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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<stdio.h>
#define SQ(x) ((x)*(x))
typedef long long ll;
struct Point{
	int x,y;
}pt0, pt1;
Point vector(Point p1,Point p2){
	Point res;
	res.x = p2.x-p1.x;
	res.y = p2.y-p1.y;
	return res;
}
int dot(Point p1, Point p2){ return p1.x*p2.x+p1.y*p2.y; }
int cross(Point p1, Point p2){ return p1.x*p2.y-p1.y*p2.x; }
ll lengthSQ(Point vec){ return SQ(ll(vec.x))+SQ(ll(vec.y)); }
void solve(Point pt2){
	const Point vec1 = vector(pt0, pt1);
	const ll vec1Len2 = lengthSQ(vec1);
	Point vec2 = vector(pt0, pt2);
	int cro = cross(vec1, vec2);
	if(cro>0) printf("COUNTER_CLOCKWISE\n");// a×b>0，则b在a的逆时针方向
	else if(cro<0) printf("CLOCKWISE\n");// a×b<0，则b在a的顺时针方向
	else if(dot(vec1, vec2)<0) printf("ONLINE_BACK\n");// a×b=0, a·b<0，则a、b方向相反
	else if(lengthSQ(vec2)>vec1Len2) printf("ONLINE_FRONT\n");
	else printf("ON_SEGMENT\n");
}
int main(){
	Point temp;
	int x, y, q;
	scanf("%d%d%d%d%d", &pt0.x, &pt0.y, &pt1.x, &pt1.y, &q);
	while(q--){
		scanf("%d%d", &temp.x, &temp.y);
		solve(temp);
	}
}
