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;
typedef complex<double> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> Poly; 
const double EPS = 1e-10;

// 内積 |a||b|cosθ
double dot(Point a, Point b){
	// return a.real() * b.real() + a.imag() * b.imag();
	return (conj(a) * b).real();
}
// 外積、符号付面積の２倍　|a||b|sinθ
double cross(Point a, Point b){
	// return a.real() * b.imag() - a.imag() * b.real();
	return (conj(a) * b).imag();
}
int in_poly(Poly p, Point x){
    int n = p.size();
    bool in = false;
    for(int i = 0; i < n; i++){
        Point a = p[i] - x, b = p[(i + 1) % n] - x;
        if(imag(a) > imag(b)) swap(a, b);
        if(imag(a) <= 0 && 0 < imag(b)){
            if(cross(a, b) < 0) in = !in;
        }
        // 線上判定
        if(cross(a, b) == 0 && dot(a, b) <= 0) return 2;
    }
    return in;
}

int main(){
    int n;
    cin >> n;
    Poly pl(n);
    for(int i = 0; i < n; i++){
        double x, y;
        cin >> x >> y;
        pl[i] = {x, y};
    }
    int q;
    cin >> q;
    for(int i = 0; i < q; i++){
        double x, y;
        cin >> x >> y;
        int res = in_poly(pl, {x, y});
        if(res == 1) res++;
        else if(res == 2) res--;
        cout << res << endl;
    }
}
