<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<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstdio>
#include<complex>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define it ::iterator
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double EPS=1e-10;
using namespace std;
typedef complex<double> P;
P pin(){
	double x,y;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int i){
	rep(i,2)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
string s[5]={"COUNTER_CLOCKWISE","CLOCKWISE","ONLINE_BACK","ONLINE_FRONT","ON_SEGMENT"};
int main(){
	P a[2];
	PIN(a,2);
	int n;
	cin>>n;
	rep(i,n){
		P p=pin();
		P b=a[1]-a[0];
		P c=p-a[0];
		int ans;
		if(cross(b,c)<0)ans=1;
		else if(cross(b,c)>0)ans=0;
		else if(dot(b,c)<0)ans=2;
		else if(abs(b)<abs(c))ans=3;
		else ans=4;
		cout<<s[ans]<<endl;
	}
}