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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 <iostream>
#include <cmath>
using namespace std;

struct vec{
    double x,y;
};

const double eps = 0.00001;

int main(void){
    vec p0,p1,p2,v1,v2;
    int q;
    cin >> p0.x >> p0.y >> p1.x >> p1.y;
    v1.x=p1.x-p0.x; v1.y=p1.y-p0.y;
    cin >> q;
    cout.precision(10);
    cout.setf(std::ios::showpoint);
    for(int i=0;i<q;i++){
        cin >> p2.x >> p2.y;
        v2.x=p2.x-p0.x; v2.y=p2.y-p0.y;
        double p=v1.x*v2.y-v1.y*v2.x;
        if(p>eps){
            cout << "COUNTER_CLOCKWISE" << endl;
            continue;
        }
        if(p<-eps){
            cout << "CLOCKWISE" << endl;
            continue;
        }
        if(v1.x!=0){
            p=v2.x/v1.x;
        }else{
            p=v2.y/v1.y;
        }
        if(p<0){
            cout << "ONLINE_BACK" << endl;
        }else if(p<=1){
            cout << "ON_SEGMENT" << endl;
        }else{
            cout << "ONLINE_FRONT" << endl;
        }
    }
    return 0;
}