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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 <stdio.h>

int main(void) {
  int x1, y1, x2, y2, xp, yp;
  scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
  x2 -= x1, y2 -= y1;
  int q, i;
  scanf("%d", &q);
  for(i = 0; i < q; ++i) {
    scanf("%d%d", &xp, &yp);
    xp -= x1, yp -= y1;
    int op = xp * y2 - yp * x2;
    if(op > 0) printf("CLOCKWISE\n");
    else if(op < 0) printf("COUNTER_CLOCKWISE\n");
    else {
      if(xp * x2 < 0 || yp * y2 < 0) printf("ONLINE_BACK\n");
      else if((x2 > 0 && x2 >= xp) || (y2 > 0 && y2 >= yp) || (y2 < 0 && y2 <= yp) || (x2 < 0 && x2 <= xp)) printf("ON_SEGMENT\n");
      else printf("ONLINE_FRONT\n");
    }
  }
  return 0;
}
