#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double pi = acos(-1.0);const double eps = 1e-10; ///这题不卡精度const int mx = 100005;struct P{double x, y;P(double x = 0.0, double y = 0.0): x(x), y(y) {}bool read(){return ~scanf("%lf%lf", &x, &y);}///由于比较的是原始输入数据，所以不用epsbool operator < (const P& p) const ///加cosnt和引用以便sort调用，其他函数不加{return x < p.x || x == p.x && y < p.y;}P operator + (P p){return P(x + p.x, y + p.y);}P operator - (P p){return P(x - p.x, y - p.y);}P operator * (double d){return P(x * d, y * d);}P operator / (double d){return P(x / d, y / d);}double dot(P p){return x * p.x + y * p.y;}double det(P p){return x * p.y - y * p.x;}///根据凸包构造之后的点的顺序，规定角度范围为(-pi/2, 3/(2pi)]double getA(){double A = atan2(y, x);if (A < -pi / 2.0 + eps) A += pi * 2;return A;}} p[mx], ans[mx], q1, q2;int n, len;double A[mx];///求凸包void convex_hull(){sort(p, p + n);len = 0;int i;for (i = 0; i < n; ++i){while (len >= 2 && (ans[len - 1] - ans[len - 2]).det(p[i] - ans[len - 1]) < eps)--len;ans[len++] = p[i];}int tmp = len;for (i = n - 2; i >= 0; --i){while (len > tmp && (ans[len - 1] - ans[len - 2]).det(p[i] - ans[len - 1]) < eps)--len;ans[len++] = p[i];}--len;ans[len] = ans[0];///这句话直接写在凸包里for (i = 0; i < len; ++i)A[i] = (ans[i + 1] - ans[i]).getA();}inline bool dcmp(double a, double b){return a + eps < b;}///二分找第一个角度大于x的边inline int Find(double angle){return upper_bound(A, A + len, angle, dcmp) - A;}inline bool judge(P q1, P q2){int i = Find((q2 - q1).getA());int j = Find((q1 - q2).getA()); ///直线两侧各找一点，若两点在直线两侧，则直线必过凸包if ((q2 - q1).det(ans[i] - q1) * (q2 - q1).det(ans[j] - q1) > -eps) return true;else return false;}int main(){scanf("%d", &n);for (int i = 0; i < n; ++i) p[i].read();if (n > 1) convex_hull();while (q1.read()){q2.read();puts(n <= 1 || judge(q1, q2) ? "GOOD" : "BAD");}return 0;}
