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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;

#define db double
const db eps = 1e-6;

struct poi {
	db x, y;
	void r() { cin >> x; cin >> y; }
	void w() { cout << x << ' ' << y << '\n'; }
	poi operator -(poi p) {
		return {x - p.x, y - p.y};
	}
	poi operator +(poi p) {
		return {x + p.x, y + p.y};
	}
	poi operator *(db d) {
		return {x * d, y * d};
	}
	db dot(poi p) {
		return x * p.x + y * p.y;
	}
	db cross(poi p) {
		return x * p.y - y * p.x;
	}
	db len2() { return x * x + y * y; }
	poi proj(poi p, poi q) {
		db s = (*this - p).dot(q - p) / (q - p).len2();
		return p + (q - p) * s;
	}
};

int ccw(poi a, poi b, poi c) {
	poi u = b - a, v = c - a;
	if(u.cross(v) > eps) return 1;
	if(u.cross(v) < -eps) return -1;
	if(u.dot(v) < -eps) return -2;
	if(u.len2() + eps < v.len2()) return 2;
	return 0;
}

int main() {
	ios :: sync_with_stdio(false);

	poi p, q; p.r(); q.r();
	int T; cin >> T;
	while(T --) {
		poi t; t.r();
		int a = ccw(p, q, t);
		//cout << a << '\n';
		cout << (a != -2 ? a != -1 ? a != 0 ? a != 1 ?
			"ONLINE_FRONT" :
			"COUNTER_CLOCKWISE" :
			"ON_SEGMENT" :
			"CLOCKWISE" :
			"ONLINE_BACK") << '\n';
	}
	return 0;
}