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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<cstdio>
using namespace std;
const double eps=1e-10;
struct point{
    double x,y;
    point(double x=0,double y=0):x(x),y(y){}
    point operator +(point a){
        return point(x+a.x,y+a.y);
    }
    point operator -(point a){
        return point(x-a.x,y-a.y);
    }
};
typedef point vector;
point p0,p1;
double dot(vector a,vector b){
    return a.x*b.x+a.y*b.y;
}
double cross(vector a,vector b){
    return a.x*b.y-a.y*b.x;
}
double norm(vector a){
    return a.x*a.x+a.y*a.y;
}
void ccw(point p1,point p2,point p3){
    vector a=p2-p1,b=p3-p1;
    if(cross(a,b)>eps){
        printf("COUNTER_CLOCKWISE\n");
        return;
    }
    if(cross(a,b)<-eps){
        printf("CLOCKWISE\n");
        return;
    }
    if(dot(a,b)<-eps){
        printf("ONLINE_BACK\n");
        return;
    }
    if(norm(a)<norm(b)){
        printf("ONLINE_FRONT\n");
        return;
    }
    printf("ON_SEGMENT\n");
}
int main(){
    int n;
    double x0,y0,x1,y1;
    scanf("%lf%lf%lf%lf",&x0,&y0,&x1,&y1);
    p0=point(x0,y0),p1=point(x1,y1);
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%lf%lf",&x0,&y0);
        ccw(p0,p1,point(x0,y0));
    }
    return 0;
}
