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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<string.h>
#include<algorithm>
#include<stdio.h>
#include<complex>
using namespace std;
typedef complex<double> qua;
void solve(qua a,qua b,qua c)
{
    b-=a,c-=a;
    qua ans=c*conj(b);
    if(ans.imag()>0) printf("COUNTER_CLOCKWISE\n");
    else if(ans.imag()<0) printf("CLOCKWISE\n");
    else if(ans.real()<0) printf("ONLINE_BACK\n");
    else if(abs(c)>abs(b)) printf("ONLINE_FRONT\n");
    else printf("ON_SEGMENT\n");
}
int main()
{
    double x1,y1,x2,y2,x3,y3;
    int m;
    scanf("%lf%lf%lf%lf%d",&x1,&y1,&x2,&y2,&m);
    while(m--)
    {
        scanf("%lf%lf",&x3,&y3);
        solve(qua(x1,y1),qua(x2,y2),qua(x3,y3));
    }
}

