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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 <bits/stdc++.h>
    using namespace std;
    #define Int int64_t
    #define dump(x) cout << (x) << endl
    #define fi first
    #define se second
    Int mod = 1e9+7;
    Int INF = 1e18;
    double EPS = 0.00000001;
    //Int mod = 1e6+3;

    int main() {
        double ax, ay, bx, by;
        cin >> ax >> ay >> bx >> by;
        complex<double> a(ax, ay), b(bx, by);
        int q;
        cin >> q;
        for (int i = 0; i < q; i++) {
            double px, py;
            cin >> px >> py;
            complex<double> p(px, py);
            complex<double> ab = (b - a), ap = (p - a);
            complex<double> e = ab / abs(ab);
            ap = ap / e;
            if (imag(ap) - EPS > 0) {
                dump("COUNTER_CLOCKWISE");
            }
            else if (imag(ap) + EPS < 0) {
                dump("CLOCKWISE");
            }
            else {
                if (real(ap) + EPS < 0) {
                    dump("ONLINE_BACK");
                }
                else if (real(ap) - EPS > abs(ab)) {
                    dump("ONLINE_FRONT");
                }
                else {
                    dump("ON_SEGMENT");
                }
            }
        }
        return 0;
    }
