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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
import cmath
EPS = 1e-4

#外積
def OuterProduct(one, two):
	tmp = one.conjugate() * two
	return tmp.imag

#内積
def InnerProduct(one, two):
	tmp = one.conjugate() * two
	return tmp.real

#点が線分上にあるか
def IsOnSegment(point, begin, end):
	if abs(OuterProduct(begin-point, end-point)) <= EPS and InnerProduct(begin-point, end-point) <= EPS:
		return True
	else:
		return False

#3点が反時計回りか
#一直線上のときの例外処理できていない→とりあえずF
def CCW(p, q, r):
	one, two = q-p, r-q
	if OuterProduct(one, two) > -EPS:
		return True
	else:
		return False

def solve(p, q, r):
	if abs(OuterProduct(q-r, p-r)) <= EPS:
		if InnerProduct(q-r, p-r) <= EPS:
			return "ON_SEGMENT"
		elif abs(p-r) < abs(q-r):
			return "ONLINE_BACK"
		else:
			return "ONLINE_FRONT"
	elif CCW(p, q, r):
		return "COUNTER_CLOCKWISE"
	else:
		return "CLOCKWISE"

a, b, c, d = map(int, input().split())
p, q = complex(a, b), complex(c, d)
n = int(input())
for _ in range(n):
	x, y = map(int, input().split())
	r = complex(x, y)
	print(solve(p, q, r))
