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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>
#define N 50005
#define re register
#define ll long long
#define P 998244353
using namespace std;
bool m1;
int n,m,K;
inline void Rd(int &res){
	re char c;res=0;
	while(c=getchar(),c<48);
	do res=(res<<3)+(res<<1)+(c^48);
	while(c=getchar(),c>47);
}
struct Point{
	double x,y;
	Point operator+(const Point&a)const{return (Point){x+a.x,y+a.y};}
	Point operator-(const Point&a)const{return (Point){x-a.x,y-a.y};}
	Point operator*(const double&v)const{return (Point){x*v,y*v};}
	double operator*(const Point&a)const{return x*a.x+y*a.y;}
	double operator^(const Point&a)const{return x*a.y-y*a.x;}
	double len(){return sqrt(x*x+y*y);}
}p1,p2,p,p0;
bool m2;
int main(){
	scanf("%lf%lf%lf%lf",&p0.x,&p0.y,&p1.x,&p1.y);
	Point p3=p1-p0;
	double l=p3.len();
	int q;Rd(q);
	while(q--){
		scanf("%lf%lf",&p2.x,&p2.y);
		double l1=(p3^(p2-p0))/l;
		if(fabs(l1)<1e-7){
			Point p01=p1-p0;
			Point p02=p2-p0;
			double res=p01*p02;
			if(res+1e-8<0){puts("ONLINE_BACK");continue;}
			Point p10=p0-p1;
			Point p12=p2-p1;
			res=p10*p12;
			if(res+1e-8<0)puts("ONLINE_FRONT");
			else puts("ON_SEGMENT");
		}
		else if(l1>0)puts("COUNTER_CLOCKWISE");
		else puts("CLOCKWISE");
	}
	return 0;
}
