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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 <cstdio>

template <class T>
T abs(T x) {
    return (x > 0) ? x : -x;
}

int main(int argc, const char * argv[]) {
    long xp0, xp1, xp2, yp0, yp1, yp2;
    int q;
    
    scanf("%ld %ld %ld %ld", &xp0, &yp0, &xp1, &yp1);
    scanf("%d", &q);
    
    long v1x = xp1 - xp0;
    long v1y = yp1 - yp0;
    
    for (int i = 0; i < q; ++i) {
        scanf("%ld %ld", &xp2, &yp2);
        long v2x = xp2 - xp0;
        long v2y = yp2 - yp0;
        
        long crossProd = v1x * v2y - v1y * v2x;
        
        if (crossProd > 0) {
            puts("COUNTER_CLOCKWISE");
        } else if (crossProd < 0) {
            puts("CLOCKWISE");
        } else if (v2x * v1x < 0 || v2y * v1y < 0) {
            puts("ONLINE_BACK");
        } else if (abs(v1x) < abs(v2x) || abs(v1y) < abs(v2y)){
            puts("ONLINE_FRONT");
        } else {
            puts("ON_SEGMENT");
        }
    }
    
    return 0;
}

