<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<bits/stdc++.h>
using namespace std;
#define ll long long
#define x real()
#define y imag()
typedef complex<double> point;


string Farhad(point a, point b, point c)
{
   point k=conj(b-a)*(c-a);
   if(k.y>0) return "COUNTER_CLOCKWISE";
   if(k.y<0) return "CLOCKWISE";
   if(k.x<0)return "ONLINE_BACK";
   if(abs(b-a)<abs(c-a) )return "ONLINE_FRONT";
   return "ON_SEGMENT";
}

int main()
{
    ll m,n;
    double  ax, ay, bx,by, cx,cy;
    cin>>ax>>ay>>bx>>by>>n;
    while(n--)
    {
        cin>>cx>>cy;
        cout<<Farhad(point(ax, ay), point(bx,by), point(cx, cy) )<<endl;
    }

}


