)
  O_c = vector_and_scalar_product(alpha_C, C)
  O = vector_sum(vector_sum(O_a, O_b), O_c)

  return O


eps = 0.00001

def vector_angle(a, b) :
  s = scalar_product(a, b)
  a = scalar_product(a, a) ** 0.5
  b = scalar_product(b, b) ** 0.5

  cos_a = s / a / b

  return math.acos(cos_a)

A = tuple(rvalues(float))
B = tuple(rvalues(float))
C = tuple(rvalues(float))

O = triangle_center(A, B, C)

OA = vector_diff(A, O)
OB = vector_diff(B, O)
OC = vector_diff(C, O)


alpha = [
  vector_angle(OA, OB),
  vector_angle(OA, OC),
  vector_angle(OB, OC),
]

R = triangle_R(A, B, C)

for n in range(3, 100 + 1) :
  step_alpha = 2.0 * math.pi / n
  k_alpha = [a / step_alpha for a in alpha]
  
  k_diff = [abs(k - int(k + eps)) < eps for k in k_alpha]

  if all(k_diff) :
    break

S = n * (R ** 2) * math.sin(math.pi * 2.0 / n) * 0.5
print S