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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;
struct point{
	double x,y;
	point(){}
	point(double _x,double _y){
		x=_x;y=_y;
	}
	point operator -(const point &b) const{
		return point(x-b.x,y-b.y);
	}
	double operator ^(const point &b) const{
		return (x*b.x)+(y*b.y);
	}
	double operator *(const point &b) const{
		return x*b.y-y*b.x;
	}
};
double len(point p){
	return (p.x*p.x+p.y*p.y);
}
int solve(point p0,point p1,point p2){
	if((p1-p0)*(p2-p0)>0) return 1;
	if((p1-p0)*(p2-p0)<0) return 2;
	if(((p1-p0)^(p2-p0))<0) return 3;
	if(len(p1-p0)<len(p2-p0)) return 4;
	return 5;
}
int main(){
	point p0,p1;
	int q;
	point p2;
	cin>>p0.x>>p0.y>>p1.x>>p1.y;
	cin>>q;
	for(int i=1;i<=q;++i){
		cin>>p2.x>>p2.y;
		switch(solve(p0,p1,p2)){
			case 1: puts("COUNTER_CLOCKWISE");break;
			case 2: puts("CLOCKWISE");break;
			case 3: puts("ONLINE_BACK");break;
			case 4: puts("ONLINE_FRONT");break;
			case 5: puts("ON_SEGMENT");break;
		}
	}
	return 0;
}
