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 PYTHON3):

Solution:
import sys, math

# Get side lengths from point set
def sidelengths(points):
    sides = []
    for i in range(0,len(points)):
        pt0 = points[i]
        pt1 = points[(i+1)%len(points)]
        x = pt0[0]-pt1[0]
        y = pt0[1]-pt1[1]
        sides.append(math.sqrt(math.pow(x,2)+math.pow(y,2)))
    return sides

# Heron's formula
def area(triangle):
    a = triangle[0]
    b = triangle[1]
    c = triangle[2]
    s = (a+b+c)/2
    return math.sqrt(s*(s-a)*(s-b)*(s-c))

# Use A=abc/4R, rearranged into R=abc/4A
def circumradius(triangle):
    a = triangle[0]
    b = triangle[1]
    c = triangle[2]
    A = area(triangle)
    return (a*b*c)/(4*A)

# s is opposite side, s1,s2 are adjacent
def angle(s,s1,s2):
    # Rearranged law of cosines
    return math.acos((math.pow(s1,2)+math.pow(s2,2)-math.pow(s,2))/(2*s1*s2))

# Given an angle of the form k*pi/n, find a reasonable value of n
def angle_denom(angle, max_n=100, tolerance=0.00001):
    r = angle/math.pi
    for n in range(1,max_n+1):
        if abs(r*n-round(r*n)) < tolerance:
            return n
    return max_n # if all else fails

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

# Input
points = []
for i in range(0,3):
    points.append(list(map(float,sys.stdin.readline().split())))

# Get sidelengths and circumradius
sides = sidelengths(points)
c = circumradius(sides)

# Since we have circumradius, it is enough to find the number of sides in the
#  original polygon.
# It is enough to look at the inscribed angles: if the original polygon had
#  n sides, then all the inscribed angles must be multiples of pi/n.
angle1 = angle(sides[0],sides[1],sides[2])
angle2 = angle(sides[1],sides[2],sides[0])
angle3 = angle(sides[2],sides[0],sides[1])
n1 = angle_denom(angle1)
n2 = angle_denom(angle2)
n3 = angle_denom(angle3)
n = lcm(n1,lcm(n2,n3))

# With the number of sides and the circumradius, calculating the area is now
#  possible: split into wedges and use sine area formula
a = 0.5 * n * math.pow(c,2) * math.sin((2*math.pi)/n)
print(str(a))