<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

//be naame khodaa
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef complex<ld> PT;
#define F first
#define S second

inline int in(){int x, y; y = scanf("%d", &x); return x; }

const int N = -1;
const ld EPS = 1e-12;

istream& operator>> (istream& is, complex<ld>& p){
	ld val;
	is >> val;
	p.real(val);
	is >> val;
	p.imag(val);
	return is;
}

PT proj(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + b*real(c/b);
}

PT reflect(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + conj(c/b)*b;
}

string pos(PT a, PT b, PT c){
	b -= a, c -= a;
	c /= b;
	if (c.imag() > EPS)
		return "COUNTER_CLOCKWISE";
	if (c.imag() + EPS < 0)
		return "CLOCKWISE";
	if (c.real() + EPS < 0)
		return "ONLINE_BACK";
	if (c.real() > 1 + EPS)
		return "ONLINE_FRONT";
	return "ON_SEGMENT";
}

int main(){	
	PT a, b;
	cin >> a >> b;
	cout << setprecision(10) << fixed;
	for (int i = in(); i; i--){
		PT p;
		cin >> p;
		cout << pos(a, b, p) << '\n';
	}
	return 0;
}
