There is a triangle formed by three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ on a plain.

Write a program which prints "YES" if a point $P$ $(x_p, y_p)$ is in the triangle and "NO" if not.

Constraints

You can assume that:

* $ -100 \leq x_1, y_1, x_2, y_2, x_3, y_3, x_p, y_p \leq 100$
* 1.0 $\leq$ Length of each side of a tringle
* 0.001 $\leq$ Distance between $P$ and each side of a triangle

Input

Input consists of several datasets. Each dataset consists of:

$x_1$ $y_1$ $x_2$ $y_2$ $x_3$ $y_3$ $x_p$ $y_p$


All the input are real numbers. Input ends with EOF. The number of datasets is less than or equal to 100.

Output

For each dataset, print "YES" or "NO" in a line.

Example

Input

0.0 0.0 2.0 0.0 2.0 2.0 1.5 0.5
0.0 0.0 1.0 4.0 5.0 3.0 -1.0 3.0


Output

YES
NO
def stl1(x):
    if a == e:
        y = b
    else:
        y = (f-b)*(x-a)/(e-a) +b

    return(y)

def stl2(x):
    if c == e:
        y = f
    else:
        y = (d-f)*(x-e)/(c-e) +f

    return(y)

def stl3(x):
    if c == a:
        y = b
    else:
        y = (d-b)*(x-a)/(c-a) +b
        
    return(y)

while True:
    try:
        lst = list(map(float, input().split()))
        point1 = [lst[0], lst[1]]
        point2 = [lst[2], lst[3]]
        point3 = [lst[4], lst[5]]
        xp, yp = lst[6], lst[7]

        pointlst = [point1, point2, point3]
        pointlst.sort()
        a, b, e, f, c, d = pointlst[0][0], pointlst[0][1], pointlst[1][0], pointlst[1][1], pointlst[2][0], pointlst[2][1]

        if a <= xp and xp <= e:
            bdr1 = stl1(xp)
            bdr3 = stl3(xp)

            if min(bdr1, bdr3) <= yp and yp <= max(bdr1, bdr3):
                print("YES")

            else:
                print("NO")
        
        elif e < xp and xp <= c:
            bdr2 = stl2(xp)
            bdr3 = stl3(xp)

            if min(bdr2, bdr3) <= yp and yp <= max(bdr2, bdr3):
                print("YES")
            
            else:
                print("NO")

        else:
            print("NO")

    except EOFError:
        break
