<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<cmath>
using namespace std;

int xp0,yp0,xp1,yp1,x,y;

struct bec{
    int x,y;
};

void judge(int x,int y){
    bec p1,p2;
    p1.x= xp1-xp0; p1.y= yp1-yp0;
    p2.x= x-xp0; p2.y=y-yp0;
    double d1,d2;
    d1=sqrt((double)p1.x*(double)p1.x+(double)p1.y*(double)p1.y);
    d2=sqrt((double)p2.x*(double)p2.x+(double)p2.y*(double)p2.y);
    if((p1.x*p2.y-p1.y*p2.x)>0) cout<<"COUNTER_CLOCKWISE"<<endl;
    else if((p1.x*p2.y-p1.y*p2.x)<0) cout<<"CLOCKWISE"<<endl;
    else{
        double c = ((double)p1.x*(double)p2.x+(double)p1.y*(double)p2.y)/(d1*d2);
        int cc=round(c);
        if(cc==1||d2==0){
            if(d1<d2) cout<<"ONLINE_FRONT"<<endl;
            else cout<<"ON_SEGMENT"<<endl;
        }
        else cout<<"ONLINE_BACK"<<endl;
    }
}

int main(){
    cin>>xp0>>yp0>>xp1>>yp1;
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>x>>y;
        judge(x,y);
    }
    return 0;
}
