<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 <iostream>
#include <complex>
#include <string>
#define x real() 
#define y imag() 
using namespace std;
typedef complex <double> point;
typedef complex<double> vectol;

string ccw(point a,point b,point c) {
  vectol ab=b-a, ac =c-a;
  vectol d =ac*conj(ab); 
  if(d.y > 0) return "COUNTER_CLOCKWISE";
  if(d.y < 0) return "CLOCKWISE";
  ab = ab*conj(ab);
  if(ab.x<d.x) return "ONLINE_FRONT";
  if(0>d.x) return "ONLINE_BACK";
  return "ON_SEGMENT";
}

int main() {
  double x0,y0,x1,y1;
  cin>>x0>>y0>>x1>>y1;
  point a = point( x0, y0);
  point b = point( x1, y1);
  int n;
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> x0 >> y0;
    cout << ccw(a,b,point(x0,y0)) <<endl;
  }

  return 0;
}