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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
pp = list(map(int, input().split()))
n = int(input())

def cross(a, b):
    return a[0]*b[1]-a[1]*b[0]
        
def dot(a, b):
    n = len(a)
    if n != len(b):
        return None
    ans = 0
    for i, j in zip(a, b):
        ans += i*j
    return ans
    
for i in range(n):
    bb = list(map(int, input().split()))
    a = (pp[2]-pp[0], pp[3]-pp[1])
    b = (bb[0]-pp[0], bb[1]-pp[1])
    if cross(a, b) > 0:
        print("COUNTER_CLOCKWISE")
    elif cross(a, b) < 0:
        print("CLOCKWISE")
    else:
        if dot(a,b)<0:
            print("ONLINE_BACK")
        elif dot(a, (b[0]-a[0],b[1]-a[1]))>0: 
            print("ONLINE_FRONT")
        else:
            print("ON_SEGMENT")
