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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 <stdio.h>
#include <math.h>

#define TOL 1e-10

using namespace std;
int main(){
    double x[3],y[3];
    scanf("%lf %lf %lf %lf",&x[0],&y[0],&x[1],&y[1]);
    double v1_x,v2_x,v1_y,v2_y;
    double n1,n2,i_v,o_v;
    v1_x = x[1]-x[0];
    v1_y = y[1]-y[0];
    n1 = v1_x*v1_x + v1_y*v1_y;
    int q;
    scanf("%d",&q);
    int i;
    for (i=0; i<q; i++){
        scanf("%lf %lf",&x[2],&y[2]);
        v2_x = x[2]-x[0];
        v2_y = y[2]-y[0];
        n2 = v2_x*v2_x + v2_y*v2_y;
        i_v = v1_x*v2_x + v1_y*v2_y;
        o_v = v1_x*v2_y - v2_x*v1_y;
        if(o_v > TOL){
            printf("COUNTER_CLOCKWISE\n");
        } else if(o_v < -TOL) {
            printf("CLOCKWISE\n");
        } else if(i_v < 0){
            printf("ONLINE_BACK\n");
        } else if(n1 < n2){
            printf("ONLINE_FRONT\n");
        } else {
            printf("ON_SEGMENT\n");
        }
    }
    return 0;
    
}
