AcCode:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn = 25;
const int maxm = 100000;
const double EPS = 1e-10;
double add(double a, double b)
{
if(abs(a+b) < EPS*(abs(a)+abs(b)))
return 0;
return a+b;
}
typedef struct P{
double x, y;
P(){}
P(double x, double y): x(x), y(y) {}
P operator + (P p){
return P(add(x, p.x), add(y, p.y));
}
P operator - (P p){
return P(add(x, -p.x), add(y, -p.y));
}
P operator * (double d){
return P(x*d, y*d);
}
double dot(P p){ //点乘积
return add(x*p.x, y*p.y);
}
double det(P p){ //叉乘积
return add(x*p.y, -y*p.x);
}
};
//判断点p 是否在线段p1 - p2上
bool on_seg(P p1, P p2, P q)
{
return (p1-q).det(p2-q) == 0 && (p1-q).dot(p2-q) <= 0;
}
//求出p1p2, q1q2的交点
P intersection(P p1, P p2, P q1, P q2)
{
return p1+(p2-p1) * ((q2-q1).det(q1-p1) / (q2-q1).det(p2-p1));
}
int n, m;
P p[maxn], q[maxn];
int a[maxm], b[maxm];
bool g[maxn][maxn];
bool input()
{
scanf("%d", &n);
if(n == 0) return false;
for(int i = 0; i < n; i++)
scanf("%lf %lf %lf %lf", &p[i].x, &p[i].y, &q[i].x, &q[i].y);
m = 0;
int aa, bb;
while(scanf("%d %d", &aa, &bb)){
if(aa == 0 && bb == 0) break;
a[m] = aa, b[m] = bb;
m++;
}
return true;
}
void solve()
{
for(int i = 0; i < n; i++){
g[i][i] = true;
for(int j = 0; j < i; j++){
if((p[i]-q[i]).det(p[j]-q[j]) == 0){
g[i][j] = g[j][i] = on_seg(p[i], q[i], p[j]) || on_seg(p[i], q[i], q[j]) || on_seg(p[j], q[j], p[i]) || on_seg(p[j], q[j], q[i]);
}
else {
P r = intersection(p[i], q[i], p[j], q[j]); //求出交点
g[i][j] = g[j][i] = on_seg(p[i], q[i], r) && on_seg(p[j], q[j], r); //交点在两条线段上
}
}
}
for(int k = 0; k < n; k++){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++)
g[i][j] |= g[i][k] && g[k][j];
}
}
for(int i = 0; i < m; i++){
if(g[a[i]-1][b[i]-1]) cout << "CONNECTED" << endl;
else cout << "NOT CONNECTED" << endl;
}
}
int main()
{
while(input()){
solve();
}
return 0;
}