<image>

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
#include <bits/stdc++.h>
using namespace std;

constexpr double EPS = 1e-10;
typedef complex<double> Point;
typedef Point Vector;
typedef pair<Point, Point> Segment;
typedef Segment Line;
typedef pair<Point, double> Circle;
typedef vector<Point> Polygon;
double dot(Vector a, Vector b){
	return real(a) * real(b) + imag(a) * imag(b);
}
double cross(Vector a, Vector b){
	return real(a) * imag(b) - imag(a) * real(b);
}
Point project(Line l, Point p){
	Vector base = l.second - l.first;
	base /= abs(base);
	double r = dot(p - l.first, base);
	return l.first + r * base;
}
Point reflect(Line l, Point p){
	return p + (project(l, p) - p) * 2.0;
}
int ccw(Point p0, Point p1, Point p2){
	Vector v = p1 - p0, w = p2 - p0;
	if(cross(v, w) > EPS) return 1;
	if(cross(v, w) < -EPS) return -1;
	if(dot(v, w) < -EPS) return 2;
	if(norm(v) + EPS < norm(w)) return -2;
	return 0;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	Point p0, p1;
	double x, y;
	cin >> x >> y;
	p0 = Point(x, y);
	cin >> x >> y;
	p1 = Point(x, y);
	int q;
	cin >> q;
	while(q--){
		cin >> x >> y;
		Point p2(x, y);
		cout << vector<string>{"ONLINE_FRONT", "CLOCKWISE", "ON_SEGMENT", "COUNTER_CLOCKWISE", "ONLINE_BACK"}[ccw(p0, p1, p2)+2] << "\n";
	}
}
