<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 <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;

int main(){
	double x1,x2,xp,y1,y2,yp1,yp,naiseki,norm1,norm2,gaiseki;
	int q;

	scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
	scanf("%d",&q);

	for(int i=0; i < q; i++){
		scanf("%lf %lf",&xp,&yp);
		norm1 = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
		norm2 = sqrt((xp-x1)*(xp-x1)+(yp-y1)*(yp-y1));
		naiseki = (xp-x1)*(x2-x1)+(yp-y1)*(y2-y1);
		gaiseki = (x2-x1)*(yp-y1)-(xp-x1)*(y2-y1);
		if(fabs(naiseki/(norm1*norm2)+1) < 0.00000001){
			printf("ONLINE_BACK\n");
		}else if(fabs(naiseki-norm1*norm2)< 0.00000001){
			if(norm1 >= norm2){
				printf("ON_SEGMENT\n");
			}else{
				printf("ONLINE_FRONT\n");
			}
		}else if(gaiseki > 0){
			printf("COUNTER_CLOCKWISE\n");
		}else{
			printf("CLOCKWISE\n");
		}
	}

    return 0;
}