<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 <vector>
#include <complex>
using namespace std;
 
#define X real()
#define Y imag()
#define EPS (1e-10)
 
typedef complex<double> P;
typedef P Vec;
 
double dot(Vec v1, Vec v2) {
  return (v1*conj(v2)).X;
}
 
double cross(Vec v1, Vec v2) {
  return (v1*conj(v2)).Y;
}
 
enum { COUNTER_CLOCKWISE, CLOCKWISE, ONLINE_FRONT, ONLINE_BACK, ON_SEGMENT};
 
int ccw(P p0, P p1, P p2) {
  Vec a = p1 - p0;
  Vec b = p2 - p0;
  if(cross(a, b) > EPS) return CLOCKWISE;
  if(cross(a, b) < -EPS) return COUNTER_CLOCKWISE;
  if(dot(a, b) < -EPS) return ONLINE_BACK;
  if(norm(a) < norm(b)) return ONLINE_FRONT;
  return ON_SEGMENT;
}
 
void putResult(int x) {
  if(x == COUNTER_CLOCKWISE) cout << "COUNTER_CLOCKWISE" << endl;
  if(x == CLOCKWISE) cout << "CLOCKWISE" << endl;
  if(x == ONLINE_FRONT) cout << "ONLINE_FRONT" << endl;
  if(x == ONLINE_BACK) cout << "ONLINE_BACK" << endl;
  if(x == ON_SEGMENT) cout << "ON_SEGMENT" << endl;
}
 
int main() {
  double x, y;
  P p0, p1;
   
  cin >> x >> y; p0 = P(x, y);
  cin >> x >> y; p1 = P(x, y);
   
  int n;
  cin >> n;
  for(int i=0; i<n; i++) {
    cin >> x >> y;
    P p2 = P(x, y);
     
    putResult( ccw(p0, p1, p2) );
  }
   
  return 0;
}