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For given three points p0, p1, p2, print


COUNTER_CLOCKWISE


if p0, p1, p2 make a counterclockwise turn (1),


CLOCKWISE


if p0, p1, p2 make a clockwise turn (2),


ONLINE_BACK


if p2 is on a line p2, p0, p1 in this order (3),


ONLINE_FRONT


if p2 is on a line p0, p1, p2 in this order (4),


ON_SEGMENT


if p2 is on a segment p0p1 (5).

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p0 and p1 are not identical.

Input


xp0 yp0 xp1 yp1
q
xp20 yp20
xp21 yp21
...
xp2q-1 yp2q-1


In the first line, integer coordinates of p0 and p1 are given. Then, q queries are given for integer coordinates of p2.

Output

For each query, print the above mentioned status.

Examples

Input

0 0 2 0
2
-1 1
-1 -1


Output

COUNTER_CLOCKWISE
CLOCKWISE


Input

0 0 2 0
3
-1 0
0 0
3 0


Output

ONLINE_BACK
ON_SEGMENT
ONLINE_FRONT
def dot(a,b):
    return a.real*b.real+a.imag*b.imag

def cross(a,b):
    return a.real*b.imag-a.imag*b.real

x0,y0,x1,y1=map(int,input().split())
q=int(input())
p0=x0+y0*1j
p1=x1+y1*1j
a=p1-p0

for x2, y2 in (map(int, input().split()) for _ in [0]*q):
    p2=x2+y2*1j
    b=p2-p0
    S=cross(a,b)
    if S>0: #符号付面積が正だったら
        print("COUNTER_CLOCKWISE")
    elif S<0: #符号付面積が負だったら
        print("CLOCKWISE")
    else:               #符号付面積が0だったら(一直線上に存在する)
        if dot(a,b)>=0:
            if dot(a,a)<dot(b,b):
                print("ONLINE_FRONT") 
            else:
                print("ON_SEGMENT") 
        else:
            print("ONLINE_BACK")
