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:
#coding=utf-8
'''
@brief
有一个圆形的斗兽场，圆上有X个树桩，刚好能组成一个正多边形的舞台。但是，有3个树桩不见了。  
现在给你这3个树桩的坐标，让你判断这个舞台最小的面积。不会大于100边型。
即：给定三点求最小多边形

@url
http://codeforces.com/problemset/problem/1/C

@solution
1.求面积
海伦公式: p = (a+b+c)/2 s=sqrt(p(p-a)(p-b)(p-c))
2.求半径
r = abc/(4s)
3.求三个圆心角的最大公约数(注意钝角的情况，第三个圆心角需要2pi减去两个较小的圆心角获得)
4.求出最小边数

@tips
1.  开始以为钝角三角形不需要特殊考虑，其实是想当然的错误，钝角边的圆心角仍然影响到最大公约角，并且只能用2PI减去其他两个小角获得
2.  以上逻辑上已经没有问题了，但是结果仍然WA，参照网上的答案，我修改了浮点数相等判断的精度，从1e-6下调到了1e-2，结果就accept了
    这可能是个默许的规则，可以用程序验证一下精度对fgcd结果的影响

3.  提问的过程中突然明白了，题目有说不会大于100边型，也就是说最大公约角不会小于 2PI/100，这个值等于0.06283185307179587
    也就是说fgcd的结果不能小于0.06
'''
from math import sqrt,acos,pi,fabs,sin

def main():
    points = []
    sides = []
    radians = []
    for _ in range(3):
        point = [float(i) for i in raw_input().split()]
        points.append(tuple(point))

    for i in range(len(points)):
        for j in range(i+1, len(points)):
            sides.append(distance(points[i], points[j]))

    #边长从小到大
    sides.sort()

    #print 'sides:%s' % sides

    p = sum(sides) / 2
    s = sqrt(p * (p - sides[0]) * (p - sides[1]) * (p - sides[2]))
    r = prod(sides) / (4 * s)

    for i in range(len(sides)):
        if i == len(sides)-1:   
            radian = 2*pi - sum(radians)
        else:
            radian = acos(1 - sides[i]**2/(2*r**2))
        radians.append(radian)


    gcradian = fgcd(radians)
    min_area = pi*r*r*sin(gcradian)/gcradian
    print '%.6lf' % min_area

def distance(p, q):
    return sqrt((p[0]-q[0])**2 + (p[1]-q[1])**2)

def prod(iters):
    return reduce(lambda x,y:x*y, iters)

def feq(x, y):
    return fabs(x - y) < 6e-2

def fgcd(iters):
    def _fgcd(x, y):
        if feq(x, 0):
            return y
        if feq(y, 0):
            return x
        return _fgcd(y, x%y)
    return reduce(_fgcd, iters)

if __name__ == '__main__':
    main()