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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
x0,y0,x1,y1 = map(int, input().split( ))
p0 = [x0,y0]
p1= [x1,y1]
v1 =[p1[0]-p0[0],p1[1]-p0[1]]
q = int(input())
def cross(a,b):
    crs = a[0]*b[1] - a[1]*b[0]
    return crs
def dot(a,b):
    dt = a[0]*b[0] + a[1]*b[1]
    return dt
eps = 0.000001
for i in range(q):
    p2 = list(map(int, input().split( )))
    v2=[p2[0]-p0[0],p2[1]-p0[1]]
    if cross(v1,v2) >eps:
        print("COUNTER_CLOCKWISE")
    elif cross(v1,v2) <-eps:
        print("CLOCKWISE")
    elif dot(v1,v2)<-eps:
        print("ONLINE_BACK")
    elif v1[0]**2+v1[1]**2 < v2[0]**2+v2[1]**2:#abs(v1[0])<abs(v2[0]):はx成分0でおかしい
        print("ONLINE_FRONT")
    else:#p1=p2はこっち
        print("ON_SEGMENT")
