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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{
	int x,y;
	point operator - (const point p){
		point pp;
		pp.x=x-p.x;
		pp.y=y-p.y;
		return pp;
	}
}p0,p1,p2;
double waiji(point p1,point p2){
	return p1.x*p2.y-p1.y*p2.x;
}
double neiji(point p1,point p2){
	return p1.x*p2.x+p1.y*p2.y;
}
void solve(){
	double xc=waiji(p1-p0,p2-p0);	
	if(xc>0){cout<<"COUNTER_CLOCKWISE"<<endl;return;}
	if(xc<0){cout<<"CLOCKWISE"<<endl;return;}
	if(neiji(p1-p0,p2-p0)<0){cout<<"ONLINE_BACK"<<endl;return;}
	if(abs(p1.x-p0.x)<abs(p2.x-p0.x)||abs(p1.y-p0.y)<abs(p2.y-p0.y)){cout<<"ONLINE_FRONT"<<endl;return;}
	cout<<"ON_SEGMENT"<<endl;
}
int main(){
	cin>>p0.x>>p0.y;
	cin>>p1.x>>p1.y;
	int T; cin>>T;
	while(T--){
		cin>>p2.x>>p2.y;
		solve();
	}
	return 0;
}

