# ヘロンの公式
# ループをどうまわすと効率的か

# ヘロンの公式より外積の方が楽かつ誤差小さい


xy= list(map(int, input().split()))


def cal(z):
    e12 = ((z[4] - z[2]) ** 2 + (z[5] - z[3]) ** 2) ** 0.5
    e20 = ((z[0] - z[4]) ** 2 + (z[1] - z[5]) ** 2) ** 0.5
    e01 = ((z[2] - z[0]) ** 2 + (z[3] - z[1]) ** 2) ** 0.5
    s = (e12+e20+e01) / 2
    ret = (s * (s-e12) * (s-e20) * (s-e01)) ** 0.5
    return ret 

def cal1(z):
    a = z[2] - z[0]
    b = z[3] - z[1]
    c = z[4] - z[0]
    d = z[5] - z[1]
    ret = abs(a*d-b*c)//2
    return ret

ans = 0
for i0 in range(2):
    for diff0 in [-1,1]:
        for i1 in range(2,4):
            for diff1 in [-1,1]:
                for i2 in range(4,6):
                    for diff2 in [-1,1]:
                        xy[i0] += diff0
                        xy[i1] += diff1
                        xy[i2] += diff2

                        ret = cal1(xy)
                        ans = max(ans,ret)

                        xy[i0] -= diff0
                        xy[i1] -= diff1
                        xy[i2] -= diff2


print(ans)