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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>
#include <cmath>
using namespace std;

int main(){

	int p0x,p0y,p1x,p1y,p2x,p2y;
	int q,cr;

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

	for(int i=0;i<q;i++){
		cin>>p2x>>p2y;
		cr=(p2x-p0x)*(p1y-p0y)-(p1x-p0x)*(p2y-p0y);
		if(cr==0){
			if(((p2x-p0x<0)==(p1x-p0x<0)&&(p2y-p0y<0)==(p1y-p0y<0))||((p2x==p0x)&&(p1x==p0x)&&(p2y-p0y<0)==(p1y-p0y<0))||((p2x-p0x<0)==(p1x-p0x<0)&&(p2y==p0y)&&(p1y==p0y))||(p0x==p1x&&p1y==p0y)||(p2x==p0x&&p2y==p0y)){
				if(pow(p2x-p0x,2.0)+pow(p2y-p0y,2.0)>pow(p1x-p0x,2.0)+pow(p1y-p0y,2.0)){
					cout<<"ONLINE_FRONT"<<endl;
				}
				else cout<<"ON_SEGMENT"<<endl;
			}
			else cout<<"ONLINE_BACK"<<endl;
		}
		else if(cr>0) cout<<"CLOCKWISE"<<endl;
		else cout<<"COUNTER_CLOCKWISE"<<endl;
	}

	return 0;
}