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 <bits/stdc++.h>
using namespace std;
#define EPS 1e-8  
struct Point{
	double x,y;
};


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

double dot(Point a,Point b){
  return a.x*b.x+a.y*b.y;
}

int contains(vector<Point> g,Point p){
  int n=g.size();
  bool x=false;
  for(int i=0;i<n;i++){
  	Point a,b;
	a.x=g[i].x-p.x;
	a.y=g[i].y-p.y;
	b.x=g[(i+1)%n].x-p.x;
	b.y=g[(i+1)%n].y-p.y;
    if(fabs(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(){
	int n;
	cin>>n;
	vector<Point> p(n);
	for(int i=0;i<n;i++) {
		cin>>p[i].x>>p[i].y;
	}
	int q;
	cin>>q;
	while(q--){
		Point r;
		cin>>r.x>>r.y;
		cout<<contains(p,r)<<endl;
	}
}
