For a given polygon g and target points t, print "2" if g contains t, "1" if t is on a segment of g, "0" otherwise.

g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n-1) are sides of the polygon. The line segment connecting pn and p1 is also a side of the polygon.

Note that the polygon is not necessarily convex.

Constraints

* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* No point of the polygon will occur more than once.
* Two sides of the polygon can intersect only at a common endpoint.

Input

The entire input looks like:


g (the sequence of the points of the polygon)
q (the number of queris = the number of target points)
1st query
2nd query
:
qth query


g is given by coordinates of the points p1,..., pn in the following format:


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points. The coordinate of a point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them.

Each query consists of the coordinate of a target point t. The coordinate is given by two intgers x and y.

Output

For each query, print "2", "1" or "0".

Example

Input

4
0 0
3 1
2 3
0 3
3
2 1
0 2
3 2


Output

2
1
0
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
template<class T> 
inline void read(T& x)
{
	char c = getchar(); T p = 1, n = 0;
	while(c < '0' || c > '9'){if(c == '-') p = -1; c = getchar();}
	while(c >= '0' && c <= '9'){n = n * 10 + c - '0'; c = getchar();}
	x = p * n;
}
template<class T, class U>
inline void read(T& x, U& y){read(x), read(y);}
template<class T, class U, class V>
inline void read(T& x, U& y, V& z){read(x), read(y), read(z);}
const double eps = 1e-10;
const int maxn = 1e3;
struct Point
{
	double x, y;
	Point(int a, int b){x = a, y = b;}
	Point(double a = .0, double b = .0){x = a, y = b;}
	Point operator+(const Point& r){return Point(x + r.x, y + r.y);}
	Point operator-(const Point& r){return Point(x - r.x, y - r.y);}
	Point operator*(double r){return Point(r * x, r * y);}
	double operator*(const Point& r){return x * r.y - y * r.x;}
} g[maxn];
int n; 
typedef Point Vector;
inline double dot(Vector l, Vector r){return l.x * r.x + l.y * r.y;}
int contains(Point p)
{
	bool x = false;
	for(int i = 0; i < n; ++i)
	{
		Vector a = g[i] - p, b = g[(i + 1) % n] - p;
		double cross = a * b;
		if(abs(cross) < eps && dot(a, b) < eps) return 1;
		if(a.y > b.y) swap(a, b), cross = a * b;
		if(a.y < eps && eps < b.y && cross > eps) x = !x;
	}
	return x ? 2 : 0;
}
int main()
{
	read(n);
	for(int i = 0, x, y; i < n; ++i) read(x, y), g[i] = Point(x, y);
	int q; read(q);
	while(q--)
	{
		int x, y; read(x, y);
		printf("%d\n", contains(Point(x, y)));
	}
	return 0;
}