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 <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#define EPS 1e-10
using namespace std;

double add(double a,double b){
	if(fabs(a+b)<EPS*(fabs(a)+fabs(b)))return 0.0;
	return a+b;
}

class P{
public:
	double x,y;
	P(double xx=0.0,double yy=0.0){
		x=xx;
		y=yy;
	}
	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);
	}
	P operator /(double d){
		return P(x/d,y/d);
	}
};

typedef P Vector;
typedef vector<P> Polygon;

double norm(Vector p){
	return add(p.x*p.x,p.y*p.y);
}

double abs(Vector p){
	return sqrt(norm(p));
}

double dot(Vector a,Vector b){
	return add(a.x*b.x,a.y*b.y);
}

double cross(Vector a,Vector b){
	return add(a.x*b.y,-a.y*b.x);
}


int contains(Polygon g,P p){
	int n=g.size();
	bool x=false;
	for(int i=0;i<n;i++){
		P a=g[i]-p,b=g[(i+1)%n]-p;
		if(abs(cross(a,b))<EPS && dot(a,b)<EPS)return 1;
		if(a.y>b.y)swap(a,b);
		if(a.y<EPS && EPS<b.y && cross(a,b)>EPS)x=!x;
	}
	return (x?2:0);
}

int main(void){
	int n;
	scanf("%d",&n);
	Polygon g;
	for(int i=0;i<n;i++){
		P p;
		scanf("%lf %lf",&p.x,&p.y);
		g.push_back(P(p.x,p.y));
	}
	int q;
	scanf("%d",&q);
	for(int i=0;i<q;i++){
		P p;
		scanf("%lf %lf",&p.x,&p.y);
		printf("%d\n",contains(g,p));
	}
	return 0;
}