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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 <iostream>
using namespace std;

int main()
{
	int p0x = 0, p0y = 0, p1x = 0, p1y = 0, q = 0, px = 0, py = 0;
	int a = 0, b = 0;
	long y = 0;
	double t = 0;

	cin >> p0x >> p0y >> p1x >> p1y;
	cin >> q;

	a = p1x - p0x;
	b = p1y - p0y;

	for (int i = 0; i < q; i++) {
		cin >> px >> py;

		y = a * (py - p0y) - b * (px - p0x);

		if(y > 0)
			cout << "COUNTER_CLOCKWISE" << endl;
		else if(y < 0)
			cout << "CLOCKWISE" << endl;
		else {
			if (a != 0)
				t = (double)(px - p0x) / (double)a;
			else
				t = (double)(py - p0y) / (double)b;
				
			if (t < 0)
				cout << "ONLINE_BACK" << endl;
			else if (t > 1)
				cout << "ONLINE_FRONT" << endl;
			else
				cout << "ON_SEGMENT" << endl;
		}
	}
		
	return 0;
}