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:
x1, y1 = map(float, raw_input().split())
x2, y2 = map(float, raw_input().split())
x3, y3 = map(float, raw_input().split())
import math
##x1 = 71.756151; y1 = 7.532275
##x2 = -48.634784; y2 = 100.159986
##x3 = 91.778633; y3 = 158.107739

##x1 = 0.0; y1 = 0.0
##x2 = 0.0; y2 = 1.0
##x3 = math.sqrt(3); y3 = 1.0

EPSILON = 0.00001
def midPoint(x1, y1, x2, y2):
    mid_x = (x1 + x2) / 2
    mid_y = (y1 + y2) / 2
    return (mid_x, mid_y)

def slope_mid_perpend(x1, y1, x2, y2):
    return -1 * (x2 - x1) / (y2 - y1)

def distance(x1, y1, x2, y2):
    return math.sqrt((x1-x2) ** 2 + (y1-y2) ** 2)

def circleCenter(x1, y1, x2, y2, x3, y3):
    x12, y12 = midPoint(x1, y1, x2, y2)
    x13, y13 = midPoint(x1, y1, x3, y3)
    if y1 == y2:
        k_mid_13 = slope_mid_perpend(x1, y1, x3, y3)
        x0 = x12
        y0 = k_mid_13 * (x0 - x13) + y13
    elif y1 == y3:
        k_mid_12 = slope_mid_perpend(x1, y1, x2, y2)
        x0 = x13
        y0 = k_mid_12 * (x0 - x12) + y12
    else:
        k_mid_13 = slope_mid_perpend(x1, y1, x3, y3)
        k_mid_12 = slope_mid_perpend(x1, y1, x2, y2)
        x0 = (y13 - y12 + k_mid_12*x12 - k_mid_13*x13) / (k_mid_12 - k_mid_13)
        y0 = k_mid_13 * (x0 - x13) + y13
    return x0, y0
x0, y0 = circleCenter(x1, y1, x2, y2, x3, y3)
def angle(distance1, distance2):
    angle = math.asin((distance1/2) / distance2)
    return 2 * angle
d_12 = distance(x1, y1, x2, y2)
d_13 = distance(x1, y1, x3, y3)
d_32 = distance(x3, y3, x2, y2)
##print d_12, d_13, d_32
r = distance(x1, y1, x0, y0)

angle1O2 = angle(d_12, r)
angle1O3 = angle(d_13, r)
angle3O2 = angle(d_32, r)
##print angle1O2, angle1O3, angle3O2

##print angle1O2, angle1O3, angle3O2
##
##print angle(math.sqrt(2), 1)

def isRemainder(angle, angle_N):
    remainder = angle % angle_N
##    print remainder
    
    if (abs(remainder - 0)<= EPSILON) or (abs(remainder - angle_N) <= EPSILON):
        return True
    else:
        return False
        
def find_N(angle1, angle2, angle3):
    N = 3
    while 1:
##        print angle1, angle1, angle3
        angle_N = 2 * math.pi / N
##        print 'angle_N', angle_N
##        remainder1 = angle1 % angle_N
##        remainder2 = angle2 % angle_N
##        remainder3 = angle3 % angle_N
##        print remainder1, remainder2, remainder3
        if  (isRemainder(angle1, angle_N) and  isRemainder(angle2, angle_N) and
             isRemainder(angle2, angle_N)):
            break
        N += 1
        
    return N
N = find_N(angle1O2, angle1O3, angle3O2)
def area_N(N):
    angle = 2 * math.pi / N
    area = N * 0.5 * r * r * math.sin(angle)
    return area

print area_N(N)
            
    


##print angle1O3, 2 * math.pi / 6
##print angle1O3 % (2 * math.pi / 6)




## test
##print circleCenter(x1, y1, x2, y2, x3, y3)
##
##import math
##
##x0, y0 = circleCenter(x1, y1, x2, y2, x3, y3)
##print distance(x0, y0, x1, y1), distance(x0, y0, x2, y2), distance(x0, y0, x2, y2)