<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>

using namespace std;

const double eps = 1e-12;
typedef complex<double> Point;

ostream& operator<<(ostream& os, const Point& p){
    return os <<  p.real() << " " << p.imag();
}

inline double cross(const Point& a, const Point& b){
    return imag(conj(a) * b);
}

inline double dot(const Point& a, const Point& b){
    return real(conj(a) * b);
}

int ccw(Point a, Point b, Point c){
    b -= a; c -= a;
    if(cross(b, c) > eps) return  1;
    if(cross(b, c) < -eps)return -1;
    if(dot(b, c) < 0)     return  2;
    if(norm(b) < norm(c)) return -2;
    return 0;
}

int main(){
    int x0, y0, x1, y1;
    cin >> x0 >> y0 >> x1 >> y1;
    Point p0(x0, y0), p1(x1, y1);
    auto state = [](int x){switch(x){
        case  1: return "COUNTER_CLOCKWISE";
        case -1: return "CLOCKWISE";
        case  2: return "ONLINE_BACK";
        case -2: return "ONLINE_FRONT";
        case  0: return "ON_SEGMENT";
    }};
    int q;
    cin >> q;
    while(q--){
        int x2, y2;
        cin >> x2 >> y2;
        cout << state(ccw(p0, p1, Point(x2, y2))) << '\n';
    }
    return 0;
}