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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 <iostream>
#include <complex>
using namespace std;
typedef complex<int> point;

string ccw(point p0,point p1,point p2){
  point temp;
  p1-=p0;
  p2-=p0;
  temp=p1;
  p1*=conj(temp);
  p2*=conj(temp);
  if(abs(temp)==0) return "ON_SEGMENT";
  p1/=abs(temp);
  if(p2.imag()>0) return "COUNTER_CLOCKWISE";
  if(p2.imag()<0) return "CLOCKWISE";
  p2/=abs(temp);
  if(p2.real()<0) return "ONLINE_BACK";
  if(p1.real()<p2.real()) return "ONLINE_FRONT";
  return "ON_SEGMENT";
}

int main(){
  int p0x,p0y,p1x,p1y,q,p2x,p2y;
  cin>>p0x>>p0y>>p1x>>p1y;
  point A = point(p0x,p0y);
  point B = point(p1x,p1y);
  cin>>q;
  while(q--){
    cin>>p2x>>p2y;
    cout << ccw(  A, B , point(p2x,p2y) ) <<endl;    
  }
  return 0;
}