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

typedef long long ll;
const ll INF = 1e18;
const int BIG = 1e9 + 555;

struct Point {
	double x, y;
	Point() {};
	Point(double _x, double _y) : x(_x), y(_y) {};
};

int n, q;
Point p[4];

Point operator - (const Point &a, const Point &b){
	return Point(a.x - b.x, a.y - b.y);
}

double dot(Point a, Point b){
	return a.x * b.x + a.y * b.y;
}

double cross(Point a, Point b){
	return a.x * b.y - a.y * b.x;
}

const double EPS = 1e-10;

double dist(const Point &p, const Point &q){
	double dx = p.x - q.x;
	double dy = p.y - q.y;

	return sqrt(dx * dx + dy * dy);
}

bool eq(double a, double b){
	return fabs(a - b) < EPS;
}

void direction(Point a, Point b, Point c){
	double crs = cross(b - a, c - a);
	double ab = dist(a, b);
	double ac = dist(a, c);

	if(eq(crs, 0)){
		double cosAlpha = dot(b - a, c - a) / ab / ac;
		if(eq(cosAlpha, -1)){
			cout << "ONLINE_BACK\n";
		} else {
			if(ab < ac){
				cout << "ONLINE_FRONT\n";
			} else {
				cout << "ON_SEGMENT\n";
			}
		}
		return;
	}

	if(crs > 0){
		cout << "COUNTER_CLOCKWISE\n";
	} else if(crs < 0){
		cout << "CLOCKWISE\n";
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> p[0].x >> p[0].y >> p[1].x >> p[1].y;

	cin >> q;

	while(q--){
		cin >> p[2].x >> p[2].y;
		direction(p[0], p[1], p[2]);
	}

	return 0;
}
